Transcript - Lecture 11-15

Transcript - Lecture 11
6.002 Circuits and Electronics, Spring 2007

We have put some of the quiz stats here. The mean was about 75%. And I must tell you that that is very impressive. I guess MIT undergrads never cease to amaze me. And this was not an easy quiz. This was a relatively hard quiz.

And that average implies that you guys did well on a relatively hard quiz. Good. Let's get back to our final lecture on amplifiers and small signal circuits. And as always let me start with a review.

Very quickly -- -- we came up with a notation to represent small signals. And our notation looked like this. Our total variable was small and capital, and this was a DC bias and this was a small signal.

This is also called the operating point. And the small signal is also called the incremental signal. In general, if you have some function, some variable of interest in the circuit, say a total variable V out, let's say it relates to some input variable as F of VI.

So mathematically we can find out V out by simply finding the slope of this function at the operating point and then multiplying it by the incremental change in the input. Gold standard math. So we do the slope of this function and evaluate it at the operating point.

So this would give us the slope of the function. And multiply that by small VI, which is incremental change. This is standard math. What this will tell you is given a small change in VI this function gives you, this expression gives you the small change in V out.

And in lecture we have pretty much used this method so far, used the math to get to where we wanted it to be. And then the way we provided biasing and so on was for our amplifier in particular we had a bias voltage, some small signal value, VS.

And this was output which was also given to be some output operating point plus a small change, which was a change in the output voltage. So what we have done here is mathematically computed small V out.

And what I am showing you here is to get the same effect in a circuit is you build your circuit and replace what used to be a total variable with a DC bias plus a small change. And then you will get your output here.

And this output will relate to this input using this expression. So this is more review. To continue on with the math review, for our amplifier VO was given to be VS-K/2(vI-VT)^2 RL. So this was the output versus input relationship for the amplifier.

And mathematically I could get the small change in the output VO by simply differentiating this function with respect to VI, evaluating that function, at capital VI and multiplying by the small change in the input.

And the resulting expression that we got for small VO -- -- was simply minus K, this was our DC value, and RL times small VI. So we derived all of this the last time. So nothing new so far. So my small signal output was some function given by K(VI-VT)RL times small vi.

And notice that this is how VI relates to VO. And this is a constant with respect to VI. V capital I is a DC bias, so this is a constant. So therefore this is the linear relationship that we had set out to get.

This term here, for reasons we will see today, this term here K(VI-VT) is called gm. Transconductance. We will look at it in more detail a little later. Even more review. So I can draw the transfer function and plot VO versus VI.

Another way to graphically view what is going on is by plotting the load line curve for this circuit, so this is VI. And I said we draw that by first plotting the -- These were our MOSFET curves. And we know that at some point the MOSFET gets into saturation, so this curve was iDS=K/2 VO^2.

And to the right side of the curve the MOSFET is in saturation. And we said we will adhere to the saturation discipline and operate in this regime. When the MOSFET gets into this region it is in its triode region.

And then we could draw the load line here. The load line codified the following relationship, iDS=VS/RL-VO/RL. This was a load line. So I have superimposed a load line on the device characteristics, and I am going to show you a little demonstration based on that at this point.

So these curves were drawn for increasing values of VI. And if I choose some operating point here then this point would correspond to some bias, this bias point would correspond to some input voltage VI, a corresponding output bias VO and a corresponding current iDS.

So iDS capitals, VO capitals, VI capitals represent the operating point values for our little circuit. So far there is nothing new. One thing we stopped the last time by pointing out that the gain of our amplifier, this is the gain, -K(VI-VT)RL.

That is the gain A of the amplifier. That gain related to VI. A gain was proportional to VI-VT. So therefore if I increased VI, I would get more gain. So the question is how do we choose a bias point? And in our particular example, let's say we are free to play around with VI.

So we play around with VI and I can choose various bias points. So where do you set the bias point? What are the various characteristics of the circuit that relate to my bias point? Well, first, of course, is gain.

The gain depends on how I choose VI. I will show you that in a moment. The second important thing, in other words, if I choose a bias point that is a small VI then my gain is going to be smaller. If I choose a bias point that's at a much higher value of VI, I get a bigger gain.

The second important consideration is operating range. Notice that if I choose a bias point here then as the input changes -- Notice VI in this graph goes up or down, and I would be traversing and following different lines here in my MOSFET characteristic.

And as VI increases the operating point would come up here and so on. So if about this operating point I varied my input voltage VI then, so let's say about this operating point, if my input VI, my small signal VI varied about a small range then correspondingly the output value would vary about this part of my load line.

So notice now that the operating range, how far can VI vary before the MOSFET goes out of its saturation discipline? Well, on the low side my VI can come down to here. And we looked at the operating ranges for an amplifier.

And I can come all the way down to VT. At that point the output will come here. Similarly at the high end VI could get up to a high value. And we computed that value in the last lecture. And the corresponding value of the input would be here.

So in some sense I can traverse all the way from here to here and have the MOSFET remain in saturation. Remember we are not talking about linearity right now, just about the valid operating range based on my definition which is that the MOSFET should stay in saturation.

So if I chose my operating point here then I get this range here. And, on the other hand, if I chose my operating point to be here, for negative excursions of the input signal I have a very small amount before I hit cutoff.

So if I chose my operating point here then for negative traversals of VI about the operating point I very quickly hit cutoff. So if I want symmetric swings then this is the best that I can do in terms of the valid input operating range if I want symmetric swings given that this is my bias point.

On the other hand, if I chose my bias point somewhere here, or very carefully chose my bias point then my input can vary on a much wider region and still get symmetric swings. And so therefore the choice of bias point also influences the maximum swing range of my input signal.

I shouldn't call this operating range. I should call it input swing range. We defined the valid input operating range as the range for which the amplifier satisfied the saturation discipline. So the two key issues, gain and the input swing.

Let me show you a quick demo and try to point out on a graph some of the characteristics that relate to the matter we have been talking about so far. So what I show here are these curves for the MOSFET.

This is VO and this iDS. This is the zero point. Ignore this line down here. This line up here corresponds to the output voltage VO. What I am going to do now is, through some careful circuit hacking, I'm going to show show you a load line and show you the bias point, and show you how the bias point can be moved up and down by changing the input voltage which changes the corresponding output voltage.

It is hardly visible out there. Is it there? OK. It is not really clear, but notice that as I increase my input, I am increasing my input. My output keeps coming down. And I hope your eyesight is better than mine because I don't see a dot up there.

I am amazed. This is the first time this has happened to me. That's OK. All right. As you can see, as I change the input value the output operating point changes, and the dot out there traverses, articulates a load line.

I guess I have to believe that there is a dot out there. Next what I will do is show you some more fun stuff. What I will do is instead of having just a dot by having a DC voltage, let me apply an input sinusoid.

So if I apply an input sinusoid at some bias then I should see an articulation of the corresponding region of the load line corresponding to the input. So, as you can see here, now the bottom line, here is my input and this is my output.

And notice that this the region of the load line articulated when the input is of this magnitude. Now let's have some fun. As I increase my input, you can see that a larger portion of the load line is articulated, right? There you go.

And as I decrease my input a smaller region of the load line is articulated. Let's leave it here for a moment. And what I will do next, this is the region here that we are looking at, let me increase the bias.

If I increase the bias, if I increase VI, what do you think should happen to this line here? Well, if I increase the bias, the line should go up, right? Because remember the dot? The dot is in the middle of this thing here.

If I increase the bias this should move up here. So that line moves up. Do you expect anything else to happen to that line? Pardon? It increases, exactly. If I increase the bias point to here then this must also increase because my gain has increased.

Let me do that. So let me increase the input bias. Indeed notice that the region of the load line articulated is larger now. Let me decrease the bias. And notice that because the gain is smaller the little segment shown is also smaller.

I have shown you two things so far. One is that I as I increase my bias the line indeed rises up corresponding to a higher value for the input operating point. And the second is that I get a larger swing in the output as I increase the bias.

Just to show that for those like me who were visually challenged in terms of viewing that little dot up there, let me get some audio so you can actually hear the sinusoidal tone. It is a big annoying.

As I reduce the bias the gain is decreased. As I increase the bias you can see that the gain is increased and the tone is louder. Let's have some more fun and let's play some music now. And what I am going to show you with the music -- The reason I play the music is not just for fun.

Well, it's 85% fun and 15% learning. Can we turn it on for a second? What I would like to do is, as we play the music, the reason I am playing the music for that 15% is so you can listen to distortion.

I want you to listen to the distortion. That is when the articulation is here you are not going to get much distortion. But as I get into cutoff you should be getting a bunch of distortion. Similarly, as you get into the triode region you should also be getting distortion because the amplification from being somewhat nonlinear here becomes highly nonlinear at those two points.

So let's just play the signal. So volume increases, or rather the amplitude increases by increasing the bias. Now you should hear the volume go down and distortion. So notice now that the bias point is way down here.

So the gain is very low, and plus there is a distortion because of cutoff. Now what I will do is blast it up here, and you will see that the volume has gone up but then you see distortion again. Let's see if you can stand the volume here.

Even the CD doesn't like that. Notice that as I went up here the volume kept increasing because the gain kept increasing, but as I got into the triode region I began to lose my gain because, remember, the amplifier doesn't have gain in the triode region, MOSFET in its triode region, and we also get a bunch of distortion out there.

Finally, it turns out that as people are building amplifiers -- I think this was in the mid to late ‘50s and ‘60s and so on. They said man, electrical engineers are not going to get their thing right.

So they invented a new kind of music which was much more tolerant to distortion. And I will play that music for you. It is called hard rock. I challenge you to tell me it is distorting. Sounds good to me.

OK. All right. That'll do it. Thank you. I hope there are no hard rock musicians in here who will come and beat me up after lecture or something. All right. Believe it or not most of that was review.

There is nothing new today besides some fun and games and so on. I will give you a breather for five seconds before jumping into something even more fun. I want you to look at the middle board here.

And, as I told you in the beginning of 6.002, engineering is about building useful systems. Engineering is not about showing off at math or saying man, I am really cool in math and stuff. Engineering is about building useful systems, and you want to find the simplest, easiest, cheapest way to get there.

Unlike deep areas of math and theory and so on, the beauty is in the simplicity. So the aesthetics are in how simply can we make things and still get to where we want to be? All through the course what you will be seeing happening again and again and again is when things begin to get too grovelly in terms of math, we will step back and say oops, we are engineers, remember? Let's find a much simpler way to do it and use intuition.

So time and time and time again, I am going to take you on a simpler path where you can solve things by inspection by pure intuition. Most circuit designers do that. So take a look at this. I don't like this nasty differentiation here.

That's getting into late high school calculus and so on. Let's avoid the math and let's see if you find some way of doing it that is even much more simpler. And that is what I will do next and show you what is called the small signal circuit view.

A purely circuit way of developing the small signal model. So let me just start by drawing the large signal equivalent circuit for you. I will draw it here for reasons that will be obvious at the end of the class.

All right. This is the large signal equivalent circuit model for our MOSFET amplifier. VS and here is my current source. iDS relates to the square of VI minus VT. So stare at that for a second. And that is a nonlinear circuit.

iDS relates to the square of VI minus VT. Let me start by making the following claim. Let me shoot from the hip here and make the following grand claim, and then I will show you how I can prove that claim.

The grand claim I am about to make says the following. A bunch of little devices here. It is a nonlinear circuit. Just suppose for a moment we do a Gedanken experiment. Suppose I replace each of my circuit elements here with its linearized element equivalent.

In other words, here is a VS source, here is a dependent current source, let me replace them with their linear equivalent circuit models. In other words, with their corresponding small signal element models.

And I will show you what those are in a second. The resistor has a corresponding small signal element. The dependent current source has a corresponding small signal behavioral element model. And what I am going to do is keep the same circuit connections and simply pull out the large signal model for the element and replace it with a small signal element model.

And by the nature of the small signal model they are all going to be linear. So what I am going to be left with is a linear circuit with simple linear circuit elements in there. And then once I have a linear circuit, I should be able to analyze that linear circuit using methods 1, 2 and 3, superposition, Thevenin, node method and so on.

And certainly the intuitive methods like superposition and Thevenin, which make life a lot easier for me with linear models, and thereby get the function that I am looking for very quickly. Again, my claim is that I can replace each of these large signal models by just small signal equivalents and then just analyze the resultant circuit.

And I claim that I should be able to get the same answer. That's a claim. All right? So what I will do is give you an informal proof for why I can do that. And I also ask you to refer to Section 8.2.1 of the course notes to go through the foundations of the small circuit model in more detail.

The intuition is that, remember KVL and KCL? I can write down KVL and KCL for every loop in that circuit and every node in that circuit. If I do KVL and KCL, I will end up with something like this.

For the input loop I get VI something or the other applying KVL. For the output loop I get V out something or the other. And then applying KCL I get some other equation in iDS. So here are my KVL and KCL equations for that circuit.

Now, KVL and KCL are simply a different representation of the circuit because within those KVL and KCL is encoded the topology of the circuit. Remember each KVL equation represents a loop and each KCL equation represents how nodes are connected together.

So KVL and KCL equations encode within them the topology of my circuit. What I do next is, say, I replace each of these with the bias plus the small signal, so I get the bias plus the small signal and keep the equations the same.

All I have done in my big set of KVL, KCL equations, I have simply replaced the total variable with the large signal variable and the small signal quantity. Then comes a key trick. The key trick is that because the bias point variables, they are a valid solution to the circuit.

The circuit is in this quiescent state, and those are valid solutions to circuit. So therefore I can cancel them out. So the VI, the large signal values can be cancelled out leaving just small signal variables in there.

So from the KVL, KCL equations I can cancel out the large signal values, the DC bias points because they satisfy the KVL and KCL themselves. In other words, I could have written VI plus V out and so on.

Since they are satisfied I just strike out the large signal variable from both sides of each of these equations, so what is left is the same KVL, KCL equations but with small variables in place of the big variables.

What that should tell you, this informal proof should tell you is that the small signal variables should then satisfy the same form of the KVL, KCL equations that the total variables satisfy. And because the KVL, KCL equations are a reflection of the topology of the circuit, what that says is that the small signal variables must also satisfy KVL and KCL.

And since these arrive from the small signal elements that says that I can replace the big elements with the small elements and KVL and KCL will hold for the resulting circuit. This is a very quick breeze through, an informal proof to show that I can replace the big elements with the corresponding little element models and then simply apply linear techniques.

Refer to Section 8.2.1 for more foundations and more discussion about the foundations for why we can do this. That brings up the small signal circuit method. The circuit method for small signal analysis has three steps.

The first step is find operating point by using LS. First you analyze your large signal circuit and find the operating point. You have to do this, because remember, the small signal models depend on the operating point values.

Remember the gain of our amplifier depended on the bias point. Second step is develop small signal models of elements. Second step is take each of the elements in your circuit and find their equivalent small circuit model for each of the elements.

Third step is replace original elements with their small signal model elements. Third step is simply take the large elements and replace them with their small signal equivalent models. Then analyze resulting circuit, and that circuit will be a linear circuit.

So let's do an example. I will just use the amplifier as an example of this method. And convince you that you are going to get the same expression at the end, but just so, so simply without even the smallest amount of grubby math.

Three steps. The first step is to find the operating point using the large signal model. And let me just do that here. I get my V out = VS-K/2(VI-VT)^2 RL. Let me just write down that out here. Don't worry about copying that down.

It is on the last page of your notes. The first step of the method simply applies the large signal model and finds out the behavior of that circuit to find out what the bias point values are. The second step is to develop the small signal model of my elements.

How do I go about developing the small signal models of elements? Let's start with the MOSFET. The large signal model for the MOSFET looks like this. Here is my Vgs. This is my gate. This is my drain.

This is my source. And I know my iDS to be K/2(Vgs-VT)^2. So this is the large signal model for the MOSFET, again in saturation. I am talking about all of these models are under the saturation discipline.

So Vgs relates to iDS in the following way for the MOSFET. That is iDS, is K/2 and that is my square law relationship. So what is a corresponding small signal model? I go ahead and start with this.

The corresponding small signal model simply says that iDS relates to Vgs in the following way. All I have to do is find a small signal equivalent where I need to find out, given a small change in the input Vgs, what is the small change in the iDS? So I can apply my standard trick to a much simpler expression here, which is iDS simply, I differentiate this function with respect to Vgs.

So I don't completely eliminate the math here, but it is a much simpler problem here. At Vgs equals the bias point times small vgs. I can find the small change in iDS corresponding to a small change in the input using this expression.

That gives me iDS as simply K(Vgs-VT) vgs. I call this gm, and I will tell you why in a second. So what does this expression say? This expression says that if I have a small change in Vgs then this will be my small change in iDS.

Notice that the resulting small signal model is also a dependent current source. It is a voltage controlled dependent current source. So the output is the current, and it is a dependent current source and it depends on the input voltage.

The good news is that notice that this one, this expression here gm is a constant related to the bias point values. Therefore, notice that the small signal model for the MOSFET in saturation, not surprisingly, is a linear voltage controlled current source according to the following expression.

So iDS=gm Vgs. Gm is a representation for K(Vgs-VT) and is called a transconductance. It is called a transconductance because it, in some sense, deflects the conductance properties of this based on the input.

So it is a transconductance. So this value is called Vgs. Therefore, I can build the small signal model as follows. Vgs is a voltage controlled current source and iDS is simply gm Vgs. So this is my gate, drain, source.

So that is the small signal model for my MOSFET. As a next step what are the other elements in my circuit? Let's see. I have a voltage source and I have a resistor, so let me find out the corresponding small signal model for a DC supply VS.

This is Page 7. I will do it mathematically for you, but often times it is always good to do a sanity check using intuition. Let me ask you, the large signal for a DC supply looks like this. The element law for a voltage source is VS equals some capital VS.

It is a constant voltage. So what do you expect to be the small signal model for a voltage source? In other words, for a small change, suppose I have a small change in the current, by how much should the output VS change? It shouldn't change.

It is a voltage source. So what does intuition tell you is a small signal model for the voltage source? A short. So the key here is that a voltage source behaves like a short circuit for small perturbations.

In other words, if I change the current flowing through it by a small amount somehow, the output is still going to held at VS. In other words, small signals are simply going to scoot through this voltage source without having any impact whatsoever on the voltage.

Or mathematically I could also do small vs is del by del IS of VS evaluated at IS equals some capital IS times small IS. And therefore VS equals zero. What that means is that the small signal model for my voltage source is simply a short circuit.

So in a small circuit voltage sources appear like a short circuit. Finally, I have a resistor, my resistor R. Let me find out its corresponding small signal model. The large signal model looks like this R, VR, IR.

And I know that VR is simply RIR. And to find the small signal equivalent I do del of IRR divided by del IR for IR calculated at some constant value times small IR. What I am looking to do is to find out what is the change in the voltage across R for a small perturbation in the current? Again, let me exhort you to rely on intuition to at least sanity check your answers.

So what do you think this should look like? It's a resistor and I have a small change in the current, by what do you expect the voltage to change? Think about, for the next five seconds, what the small signal model for this should look like and then I will go ahead and write down the answer.

So differentiating I simply get RIR. In other words, for a resistor the small signal model is the resistor itself. So what I have done so far, let me just take you through where we are right now, give you the big picture there.

I began by suggesting that looking to find an even simpler way to do small signal analysis. I gave you an informal proof to show that if I had small signal element models for all of my elements, I could simply replace them in the circuit and then do a corresponding linear circuit analysis phase to get the result I am looking for.

There are three steps to the method. As a first step we began by finding small signal models for each of our elements. For the nonlinear MOSFET the small signal model was a linear dependent current source.

For a voltage source the corresponding small signal model was a short circuit. Again, that makes sense intuitively if I change the current through a voltage source by a small amount. By how much does the voltage change? It is a voltage source, silly.

The voltage doesn't change. So the small signal V, the small change in the voltage is zero, and that is the same thing as a short circuit. For a resistor by how much does the voltage change if I change the current by a small amount? Well, it will change by R times the current change, and that is the property of a resistor, R.

As a final step what I would like to do, on Page 8, I'd like to very quickly draw for you the small signal circuit and then analyze it. This is the large signal circuit. That is a large signal circuit.

And let me draw the small signal circuit. And the method says simply pluck out, gouge out each of these elements. And simply replace each of these nasty nonlinear elements with the corresponding small signal linear equivalents.

So let's do that. Remember, for the input you replace input with its small signal voltage because I am telling you that it's sourcing a small change in VI. So that is VI. And then I replace a short for VS.

I replace an R for RL because it is an RL itself for the small signal model. And then for the dependent source, we discovered that the dependent source was a linear dependent source given where ids=gmvi.

Remember, this was my small VO. Here you go. I have a small signal circuit here where I have simply created that by replacing each of the big elements by little rinky-dink elements. Now these are all linear elements so I can do a really simple linear analysis.

What method shall we use? Well, this is so simple. I will just go ahead and use the node method. So applying the node method at the node with voltage VO, what I will do is the current going up, VO divided by RL equals the current going down iDS.

And so the current going up is VO divided by RL and the current going down is -- Oops, I should have done this. The total current going out is zero, so the sum of these two is zero. That is my good old node method here.

And I know that iDS is simply gmvi equals zero. So right there I have the relationship between VO and VI. So VO is simply minus gmviRL. And remember gm was simply K VI minus VT. We are done, OK? What have we here? I created a linear circuit which simply comprised small signal models for each of my big elements.

And then I simply did a straightforward linear analysis using any one of the linear techniques I knew about. This is simple enough so I apply the node method. And I've got the equation at this node, simplified it and I directly got the answer.

In one or two steps I directly gave you the output as a function of the input. It can't get any simpler. Thank you.

Transcript - Lecture 12
6.002 Circuits and Electronics, Spring 2007

Good morning. Today we move in the direction that takes a big turn from the direction we have been going in so far. All the devices we have had up until now, resistors and voltage sources, and even your digital devices like the AND gate or the inverter and so on had a very specific property.

We didn't dwell on that property, but that property was that these were not what are called memory devices. In other words, the outputs at any given time are a function of the inputs alone. In other words, if you took your inverter or your NAND gate for that matter and you build a circuit comprising 50 NAND gates connected in structures that we have talked about, you apply an input and boom you get an output.

And your output is a function of the inputs alone, right? The same thing with your resistors and voltage sources. At any given point in time your output VO of T was some function of the input VI of T.

What we are going to do today is discuss a new element which will introduce a whole new class of fun stuff for all of us to deal with. And that is called storage. In other words, the output of a circuit is now going to depend not just on the inputs but it is going to depend on the background or it is going to depend on where the circuit has been in the past.

So past is going to matter. It is a very fundamental difference. And what I would like to do is start by giving you folks a little bit of a surprise. I am going to do a little demo taking two of your inverter circuits.

I am going to start by taking a couple of inverters. Remember, I am using this structure here as an inverter. And I am going to couple this to another inverter and take an output C, some VS, some load resistance RL, my B terminal and my A terminal.

So I'm going to apply some input between ground and my A terminal. And for fun I want to apply a square wave at the input. A square wave between zero and 5 volts. And this is how my time goes. Let's assume that VS is 5 volts.

So what I am going to do is plot for you the behavior of this inverter. I am going to plot for you A, which would look like this. I am going to plot for you B, which would be the inverted wave form.

And then plot C, which would be a wave form that looks like this again. Let me do a plot here. So this is A. -- and so on. Time goes this way. And let's say this is between zero and 5 volts. And B should be an inverted wave form that should look like this.

If all that we believe of the world so far is true then this is how things should behave, so C should look like this. This is what the world should look like and if everything that you learned about is true and correct and all of the good stuff.

Let me show you a little demo and see if I can try to pull the rug out from under all that you have learned so far and show you some surprising stuff. Here are the three wave forms that I showed you up here.

This is my A. This is my A wave form. This is the B wave form. Notice that B, as you expect, is an inverted form of A. And this is C. We all expect this, correct? But what I am going to do is let me expand the time scale on this so that I can look at these transitions a little bit more carefully.

I am just going to expand the time scale. There you go. All I have done is expanded the time scale and spread that out a little bit. And what you see there is quite different from what you expect.

A is a square wave as expected, but B is stunningly different. It is a zero as expected because this is a one. But here I get some really strange behavior, behavior that is like nothing on earth. Like nothing you have seen before.

And then, of course, it becomes a one eventually, but there's some really, really shady stuff going on here. And so far you are not prepared to deal with this. We have not given you the facility to deal with his issue.

What is the problem with this? We could say who cares? What is the problem with this? Let's look at the result. I am looking at this, I am focusing on this piece here. And notice that instead of being a sharp rise it looks like this.

It is going up a little bit more slowly. What kind of problem would that create? The problem that it creates is the following. Let me play around with this graph a little bit more. What I am going to do is just take this output here, the C output and line it up against the A output.

And so I am going to line up the C wave form on top of the A wave form. So you can see for yourself if something really, really strange and nasty is happening, I am just going to move up the C wave form and line it up.

What is happening out there? If you look carefully, what you observe is that the C wave form transitions just ever so slightly later than the A wave form. Look here. And I claim that it is because of this.

Because of this, the C wave form falls just a little bit later, and that little thing we see out there is a delay. So nothing you have learned so far prepares you for this. Suddenly, instead of the output exactly following the input, my output is following the input but a little bit later.

And it is this fact of life that things happen a little bit later, is really the reason why each of you and all of us needs to buy new computers every couple of years. This simple basic fact. If this fact of life didn't exist, you would buy one computer and be done with it for life.

Intel would make gobs of money one year, and so would Dell and Gateway and so on, and then no more. That's it. This is it. But because of this a little itty-bitty difference here the entire semiconductor technology is charging along trying to do something about that.

You buy newer and newer computers each year. It turns out this little itty-bitty thing here, that is called the delay, the inverter delay. And it happens because of a specific element that has been introduced here that we have not shown you so far.

And a large part of the semiconductor industry and follow-on courses and design and so on focuses on how could I make my delay smaller, how can I get to be faster and faster and faster? This relates to how fast we can clock your Pentium IV.

Remember it came all the way to 1.3 gigahertz? What's the fasted Pentium money can buy today? What is the fastest P4? Oh, 3.2 have come out? I don't know. Ken claims 3.2. But, yeah, there you go, 3.2 gigahertz.

It all has to do with this little itty-bitty thing. You saw it for the first time here. When some of you become CTOs at Intel and so on, just remember that it all began on October 16th with this little rinky-dink thing here.

What you are going to learn now is some really cool stuff that has huge implications for life. So why does that happen? Why did this transition happen just a little bit later? The reason is that remember when this wave form reaches VT, the threshold voltage of this MOSFET, this guy is going to switch, right? So because of the slower rise of the voltage, the VT is going to be reached a small amount of time later.

So I am going to hit VT slightly later. And because of that this guy is going to transition just a bit later because this intermediate wave form B is slower. It hits VT just a little bit later than if it would have made an instantaneous transition.

And therefore my output falls just a little bit later and this gives rise to my delay in the inverter. We can call that d if you would like, some delay. In your course notes, this material is covered in Chapters 9 and 10.

That was to kind of motivate why we are going to be doing all that you we will be doing. Don't anybody come within a foot of this even by mistake. I mean it. It is pretty deadly stuff. Today we will talk about the capacitor.

And in the next couple of lectures I am going to tie it all together and show you how this relates to that. I will show you exactly how the delay happens. You can compute it based on some simple principles that you will learn about in the next couple of lectures.

What I am going to do is first of all show you, I claim that that delay happens because of the presence of a capacitor somewhere in there. What I will do now is take you into a closer look, take a closer look at the MOSFET and show you were the capacitor is.

This is the MOSFET that you have seen so far, drain, gate and source. This is called an n-channel MOSFET. And what I am going to do is dissect this and show you what is actually happening, what this looks like on silicon.

So here is my slab of silicon. It is very thin. And let's say this is, I won't go into details here. You will learn a lot more about this in future device classes like 301 and so on, but suffice it to say I will just introduce it here to give you a sense of where the capacitor is.

This is p-type silicon. And the way you build a MOSFET is you create a couple of tubs in which you dope to be n-type. The basic silicon is dope p-type. And this guy here is n-type. And what you do is a thin oxide layer is placed on top of that and then on top of that a thin metal layer.

This is a metal layer. This is a thin piece of oxide, silicon dioxide. And this is my P substrate. Now this is a little metal layer that is really a wire on top of the silicone. This metal layer could be some sort of a wire that meanders around on the surface of silicone.

And this is a wire that connects to the gate. This is the gate of my MOSFET. And this guy here is the drain. And this guy here is the source. And this is my gate. So there is a little piece of metal here.

This is this piece of metal here. And there is a piece of oxide and then my silicone substrate. Notice that this is my oxide. When I apply a positive voltage to the gate here with respect to the substrate, what happens is that I draw up negative charges.

I draw up electrons here into this channel region and I have corresponding plus type out here so that I get a view here that looks like a couple of plates. And I end up with an oxide in the middle. There is no connection.

Two plates separated by a small distance with plus q and minus q on the plates. And, because of that, what ends up happening here is that this piece behaves like a capacitor. So a capacitor has two plates with a thin insulating material in the middle with some permittivity epsilon.

And so I get a little piece of a capacitor here. That is the capacitor that is forming. I did not set out to build that capacitor, but there is a capacitor nonetheless. So when I apply a positive voltage at the gate, negative electrons are pulled up here which forms a channel, and then a current can then flow.

And that is how the MOSFET turns on. So n-type electrons back to n-type, and I get electron flow here and that gives me my channel. This is just kind of devices in four minutes or less. You will do an entire course on this, if you like, if you take 301.

What we do is to be able to capture the behavior that we just saw, the funny delayed behavior, we have to augment our model. We have to introduce a new element. So what we do is here is a MOSFET, gate, drain and source.

And notice here we model this by putting a little capacitor, CGS between our gate and the source. So this becomes a simple model for our MOSFET device which is the good old gate drain source device from the past with a little capacitor CGS having some value for CGS in maybe ten to the minus 14 or thereabouts farads.

So that is a little capacitor that has come about in this device that we fabricated here. It is that capacitor that is at between node B and ground because it is between the gate and the source of the second inverter.

And it is that capacitor that is playing the games that we saw out there. So let's look at some of the behavior of an ideal linear capacitor. A capacitor, as I said, has a couple of plates. There are a couple of plates.

Between the plates is some dieletric, permittivity epsilon. Let's say the area of the plates is A, and let's say the plates are separated by a distance D. I get some charge here, let's say q. So q and minus q on the capacitor.

And the capacitance C is given by epsilon A divided by D. Epsilon, as I said, is the productivity of the dielectric. So if it is free space then it would be epsilon zero which is the permittivity of free space.

That is the capacitance in farads. And the symbol looks like this. Capacitor C. Voltage v. Current i. So this, much like the resistor, voltage source and so on, this now becomes a primitive element in your tool chest of elements like the voltage source and so onn.

Capacitance with the voltage v across it and a current i. And I have assigned the associated variables here according to the associated variable discipline. A question to ask ourselves is remember we said we are all now in a playground from all of nature, in this playground where the lumped matter discipline holds? And also remember that we said that for the lumped matter discipline to hold we have to make a couple of assumptions.

One of those assumptions was that dq/dt, for all their elements should be zero for all time. So right now what about the capacitor? It has got some charge q. So charge must have built up somehow. Does that mean that I lied all along, that we are no longer in this playground, that we have been ejected from the playground because of the capacitor, or are we still in the circuits playground in which the lumped matter discipline holds and all good things happen and so on? It seems like a contradiction, doesn't it? I took you from Maxwell's playgrounds to the EECS playground where I said the lumped matter discipline holds.

And one of the foundations of the LMD was that dq/dt should be zero for all time inside the elements that we are going to deal with. And right now boom, it's not four weeks into the course and Agarwal introduces an element and it has q in it.

It turns out that the capacitor also adheres to the lumped matter discipline. Remember the discipline says that dq/dt is zero for all time within elements. So I am going to be clever. What I am going to do is I want to choose element boundaries in a very cleaver way.

Notice that if I have q here on this plate then I get minus q on the other plate. So if I take the whole element, the element as a whole, if I am careful in terms of how I package my boundaries, if I put both my plates inside my element boundary then I still do get the net charge being zero.

So dq/dt is indeed zero for all time provided I make sure that my element has both the plates. Therefore, if you come across somebody else that gives you an element that says I have an idea. Let's create a new branch of electrical engineering in which we model the capacitor not as one element for two plates, but let's build a capacitor by combining two new elements, two garbage elements called G1 and G2.

G1 is like the top plate. G2 is the bottom plate. I put them together and I get a capacitor. But notice if I just pick one plate then the element G1 will not adhere to the LMD. It adheres to the LMD because I choose my element boundaries in a way that both plates come within it.

So it is very fundamental and key. And you can read a lot more about it in the course notes. I purposely dwelt on that simple point because I think it is foundational and important. And you really need to understand that the capacitor does satisfy LMD.

We are still in the good old playground. A few simple facts here. These are in the notes. And you have also seen this before, I am sure. I can relate the charge to the capacitance and the voltage as q is equal to Cv.

And q is in coulombs, this is in farads and this is in volts. So there is some charge q stored on the capacitor and it is in coulombs and q is equal to Cv. So I can differentiate this with respect to time to get the current, and that becomes i=dq/dt.

So the current at any given time is dq/dt. And so I substitute for q in terms of Cv here. That is what I get. So the current i=d(Cv)/dt. A 6.002 assumption, capacitance in general can be time-varying.

I can get time-varying capacitors. In fact, there are some sensors which are capacitive. And, as I talk, my sound waves can change the pressure on the top plate of the capacitor. And move the top plate of the capacitor, thereby changing the capacitance by moving the plate.

Remember d here, as the plate moves closer I get a higher capacitance. So we won't be dealing, unless explicitly said so, with time-varying capacitances. So what we can do is 6.002 allows us to write Cdv/dt.

So my current source capacitor is Cdv/dt. I can also write down the energy, capacitors store energy. E=1/2Cv^2. I am sure you have seen all this before in physics and so on. That is the amount of energy stored in the capacitor if it is holding a charge q.

Let me do a little demonstration for you. They don't make glasses like they used to. Our friend Lorenzo has charged up this capacitor. It is a huge capacitor. It is a 250 volt capacitor so it is nasty.

He has charged it up and has kept it there. And to show you that it does contain stored charges it has been sitting there holding charge. Maybe the first row should go backwards, just step back for a second.

I think you guys would be safe but I just don't want to take any chances. This is holding a bunch of charge. It is kind of sitting there. If I short the terminals it should try to say oh, I've got a path, let me get my charge out.

All right. Let's do it. This is always a scary moment for me. And I say a little prayer before I do this. Good? OK. Gee, you guys would love to see me getting fried, huh? All right. Let's see.

So it did contain charge. So there is a reason why Lorenzo puts one hand inside his pocket when he shorts it, because there is a natural tendency to hold the wire with both hands, and la, la, la, la, la and put it across the capacitor.

By doing this you are guaranteed that you will just be touching it with one hand. Hopefully you folks will remember for life that a capacitor can sit around and hold its charge for a while. All right.

That is enough of fun and games. Let's get on with our business of building circuits. What I am going to do is, as I promised you, I am going to close the loop on that example by halfway through the next lecture.

I'm going take you on a bit of a journey involving capacitors and resistors and involving some analysis, and then we will close it all up for you at about the middle of next lecture. What I would like to do next is here is a new element.

And let's do some fun stuff with elements. Well, you know about voltage sources, you know about resistors, let's put them together and see how they behave. Let's have a capacitor here, C, vc(t) and some current i.

What I am going to do, in general, whenever I have something new or something strange, let's say like a capacitor or some other device. It is interesting to model the rest of the circuit behind it if it contains only resistors and voltages and linear elements as a Thevenin equivalent.

So let me do that. This is R and this is vi. This stuff in the back is my standard pattern, voltage source in series with a resistor, and I connect that across my capacitor. But remember, although you saw those funny wave forms and so on, the capacitor is a linear device.

Because you can see from here that the current relates to dv/dt. That is a linear operation. You don't see V squareds and Vis and things like that in there. It's is a linear device. Let's go back to our trusty old method, the node method.

If you just blindly apply the node method and simply grunge through a bunch of math, you should be able to get to the answer, that is for some voltage v or some form of voltage vi, I should be able to figure out what vc looks like.

So let's do that. This is the node that is of interest here with the unknown node voltage vc. So let me apply the node method. (vc-vi)/R is the current going this way. That plus the current through the capacitor should equal zero.

And what is the current through the capacitor? The node method tells me that, get the current in terms of the element values. We know that the current is given by CdvC/dt.=O. Just shuffling things around a little bit, I can write RC dvc/dt+vc=vi.

We are writing the node equation and then getting the equation that characterizes this little circuit. Notice here that this has units of volts. And since I have time here, this also must have units of time.

Let's go about solving this little circuit and understanding how it behaves. The specific example that we will look at looks like this. Let's say the capacitor voltage at time T=0 is V0. This is given.

So at time T=0, I am telling you that the capacitor contains a charge. And because of that there is a voltage V0 across it. That capacitor had a voltage of 250 volts across it and most of the devices we deal with in laptops and so on today, like the Pentium IV, voltages are on the order of 1.5 volts, very small voltages.

So that is the value in the capacitor, the voltage. That is called a state. This is called the state, capacitor state. It is the state of the capacitor. And I also give you that vi(t)=VI. So my voltage is VI.

And somehow, I am not telling you how, but some how it arranged to have the capacitor voltage be V0 at time T=0. Now I want to look to the solution to this for t greater than or equal to zero. And in that time my voltage vi is at some capital VI, some DC voltage VI.

So I am going to solve the differential equation RC dvc/dt+vc=vi given these two values. Input is DC voltage VI and VC0 is V0, the initial charge in the capacitor. So from now until almost to the end of the lecture, it is just going to be math by solving this very simple first order differential equation.

And the key here will be that throughout 6.002 we will be following one method to solve these. There are many methods to solving differential equations, and we will follow one method. That method is called the method of homogenous and particular solutions.

In 1802, I believe, you would have learned maybe this, and certainly other methods. You can use any method to solve it. We will just stick to one method. And this is also used in the course notes.

In this method what we do is take the solution VC by finding two other components. One is called the homogenous solution. And summing that up with the particular solution. And that is the total solution.

So total solution is the sum of the homogenous and the particular solutions. And the method has three steps. As I said before, we will be using this method again and again with every differential equation that we encounter in this course.

And you won't encounter a while lot. The first step we find the particular solution. The second step, find the homogenous solution. The total solution is the sum of the two. And then find --- There will be some unknown constants depending on the equation that you have.

And in the end we simply find the unknown constants by applying the initial conditions that we have. Boom, boom, boom. Particular. Homogenous. Find constants. Three things. So let's go about solving this equation and apply those three conditions.

Again, remember, what I am doing now for the next 10 minutes or 15 minutes is using math that you know about to simply solve this first order of differential equations. There is nothing really new that I am going to talk about here.

One is to find the particular solution vCP, which will then be added into the vCH to get me the solution. So the way you find the vCP is you find any solution that satisfies this equation. This is the equation.

You find any solution that satisfies it. And find the simplest possible solution that money can buy. Find it. That's the particular solution. Any solution is fine. In this case, a really simple one would be vCP equals VI.

Let's see if a constant works. One thing you will realize in differential equations is that they are actually much simpler than they seem. And the reason is that almost every time you have to assume you know the answer, and then you are checking to see what you assumed was correct.

Assume the answer is this like you are really smart, and then check it out and say oh, yeah, that must have been the answer. So here we assume that I think VI is going to work so let's try it out. Substituting in here.

RC dvc/dt is 0. vi is a constant. So I get vi equals vi, so therefore this is a particular solution. Done. I substitute vi here. So dvi/dt=0. This vanishes and vi=VI. Bingo. Therefore, VI is a solution to this equation.

So I am done with my vCP. And in general what you have to do is use trial and error. By trial and error try out a bunch of solutions until you get lucky. In general, again, in all of 6.002 for many of the excitations a simple constant usually suffices.

Our second step is to find the homogenous solution. And we can also do that very quickly. And to do that we have to find a general solution to the homogenous equation. The homogenous equation is the same differential equation but with the drive set to zero.

We want to follow a set pattern to solve the differential equations here, and the set pattern is find vCP, vCH, find constants. And to find vCH we are also going to follow a set pattern to find the homogenous solution.

So we set the drive to zero, so vi is set to be zero. And I need to find a general solution to this. As I promised earlier, diff equations are really, really simple because the way we are going to solve them is we are going to assume we know the answer and then go check it.

So let's try Ae^st. Let's try and see if this can solve this particular equation for some values of A and S. I am telling you that the solution is going to be of this form. Assume it. And then simply go ahead and find me A and S, and do that by substituting it back into the equation and find out the corresponding As and Ss.

So let's go ahead and do that. I get RC. I substitute this back up so I get dAe^(st)/dt+Ae^st=0. And let me plug that in and see what comes. I get RCAse^st+Ae^st=0. I want to discard the trivial solution of A being 0.

That is a trivial solution so I will discard that. And what I will do is cancel out the As from here, assuming A is not zero, and cancel e^st here. And what is left is RCs+1=0. What this is saying is that if I can find an S such that this is true then Aest is a general solution to my homogenous equation.

This is easy enough. And so S=-1/RC. If I choose my S to be -1/RC then the simple math that I have gone through shows me that this must be the solution to the homogenous equation. Or in other words vCH=Ae^(-t/RC).

All this is saying is that Ae^(-t/RC) is a solution to my homogenous equation. A is an unknown constant. A is some constant. I don't know what that is yet. Notice RC has popped up again. And the cool thing about RC is that, this is time, this also has units of time.

We commonly represent RC as some time constant tau, as units of time. Associated with that circuit is the time constant tau, which is simply RC. I commonly write this as Ae^(-t/tau). I am very the end here.

I have the particular solution here. I have got the homogenous solution there. I need to tell you about something else. The way I found the homogenous solution was in four steps. I assumed a solution of the form Ae^st.

I created this equation here in S. This is called the characteristic equation for that circuit. We will see this time and time again for RC and other forms of circuits. Assume a solution of this form.

Construct the characteristic equation. Find the roots of the characteristic equation. In this case it is an equation in S. So this is the root. And then form the solution based on that root. Four steps.

Ae^st, characteristic equation, root and then write down the general homogenous solution. Four steps there. And finally I want to write down the total solution. And the total solution is simply vCP+vCH.

And vCP was VI and vCH was Ae^(-t/tau). tau was simply RC. That is my solution. Now, remember the last step. The last step was form the total solution and find out the remaining constants. Find out the remaining constants by using my initial conditions.

At t=0, I know that vC=V0. I know that. And so therefore I can substitute t=0 to find the constant. So I know that VO=VI+A. t=0, this thing becomes 1, and so I get this equation from which I get A=V0-Vi.

In other words, my solution vC is simply VI+(VO-VI) e^(-t/tau). So the last 15 minutes have just been math. No electrical engineering here, but electrical engineering stopped at the point where you wrote this differential equation down, went through a bunch of math and came up with a solution.

Purely mathematically. So here I simply used math to get you the solution. And, as I have been promising you throughout this course, in the next lecture I will give you an intuitive EE method of doing it.

Real electrical engineers, real EECS folks don't do it this way. Real EECS folks do it intuitively. And I will show you how to do it in four easy seconds in the next lecture. But you need to understand the foundations of how this comes about, and so this is the answer.

You can also get the current iC is simply Cdvc/dt. I won't do that for you, but you can simply differentiate it and get the current. So I can plot for you vC, time t, vC. The intuitive way of looking at this is I have VI which is the final value of the voltage.

When t is infinity this part goes to zero so the vC is simply VI. And then there is a component V0-VI which decays according to this starting out at an initial value of V0. Notice when t is zero vC is V0, you can see that in the equation, and so it starts out at V0 and ends up at VI.

I start here, I end up here. And this portion V0-VI decays out over time like this. And this decay is governed by the RC time constant or tau. I am going to show you very quickly a couple of examples of wave forms, one that goes like this and one that looks like this.

This is when I start with some value V0 and I don't apply any input, it should decay down to zero, t, t, vC, vC. If I apply zero for VI then this should simply decay down to nothing over time. And if I apply some VI but there is no state in the capacitor then that same equation is going to look like this.

You can go and confirm for yourselves that when I apply some input but the capacitor has zero state, I start at zero, I finish up at VI and my wave form looks like this. There you go. That's the first one.

The second one where I have 5 volts on the capacitor and no input. Assume that at time equals zero I take away an input, short the input voltage to ground for example, apply zero volts. You will see the decay from 5 volts to 0 volts.

And in the first case I start with zero volts in my capacitor, I apply input of 5 volts, and notice that at t=0 the capacitor rises up to that level. So notice that these circuits with capacitor and resistors are typified by wave forms that are exponential rises and exponential decays.

We will see more of that next time.


Transcript - Lecture 13
6.002 Circuits and Electronics, Spring 2007

OK. Good morning. In the last lecture I did a little demonstration for you where I showed you a pair of inverters. And showed you that the output of the first inverter looked weird, certainly not like anything we have seen thus far.

It looked like a slow rising transition like this. And using that motivation we have begun our study of RC circuits. And in particular for today the lecture is titled "Digital Circuit Speed". We are going to look at the fundamentals of digital circuit speed.

And it all boils down to an RC delay. By the end of the lecture, I am going to show you two numbers that you can look at a circuit and obtain by observation, multiply them out and you will get a good idea of the speed at which a circuit will run.

It is pretty amazing. So as a quick review -- The relevant section for this is Chapter 10.4. As a review, we said to understand things like this we need to develop the foundations for RC circuits.

And the example I covered was that of a very simple circuit that looked like this -- An RC circuit of this form. And I also showed you that for an input of the form, input that steps from zero to VI at time T equal to zero.

And assuming that the capacitor state at time T equals zero was zero. What this means is that the capacitor starts from rest, so at time T=0, oops, this is VI, I'm sorry. So we assume that the capacitor starts from rest.

At time T=0 I apply a VI step, capital VI. And then I want to look at how the voltage across the capacitor behaves. And we did a bunch of analysis. And at the end of the day, in the final demo in the lecture last time I showed you that the capacitor would behave like this.

It would start off at, oops. I am sorry. This should be, let's assume that started off at VO. We get a different equation for zero. So let's say the capacitor started off at VO, in which case VC at time T=0 is VO as expected.

And we showed that the output would look something like this. After a long period of time this would come up to VI and this rise had a time constant of tau=RC. So we wrote the equation for this waveform.

And this is the case when VI is greater than VO. I would like you to stare at the circuit and this result here to get more intuition on what is going on. At time T=0, VC starts off at VO as expected because I am telling you that is the case, that is initial condition.

It starts off at VO. Then this one steps to VI. There is no infinite transition anywhere here, and so the capacitor holds its voltage at VO, at time T=0. And then the VI here, which is greater than VO, begins to charge the capacitor up, charge it through this resistor.

And so therefore the capacitor charges up. After a long period of time, from the basic foundations of capacitors, we know that the capacitor appears like a long-term open circuit to DC. This is a DC voltage VI.

So it appears like an open circuit. So after a long period of time VI appears at the end. And from here to here I have an exponential rise that is typified by an equation of the form -t/RC. This kind of waveform rising from a smaller value to a higher value is typified by this expression.

We saw the expression when we developed the equations last time. On the other hand, if the input was such that VI was smaller than VO, so let's say VI was smaller than VO then what will happen is that the capacitor voltage would start off at VO, because I am telling you that is the initial condition, and would then decay in this manner to the final value of VI which is the input.

Instead of going up this way it decays down to the final value applied to the circuit. Again, the time constant is RC. But this is typified by a form, this is exponential rise and this guy e^-t/RC is an exponential decay.

The key thing to remember is that when you have RC circuits of this form, the waveforms that you get are either each of the e^-t/RC or 1-e^-t/RC. So you can now begin to see how waveforms such as that come about.

We will do an example and sit down and compute the inverter delay. And notice that this waveform here is very typical or corresponds to this waveform that we see here. Here I am starting at VO. And assuming this axis starts off at zero, this one starts very close to zero and then rises up to some final value.

So far I have reviewed some material for you that I covered the last time. As a second step, I would like to give you a much more intuitive approach -- -- that doesn't involve solving any differential equations.

And the reason I do this is that most experienced circuit designers do not sit down and write differential equations each time they see an RC circuit. When you are starting out and you see an RC circuit, you say node method and you write the differential equation, but experienced people don't do that.

They look at it and they can sketch the waveform out by inspection. And I will show you how to do that. It is indeed incredibly simple once I give you some intuition. Throughout the rest of this course, I will be showing you many such examples where initially I develop the foundations of stuff and then show you an intuitive approach that very quickly lets you either get the final answer or at least sanity check the answer that you have gotten.

And this is how experienced circuit designers deal with stuff. How many people here have seen this movie Bend it Like Beckham? So you know this Beckham character doesn't think about how he is going to curve the ball.

He just does it and it happens. He doesn't sit down writing differential equations to find out the projectile trajectory and all of that stuff. You just kind of do it. These series of intuitions I am going to give you is going to be in line with the Bend it Like Beckham kind of intuition.

And this one in particular I would like to do in honor of one of your recitation instructions Professor David Perreault. And so this piece of intuition is going to be termed "Practice it Like Perreault".

Watch what I do with the other names. Professor David Perreault is really a world expert in designing really incredible power supplies for very, very small chips and so on. He doesn't start writing differential equations to do this stuff.

He looks at it and sketches it out. Let me show you how he would do this. Suppose I have my circuit like before, VI, R and C, and I am telling you that VC(0)=VO. And my input VI is a step that looks like this.

VI is a step. How would Professor Perreault do this? Let's do it completely by intuition. No math here. All right. We know that I have told you that this guy starts off at VO. I am telling you that.

You know it is going to start at VO. And there is no impulse or huge infinite transition, and so the capacitor starts off at VO. We also know from basic capacitor properties that after a long period of time, in the steady state, this is but a DC voltage.

If you apply a DC and here is my capacitor. After a long period of time this guy is going to look like an open circuit. It is going to charge up to some value and then is going to look like an open circuit.

Because if it didn't, you would keep charging it and its voltage would keep increasing. That doesn't happen, it looks like an open circuit. So it looks like an open circuit in the long run. The voltage across it must be capital VI.

If I don't have current flowing in the circuit then the only way that can happen is -- This open circuit. Capital VI appears across the capacitor. Well, after a long period of time I know that the output must look like this.

In this case, I have assumed VI is greater than VO. So you have two points of your curve, VO and VI after a long period of time. And, as I told you earlier, with capacitors you get two kinds of curves.

Two things. What you do is go zoop. There you go. You're done. And this has an exponential rise. This is with the form 1-e^-t/RC. So we can write an equation for that as follows. VC we know has something to do with minus t/RC.

This is of that form, so there has to be that term in there somewhere. And I start off with VO. At time T=0 this is one and this is one, so this term becomes a zero. At time T=0 that becomes a zero so I get VO here.

I am going to make sure this stuff stays zero at time T=0, so I start off with VO. Now, as time wears on what happens here? This voltage here, VI-VO, if you look at this difference. That is exponentially decaying over time.

And so therefore all I have to do here is write VI-VO. There is the answer. I know the form of the curve. I am just fitting an expression that meets this form. This starts off at VO. When time T=0 this second expression is zero and so it is VO.

And this difference here decays down to zero. And this difference here, VI-VO is multiplied by this term here and that is what I get. And you can confirm this. At time T=0 this is zero. At time T infinity this goes to zero, this goes to zero leaving a one, and VO and minus VO cancel and I get a VI.

Virtually any such simple voltage source, current source, resistor, capacitor, circuit for most inputs like steps and so on can be analyzed in this manner. Initial value, final value, it's simple. And just to show you that this is simple, I am going to label this expression this way.

It is of the form 1-e^-t/RC. Just remember that. Now, by the same token, what if VI had been smaller than VO? Then that is simple, too. I would have had my VI being here. VI would have been here.

And that is of the form. In this particular situation, here is my VI, my starting value and I do this. And just to label that, let me label that this way. I just told you that for RC circuits you go this way or you go this way.

So it is down here. I get some kind of an exponential decay. And, like before, think of this one. This one has VI as a base value here. And the difference between the two is VO minus VI. And that difference decays.

So I have a VI out here, and this difference decays so I get VO-VI and that decays in this form. So I get an exponential decay of this difference here. Just stare at it for a while longer. You should be able to just go and knock it off like this, just like Professor Perreault would.

No differential equations. Just write it down by looking at the curve. Let's keep these two in mind, OK, these forms? One is the 1-e^-t/RC form and the e^-t/RC. Both have a time constant RC. Let me just make this a dashed line just to be on the safe side here.

That is our first piece of intuition. And, as I pointed out before, in problems you face in life or in ones that we give you, feel free to use the intuitive method. Or what you can do is apply the mathematical method and then check your answer by using your intuition.

What I would like to do next is apply what you have learned so far to figure out what we set out to figure out, which is the delay of my inverter. I had promised you that by the end of this lecture I was going to close the loop on that little demo.

I was going to close the loop for you on this little circuit that we had looked at, one inverter driving another inverter. This was A, this was inverter X, and this was my node B. The green curve you see out there, the middle one has a transition shown up there.

And what I am going to do next is use the results we have gotten so far to compute a number. We are going to compute a delay number both for a rising transition. We will call that delay DR for rising transition.

And we will compute a delay for the falling transition DF. Remember, that this is the input that falls down sharply. The intermediate node B rises much more slowly. And because this rises much more slowly this guy here falls a little after this transition here, and so there is a delay.

And I am going to apply what we have learned so far and do an example for you and figure out what that delay is. This is an absolute foundational calculation done in building digital circuits all the time.

It is remarkable that something so simple is used in designing even the most complex of circuits to obtain very quick ideas of what my delay will look like when I have some subcircuit driving some other piece of subcircuit.

Let me just draw a few equivalent circuits for you. The internal circuit looks like this. This is my inverter X, A, my node B. And notice that I have this capacitor CGS. Since I am interested in this node, let me show you that, this capacitor explicitly, it is because of this capacitor here that arises because of this MOSFET here between the gate and the source.

And that capacitor gives rise to this RC thing that we are seeing. This is RL, this is RL, VS, VS. And let's say, just as up there, at time T=0 I get a transition like so, a falling transition from say 5 volts to 0 volts at the node A.

This is VA here. That is shown up there. And VB -- We had expected that VB would look like this. We expected VB to be instantaneous and looking like that, but instead because of the capacitor VB looks like this.

And remember, again, this is of the form 1-e^-t/RC. And we will write down the answers by inspection. From this let me draw the connection to circuit delay by showing you another little graph here t, VB, zero.

And what I am going to show you, this is 5 volts. And so the output goes like this from close to zero to 5 volts. It is close to zero. Because, at least with the inverters we have been seeing in lab and so on, the RON for the inverter is very, very small compared RL.

So it is virtually zero down here. And so what is the delay? I mentioned there are two delays of interest. One is the rising delay. That is the logical value at the end, if I wait a long enough period of time, is a logical one.

Delay is simply defined as starting from here how long does this output take to get to a valid one? At what voltage here can I say that this transition corresponds to a logical one? At what voltage here can I say that that represents a valid one? Any ideas? Yes.

It depends on the discipline, bingo. So it depends on the discipline. Now let's get more specific. Since it depends on the discipline, at what value based on something in the discipline can I say this thing is a logical one? This is an output remember.

VOH, bingo. There is some VOH somewhere. And it takes some amount of time to get to a valid logical one output, ergo there is your delay. This is tR. And I call this the rising delay of the inverter X.

It is interesting that the rising delay of inverter X, based on our model, depends on the parameters of this inverter and the parameters of whatever it is driving. So remember that the delay is not necessarily just the property of the inverter itself, but it depends on the context.

If I stick my inverter before another inverter like this, it is the capacitance on that inverter by our model that tells me what the delay is going to look like, of course in addition to RL. And we will do the math in a few seconds.

By the same token, if I had this wire connecting not to one inverter but going to ten other inverters, I expect to have a capacitance equal to ten times CGS. And so therefore this thing should rise even more slowly, correct? The more capacitance on here the slower it rises up.

Simple. If I put more and more load on this line by putting more and more MOSFETs on that line, more and more inverters this will rise slower. In our example I just have one, so let's go ahead and compute the delay.

This is called the rising delay of X. That says that for this node here to go from its output value to a valid one, which is VOH how long does it take? Notice that if this capacitor was zero then you would have seen an instantaneous transition.

If you have an instantaneous transition then notice that the rising delay was zero. That was the model we had looked at up until learning about capacitors. So let's go ahead and compute the number.

I can draw an equivalent circuit for computing a rising delay. The equivalent circuit for the rising delay looks like the following. The VS voltage source, with a resistor RL and a capacitor CGS, because when I turn this guy off, this guy has gone off, and so as far as the rise time of this node is concerned I can look at this circuit, ground through CGS through RL through VS back to ground.

And just for simplicity, let me draw this in a form that we understand. CGS. Let me use this as my ground node. And this is the voltage VB. And this is RL. And V is simply VS once that transition happens.

My other equations here, VI=VS. And what is VB(0)? VB(0) is at what value does this node start out? Notice that for simplicity here if this RON is much, much smaller than RL, then this node would be very close to ground.

So I will just go ahead and say that VB at T=0 is approximately zero. And then what I want to find out is what does the value look like for time starting from zero and then going forward? Well, we have become experts at this now.

Let's do the intuition here. Start off with zero. That's good. Because my initial value is zero, I start off here. What is the final value? After a long time, since this is a DC voltage, what would be the value at VB after a long time? Pardon? VS.

If I wait long enough then it is going to be at VS. This is greater than the initial value, so we're done. That is my 1-e^-t/RC form. It took me three seconds there. It's pretty cool. We could add the expression for this.

And the expression was I take my starting value, which is zero, and I add to that this difference VS and I multiply that by this form. There we go. And remember I get this from that rising form up here.

V0=0, this is zero, so it is simply VI times that, and VI=VS. I really would like you to get this intuition. If I had two choices, one is that you understand the intuition and are able to sketch that versus in your sleep be able to solve the differential equation and get to the answer.

I would much rather you get the intuition, if it is one or the other. It is very simple. Start off at zero, I go chuck, and boom, I get to VS and this is my 1-e^-t/RC form. I need to compute tR. And tR is the time that this takes to get to VOH.

For what value of time, for what T, does VB reach VOH? I want to find tR. What's tR? From that equation, that simply tells me the trajectory of VB as a function of time. And so I need to find out what is T for which VB is VOH? I write VOH=VS (1-e^-t/RC).

So after a rise time my output is going to be VOH. And so let me go ahead and find tR. Let's see. I bring this to this left-hand side and divide VOH by VS, and then I move things around and what I end up getting is -tR/RC and on the other side I get ln(1-VOH/VS).

Divide VOH by VS, that is this, move this to the other side, and move e^-t/RC to this side. And take logarithms on both sides. This is what I get. tR is therefore -RLCGS ln(1-VOH/VS). That is my rise time.

You can just do this by inspection. It is just so awfully simple. Just to give to some intuition with numbers and so on. Let's say that RL=1K, VS=5 volts, VOH=4 volts, CGS=0.1 pF. This happens so often that we often time call it "puff".

0.1 puff. It is pF, it's called puff. If it is nF, I don't know why they didn't call it "nuff". They just call it nanofarads. TR for these numbers gets to be one times ten to the three times point one times ten to the minus twelve for pico-farads ln(1-4/5).

And if you do the math you get this down to 0.16 nanoseconds. This means that if I had an inverter like that droving another inverter then my output transition would be delayed by 0.16 nanoseconds.

Trust me, when Intel builds microprocessors or when Broadcom builds its cable modem chips, they have to do this one way or the other using a computer tool or by hand for virtually every little subcircuit in their chip.

That is how you get the delays or some approximation thereof. What I want you also to do is, for no particular reason, I will just compute for you the following quantity RLCGS. The time constant of that circuit for no reason at all.

I am just going to compute it and stick it here. And RLCGS 1 K times 1 pF is simply 0.1 nanoseconds. I am just writing it and sticking it there for no particular reason. The next step let's do the falling delay, DF.

That is the rising delay. And, although I didn't show this to you in the demo, there is a corresponding delay of the fall time. It doesn't fall instantly, but rather it falls rather slowly. Let's draw the equivalent circuit for when the node X falls.

Notice that in my inverters here, this node starts off being at VS. This is high. And this is going to fall because when I turn this transistor on it is going to pull this node to ground or it is going to fall down.

And what is the equivalent circuit? The equivalent circuit is that ground through capacitor to this node. At this node I have RON connecting to ground and I have RL connecting to ground through VS.

Let me draw that little circuit for you. Remember life begins and ends on storage elements, so I will draw them first. My storage element CGS. That is VB. And, as I said, this is node X, it goes from RON to ground, and it also goes through RL through VS to ground.

And in this particular situation VB of zero for the following delay, VB starts off at VS so VB of zero is VS. And the final output I am not sure yet. What is the final value of the voltage at this node? I don't know that yet.

I need to compute that. So what I will do is whenever you see something like this, a capacitor connecting to linear stuff, or a nonlinear element connecting to linear stuff. For no apparent reason you should at least think about what? Think Thevenin, exactly.

And then see if you can use the Thevenin method to simplify your life. Capacitor, a bunch of stuff here, I need to find out the initial value. Oh, I know that. That is VS. Done. I need to find the final value using my intuitive method.

For the final value, I could do it just by looking at this, but I wanted to throw in Thevenin. Hey, let me try to the Thevenin equivalent and see if that makes my life any easier. VTH. The Thevenin method says that you can replace this circuit here with a Thevenin equivalent of the sort for the purpose of determining what happens at this node given that that is linear.

So I need to find out that for the purpose of determining what happens at the node X. I have to replace this with its Thevenin equivalent. And I now need to find out RTH and VTH. So I get RTH by looking in here, shorting this guy and looking at the resistance.

So I look in like this, then I short this guy here and I get RL in parallel with RON because this one shorts to ground. So RTH is simply RL in parallel with RON. This is a convenient notation for RL being in parallel with RON.

And you all know the value of that. It is another one of our very simple patterns like voltage divider and so on. Resistances in parallel can be computed as RL RON divided by RL plus RON. What is VTH? VTH is the open circuit voltage here.

If I take out this capacitor, I want to find out what the voltage here is. Ah-ha, voltage divider. VS, the voltage divider here, RL and RON. I could write this down as VS times RON/(RL+RON). Remember you will see again and again and again and again in 6.002 or any circuit stuff that you do, you will see them all over Thevenin.

Voltage dividers, current dividers, resistances in series, resistances in parallel, RC thing-a-ma-jigs like this. So if you just remember those 10 to 15 intuitive patterns then you are pretty much set for life.

It just comes on again and again and again. Parallel resistors. Voltage dividers. You should be able to write down a voltage divider in your sleep. So this is what I have. Let me now write down intuitively what I expect the node X to do just by inspection.

Let's see. What is the initial value of the voltage across the capacitor, intuitive method? This is how Professor Perreault would do it, remember? He would start off by saying ah-ha, initial value is VS because I am told it is VS.

I start off with VS. And so I start off here. What is the value after a long, long time based on this circuit here? V Thevenin. After a long time this is a DC voltage because that is a DC voltage.

The capacitor looks like an open circuit after a long time. And VTH appears there so it is simply V Thevenin. And then when you see those two, boy, I love doing this, you go like this. That is the coolest part.

And then I am done. It is so simple. Three seconds or less, I am able to tell you what the delay of an inverter is purely by intuition, completely intuitively. I mean I haven't done any solving. It is just by observation.

Took this circuit, made my life easy, Thevenin, looked at RTH, VTH and then sketched it by inspection. Again, if you find that things are really, really, really simple don't be surprised. Once you get some conceptual understanding things are indeed very simple.

You can eliminate a lot of math just by staring at things attempting to build up the intuition. As a next step what I can do is write down the expression for VB. And I write down the expression from a falling transition.

How do I do it? What was it? What is the method? I take the lowest value of interest here. That is VTH. And then I add to that this difference decaying exponentially. And that difference is simply VS-VTH.

And that decays exponentially. This form is the e^-t/RC form. And, boom, I am done. Many of you are wondering, Professor Agarwal, if life was so simple, why on earth did you have us mess around with those differential equations to get here? You show us differential equations and then you don't use them anymore.

Well, that is a good question. The answer to that is that you need to understand the foundations. Once you understand the foundations you can find simplifying techniques to get to where you need to be, but you need to understand the foundations.

You need to at least see why things are the way they are at least once. Understand the foundations and then find intuitive ways of getting your answers. So now my falling delay here is, I start off with VOS and I need to get all the way down to what value to compute.

At some point here, this is a valid one, at some point VB becomes a valid zero for the output. And that is when I stop my tF block. What is the value here for this to be a valid zero? Don't all yell at once.

VOL. I simply had to figure out what is the value of time, this is Page 7, for which this expression decays down to VOL. So it is VTH+(VS-VTH) e^-tF/RC. Then I simplify this. How do I do that? VOL-VTH.

Then I divide that by VS-VTH. So VOL-VTH. Divide that by VS-VTH. Take logarithms on both sides and then multiply by RC. So I get tF is -RC log of that. This is R Thevenin and this is CGS. How did I get this? VOL-VTH divided by VS-VTH.

Take logs on both sides. And then multiply throughout by -1/-RC and I get my tF. Done. Let's do it for the same set numbers, just that we add an RON of 10 ohms. I will do this for RON of 10 ohms and compute the value for you.

tF=-RTH. RTH is RON parallel RL. This is 10 ohms. That is 1K. So 10 ohms in parallel with 1K is approximately 10 ohms. So let me just use approximately 10 ohms. 1 pF, that is RC times ln of VOL.

Oh, I need to give you a VOL. Let's say my discipline has VOL being 1 volt. And so therefore I end up getting a VOL-VTH divided by VS-VTH. Since RON is much, much, much smaller than RL, since RON is 10 ohms and this is 1K, most of VS will drop across RL.

This is a hundred times smaller. Compared to VOL, which is 1 volt, VTH is very, very small. VTH will be on the order of 0.05, and so therefore I simply write down VOL here and say VTH is approximately zero, and I get VS-VTH.

This is approximately 5. So let me just say this is approximately. And if you do it you will get 1.6 pico-seconds. Again, just for fun, let me write the corresponding RC time constant for the circuit, which is RTHCGS.

So RTH is approximately 10 ohms and CGS is 1 pF, so this is 1 picosecond. Now you will understand why I have been writing this time constant down. It turns out that the time constant is a very, very important number.

So you see an RC circuit, and you compute its time constant for an RLC connection like this, it is the series resistance times the capacitor. The time constant is a very important number. And usually the circuit delays are in the neighborhood of the time constant value.

In this case this is 1 pS. That is 1.6 pS. And in this case we had 0.1 nS and 0.16 nS. So the time constant itself is a good indicator of what your delays are going to be like. If you have no time, you are sloshing your cereal down in the morning and you need to know how long the delay of the inverter very quickly, you have three seconds.

Just do the RC and that is a good first approximation. What I would like to do next in the last three or four minutes is set up a little demo for you for your recitation, and then your recitation will cover it.

This is a true story. This really, really happened. In this West Coast school, which shall remain nameless, they had a chip, they built a chip. And the chip had a bunch of pins, as you might imagine.

And the pin, as you have a trace on a board, a wire on a board there are some capacitance attached to wires, between the wire and ground. And that is a capacitor. And they just called it a load capacitance.

It could have been 0.1 pF or 0.01 pF or something like that. What they found when they built this chip -- What they found was that the voltage here they expected to look like this, this computer science abstraction and so on, zero to one transition, boom, it should look like this.

But for the reasons we saw today the observed transition was much slower and looked like this. So the students said ah-ha, let's speed up this chip. We can speed up the chip by looking at the RL and RON of my driving inverters.

And if I make RL small -- Notice if I make RL small my delay is small. If I make RON small my falling delay is small. So let's make really small RLs and RONs and let's all have fun. Unfortunately, what they observed was that by making RL and RON both small, the RC time constant small they expected to see a much sharper rise time.

And this was the original. But what really happened was -- They expected this to get faster and kind of look like this, but what happened was disaster struck. What they observed was something like that.

This is a real-life story. And so instead of getting something like this they go something like this. And why is that a problem? That is a problem because notice when I expect to be at a zero, I got some spikes that went higher than VIL into the forbidden region and did bad things to me.

So let me show you a little demo and show you that that's exactly how the circuit is behaving. Notice that this is what I expect but this is what I see. Look at the purple curve here. Notice these spikes that are showing up there.

This is true. They saw it happen. And why is this happening? It turns out that what was happening was that the two pins were next to each other. And I will show you a little demonstration here. Let's see if you can figure out why this was happening.

Think of these as two pins and the pins are close together. I am just modeling the two pins with a role of wire. And what I am going to do is -- I am going to separate the wires and keep them far apart.

It is like keeping my pins far apart. Hey, guess what happened? Those nasty spikes went away. But then I cannot keep my pins 1 meter apart on a chip. Your laptops are going to look 20 yards long.

You want the pins to be very close to each other so that you can have many pins on chips and therefore have very small systems. But then look, I get the spikes. Any idea why that is happening? Why is that when the pins are close together I get those spikes? Any ideas? Somewhat? We just learned about capacitors, so this must have to do with capacitors.

There is this parasitic capacitor between the pins, exactly. Here is what is happening. Here is what I expect. I expect a nice square wave at the output. But instead I have a pin next to me. And I have a faster wave form driving it.

And so therefore there is a parasitic capacitor here. And because of that I get something called "crosstalk". And the model for crosstalk is some resultant resistance with the parasitic capacitor and I get those spikes.

And the 6.002 experts saw the solution. They said how do we fix this problem? 6.002 experts said the way we fix this problem if it is slow it may be better. Instead of having sharp transitions let me drive it with slower transitions.

Let's switch to the demo again. You will see this in recitation, but I will show you the demo very quickly. I have a sharp transition of the input, which is that yellow thing out there. I am going to make the transition slower.

Switch to a triangular wave. And you will notice the spikes go away. Oh, no. That is the wrong one. The other one. There you go. The moment I switch to a slower transition boom, the spikes go away.

You want to switch back to square? There you go. The 6.002 experts saw the solution. Slower transitions. And you will do this example in detail in Section tomorrow. Thank you.

Transcript - Lecture 14
6.002 Circuits and Electronics, Spring 2007

I will be replacing Professor Agarwal today because he is away. I am one of the recitation instructors for those of you who have not seen me. We will talk today about a neat application of RC networks and expand those to application in MOS memory systems.

To connect with everything, we will get back to the basic circuit that we have been discussing so far. And you recall the circuit that we have been studying, the canonical RC with an input voltage function of t.

And we had specified that we solved this problem for the case of a step input or a condition in which a t=0. At t greater or equal to zero vI is equal to some capital VI value that for now on is constant.

And the other condition that we discussed was the value of the voltage on the capacitor that would exist at time t=0. Let's call that vc(0). And in general there is some finite value here. It can be zero or it can be different from zero.

Given that, we learned how to write down directly, without messing around with differential equations, the answer for the voltage on the capacitor vc(t), let me define also my vc right here, is equal to VI, the final value, plus vc(0), the initial value on the capacitor, minus the final value, e^–t/RC.

This is our standard equation to which we plug in, and it's either a rising exponential if VI is larger than VC or a decaying exponential if VI is a smaller value than VC. This should all be familiar.

And, again, as pointed out in the notes, the reading for today is 10.3 and for the new material you should look at Chapter 11 where we discuss memory. This is where we stood as of last time. Now, I would like to discuss a little bit more about the storage of charge in capacitors.

And how we can take advantage of that for storing logic state. One of the things that I am sure you must be aware of is that one of the perhaps most massively produced chips is actually the so-called DRAM which you find in every PC and every computer that exists anywhere.

This DRAM is dynamic random access memory in which we can store a state and come back and look at it at any time later, provided we don't power off our machine. The logic state in the basic memory elements, of which instead there are close to 1 giga elements per chip, are stored on capacitors.

And so we will play a little bit with that concept today. And, although we're not going to discuss the specific example of the DRAM, the basic elements of the DRAM you will see actually in a demo shortly.

So that's the general response of this network that I have here to an input VI that happens at t=0. Now, the one thing that you recognize immediately is that it really doesn't matter what the value of VI was for t less than zero.

What really counts is the value of VI at t=0. And that's the value that we're interested in. Now, there is an implicit statement in that. And that statement is that somehow that network appears like this at t=0.

So, there has to be some switch there, and you will see that, that basically starts my condition to that at t=0. And so the history of VI really doesn't matter. The response following that equation that we have there will depend on the initial value which is vc(0) here.

Now that is the voltage on the capacitor at that time. And then assuming that VI is a value that is larger than vc(0) will have a rising exponential that will come to this value. And this is the time constant RC and this is time.

So, the capacitor starts with some voltage here and goes to a new voltage that is imposed by the input for time greater than zero. We can define at any one time, say this time, this time, this time, this time the state of the capacitor.

The state. What is the state of the capacitor? The state is the summary of all inputs that are relevant to predicting the future. If I know the state of the capacitor this time, I can predict what it is going to go given a response VI here in the future.

So, predicts the future. Now, what is the state variable on the capacitor? What is actually stored on the capacitor? You can say, well, what is stored is voltage. The real physical quantity that is stored is the charge q which is for linear capacitors related to the voltage, let me actually write it correctly, vc like this.

So, the real state variable is this. But for a linear capacitor, since there is one-to-one relationship between the two, v is also a state variable. OK, so let's then go back to our original circuit.

What we have is -- -- vc(t), so that's the future value of the voltage on the capacitor, is a function of vc(0), the initial value and the variable input now in the future time. And for the case of vI(t) being constant VI for t greater or equal than zero we have the equation that we just described.

Nothing new. All the past inputs to the capacitor for time t less or equal to zero is summarized in this value. And vi being constant the future is predicted from that. So, that's the concept of the state.

There is an initial state on the capacitor. And then there is a final state that will be reached when equilibrium actually is achieved. There is a fair amount of discussion in the text, and we don't go in great detail here, but it is both convenient for analysis and also it's interesting in many cases to look at the response of a linear network for two different conditions.

So, we're interested in two cases. One is the so-called zero state response. Now, what is the zero state response? It's the response to a condition in which we impose an input and impose also the condition that the initial value, initial state of the capacitor is zero.

So then we ask how does it respond to vi(t)? So, starting with a capacitor at zero state what is the response? It allows us to decouple the initial conditions from the response to the input. Now, you will see that this is actually very useful.

The second condition to which we're also very interested is the so-called zero input response. What is that? That is vi(t)=0. Now, it's the condition under which there is no input. vi(t)=0. The question here is how does it relax? We're starting with an initial state.

So, how this state relaxes out in the circuit. Now, the zero state response, this one here is Z so called SR for our case, which I will write like this, vC, ZSR is simply a rising exponential. We start from zero and we go to VI.

So, it's VI-VI e^–t/RC. So, that's the ZSR. The ZIR, the zero input response is like this. It's the decay of the initial voltage on the capacitor to zero or to equilibrium. Starting from vC(0) we're decaying like this.

Now, do you see something that's rather obvious from what's on the board in terms of ZIR and ZSR and the final complete answer which is there? They are specific cases, but how do they relate to the full answer? It's the sum.

It's the superposition of the two. What basically we see here -- And that's actually a general statement, is that vC = vC,ZSR + vC,ZIR. Now, you may say this is trivial because we started from that, ended back in that from some very simple observations.

However, we are not always solving networks for responses that are steps. The input voltage may be a ramp. We did that in recitation. Or, it could be an impulse. Or, it can be a more complicated function.

Having this observation in place actually allows us to solve the problem rather neatly. If I have time at the end, I might come back to this. So, this is the same equation as I started with, arrived at from a principle of superposition of two different solutions.

One application of state which can be, since we have energy storage element here, the capacitor, which can be stored on the capacitor is in memory. And you may ask, so why do we need a memory node to perform logic? Well, there are cases in which a result depends on previous results.

So, a computation proceeds in time. In order to do that, we need to store intermediate results and proceed forward. One good example is if you're doing a continuous summation, say, on your calculator, you keep putting things in the memory.

The M+ button, right? And you keep adding a series of numbers. Every time we store the sum of the previous operation we add another number and so on. Clearly we need some way of storing state. For a complete computing system, we need combinational logic and we need memory.

In fact, these are the two basic elements that are essential for any kind of computing system. We need to remember intermediate results. We need to remember transient inputs. And that's the role that all these enormous amount of memory that comes to play in computers is doing.

The basic memory abstraction is as follows. Imagine a block which needs to be populated by transistor, resistor, capacitor, whatever elements. And it has a control input, which we will call the store.

It has a state input that we will call dIN and has an output dOUT. When we're telling this element, OK, now it's time to store, it looks at the input dIN and stores it for, in principle, an infinite amount of time.

If we were to make a drawing of this, of what this looks like, let's suppose, let me do all this in one axis. So, time moves this way. Let's suppose that we have an input dIN that looks like this, and the store command comes in the form of a logic.

Let's actually suggest here this is logic one, this is logic zero. And, although this is not absolutely necessary, let's also define that the store command comes in the form of a logic one at this store input.

Store, let's say, looks like this. What does the output look like then in this particular case? Assuming that the output was dOUT, the stored element was zero prior to the store, then the output would look like this.

This is dOUT. As you can see, it would remember the one that it saw at this point. In fact, it would do that irrespective of what was stored in this memory cell. For example, suppose it was storing one and the output didn't change, it's still one.

If it was storing a zero, it would flip to a one. If we had another store, let's say here, then what happens? Then it would go back down to zero because now we sampled an input that is zero and we flipped the state.

That's what a memory -- -- element or cell would do for us. It would remember the output state. And, not only that, but in principle it should be undisturbable. In other words, I may do something to this dOUT but it should not flip the state.

And that comes about quite a bit. Because in actual integrated circuit memory there is lots and lots and lots of nearest neighbors to this cell which, when they're flipped, have a cross-coupling to the cell.

The cell must be designed robust enough that it doesn't flip, that no coupling actually occurs. All right. Now we're going to try to apply what we've learned so far to invent a basic memory element.

And, believe it or not, this is the key to the DRAM. Let's implement this in a circuit. Suppose I have a switch here like this. And I will put a capacitor. I take my dOUT here. This is dIN. And the switch is operated by a command here that we will call store.

When store is one it goes up. When store is zero it is down here. That's capacitor C. This is the storage node. What are we actually storing in this case? Let's suppose that this voltage here is 5 volts.

I flip the switch up to one and I flip it back down to zero. What's the voltage in this capacitor here? 5 volts. Now the capacitor is at 5 volts, I put dIN to ground, flip the switch back up and then back down to its known storing condition.

What's the voltage in the capacitor? It's zero, exactly. So, it does store the value of the voltage that it saw, five or zero, high and low. It stores it because it stores charge. That's actually the physical quantity that's stored.

It's manifested as a voltage, which we see. All right. Now, is this, oh, before I move from here. What is the basic cell in a DRAM, one that you go out and buy by the billions of cells? It's actually this.

The only difference is that this switch here is replaced with a MOSFET. And that's all it is. So, a MOSFET plays the role of the switch. When the gate is high this is a resistor and connects the input to the capacitor.

And when the gate voltage is below the threshold voltage this is an open, as we've seen, and it isolates the transistor from the output. So, that's the basic memory element. And, as I said, it's the key to a DRAM.

OK. Now let's consider a little bit the conditions of operation of this thing. Let me draw the circuit in two conditions. One in which it is storing, one in which it is sampling and one in which it is storing.

Not to redraw this thing. Assuming that I have a MOSFET there, I would have the on resistance in place here when store=1. Now, in principle, the output is connected to -- -- some load resistance. We'll talk a little bit more about this load resistance in a minute.

This is the situation when we are at store=1 situation. For example, let's suppose that dIN is 5 volts. Now, what is the situation for store=0? It's very simple. We have the capacitor C and dOUT and here we have a resistance.

The switch is open. This is store=0 condition. What we have in this case is we have a problem similar to what I was discussing earlier. It is a ZIR, if you like, situation. And this you can think of as a ZSR if we're starting with zero charge on the capacitor, but I'm interested in this part.

In this case, I am starting with a vC(0)=5 volts. And I'm asking myself how long will this cell hold the value? And, in fact, that is actually what happens in a dynamic RAM. The value on the capacitor is not stored forever.

In fact, that's why we call it dynamic because we have to come back and restore it every once in a while. For how long are we going to store the charge? What's the response of vc for t greater than zero after the switch flicked? It's very simple.

It's vc is equal to 5 volts e to the minus t over RC, right? That's the response. We have a decay. And applying to the things we know. We start from 5 volts, let's say here, I have a decay going down towards zero, at some point we are going to cross the threshold for high.

The only period in which I have a valid output, if the capacitor was storing a one, is this period here. This is the only period in which I have valid stored one because, once I go beyond capital T here, I have crossed the legal limit, threshold for discriminating a high output.

And from then on the output is no longer valid. So, this memory is good provided time is less than capital T. It's not a case in which the capacitor can hold charge forever. In fact, we can calculate, that is we can solve for T in this particular case.

It's in your notes. Nothing really profound. T is equal to minus RC log VOH over 5 volts. So, this is basically what the response is going to be. Now, there is an implicit assumption here, which is that the store pulse width is much, much larger than RON C.

In other words, when we want to store a one here starting from zero, we better charge it all the way up to 5 volts in the time that our switch is connected here. And what is the relevant time constant? It's going to be the RON C.

In fact, it's actually the RON parallel RL with C. But typically RON is much, much less than RL so we don't have to worry about that. Dominant time constant is RON C. So, provided these things are happening, we have a memory.

Now, we can try to improve things a little bit. We see here that we will have a decay to an invalid state in time T. How can we improve things? One way to improve things are the buffer. Here is our memory element again.

Here is the capacitor. This is the storing node. Now I am going to put the buffering effect. I am going to put two buffers here. Two invertors, I should say, because if I am storing a one here I want to be able to see a one here as well.

And, in this case, what I am looking at is the RIN of the buffer. And, in principle, I have out here the RL. Now, this is better because if RIN is much larger than RL then the time T, in this case, is much larger than the case without buffer.

So, we buffer the effect of VL. This could be one of these neat circuits we saw in recitation like a source faller, for example, or it can be just an inverter in which case you just see the input of a transistor.

So, now this condition can be satisfied. Let me give you some cases which are some numbers that are typical for a dynamic RAM. Typical times we're talking about is RIN on order of 1 gigaohm and storage node capacitor on order of 1 femtofarad to one picofarad.

Now, if you can do the math in your head, which is just multiplication, you will see that the time constant, the RC is between 1 millisecond to 1 microsecond. And for DRAMs, actually, we try to be in the order of milliseconds.

These are the times we're talking about. If I have this kind of circuit, somehow there has got to be additional circuitry that comes back, samples the voltage here and restores it. And that is actually what is happening in a DRAM.

And my laptop is working there and its DRAM keeps getting refreshed every, say, millisecond or whatever the condition is. But, in our case, we are going to do a slightly different case in which we will create a static memory.

Let's actually look at, first of all, the case of the discharge. Pay attention to, let me actually break the loop here. This is my capacitor. This is a resistor that is in series with a capacitor like you see here.

Actually, I am going to keep that resistor in series with the capacitor, even in this case, because I have it for my second part of my example. I charge the capacitor to 5 volts. And you can see here this lights up, I hope everybody can see it, proportional to the voltage that I have here.

From here on it's all logic levels. So, the intensity of light here will always be the same. It's either lit or it's not lit. Right now I am charging the capacitor. In fact, let's see. Maybe I can discharge the capacitor first.

Here the capacitor is discharged. As you can see, the input is zero, the output is a one, and then the output of this inverter here is a one. I have two inverters in series. And I am going to charge the capacitor.

I charged it to 5 volts and this lit up, this is off of course, that's an inverter, this is a valid zero, produce a valid one. And now I am going to take the input out. As you can see it's stored. In fact, we have to wait for a very long time.

We don't have enough time to wait for this to discharge, so instead what I am going to do now is I am going to add also the resistor. Now I am going to flip the resistor in parallel with the capacitor to imitate what happens when we have an input resistance.

You saw that there was a discharge of the capacitor. This input level went down. Voltage here flipped over to a one. Let me do it again now with a resistor in place. Storing charge on the capacitor.

That's the store command. Now, don't store. I have less, about a second. The element here is 20,000 microfarads and 100 ohms which gives me a time constant of two seconds. Assuming a VOC of the order of, let's say, I don't know what it is for this case, 2.5, the log would be about 0.5, so it cuts basically the time to about one.

So, it lasts about one second, if my math is all correct. It's actually a little longer than a second, excuse me, but the point is that the charge is gone. Now, notice, however, that there is something I can do here, which is that suppose I take the switch or a switch and bring it back and provide a path from the output to the input here.

And this switch is open when this is closed and closed when this is open. So, this basically is the compliment of store. What I am doing now is I put a charge here, it produces a valid one at this point, and then I am feeding this valid one back to the input.

As you can see, this will now allow me, even though I have a high resistance, to store the value for a long time. In this case, what I am going to do is I am going to connect the output, as you can see here.

And I have my resistor in. And I am storing zero here, storing 5 volts. Now I am going to flip the switch. Basically, I mean the don't store, don't look case. You notice this dims a little bit. Sorry.

No, I want the resistor in. There. Yes. OK, so the output remain value. This dimmed a little bit but the output has remained OK. All right. So, we've provided a feedback. Now we've created a static memory.

This will hold charge for as long as the circuit is powered up. Now, there is still one little problem that I have with this kind of configuration. And that is if I disturb this output the charge may, the state may change.

So, for example, let's say that I have -- I disturbed it by coming close to it, so let's charge it again. OK. I flipped the switch. I flipped the state from the output. That is an invalid condition.

I shouldn't be able to do that. How do I avoid that? How can I avoid this problem that you just saw? Well, I need yet another buffer. The answer is in your notes. If I don't take the output here but rather take the output here, or if I don't want an inverted output, if I don't want an inverted output, I could put yet another element there.

Then the situation would be fine. In this case, let me do it again. Charge. Why isn't this lit? A bad one? Now, of course we disturbed the input. Now, of course I can do anything I want here. Nothing happens, but you may say this is a trivial case because this is already zero.

So, I am going to change the state. Here's is the changed state. See. I can show this. Nothing happens up there. So, this is an interesting situation in which I am buffering the output so that the output does not feed back to the input.

And, by and large, in designing circuits this is something that we do. Now, in the remaining three minutes there is an example that we have. Can we put the laptop here? OK, so here is an example of how memory can be put together now to create something a little bit more complicated.

And you can see the memory cells that we were discussing here. There's four of them, so this is a four bit memory. There is a decoder at the beginning here which decodes the address of each cell, so the input here will tell me which cell I need to address.

Let's look at the truth table. This is the truth table for the decoder. As you can see, depending on the address that I have here, this is zero, one, two and three in a binary system, only A, B, C or D is up, is high.

Which means that this end operation here only allows the input that is presented to all of the cells, what is going through the AND gate here to appear at the output. If, for example, we have a one, zero, the only end input that is going to be high is going to be this one.

And that means the only cell that will look at the input when the store comes up is going to be this one here. At that point it will store whatever is on the input cell because that's an AND operation.

That is a simple example of a memory. And following that simple arrangement you can build incredibly large memory systems. So, that's all I had for today. And I will see you on Tuesday.

Transcript - Lecture 15a
6.002 Circuits and Electronics, Spring 2007

All right. Good morning, all. So we take another big step forward today and get onto a new plane of understanding, if you will. In the last week and a half, our focus was on the storage element or storage elements called inductors and capacitors.

And capacitors stored change and inductors essentially stored energy in the field, the magnetic flux. And the state variable for an inductor was the current while that for a capacitor was the capacitor voltage.

We also looked at circuits containing a single storage element, we looked at RC circuits and we also looked at circuits containing a single inductor. And this was a single inductor with a resistor and a current source or a voltage source and so on.

What we are going to do today is do what are called "second-order systems". So they are on the next plane now. And with this second-order of systems, they are characterized by circuits containing two independent storage elements.

They could be an inductor and a capacitor or two independent capacitors. And you will see towards the end what I mean by two independent capacitors. If I have two capacitors in parallel, they can be represented as a single equivalent capacitor so that doesn't count.

It has to be two independent energy storage elements and resistors and voltage sources and so on. And what we end up getting is what is called "second-order dynamics". And much as first order circuits were represented using first order differential equations, this kind you end up getting second-order differential equations.

Before we go into this, I would like to start motivating this and give you one example of why this is important to study. There are many, many examples but I will give you one. What I would like to do is draw your attention to our good old inverter driving a second inverter.

The same circuit that we used to motivate RC studies, one inverter driving another. So let me draw the circuit. Here is one inverter. This is, let's say, 5 volts and this is, let's say, 2 kilo ohms.

And I connect the output of this inverter to a second inverter. And what we saw in the last few lectures was that in this specific example there was a parasitic capacitor or a capacitor associated with the gate of this MOSFET.

And that could be modeled by sticking a capacitor CGS between the gate of the MOSFET and ground. And we saw that the waveforms here, if I had some kind of step here. Let's say, for example, a step that went from high to low.

Then out here I would have a transition that instead of going up rapidly like this would transition a little bit more slowly. And this transition was characterized by an RC time constant. And this is what gave rise to a delay in the eventual output.

So that is what we saw previously, single energy storage element. Today what we are going to do is we are going to look at the same circuit, the exact same circuit, and have some fun with it. What we are going to say is look, this thing is pretty slow, so what I would like to do is -- why don't we go ahead and put that up.

What we are going to see is that the yellow waveform is the waveform at the input here. And the green waveform here is the waveform at this intermediate node. And notice that this waveform here is characterized by the slowly rising characteristics that are typical of an RC circuit.

There are some other weirdnesses and so on going on here like a little bump and stuff like that. You can ignore all of that for now. It happens because of certain other very subtle circuit effects that you won't be dealing with, called Miller effects and so on that you won't be dealing with in 6.002.

So focus then on this part here. It is pretty slow. And because of that slow rising, I get a very slow transition and I get some delay in my inverter. So you say ah-ha, we learned about this in 6.002, I can make it go faster.

How can you make the circuit go faster? What could you do? This is rising very slowly. How can you make it go faster? Anybody? You have multiple choices, actually. What are your choices here? Pardon.

Decrease the time constant. And how would you decrease the time constant? The capacitance is connected to this MOSFET gate here. I didn't want it in the first place but it is there, I cannot help it, so I can decrease the resistance.

Good. Let me go ahead and do that. What I will do is I am going to knock this sucker out and stick in a new resistance that is say 50 ohms, a much smaller resistance. That should speed things up, right? That should make things go much faster because this is a smaller time constant because R is smaller, correct? OK, let's go do it.

And let's see if we get what we expect. I have a little switch here. And using that switch, I am going to switch in this little resistance. Whoa, what on earth is happening out there? This is so much fun.

What I did is I switched in a small resister here to decrease the time constant, but it looks like I got a whole bunch of crapola that I did not bargain for. This is certainly very fast, it goes up really fast, but I am not sure where it is going, though.

Let's stare at that a little while longer. Let me expand the time scale for you. Look at this. Instead of a nice little smooth thing going up. I get something that looks like this. It looks something like a sinusoid.

It looks sinusoidal, but then it is a sinusoid that kind of gives up and kind of gets tired and kind of goes away. Right? It kind of dies out. So nothing that you have learned so far has prepared you for this.

And, trust me, when I first did some circuit designs myself a long, long time ago I got nailed by that. I looked at my circuit, and what ended up happening was I was noticing these sharp lines at all my transitions.

When I looked at my scope, I expected to see nice little square waves but I saw these little nasty spikes sitting out there. And then when I stared at it more carefully, those spikes were really sinusoids that seemed to kind of get tired and kind of go away.

So those are nasty, those are real and they happen all the time. And what we will do today is try to get into that and understand why that is the case. We will understand how to design that away. And that is a real problem, by the way.

And the reason that is a real problem is the following. Look at this. Look down here. Because this intermediate voltage is meandering all over the countryside here, at this particular point the intermediate voltage dips quite low.

And because it dips quite low look at the output. The output has a bump here. And it is quite possible for this output bump to now go into the forbidden region. Or worse. If this swing here was higher, this could have actually gone onto a one, so I would have gotten a false one pulse here.

Instead of having a nice one to zero transition, I would have gotten a one to zero, oh, back to one, oh, back to zero and then back down to zero. So this is nasty stuff, really, really nasty stuff.

What we will do is understand why that is the case today and see if we can explain it. What is going on here? What is really going on here is take a look at this circuit here. I will take a look at this path here.

So this is your VS voltage source. Path kind of goes like this and around. It turns out that this circuit is a loop here. And when there is current flow, going down to basic physics you remember that I also enclose some amount.

So there is a current flowing in a loop. And because of that there is an effective inductance here. And, in fact, any current flowing through a wire above a ground plane, for that matter, can be characterized by the inductance.

So I can model that by sticking a little inductor here. So my real circuit is not exactly a resistor and a capacitor, but my real circuit is an inductor as well that comes into play because of this wire.

Every wire, when there is a current flow, has an inductance associated with it. And because of that the real circuit is resistor, inductor and capacitor. So I end up with two storage elements now, and the dynamics of that are very different from that with a single storage element.

That is just a bit of motivation for why our study of inductors is important. And I can draw a quick circuit here. If you look at the circuit, start from ground, the voltage VS and there is a resistor here.

And then I have an inductor and then I have a capacitor. So it is a voltage source, resistor, inductor and capacitor. For this whole week we will be looking at circuits like this. Today what I would like to do is start very simple, start with the simplest possible form of this so that you can begin building up your insight and then go into more complicated cases.

Today what I will do is simply begin with a case where I don't have a resistor here and simply study a voltage source, an inductor and a capacitor and understand what the voltage looks like out here.

So we look at the dynamics of a little system like this. Before we go on, I want to caution you about something. It is just happenstance that I have introduced for you capacitors based on the parasitic capacitance here and inductance based on parasitic inductance.

I would hate to leave you with the impression that inductors and capacitors are "bad". Because when you think of a parasitic, you know, parasites. These are parasitic. You didn't expect them there, didn't expect this here and we got the weird behavior.

So parasitics have a bad connotation to them. I do not want to leave you with a bad taste in your mouth about capacitors and inductors that these are just bad things. We just have to deal with them and deal with second-order differential equations and all that stuff because they're just bad stuff and we just have to deal with them.

I don't want you to end up going through life hating capacitors and inductors. Just because of my choice of examples, it just happened to be introducing them as capacitors. I want to point out that these are fundamental lumped elements in their own right.

They are very, incredibly important and useful circuits where we designed capacitors and inductors because we want to have them in there. There are many circuits that we will look at where we really want the inductor in there.

We will design an inductor by wrapping wire around in a coil and get bigger inductances and so. Just remember that this can be parasitic in some cases, but in many cases it's good, inductors are good, so just stick with that thought.

These are mostly good so don't go around hating them. All right. Let's go on and analyze a basic circuit like this. And what I would like to cover in the next hour are the foundations of something like that.

I will take you through the foundations so you understand how it works. And, as always, what I am going to end up with is build up the foundations, help you understand why we got where we were and then help you build intuition.

And then show you a really, really simple intuitive way of doing things in terms of how experts do it. And the real cool thing about EECS is that the way experts do things, things are really, really very simple in the end.

But you need to build up some intuition to get there. So our circuit looks like this in terms of my two storage elements. I have a voltage vI, inductor L, capacitor C and I am going to look at the voltage across the capacitor and my current through the capacitor.

So v(t) is the voltage across the capacitor and my current is the current through this loop here, which is the same as the current through the capacitor or the current through the inductor. And we are going to proceed in exactly the same manner as we did for first order differential equations, write the equations down and just boom, boom, boom, boom, go down the same sets of steps but just get to some place different.

We are going to start by writing a node equation for this node here. That's the only node for which I have an unknown voltage. The node here is vI, so I need to find this, there's just one unknown node voltage.

And I am going to need some element laws. For the capacitor I know the iV relation is given by the i for the capacitor is Cdv/dt. And just to show the capacitor I am just calling it dvc/dt. Similarly, for an inductor, L, the voltage across the inductor is given by Ldi/dt.

So this is the vI relation for the capacitor, the vI relation for an inductor. It also suits us to write this in an integral form. So if I integrate both sides of this equation and I bring L down to this side, I end up getting something like this, 1/L minus infinity to t, VLdt, and that is simply iL.

I am just simply replacing this with an integral form. So this is a VI relationship for the inductor and this is for the capacitor. So let me now go ahead and apply the node method for my circuit here.

Here, for the node method, I have to equate the currents coming into the node or sum the currents coming into the node and equate that to zero. And while I do that I simply replace the currents by the corresponding voltages using the element laws.

So what do I get? I get the current going in here to the inductor is equal to the current going through the capacitor. What is the current going the capacitor? In terms of its v relationship it is Cdv/dt.

And the current going to the inductor is given by this relation here, which is simply 1/L minus infinity to t. The voltage across the capacitor is simply (vI-v)dt. I have just written down the node quotation for this node here.

Now I will just apply a bit of math and simplify it and get the resulting equation. What I can do is simply differentiate with respect to t here. And get this to be Cd^2v/dt^2, the second derivative of v.

And here what I end up getting is 1/L(vI-v). So I just differentiated the whole thing by d/dt here. And then I just move L up here. I bring d^2v/dt^2 out here. And then I get a minus v here, and that will be equal to, oh, I'm sorry.

Let me leave this here. Bring the minus v to this side so it becomes a plus and leave vI on this side. So I end up getting LCd^2v/dt^2. I bring L up here. And then I take v to the other side. Plus v and leave vI here so I get vI.

That is second order differential equation that governs the characteristics of the voltage, v. So much as the voltage across the capacitor was a state variable in our RC circuits or the current through the inductor was a state variable in our RL circuits, out here both the current through the inductor and the voltage across the capacitor are my two state variables.

And so here I have a second-order equation in my voltage, v. Again, going through the foundations here, I am now going to go through a bunch of math. Up to here it was circuit analysis, and now I am just going to do math.

For the next three or four blackboards just math. You can solve this second-order differential equation any which way you want. But just to keep things as simple as possible, in 6.002 I solve all the differential equations, it turns out we are fortunate enough we can do that, using the exact same method again and again and again, the same thing can be applied.

And the method that we use to solve it is the method of homogenous and particular solutions. So the first step we are going to find the particular solution, vP. Second step we find the homogenous solution, vH.

And the third step we are going to find the total solution as the sum of, v is simply the particular plus the homogenous solution and then solve for constants based on the initial conditions and the applied voltage.

So let's write down initial conditions. Let's assume, for simplicity, that my initial conditions are simply the voltage across the capacitor is zero to begin and the current through my inductor is also zero as I begin life.

Now, this is what is called "zero state". v and i are both zero, and so the response of my circuit for some input is going to be called ZSR. You've probably heard this term in one of your recitations.

So zero state response simply says I start with my circuit at rest and looks at how it behaves for some given input. That is a little term you may end up using. My input next. I am going to use the following input.

vI of t is going to be a step, is going to look like this. My input is at t=0 v is going from zero to some voltage VI and then stay at that voltage. It is going to be a step. Kaboom. And you can see why I am going with this set of variables, because I want make this situation as close as possible to the funny behavior we observed there.

Remember we had a step, and because of the step we had some behavior at that node? So I will try to bring you as close to that. In tomorrow's lecture, I am going to close the loop around that and derive for you exactly the behavior we saw on the scope.

And to get there I am going to be try to be as close as possible to the constants and other parameters in the demo. So VI is a step and zero state. Just in terms of notation, this kind of a step input occurs pretty frequently.

And we just have a special notation for it. We simply call it VI is the final value here. And we call it u(t). So VIu(t), u(t) simply represents a step at time t=0, steps from zero volts to VI. That is just a little more notation that will come in handy at some point.

More math now. Three steps, particular solution, homogenous solution, total solution/constants. This is almost like a mantra here, like a chorus. Homogenous solution we compute using a four-step method.

And four-step method for homogenous solutions, it turns out that it happens to be that way for all the equations we will see in our course. The first step would be assume a solution of the form Ae^st.

Exactly as with RCs. If you close your eyes and do exactly what you did for RCs you will get to where you want to be. You assume a solution of the form Ae^st. Substitute that into your homogenous equation.

Obtain the characteristic equation. Solve for the roots. And then write down your homogenous solution. Same sort of steps again and again and again until you get bored to tears. Particular solution.

For the particular solution, I simply need to find a solution, any solution, if not the most general one but any solution that satisfies the particular equation which satisfies that equation. LCd^2vP/dt^2+vP=VI.

My input is a step and I am going to look for the solution for time t greater than zero. Notice that for time t less than or equal to zero, v is going to be zero. So I am looking for a solution greater than t=0.

Here, if I substitute vP=VI, that is a particular solution. Because if I substitute VI here this goes to zero and then I get VI=VI, so this works. I promised you this was going to be simple. You cannot get any simpler than that.

I have done my first step. I found the particular solution. And VI is a good enough particular solution so I will use it, I will take it. As my second step I am going to find vH or the solution to the homogenous equation.

And the homogenous equation is simply that equation with drive set to zero. What I get here is LCd^2vH/dt^2+vH=0. That is my homogenous equation. I simply set the drive to be zero. And to find the solution here, I go through my four-step method.

Again, in 6.002 following the kind of Occam's principle, we just show you the absolute minimum necessary to get to where you want. The absolute minimum necessary is it turns out that we can solve all our differential equations that we use here by using the methods of homogenous and particular solutions.

And every homogenous solution can be solved by a four-step method. That is about as minimal as it can get. So no extraneous stuff there. The four-step method, four steps. The first step is assume a solution of the form vH=Ae^st.

What I have noticed is that students starting out are usually scared of differential equations. I know I was when I was a student. And the trick with differential equations is that it is all a matter of psych.

Just because you see some squigglies and squagglies and a bunch of math and so on you say oh, that must be hard. But differential equations are actually the simplest thing there is because in a large majority of cases the way you solve them is you assume you know the answer, someone tells you the answer.

And then all you are left to do is shove the answer into the equation and find out the constants that makes it the answer. Just a matter of psych. Psych yourselves that this stuff is easy, because I am telling you what the solution is.

All you have to do is substitute and verify. If you think about differential equations that way or a large majority of them, it really is very simple if you can just get past the squigglies here. Just get past the squigglies and then just simply stick in some simple stuff and it works.

I mean it just cannot get any easier. I cannot think of any other field where the way you find a solution is assume you know the solution and stick it in. It has never made any sense to me but that is how it is.

So we assume the solution to the form Ae^st, you stick it in there, and you have to find out the A and s that make it so. It cannot get any simpler than that. Let's stick the sucker in here and see what we can get.

Substitute Ae^st here I get LCA, and second derivative, so it's s^2 e^st. And Ae^st on this one here. And that equals zero. And then let me just solve for whatever I can find. Assuming I don't take the trivial case A=0, I cancel these guys out.

And what I am left with is simply LCs^2+1=0. In other words, what I end up getting is B, s^2=-1/LC. My first step was, I am giving you solutions, stick them in there, assume a solution of this form.

Second step is get the characteristic equation. And the way you get the characteristic equation is that you simply stick this guy in there. And what you end up getting is some equation in s^2. Do you remember what you got for first order circuits? What s was? What is s? For first order circuits, what did you get as a characteristic equation? s+1/RC=0.

The same thing. Just remember to blindly apply the steps. It will lead you to the answer. This is called the "characteristic equation". This is incredibly important. You will see in about a couple weeks from now that once you write the characteristic equation down for a circuit, it tells you all there is to know about the circuit.

And often times you can stop solving right here. To experienced circuit designers this tells me everything there is to know. This is really key. That's why it's called a characteristic equation. I believe in problem number three of the homework that will be coming out this week, that is exactly what you are going to do.

I am going to give you a circuit, ask you to get to the characteristic equation quickly and then from there intuit the solution. Write the characteristic equation and then just intuit solution, it's that simple.

So, step A, assume a solution of the form, step B, write the characteristic equation down. And let me just simplify that a little bit. I go ahead and find my roots. And my roots here, remember that j is the square root of minus one.

And so what I end up getting is, my two roots here are, plus j square root of 1/LC and minus j square root of 1/LC. Two roots. And just as a shorthand notation, much like I had a shorthand notation for RC, what was my shorthand notation for RC? Tau.

Just as tau was big in first order, we have a corresponding thing that is big in second order and that is omega nought. Omega nought is simply square root 1/LC. Just as tau was RC, omega nought is a shorthand here.

And so s is simply plus or minus j omega nought. Notice that in this equation here, if you take the square root of LC there that has units of time, so one divided by that has units of frequency. Notice that this guy is a frequency in radians.

I end up getting my roots of the homogenous equation, and that is my third step. And as my fourth step, I simply write down the homogenous solution as substituting s with its roots and writing the most general possible form of the solution, and that would be A1e^(j omega nought t)+A2e^(-j omega nought t).

Done. Some constant times this solution plus some other constant times, the other solution. Plus zero omega nought. Remember it comes from here, Ae^st. I assume the solution of this form, so my solution in this most general case would be s being j omega nought in one case, minus j omega nought in the other case, and I sum the two to get the most general solution.

So blasting ahead. I now have my homogenous solution. And as my third step of solution to differential equations I write down the total solution, v=vP+vH, particular plus the homogenous solutions.

And v=VI, was my particular solution, +A1e^(j omega nought t)+A2e^(-j omega nought t) is my complete solution. The final step, write down the total solution and find the constants from the initial conditions.

To find the constants from the initial conditions, let's start with, the voltage is zero to begin with. This equation governs the characteristics of v, so I need to find the initial conditions. First of all, I know that know that v(0)=0.

From there I substitute t=0. And so this goes to one, this goes to one, and I end up getting 0=VI+A1+A2. That is my first expression. And then I am also given that i(0)=0. And so I can get that as well.

How do I get i? This is v. I know that i=Cdv/dt, so I can get i by simply multiplying by C and differentiating this with respect to t. I get C, this guy vanishes so I get d/dt of this. So it is CA1(j omega nought) e^(j omega nought t)+CA2(-j omega nought)e^(-j omega nought t).

From here I am given that that is zero, and so therefore this guy becomes a one, this guy becomes a one, j omega nought, j omega nought cancel out. What I end up getting is A1=A2. From the second initial condition I get A1=A2.

From these two, if I substitute here for A2, I get VI + 2A1 = 0, or A1=-VI/2. That is also equal to A2. Therefore, my total solution now can be written in terms of the actual values of the constants I have obtained.

I get VI-VI/2. So A1 and A2 are equal. I just pull them outside. I pull VI-2 outside and I stick these two guys in parenthesis in. Again, I promised you no more circuits from here on until the very last board or something like that.

It is all math, so not much else happening there. More math. If you would like, I could skip all the way to the end and show you the answer. But I just love to write equations on the board so let me just go through that.

I am going to simplify this a little further here. And we should remember this form by the Euler relation, ejx=cos x+j sin x. And by the same token, (e^jx + e^-jx)/2=cos x. You all should know this from the Euler relation.

So were are using this guy here, ej^x + e^-jx=2cos x. And so this one is 2 cosine of omega nought t, 2 and 2 cancel out, and what I am left with is v(t)=VI-VI cos( omega nought t). And the current is Cdv/dt, which is simply CVI sin( omega nought t).

Just remember that omega nought is the square root of 1/LC. We are done. In fact, I did not give that answer the importance that was due so let me just draw. There. That is better. Enough math. In a nutshell, what did we do.

We wrote the node method, it's a very simple circuit, to write down the equation governing that circuit. And then we grunged through a bunch of math. Not a whole lot here. It is pretty simple. And ended up with a relation that says the voltage across the capacitor for a step input, assuming zero state, is a constant VI-VI cos omega t.

Notice that even though I have a step input, the circuit dynamics are such that I get a cosine in there. You can begin to see where these cosines are coming from now. They come in here. And if you recall the example I showed you earlier of the inverter circuit, remember there was a cosine that decayed, that was sort of losing energy and kind of dying out? So you can see where the cosines are coming from.

And just to draw you a little sketch here. Let me draw v and i for you and let me plot omega t, pi/2, pi and so on. Let me plot VI. When time t=0, VI=0, cosine omega t is one, and so VI-VI=0. That is simply a cosine that starts out at zero here, and at pi I get cosine omega t is minus one, so I get plus VI on the other side.

So I end up at +2VI. At this point the voltage is here. And notice that this guy looks like this. It is a cosine that is translated up so that its mean value is not zero but VI. It is just a translation up of a cosine.

Similarly, in this case for the current it is a sinusoidal characteristic. And it looks something like this where the peak is given by CVI, oh, I messed up. When I differentiated this is missed the omega nought out there.

What I would like to do now -- This is the form of the output for a step input. What I would like to do next is show you a demo. But before I show you a demo, I always found it strange that I have a step input and then I have two little elements, how can I get a sine coming out of the output? I would like to get some intuition as to why things behave the way they are.

I could go and pray to find out, but let me just give you some very basic insight as to why this behaves the way it does. Let me draw the circuit for you here. And this is my inductor L and capacitance C.

Remember this is v. Let me just walk you through what is happening there and get you to understand this. Now, you have seen sines occur before. If you go and write down the equation of motion of a pendulum, you know, you have a pendulum, you move it to one side, let go.

It is also governed by sinusoidal characteristics. And you will find that the equation governing its motion is very much of the same form, and you get the sinusoid where you have energy that is sloshing back and forth between maximum potential energy to maximum kinetic energy and zero potential energy back to maximum potential energy, zero kinetic.

So it is energy sloshing back and forth. The same way here. Capacitors and inductors store energy. Let's walk through and see what happens. I start off with both of them having the stage zero, zero current, zero voltage.

I apply a step here. Boom, the step comes instanteously to VI. I notice that the capacitor voltage cannot change instantly unless there is an infinite pulse of a sort, so this guy cannot change instantly.

And so its voltage starts off being zero. So the entire voltage here, KVL must be true no matter what. They are absolutely fundamental principles from Maxwell's equations. KVL must hold, which means that the entire voltage VI must appear across the inductor.

I put a big voltage across the inductor and its current begins to build up. There you go. A voltage across the inductor, its current begins to build up. As its current begins to build up that current must flow through the capacitor, too.

And as current flows through a capacitor it is depositing charge into the capacitor. As the capacitor begins to get charge deposited on it, its voltage begins to rise. Let's see what happens here. Its voltage keeps rising.

At some point, the voltage across the capacitor is equal to VI. But then VI equals this VI here. So when the two become VI, the inductor has zero volts across it. So there is no longer a potential difference that is increasing the current in that direction.

At that point, at pi divided by 2, I have some current going into the inductor so there is no longer a pressure that is forcing more current through the inductor because this voltage reaches VI. But remember capacitors like to sit around holding voltages.

Just remember that demo. That rinky-dink capacitor sat there stubbornly holding its voltage. And it had a huge spark towards the end. It just sat there holding its voltage. In the same manner, inductors love to sit around holding a current.

They will do whatever they can to keep the current going through them. It has got the current going through. And few forces on earth can change that. And so therefore, even though the capacitor voltage is VI and the voltage drop across the inductor is zero, it still keeps supplying a current.

It has got the current. It's got inertia. It keeps going. It is like a runaway train. You may not be pushing the train from the back, but once it is running it has got kinetic energy and is going to run no matter what for a least some more time, even if you take away the force on the train.

So I have taken away the force on the punching more current through, but it has kinetic energy. It has current flowing through it so it continues to supply a current. Because it continues to supply the current the capacitor voltage keeps increasing.

This is a subtle insight which is absolutely spectacular that with zero volts across it, it still keeps pumping that current. Capacitor voltage has gone up. And guess what? The voltage on this side is higher now but this guy is still pumping a current.

Man, I have been born to do this, you know, I shall pump a current. However, because the voltage has now gone up here gradually the current begins to diminish. So the capacitor is concerned. You pump a current into me, my voltage goes up.

At some point, like a runaway train, it comes to a halt. The current through the capacitor drains and now goes to zero and the capacitor voltage reaches 2VI. So this is at 2VI now and this is at VI.

Now the situation is not in equilibrium. At this point there is zero current through it, but guess what? I have a VI pumping in this direction now. I have the same VI punching in this direction. So guess what? Its current must now build up in this direction and its current begins to build up in that direction.

That begins to discharge the capacitor and the capacitor then goes on to a negative, or the current goes down to a maximum negative current, and this process continues. What you are seeing here is energy.

It is sloshing back and forth between the two, and that is kind of a key. I will just quickly put up a demo that you can watch as you are walking out. With a step input, notice the green is the voltage across the capacitor and the orange is the current through the capacitor.

Transcript - Lecture 15b
6.002 Circuits and Electronics, Spring 2007

Before I begin today, I thought I would take the first five minutes and show you some fun stuff I have been hacking on for the past three years. This has to do with 6.002 and circuits and all that stuff, but this is completely optional, this is for fun, this is to go build your intuition, this is to check your answers, whatever you want.

This is not a required part of the course. Just for fun. There is this URL out here that I put down here. I have been hacking on this system for the past three years, and for the first time this year and very tentatively and gingerly introducing it to students.

The idea here is that it is a, that is kind of defocused. Any chance of focusing that a little bit better? The idea of this is that it is a Web-based interactive simulation package that I have pulled together.

And what you can do is you can pull up a bunch of circuits. Notice that the URL is up here. It is euryale.lcs.mit.edu/websim. And there is the pointer to it. So you have a bunch of fun things you can play with.

And we have gone through all of these things in lecture. Let's pick the MOSFET amplifier. You come to this page. This is something you have seen in class. And let's play with this little circuit.

And you see the mouse? Good. You can set up a bunch of parameters. You can set up the MOSFET parameters VT and K. You can set up the value of R for your resistor, you can establish a bias voltage, and you can have an input voltage vIN.

So you can apply a bunch of input voltages. You can apply a zero input, unit in pulse, unit step, sine wave, square waves. Or this was the part that took me the longest to get right. You can also input a bunch of music.

And so far I just have two clips, so you are going to get bored listening to them. Good. So you can also input music. And what you can do is you can watch the waveforms, you can listen to the output and do a bunch of fun stuff.

One experiment I would love for you guys to try out. Again, remember, this is completely optional. Just for fun. You can apply some input. Step input, for example, to an RLC circuit and spend 30 seconds thinking about what should the output look like.

I divine that the output should look like this and then do this and see if what you thought was correct. And it's fun to kind of play around with it. Let me start with, just as an example, let's say I input classical music.

And let us say I would like to listen to the output here that is the voltage at the drain terminal of the MOSFET. For listening it sets up a default timeframe to listen to, so you go ahead and do it.

This shows you the time domain waveform of a clip of the music and then you can listen to it. Lot's of distortion, right? As you can see, there is a bunch of distortion. And that is as you expect because the peak-to-peak voltage is 1 volt, the bias is 2.5, and so this is clipping at the lower end, plus the MOSFET is nonlinear.

You can play around with a bunch of things and you can have a lot of fun. And the reason I created this is that MIT is putting a bunch of its courses on the Web. And one of the hottest things about courses like this is the lab component.

If you are beaming a course to, say, a Third World country or something, how do you get people to set up the massive lab infrastructure? I know you hate your oscilloscopes, I know you hate your wires, I know you hate the clips, but the fact is you have them.

I know a lot of places those are way too expensive to pull together, which is why I have been creating this Web-based kind of interactive laboratory so that people can learn this stuff over the Web.

Let's go do another example very quickly. Let's say you learned about, well, let's do RC circuits. Here is the parallel RC circuit. And you can set up capacitor values, resistor values, you can set up input.

Here, let me look at the time domain waveform for the voltage across the capacitor. And this time around let me play a unit step. And let's see what the output is going to look like. You can think in your minds what should the output look like, and then you can go and plot it.

There you go. That's what the output looks like. So you can play around with it and have fun. That's all the good news. The bad news is that so far I just have one Pentium III machine behind us.

It is a Linux box, so don't all of you try it at once. However, what I have also done, and that took me another six months of hacking in the small amount of time professors have to hack on stuff, I've hacked an incredibly elaborate cashing system so that once anyone in class tries out some combination of parameters it goes and squirrels away all the outputs.

If anybody else types in the same sets of parameters it will just get all the output and play it back to you. So if enough of you play with over time, we may end up cashing all the important waveforms and music clips and all of that stuff.

I have allocated a few gigabytes of storage, so I am hoping that it may work. Go forth. Play with it. And this is completely my fault, so if there are any bugs or anything simply email them to me.

This is the first time this is coming alive so bear with it. Now let me switch back to the scheduled presentation for today. All right, hope and pray that this works. Yes. Good. I am going to do today's lecture using view graphs.

And the reason I am going to do that and not do my usual blackboard presentation which I way, way, way prefer to a view graph presentation. The only reason I am going to do this for today, and maybe one more lecture, is that there is just a huge amount of math grunge in this lecture.

What I want to do is kind of blast through that, but you will have it all in the notes that you have, so that you don't waste time in class as you watch me stumbling over twiddles and tildes and all that stuff.

The key thing here is that the insight is actually very simple. The beginning and the end are connected very tightly and very simple. There is a bunch of math grunge in the middle that we are going to work through and, again, follows a complete old established pattern.

So, in that sense, there is really nothing dramatically new in there. Let me spend the next five minutes reviewing for you how we got here, what have we covered so far and set up the presentation. The first ten view graphs I am going to blast through and just tell you where we are in terms of LC and RLC circuits.

I began by showing you this little demo, two inverters, one driving. I can model the inductance here with a little inductor, the capacitor of the gate here. And recall that when I wanted to speed this up by introducing a 50 ohm smaller resistance, I got some really strange behavior.

Just to remind you, for Tuesday's lecture it would help if you quickly reviewed the appendix on complex algebra in the course notes. Remember all the real and imaginary j and omega stuff? It would be good to very quickly skim through that.

It is a couple of pages. Remember this demo? And the relevant circuit that is of interest to us is this one here. It is the resistor, there is the inductor and there is a capacitor. This is Page 3.

I am just going to blast through the first ten view graphs. It is all old stuff. Then we observed the following output. We applied this input at VA and we got this output, a very slowly rising waveform because of the RC transient.

And because of that you saw a delay. Notice that this delay was because of the slowly rising transient. This waveform took some time to hit the threshold of the neighboring transistor. So we say ah-ha, let's try to speed this sucker up by reducing the resistance in the collector of the first inverter.

And so I had this input. Now, to my surprise, instead of seeing a nice little much higher and much faster transitioning circuit, well, I did see a much faster transitioning circuit but I got all this strange behavior on the output that I was interested in.

And because of that, if these excursions were low enough, I could actually trigger the output and get a whole bunch of false ones here because of these negative excursions which should not really be there.

That was kind of strange. In the last lecture we said let's take this one step at a time. Let's not jump into an RLC circuit. Let's go step by step. Let's start with an LC, understand the behavior.

We started off with an LC circuit of this sort, and using the node equation we showed that this was the equation that governed the behavior of the circuit. And then we said that for a step input and for zero initial conditions, that is the zero state response, let's find out what the output, the voltage across the capacitor looks like.

And so we obtained the total solution to be this. And there was a sinusoidal term in there. And the omega nought which was one by square root of LC. And this was the circuit. And so for this step input notice that the output looked like this.

So far an input step I had an output that went like this. Notice that it is indeed possible for the output voltage to actually go above the input value VI. This is kind of non-intuitive but this can happen.

So this waveform jumps up and down. But the steady state value, on average if you will, is VI. On the other hand, it does have sinusoidal excursions and this kind of goes on because there is nothing to dissipate the energy inside that circuit.

By the way, the fact that the capacitor voltage shoots above the input voltage is actually a very important property. We won't dwell on it in 6.002, but just squirrel that away in your brain somewhere.

I promise you that some time in your life you will have to create a little design somewhere that will need a higher voltage than your DC input. And you can use this primitive fact to actually produce a DC voltage higher than you are given, and then use that somehow.

In fact, there is a whole research area of what are called DC to DC converters, voltage converters. Let's say you have 1.5 volt battery, a AA battery, but let's say a circuit needs 1.8 volts. The Pentium IIIs, for example, needed 1.8 volts.

And the strong arm is another chip that required 1.8 volts a few years ago, but the AA cell was 1.5 volts. How do get 1.8 from 1.5? Well, you have to step it up somehow. And this basic principle where the voltage can jump up above the input is actually used, of course with additional circuitry, to kind of get higher voltages.

It is a really key point that you can squirrel away. This was pretty much where we got to in the last lecture. This starts off the material for today. What we are going to do is take that same circuit, but instead we are going to put in this little resistor here.

This is what we set out to analyze. And for details you can read the course notes Section 13.6. The green curve here was the behavior of the LC circuit. And what we are going to show today is that the moment we introduce R this sinusoid here gets damp.

It kind of loses energy. And I am going to show you that the behavior is going to look like this. By introducing R this guy doesn't keep oscillating forever. Rather it begins to oscillate and then kind of loses energy and kind of gets tired and settles down at VI.

And remember the demo. This is exactly what you saw in the demo. You had a step input and you had this funny behavior. And for the RLC that is exactly what it was. So today's lecture will close the loop on what you saw in the demo and the weird behavior, and I am going to show you the mathematics foundations for that today.

Let's go ahead and analyze the RLC circuit. I purposely created the entire presentation to follow as closely as possible both the discussion of the RC networks and the LC networks so that the math is all the same.

Exactly the same steps in the mathematics are in the exposition of the analysis. What's different are the results because the circuit is different. So don't get bogged down or whatever in the mathematics.

Just remember it is the same set of steps that you are going to be applying. We start by writing down the element rules for our elements. Nothing new here. For the inductor V is Ldi/dt. The integral form which is simply 1/L integral vLdt=i.

We saw this the last time. And for the capacitor, the current through the capacitor is simply Cdv/dt. Those are the two element rules for the capacitor and inductor. The element rule for the resistor, of course, is V=iR.

You know that. And for the voltage source we know that, too, the voltage is a constant. Just follow the same established pattern. By the way, just so you are aware, I have booby trapped the presentation a little bit to prevent you from falling asleep.

You see the dash lines here? Whenever you see a dash line, that stuff needs to be copied down. Don't trip over that. Don't say I didn't warn you. We start by using the usual node method. And I have two nodes in this case.

Unlike the LC circuits, I have two unknown nodes. One is this node here with the node voltage vA and the second node is the node with voltage vT. Let me start with vA and write the node equation for that.

It is simply 1/L, the node equation for this is the current going in this direction with is vI-vA integral and that equals the current going this way which is vA-v/R, node equation. I then write the node equation for the node v, for this node here, and that is simply (vA-v)/R=Cdvdt.

And that is what I have here, two node equations. Let me summarize the results for you and then show you a view graph where I grind through the math as to how I got the result. Here is the result I am going to get.

If I take these two node equations and I massage some of the mathematics, I am going to get this result. And I will show you that in a second. By grinding through some math and solving these two equations and expressing this in terms of v, I get a second order differential equation, d^2v blah, blah, blah.

Notice that this is different from the LC in this term. Every step of the way you can check to see if I am lying or I am correct. I will indulge you, indulge myself rather with a little story here.

Richard Fineman was a known smart guy. And one of the reasons that he was that was in the middle of talks he was known to get up and ask some of the darndest, hardest questions and say ah-ha, you have a bug in this proof here or a bug in this equation that is not right.

And usually he would be correct. So his trick in doing this and which is one reason how he became a known smart guy. What he would do is, as the speaker went on talking he would kind of follow along and think of a simple initial primitive case.

In this case, I have an RLC circuit. So think of a simpler case of this. A simpler case of this is R=0. Whenever you set R to be zero, you should get exactly what we got in the last lecture, correct? That is what Fineman would do.

He would boil this down to a simpler case, make some assumptions and just follow along. And whenever he found a discrepancy between the math here and his simple case he would say oh, there is a bug there.

If you want you can catch me that way. Here, what Fineman would do is replace R being zero, and notice then this equation here is exactly what we got the last time with R being zero. Just remember that Fineman trick.

This is the equation we get, the second-order differential equation with an R term in there. And let me just grind through the math and show you how I got this from this. So the two node equations again.

And what I do is I start by taking these two equations and differentiating this with respect to t and this is what I get. And, at the same time, I have replaced (vA-v)/R here by this term. I replace this with this and differentiate.

Then I simply divide the whole thing by C. Then I take this expression here and write down vA is equal to this stuff here. Next I am going to substitute this back for vA and eliminate vA. So I take this vA, stick the sucker in here, and thereby eliminate vA and get this.

And then I simplify it and here is what I get. That is what I get. I just grind through the two equations and get that result. So like a stuck record I will repeat our mantra here, which is here is how we solve the equations that we run across in this course, the same three steps.

Find the particular solution. Find the homogenous solution. Find the total solution and then find the constants using the initial conditions. Same steps. You could recite this in your sleep. And the homogenous solution is obtained using a further four steps.

Let's just go through and apply this method to our equation and get the results. vP is a particular solution and vH is the homogenous solution. With a particular solution, oh. Before I go on to do that, let me set up my inputs and my state variables.

My input is going to be a step. Remember, I am trying to take you to the point where the demo left off. The demo had a step input, so I am going to use a step input rising to vI. And I am going to with the initial conditions being all zeros.

So the capacitor voltage is zero, inductor current, another state variable is also zero, and therefore this is also fondly called the ZSR or the zero state response because there is only an input but zero state.

Again, remember the dashed lines here. Don't say I didn't warn you. Let's start with a particular solution. This is as simple as it gets. I simply write down the particular equation and stick my specific input.

And remember the solution to the particular equation is any old solution, it doesn't have to be a general solution, any old solution that satisfies it. And I am going to find a simple solution here.

And V particular is a constant VI. It works. Because remember this has been working all along. And I am going to keep pushing this and see if this works until the end of the course. Guess what? It will.

So this is a solution. I'm done. That is my particular solution. Simple. Second, I go and do my homogenous solution. And the homogenous equation, remember, is the same old differential equation with the drive set to zero.

Remember that sometimes this equation with the drive set to zero is the entire equation you have to deal with in situations where you have zero input, for example. Or in other situations in which you have an impulse at the input.

And the impulse simply sets up the initial conditions like a charge in the capacitor or something like that. So we are going to blast through this four-step method. The method simply says that four steps, I am going to assume a solution of the form Ae^st.

And if you think you've seen that before, yes, you have seen it many times before. And you will see it again, again and again. And we need to find A and s. We want to form the characteristic equation, find the roots of the equation and then write down the general solution to the homogenous equation as this.

Same old same old. Let me just walk through the steps here. Step A, assume a solution to the form Ae^st. And so I substitute Ae^st as my tentative solution to the equation. Again, let me remind you that the differential equations that we solve here are really easy because the way you solve them is you begin by assuming you know the solution and stick it in and find out what makes it work.

I am going to stick Ae^st into this differential equation, and A comes out here. Differentiate this d squared, I get s squared down here, A s here and this simply gets stuck down here with the 1/LC coefficient.

The next step I begin eliminating what I can, so I eliminate the A's, then eliminate the e^st's, and I end up with this equation here. I end up with this equation. This is my characteristic equation.

It is an equation in s. Do people remember the characteristic equation we got for the LC circuit? Remember the Fineman trick? That's right, LC. S^2+1/LC=0. This thing wasn't there. All you do is simply follow the R.

Just follow the R. Just imagine this is a dollar sign and kind of follow it. And you will see what the differences are between the LC and the RLC. So this is the characteristic equation. What I am going to do, iss much as I wrote the characteristic equation for the LC circuit, by substituting omega nought squared for 1/LC.

Let me do the same thing here but introduce something for R and L as well. What I will do is let me give you this canonic form. The very first second-order equation I learned about when I was a kid was this one, S^2+2AS+B^2 or something like that.

Let me write it in that form where I get 2 alpha s plus omega nought squared. Again, remember the alpha comes about because of R. So omega nought squared is 1/LC and alpha is RL/2. Omega nought squared is 1/LC and R/L is equal to two alpha.

I am just writing this in a simpler form so that from now on going forward I am just going to deal with alphas and omega noughts. Once I get to this characteristic equation, after that I can give you one generic way of solving it.

And depending on the kind of circuit you have, a series RLC, which is what we have, or a parallel RLC we will simply get different coefficients for the alpha term. This is going to stay the same but this term will look different, alpha is going to look different.

There is a real pattern here. And what I am doing is simply focusing on what is important, what the differences are between the pattern. You learned the LC situation and the RLC situation. Given this I can now write down, I am just simply replacing this as my characteristic equation in dealing with alphas and omegas.

I will give you a physical significance of alpha in a little bit. Do you remember the physical significance of omega nought? That was the oscillation frequency. In other words, given an inductor and capacitor, you put some charge on the capacitor and you watch it, it will oscillate.

And its oscillation frequency will be one by a square root of LC. The magnitude of the initial conditions will determine how high are the oscillations or what the phase is in terms of when it starts, but the frequency is going to be the same.

Step three, to solve the homogenous equation, is find the roots of the equation, s1 and s2, and here are my roots. Good old roots for a second-order, little s squared equation here. Finally, given that I have the roots, I can write down the general homogenous solution.

So general solution is simply A1e^s1t, A2e^s2t. That's it. That's the solution. This looks big and corny, but we are going to make some simplifications as we go along and show that it ends up boiling down to something cos omega t.

The math is kind of involved but we get down to something very simple, a cosine. Hold this general solution. From that, as a step three of the differential equation solution, I write the total solution down.

And my total solution is the sum of the particular and the homogenous, so therefore I get this. VI was my particular and this term here is my homogenous solution. Now, if I wasn't doing circuits and simply trying to solve this mathematically here is what I would do.

I would find the unknown from the initial conditions, so I know that v(0)=0. And so therefore if I substitute zero for V(0) I get this. If I substitute zero here, t is 0, t is 0, so I simply get V1+A1+A2.

And let me just blast through because I am going to redo this differently. i=Cdv/dt. And so that's what I get. I substitute zero and this is what I would get. I hurried through this. Don't worry.

I'm going to do it again. If you just do it mathematically, you can solve this equation here and these two simultaneous equations in a1 and a2 and get the coefficients and you are done. But it doesn't give us a whole lot of insight into the behavior of these terms here.

What I am going to do for now is kind of ignore that. Ignore I did that and instead try to go down a path that is a little bit more intuitive. Let's stare at this expression we got for the total solution.

That is the expression we got. All I did is, I had alpha in there, I simply pulled out the alpha outside. So this is my total solution, V1-A1e^(-alpha t) something else and something else. Three cases to consider depending on the relative values of alpha and omega nought.

If alpha is greater than omega nought then I get a real quantity here. The square root of a positive number, I get a real number, and that number will add up to the minus alpha and I am going to get a solution that will look like, oh, I'm sorry.

Let me just do it a little differently. There are three situations here. One is alpha greater than omega nought. Alpha equal to omega nought. Alpha less than omega nought. Alpha is greater, alpha is less, alpha is equal to this term inside the square root sign.

For reasons you will understand shortly, we call this "overdamped" case, the "underdamped" case and the "critically damped" case. When alpha is greater than omega nought this term gives me a real number, and I get something as simple as this.

Remember, for the series RLC circuit, alpha was R/2L. So if R is big, in other words, if in my RLC circuit R is huge then I am going to get this situation. My output voltage on the capacitor is going to look like this, the sum of two exponentials.

And if I were to plot it very quickly for you, for a VI step, V would look like this. So v would simply look like this because it is the sum of a couple of exponentials. All right. Now, alpha is positive here.

Remember alpha1 and alpha2 are both positive. These two added up, because of this constant VI, give rise to something that increases in the following manner. Let's look at the situation where alpha is less than omega nought, where the term inside the square root sign is negative.

What I can do is pull the negative sign out and express it this way. What I am going to do is since alpha is less than omega nought, I am going to reverse these two and pull out square root of minus one to the outside.

This is what I get. I am just playing around with this so that whatever is under the square root sign ends up giving me a positive real number. So I pull the j outside and this is what I get. Now, let me blast through a bunch of math and end up with something very, very simple for this underdamped case.

Let me define a few other terms. I am going to call omega nought minus alpha squared the square root of that. I am going to call it omega d. And here is what I get. So I have defined three things for you now, alpha, omega nought and omega d.

And I get this equation in terms of alpha and omega d. And then, remember from your good-old Euler relationship? e to the j omega d is simply cosine plus a j sine. I am just going to blast through a bunch of math rather quickly.

Once I replace this in terms of a cosine and sine, cosine and a j sine and then collect all the coefficients together, I get an equation of the form VI plus some constant e to the minus alpha t, cosine, the sum of the constant e to the minus alpha t, sine.

Remember the sines and cosines are coming out, but because of my R I am getting this funny alpha here, e to the minus alpha here. So I am getting sums of sine and cosine. And K1 and K2 are some constants which I will need to determine for my initial conditions.

I am going to continue on with this and keep on simplifying it because, as I promised you, I want to get to something that is just a cosine. I want to go down this path. I am not going to cover this case, the critically damped case.

And I will touch upon it later but not dwell on it. Let me continue down the path of the underdamped case, and this is what we have. Continuing with the math, let's start with the initial conditions, v nought equals zero, and that gives me K1 is simply -VI.

So at v(0)=0 t is zero, so this terms goes away, the cosine becomes a 1, e^(alpha t) goes away, and K1=-VI. Then I know that i(0) and i is simply Cdv/dt. And I get this nasty expression. I substitute t=0 and I get something that looks like this.

I know what K1 is, and so therefore K2 is simply -V1alpha divided by omega nought. I have taken this expression where the unknowns K1 and K2 are to be found. I set the initial conditions down at t=0 and I get K1 and K2 as follows, which gives me the following solution.

This is the solution I get where I do not have any unknowns anymore. Remember that omega d and alpha are directly related to circuit parameters. Alpha was R/2L and omega d was square root of alpha squared minus omega nought squared.

** omega d = sqrt(alpha^2 - omega_0^2) ** And omega nought squared was 1 by square root of LC. So I know it all now. I still have sines and cosines here, so I am going to simplify this a little further.

Oh, before I go on to do that, let's do the Fineman trick again and notice if I am still true to the LC circuit I did the last time. Remember when R goes to zero alpha goes to zero. Because alpha is R divided by 2L.

If alpha was zero what happens? If alpha was zero, this guy goes to one, this whole term goes to zero and omega dt now ends up becoming omega nought, and I get this term here. I get VI-VIcosine(omega t), which is exactly what I expected in my equation.

This is the same as the LC case that I got. Let's go back to this situation and simply if further. If you look at Appendix B.7 in your course notes, Appendix B.7 is a quick tutorial on trig. And in that trig tutorial you will see that, and you have probably seen this before, too, multiple times, the scaled sum of sines are also sines.

This is an incredibly cool fact of sinusoids. If you take two sinusoids of the same frequency and you scale them up in any which way and add them up you also end up with a sinusoid. It is hard to believe but it is true.

It is an incredible property of sinusoids. Take any two sinusoids, scale them in any way you like, same frequency, add them up, you will get a sinusoid. What that is saying is that, look, here is a sinusoid, here is a sinusoidal function, and I am scaling them up in some manner.

So I should be able to add them up and be able to express that as single sine. And to be sure you can, look at the Appendix, and there is an expression for a1 sinX plus a2 cosX is equal to a cosine of blah, blah, blah.

This is what you get. No magic here. Just math. From here I directly get this. And look at what I have. It is absolutely unbelievable. v(t) is simply VI, there is a constant here, this an e to the minus alpha term and there is a cosine.

Again, to pull the Fineman trick, if this alpha were to go to zero here then you would end up with the expression you had for the LC situation. Let's stare at this a little while longer. There is a constant plus a minus, a cosine term, so there is a sinusoid at the output, and there is an e to the minus alpha which ends up giving you the decay you have seen before.

In other words, to a step input, the LC circuit would give you a sinusoid. That is what the LC circuit would do if alpha was zero. But because of this alpha term here, e to the minus alpha t, that gives rise to a damping effect, so this causes this thing to become smaller and smaller as time goes by until this term goes to zero at t equals infinity.

This guy damps down and so therefore you end up getting the curve that you saw like this. Twenty minutes of juggling math solving a second-order differential equation, but what ends up is the same sinusoid but it is damped in the following manner such that the frequency, rather the amplitude keeps decaying until it starts off at zero and then settles down at vI.

This is exactly what you saw in the demo that we showed you earlier. The critically damped case, I am not going to do it here. I am going to point you to the following insight. The underdamped case looked like this.

It was a sinusoid that kind of decayed out. That is the underdamped case. And then I showed you the overdamped case. The overdamped case looked like this. And, as you might expect, the critically damped case is kind of in the middle and looks like this.

So the overdamped case would look like this, underdamped like this, and the critically damped case kind of goes up and kind of settles down almost immediately. This is when alpha equals omega nought.

I won't do that case here, but I will simply point you to Section 13.2.3. Just to tie things together, recall this demo here that we showed you in class yesterday. This is exactly the kind of form of the sinusoid you saw because of that input step.

If you want to see a complete analysis of inverter pairs and look at the delays and so on because of that, you can look at Page 170 and example 898. In the next five or six minutes, what I would like to do is stare at the RLC circuit.

And much like I showed you some intuitive methods to get the RC response, what we are going to do is do the same thing for the RLC. In the RLC situation, much like the RC situation, experts don't go around writing 15 pages of differential equations and solving them each time they see an RLC circuit.

They stare at it and boom, the response pops out, the sketch pops out. This one is going to be another one like our Bend it Like Beckham series here. And this one is in honor of Leslie Kolodziejski.

And I call it "Konquer it like Kolodziejski". Again, as I said, experts don't go around solving long differential equations and spending ten pages of notes trying to get a sinusoid. They look at a circuit and sketch response.

I am going to show you how to do that, too. And what you can do is, to practice, go to Websim and try out various combinations of inputs and initial conditions and sketch it, time yourself, give yourself 30 seconds or a minute if you like, and sketch it and check it against the Websim response.

If it doesn't match either you are wrong or there is a bug in Websim. What I am going to do is, the response to the critically damped and underdamped case was very easy to sketch out. You started with an initial condition, you settled at VI and just kind of drew it like that.

The interesting case is the underdamped case, and that is what I am going to dwell on. Before we go on and I show you the intuitive method, as a first step I would like to build some intuition. Let's stare at this response here and try to understand what is going on.

This is the response that we saw. And this fact that you see an oscillation happening is also called "ringing". You say that your circuit is ringing. All right. You see some interesting facts. You see that frequency of the ringing is given by omega d.

This cosine omega d, so that is the frequency omega d. So the time is 2 pi divided by omega d. The oscillation frequency is omega d, but omega d is simply omega nought squared minus alpha squared.

Once you have a big value of R alpha becomes very small and omega d is very commonly equal to, very close to omega nought. So omega d and omega nought very commonly are very close together. And when that happens this frequency is directly omega nought.

Alpha governs how quickly your sinusoid decays. e to the alpha t here is the envelope that governs how quickly my sinusoid decays. And notice that each of these terms, alpha and omega nought, comes directly from my characteristic equation.

Which means that once you get your characteristic equation you really don't have to do much else. And up until now you still have to write the differential equation to get the characteristic equation, so you still have to do some differential equation stuff, but in two lectures I am going to show you a way that you can even write down the characteristic equation by inspection.

Look at your circuit and boom, in 15 seconds or less write down the characteristic equation. It is absolutely unbelievable. What are the other factors that are interesting here? Of course I need to find out initial values.

I start off at zero. This is my capacitor voltage. If I don't have an infinite spike or an impulse my capacitor voltage tries to stay where it is and starts off at zero. And the final value is given by VI, the capacitor is a long-term open so therefore VI appears across the capacitor.

In the long-term my final value is going to be VI. There is one other interesting parameter, which I will simply define today but dwell on about a week from today, and that is called the Q. Some of you may have heard the term oh, that's a high Q circuit.

Q is an indication of how ringy the circuit is. And Q is defined as omega nought by 2 alpha. It is called the "quality factor". And it turns out that Q is approximately the number of cycles of ringing.

So if you have a high Q you ring for a long time and if you have a low Q you ring for a very short time. That is called the quality factor defined by omega nought by 2 alpha. Notice that Q, omega nought, alpha, omega d, all of these come from the terms in the characteristic equation.

We will spend more time on Q later. With this insight here is how I can go about very quickly sketching out the form of the response. Here is my circuit. I want to sketch the form of the response for a step input at vI.

Zero to vI step input here, I want to find out what happens at this point. This is described to you in a lot more detail in Section 13.8 in your course notes. Let's go through the steps. Let's do the really simple situation first.

Let's also assume for fun that you are given that v(0) starts out being some positive value. Some v(0) which is a positive number. And, to make it harder on ourselves, let's say i(0) starts out being some negative number.

So i(0) is some negative current. The first thing I know is v(0), the capacitor voltage starts out here, which can change suddenly. And I also know that in the long-term this is an open circuit. So that this voltage vI will appear directly across the capacitor in the long-term.

So I get starting out at v(0), ending at vI, I am also half the way there. I know the initial and ending point of the curve. And then I know that somewhere in here there must be some funny gyrations here, because remember I am dealing with the underdamped case.

And you can determine that from alpha and omega nought. If alpha is less than omega nought, you know that you are in the underdamped case and this is what you get. Let's compute and write the characteristic equation down.

A week from today you will write it by inspection, but for now you will do it by writing down a differential equation. And from the characteristic equation you will get omega d, you will get alpha, omega nought and Q.

So omega d gives you the frequency of oscillations. My frequency of oscillation is now known. From Q I know how long it rings, because I know it rings for about Q cycles. I know that ringing stops approximately here.

And then I know that between that the start and end point my curve kind of looks like this, something like this. Right there we are 95% of the way there. The only question is I do not know if it goes like this or it goes like this.

I am not quite sure yet if it starts off going high or starts off going low. Not quite clear. I also do not know what the maximum amplitude is. It turns out this is rather complicated to determine so we won't deal with that.

Just simply so you can draw a rough sketch. The questions is which way does it start? I could leave it for you to think about. Yeah, let me do that. It is given on this page so don't look at it.

Think about it, and think about how you can determine whether it goes up or down. It turns out that in this case it is going to down and then ring. See if you can figure it out for yourselves and then we will talk about it next week.