Transcript - Lecture 6-10

Transcript - Lecture 6
6.002 Circuits and Electronics, Spring 2007

-- will try it again at the end of this lecture and you show you that stuff hopefully next time. For today we are going to start with nonlinear analysis. Before we do that I wanted to do a little bit of review.

I wanted to give you the past three weeks in perspective and show you how all of these things fit into the grand scheme of things. We began by building a great little playground, and within that playground we said that by enforcing upon ourselves the lumped matter discipline we created the lumped circuit abstraction.

So within that playfield we assumed that we had dq by dt and d phi by dt to be 0 so that gave us as the lumped circuit abstraction. And within that lumped circuit abstraction, within this playground we looked at several methods of analyzing circuits, including the KVL, KCL method.

We also learned the method involving composing resistors, the voltage dividers and so on and solving circuits intuitively. And we also looked at the node method, which is kind of the workhorse of the circuits industry.

So when in doubt apply the node method and it will get you where you want to go. Now, we also said that this is good, here is our playground. We said hey, if we focus on those circuits that are linear we come to the left part of our playground.

And we said that for linear circuits in this part of the playground we can further use a couple of techniques, a few techniques, superposition, Thevenin, Norton and so on. So these techniques allow you to very quickly analyze complicated circuits, especially when you're looking to find a single current, or voltage or some parameter of interest.

Whenever you see, if you see a circuit containing multiple voltage sources or two or more voltage sources or current sources, as a first step think superposition. And so these are very powerful techniques that let you analyze very complicated circuits very effectively.

After we did this we said, oh, let me draw another playground here. This is another piece of our playground. And if these are linear circuit then this half of the playground is nonlinear circuits. And we said that if you further focus on discretized values, if you discretized values and focused only on circuits that dealt with binary signals, highs and lows, then we came into this small regime of the playground.

And notice that digital circuits are, by their very nature, nonlinear. Remember the circuit, A, B, this was one of our NOR gate circuits? And if you look at transfer functions, that is if I plot, let's say for example, for some combination of input values.

Let's say I plot v in verses v out. Let's say, for example, I turned this guy off by setting B to 0 and then I simply apply a low to high transition at v in, then what I would see at the output is a transfer function of the following sort where as v in changes the output switches at some point and then stays at a low value.

So when v in is low v out is high and v in and high v out is low. So that's kind of the v out versus v in when B is set at 0. So notice that this is a nonlinear curve. This is not a straight line.

It's a nonlinear curve. And so therefore in the digital domain we see highly nonlinear functions that look like this and so on. However, take a look at this circuit. Suppose I focus on the circuit for a given set of switch settings.

Let's say, for example, I focus on the circuit when A and B are both 1s. For a given set of switch settings, notice that I'm going to be either in this region or in this region. Notice that this region is a straight line.

So if I focus on let's say both A and B at once then I get something like this. And in this situation, for a given set of switch settings, notice that my digital circuit now can be analyzed using linear techniques.

So therefore my digital gets moved into the linear domain for a given set of switch settings. So if I fix my switch settings and look at the circuit then each circuit, for a given set of switch settings, is comprised of voltage sources and some resistors and it's a linear circuit.

Again, I can go back and apply all my linear techniques to virtually all the digital circuits that you will be dealing with in 6.002. Again, remember if I fix my switch settings, if I fix the inputs then the output can be determined using linear techniques.

Because the digital circuits we're showing you in 6.002 simply comprise linear elements like voltage sources and resistors and so on. You'll see some more later. But you can apply your linear techniques and analyze them.

The cool thing here is that with just two weeks of stuff that you've learned in 6.002, you are well on our way to being able to analyze certain classes of digital circuits for a given set of switch settings and many, many, many linear circuits.

What we will do today is focus on nonlinear circuits. So we look at this space. Notice again that up until now we've dealt with these three methods, which apply to all circuits within this playground, the lumped circuit playground.

And the subset of that is the linear domain. And we can analyze linear circuits in this way. And digital circuits, for a given set of switch settings, also fall within this category. So notice that you can go ahead and analyze the digital circuits using superposition or other techniques like that.

The next big step for us is to begin our analysis of nonlinear circuits today. The important thing to remember is that nonlinear circuits are also within this big playground in which we are going under the lumped matter discipline.

So nonlinear circuits are also lumped circuits. And therefore because we are in that playground we can use any one of our techniques, KVL, KCL or the node method to analyze nonlinear circuits. So if you see a nonlinear circuit, don't get daunted.

Just remember this is meant to be simple stuff. So let me simply write down the node equation and analyze it. There is really nothing new in today's lecture. I'm just going to show you a nonlinear circuit and analyzing using techniques that you already know.

Today nonlinear circuits. And we look at several methods of analyzing nonlinear circuits. We look at the "Analytic Method". We look at a "Graphical Method". You will look at a "Piecewise Linear Method" in the book.

I won't be covering this in lecture. You can read Section 4.4 for the piecewise linear method. In this method you take your curves and you approximate them with a bunch of straight line segments, kind of like the v out, v in curve I've shown you there, and analyze the circuit using linear techniques within any given straight line segment.

We will also do incremental analysis. This is also called small signal analysis. So I will cover these two today, I will introduce this one today, and wrap that up during the next lecture. Let's start with a simple example.

So I have some voltage, V, some voltage source V. And I have some resistor, R. And I have a fictitious device here that I labeled D. Let's call this fictitious device the "Expo Dweeb". I purposely chose a funky name because this is a fictitious device.

Let's call it the Expo Dweeb. And let me write down the associated variables for this device as follows. iD is the current flowing into this terminal and vD is the voltage across this device. So this is a nonlinear device.

And this device is characterized by the following equation. Much like resistors were characterized by an iV relation, V is equal to iR, or i is equal to V/R. This device is also characterized by the following element relationship.

It's a e raised to bvD. So there is an exponentiation here. Again, this is a fictitious device. And I'll show some funky things that it does in a second. It's a very simple relation. It's an exponential relation where the current relates to the exponentiated value of the voltage vD across the element.

So I can plot iD versus vD for this element as follows. Notice that when vD is 0 iD is a, so I have a here, and it looks like this. It's a funny device, a fictitious device. So when vD is 0, I have some current flowing the device, and as vD increases I get an exponential increase in the current through that device.

This device is funny in the sense that it is not a passive device in that notice that when vD and iD are positive the product is positive, which is fine, which says that it is consuming power. On the other hand, on the left-hand side notice that the vI relation is negative, which means that when I put a negative voltage on it, it can still sustain a positive current.

This must imply that the device is producing power. But for the purpose of a nonlinear analysis we don't have to worry about that. Let's just do it mathematically and find out what it looks like. So back to this again.

I have a voltage source, a resistor and my Expo Dweeb connected in that manner. Now, again, reflect on this pattern. A voltage source or a current source, a resistor and some device. This is a very standard pattern you will see again and again and again.

In particular, if you look at this device, it's a nonlinear device here. And facing the nonlinear device is a voltage source in series with a resistor. And the reason I say that this is an incredibly important pairing is the following.

Notice that if on the left-hand side I had any linear circuit and I had a single nonlinear element in that circuit. Notice that by a Thevenin reduction that you've learned you can take this entire mess.

If all you care about is the behavior of the nonlinear device, for the purpose of analyzing this nonlinear device, you can take this entire linear circuit, no matter how complicated it is, voltage sources, current sources, resistors and a bunch of other funky stuff, you can boil all of that down to a Thevenin equivalent, a voltage and a resistor in series.

So we can trick you. We can give you a complicated circuit and say ah-ha, tell me what the current is through this device if I apply some voltage, 3 volts there. What you can do is you can say ah-ha, I don't care what happens here so I'm just going to replace the whole thing with a Thevenin equivalent.

And you've done your homework now and you can calculate Thevenin equivalents for circuits. And simply replace this and then go ahead and solve the circuit. Again, remember we are engineers. We are looking for answers.

We are looking to build interesting systems. And, in general, we like to take the simplest path possible to the solution. So simplify your lives and create a simple Thevenin coupled to a nonlinear device and then you will be rolling.

When we talk about a variety of other circuits, nonlinear circuits, time-varying circuits and so on in the rest of this course, we will look at this pattern again and again and again and again until we are blue in the face.

And, just remember, the reason we keep looking at this pattern is that whenever you have some big linear mess connected to some interesting device what you can do is if all you care about is analyzing the behavior of that device, you can take this linear mess and simply figure out the Thevenin equivalent, or the Norton equivalent if you like and replace this whole thing with its equivalent and then go ahead and analyze it.

So boil an arbitrarily circuit down to a very simple pattern of this sort. What this means is because of this brilliant Thevenin simplification, going forward through the rest of this course we will mostly deal with very simple circuits like this, voltage source, resistor and the device.

That's it. Very, very, very rarely will you see multiple sources and lots of resistors in a circuit. It's usually going to be simple stuff. And remember how we got here, by making a Thevenin simplification of a linear mess.

All right. If in homeworks or quizzes or in real life, or in many examples of real life, if you find that you have to deal with a lot of grunge and a lot of mess, step back and think a little bit. Try to use intuition and see if you can simplify things using some clever trick or method.

Method 1 of analysis. Let's go ahead and analyze this pattern here, this template circuit, if you will, a voltage source a resistor and a nonlinear device. This is the analytical method. And remember the node method applies, so let me go ahead and apply the node method.

To apply the node method, what do I do? I first have to select a ground node. Let me insulate this as my ground node. Let me label all the nodes with their voltages. So this node has voltage V and this node has label the capital D.

So let me go ahead and analyze this using the node method. So the node method says for each of the nodes in the circuit whose voltage is not known go ahead and write down KCL implicitly applying the element relationships to replace the current values with the voltage values.

Let's start with the current going in that direction. Current going from the vD node through resistor R, which looks as follows, vD – V divided by R. That's a current going that way. And the current going down is iD.

In general, when I apply the node method, I don't write iD here but I go ahead and write the element relation ae to the bvD here. Then I get an equation in vD and I just solve the mode voltage. However, just to make a couple of extra points later, let me go ahead and do that in two steps, write down this and then go ahead and write down iD separately as ae to the bvD.

Again, remember, don't get confused here. In a node method, I don't write down a second step. I directly write down ae to bvD in place of iD. I get one equation in vD, I go solve it. Just for fun today, I'm taking two steps here, writing iD and explicitly putting down iD as ae to the bvD.

Now, that's it. I mean this is all there is to it. You guys can now go ahead and analyze nonlinear circuits. You get a bunch of equations, a bunch of unknowns, go solve. I have two equations here.

vD and iD are my unknowns and I can just go ahead and solve for them. Now, in general with nonlinear circuits, often times it's hard to get a closed form solution so you may have to use a bunch of methods.

You can try a closed form solution or you can try numerical solutions or you can do trial and error. In this case, I'll just go ahead and tell you. Suppose I choose V as 1 volt, R is 1 ohm and b is 1 over volt and a is ¼ amps for those values, approximately vD is roughly 0.5 volts and iD is roughly 0.4 volts.

You can do this by using trial and error or other methods. In 6.002 we don't dwell on working too hard to solve equations of this sort. If you cannot substitute this in here and solve it directly, we don't ask you to go and learn numerical method and the techniques and so on to solve it.

But just remember that you can use trial and error or you can use back substitution and other techniques that you will learn in future numerical methods classes and apply it here. But suffice it to say that, for here we can stick with trial and error if you like.

And for these values, vD and iD are 0.5 and approximately 0.4. You're done. It's really that simple. Yes. Oh, I'm sorry. Good catch. I know there is one person that's not sleeping here. Good. So, as I said, there's not a whole lot to it.

Whether it's a nonlinear circuit or a linear circuit and as long as I am inside this playground here where the lumped circuit abstraction holds, I can apply my node equations and then go ahead and solve it.

Let me show you a few more methods so we can articulate your repertoire of tools for nonlinear circuits. And I'd like to show you a graphical technique. I personally rarely use a graphical technique to solve circuits.

And why am I sharing this with you? It turns out that often times by looking at things graphically you can get some better insights into circuit behavior. You can also show cool demos when you show graphs of responses kind of playing with each other and so on.

So this is fun for getting intuition and things like that. Graphically all I'm really going to do is solve those two equations graphically. So I'm going to plot equation one. Let me rewrite equation one as follows.

iD is -- I'm just rewriting equation one as follows. V/R – vD/R. And I can also draw the second guy -- OK, I can do this as well. I can do an iD versus vD plot. And in this particular situation, you've seen this already, that's my iD versus vD curve right there.

And I can do the same for this one here. So this equation establishes the following straight line relationship. It says that when vD is 0, iD is V/R. So that's here. And similarly when iD is 0 then vD is equal to V so I get something here.

So that's my straight line relationship corresponding to this equation here. So what I can do is I can simply solve these by superimposing the two curves on the same vD, iD template here and finding the intersection of the curves.

So I can take this curve corresponding to two and I can take this curve corresponding to one, and this is V/R and this is V, 0, and I can find the intersection point. This curve here, for reasons that will be obvious about three weeks from now, is called the load line.

It's called the load line. You will understand why that is so in a later lecture. So I've given you a template on Page 6 to boil these two down into one equation. So there, again, you can substitute the values for V is 1 volt and R is 1 and so on and so forth and get the same kind of result as you did previously.

So there is really nothing new here. All I've done in the second method is combined the two equations graphically and found the solution by looking at where the two curves intersect. At the start of the lecture I also told you that you may want to go and check out the piecewise linear technique -- -- in Section 4.4 of the course notes.

All right. For today let me do a third method called "Incremental Analysis". This technique is also called the small signal method. I'm going to show you, before I go into the method, in today's lecture what I'll do is I'll give you a motivating example for why we need the small signal approach.

I'll give you a motivating example and show you a little demo. And then I will close with showing you a problem with applying a standard approach, and I'll ask you to see if you can figure out a way to handle it in time for next lecture.

So let me give you the motivation here. So here is what I want to do. Many of you have seen one of those electric eye garage door openers, right? You have a receiver at one end and you have some kind of a light beam at the other, and when you walk through it stops, or rather it cuts the circuit and stops the door from closing.

And when no one is going through it maintains a connection and lets the door close. So what we did is we went to Home Depot, or one of those stores, and bought a very standard device that essentially produces some response when light impinges on it.

And my goal will be to see if I can send music over the light beam using a simple garage door opener device. So here is the little circuit that I will do. We actually went there and built this. I will also show you a demo.

Here is my time-varying voltage, vI(t), and this is some music signal. And get some music signal. And I want to connect this to this device, which is a device found in garage door openers. I am going to call it a LED.

If you like, you can view it as, this is very similar to our Expo Dweeb. This is called a "Light Emitting Expo Dweeb". That's why it is LED. So what the LED does is, as I apply this voltage across it, that same voltage appears across the Light Emitting Expo Dweeb.

And there is some current that flows through the device. And for our analysis we will assume that this device virtually has an identical iD characteristic to the Expo Dweeb just that it emits light.

So when I pass a current through it, it emits light. And the light intensity is proportional to the current that flows through. So it emits light and light intensity, LD, is proportional to iD. Here is my little light emitting device, which when current flows through it, itproduces light because its intensity is proportional to the current.

And what I will do is I will stick in the receiver here. Think of it as a photo resistor or some other device where I am going to connect that in a circuit. I am not going to spend too much time on this side.

I'm going to focus on the left-hand side here. And let's say I have some kind of amplifier and speakers and so on and so forth. Suffice it to say that when the light falls on this device PR that iR that goes through here is proportional to the received light intensity.

So if the current is proportional to the received light intensity then I amplify that signal in my amplifier and I get the music playing out here. And notice that the following chain of dependences apply.

So I have an input music signal VI. That gets converted to some iD. These are all time-varying signals, so VI is a time-varying signal and so is iD. And iD gets converted to light of some intensity LD.

This in turn gets attenuated somewhat and is received at the photo resistor. And I get some intensity LR impinging on that device there. And that in turn produces a current iR and then iR is amplified and goes through a speaker and so on and produces sound.

Notice that using this chain I've taken a music signal here and I am playing it here. And just imagine that this is your garage door opener device here where the light emitted is being articulated by the voltage signal VI.

And received here. So notice that if I cut this, if I stick something in here and block it then I get no response here, but if I take my hand away then I do get some response. But this is fine. This should work.

You could try this at home if you'd like. If you have a garage door opener, just stick a little circuit like this and it should simply work. We have a problem, though. The problem is that, as I said, I'm using the Expo Dweeb here, the light emitting Expo Dweeb, and its characteristics are as follows.

iD is exponentially related to the voltage vD, so this is nonlinear. And that's a real problem. Because this is nonlinear, I am going to get a distorted output. Let me show you a little wave form, a little graph to show you how the distortion happens and then show you a little demo showing you the distortion.

Let me graphically show you the kind of distortion that is happening here, and I will do it by drawing the following graph. So this is the vD, iD curve for our device. And what I'm going to plot for you is if I have a time-varying vD voltage, I just want to see what the time-varying iD current looks like.

And a trick to plot that is to take your input voltage like so. And let's say I apply a sinusoid. So I am just taking a time-varying sinusoidal voltage and rotating the plot 90 degrees like so, so I can see where these points correspond to on that curve.

So what this says is that at some point here, for example, where vI, at this point and time, vI is here. Notice vI and vD are the same thing because vI is applied across vD. vI directly applies across the device, and so vI equals vD at all time.

So this voltage here corresponds to this voltage, it corresponds to this current and then I can find out what the current is for that voltage. By using the same artifice I can plot the output current iD like so.

So for this value I get some current here. And so at time T0 I start here. And notice that as this signal moves up here, I can find out the corresponding values of iD by looking at where a straight line intersects here and plotting the values here.

I have a nice little graphical animation to show you this. Hopefully, the laptop will work tomorrow and we can check that. I am doing nothing new here. Just showing you a trick to be able to plot vI versus v out relationships, or vI or versus other relationships based on some kind of a transfer function.

So what you end up getting is something that looks like this. Why is that? Notice that this curve here corresponds to the signal. As this signal moves from here to here, this point moves from here to here and that corresponds to this iD.

When this moves from here to here that corresponds to a point moving from this part of the curve to here, and that looks like so. And then for the whole negative incursion, notice that the whole negative incursion moves here, so for that entire negative incursion I get an output that looks like this.

Notice that this device has completely cut off and hammered negative going signals. What it's done is that rather than giving me a nice little negative spike incursion here, or excursion here, what this is doing is that it is taking this excursion and simply slamming it down to this value here.

And then again, when I go back up, I get this peak here. So notice that what was a nice little sinusoid out there gets hammered and squished into this funny curve here. What this device is doing is for positive values it tends to produce exponentially greater current so I get boom, high-rising peaks corresponding to these two, and for negative going voltages it simply compresses them to a low positive value here.

And that's what I see here corresponding to negative excursion. So notice that what this will do, if I view sound, if I input sound here, and sound has negative going excursions it will simply scrunch them.

But more or less let the positive things through. And that is going to give rise to a bunch of distortion in my signal. So I would like to show you a little demo. Actually, we've gone ahead and built a little device like this.

We have an honest to goodness little device costing, I don't know, 50 cents or $1 or something, which is a little voltage, it's a device that emits light proportional to the current flowing through it.

I have a receiver. And I am going to play some music, and you will listen to the output here. And hopefully you should see a bunch of distortion because of that effect that I showed you. And what I will do is, before we do that, you will see two curves up there.

The yellow, I believe is the vI, is the input, and the green, I believe, is a signal proportionate to -- The other way around. Oh, I see. So green is the input. So green, the lower one is the input and the upper one is the distorted output.

So we are going to play some sound through it, music through it and you can listen, through a little CD player. So a couple of things. The good news is that it works. However, I doubt that music artists will come to my studio to record if this is the quality of what I produce.

Do notice that there are hardly any negative going excursions in that curve up there, right? All the negative ones have been like scrunched up down into a flat line there, and that's the reason I get this distortion.

And just to prove to you that I am indeed using a garage door opener device and not faking it here, I am going to just shut the signal off by stopping the light using a piece of paper here. So notice that this device here is the little device that has a light beam going through the center, and I am going to take this piece of paper, can you turn it up? So let's have some fun with this.

If I were to put this piece of paper halfway down, I should get half the intensity, right. So my sound should diminish in volume a little bit. Maybe that will work. Let's see if it works. Nothing to do with 002 but it's just fun.

Louder. You can make it loud. Too much coffee. My hand is shaking. I guess you did see the lowering of volume, right? OK. Just way too much coffee, and so my hand was shaking too fast imposing its own sine wave on top of the signal.

What did I show you? This was garbage, right? We had a nice little signal input, and the output was completely distorted because I was playing sound over this and this is what happened. Switch to Page 9.

Now, this is what I would have liked to have happened. On Page 9 what I would have liked to see happen is this. Suppose I had a light emitting device that looked linear, a straight line where the current was linearly related to vD.

Then what I would see, if I had a sinusoid here then I would get a sinusoid here. No distortion there, right? If only things were like I wanted them, if I had a linear device, but I don't have a linear device.

I have an Expo Dweeb. Now you know why I call it a dweeb. Well, I'd like a linear device and it's exponential. But this is what I would like. And if I had this I wouldn't show it to you today. If I had this my music would go through without any distortion and I wouldn't have to run cables through my attic.

I could just use my garage door opener to play signals from my bedroom and living room and so on, right? So the key thing here is how do I get this? And what I would like you to do is think about it yourselves.

What I am given is something like this. This about it yourselves, you know, what would you do? See if you can come to me before lecture tomorrow or Thursday and tell me the answer, OK?

Transcript - Lecture 7
6.002 Circuits and Electronics, Spring 2007

OK, good morning all. So before we begin, I just thought I'd show you a little news item that I happened to read that was very relevant to what we covered recently in 6.002. So you recall when we did the digital section a few days ago last Thursday, we talked about a switch.

We talked about the MOSFET switch, which when turned on and off, by input signals could help build gates which would then be combined in tens of millions of quantities and go into chips like the Pentium 4 and AMD Athlon 64, and so on it so forth.

So I just saw this news item that I came across, and this says they are rethinking the basic construction of the products. It talks about the semiconductor manufacturers like AMD, Intel, and others that build digital chips.

They are rethinking the basic construction of the products down to the architecture of the transistor. That's a MOS transistor, and the on/off switch inside the chip. OK, now this might imply that there is a single switch inside the chip, but no, there's tens of millions of transistors, or tens of millions of switches inside a chip.

And pretty much any advancement that can be made to the basic transistor can have a 10 million to 20 million times effect because there are that many of them on a single chip. So I thought that was very appropriate.

OK. Let's dive into a quick review. So this week, we had begun nonlinear analysis, and I just thought I'd blast through a few animations that I've created, trying to give you more insight into the behavior of some of the things that we have done.

Now first of all, as I did the last time, let me try to put it in perspective most of what you've learned thus far, and what we will be learning today. So the past week, we have been focusing on nonlinear analysis.

And as I pointed out, here is how this fits into the big picture. So, we had our 6.002 world, at what we said is that we are engineers. We are going to devise our own playground in which to play with our own rules.

And that's our playground. That's what we're going to learn about in 002, and for that matter, the rest of EECS at MIT. It's all within this playground here. And this is the playground with lumped circuit abstraction, and good old KVL, KCl, node method, your basic composition rules apply within this playground that directly come from Maxwell's equations because you have made the lumped matter discipline assumptions.

OK, so then we said a large part of the playground is linear, and some other much more intuitive techniques apply within the linear portion of that playground, techniques like the superposition, Thevenin and Norton.

In most exercises, and quizzes, and experiments, and so on that you do in real life, you can pretty much apply these simple techniques. Very rarely do you have to go into the node method for circuits that are more complicated than single source and a couple of elements.

And then, there's the nonlinear part. Remember, the reason I showed this is that this is the same playground. OK, linear and nonlinear are part of the same playground. OK, even nonlinear elements are lumped circuit elements, and they follow KVL, KCl, the node equation, and so on.

And then, last week we spent some time talking about the digital abstraction. So we focused on a smaller region of the playground. And the assumptions we made in there were even tighter. We said that it is part of the playground we shall only deal with binary values.

We'll digitize or lump values into highs and lows, and that's where our circuits are going to be. And these circuits, when looked at as a whole, were nonlinear. So, this is a simple NAND gate circuit.

And this is the input/output characteristic. So, for example, if I hold B at zero, and I apply a zero to one transition at A, then this is the output that I will see at C. So notice, this is decidedly nonlinear.

Then I said that, look, suppose we had to fix the input values at a given set. OK, so let's say, for example, I fix A at one, and B at one. OK, and then look at the circuit in this situation. What do I find? What I find is that the entire digital set of circuits that we were looking at move over into the linear space for a given set of switch settings, OK? So, when I set A 1 and B 1, A equal to one and B equal to one, my NAND gate becomes like this.

OK, it's a simple resistive network with a voltage source, VS. So, for a fixed set of inputs, for a given set of inputs, if I don't change my inputs, then my circuit looks like a linear circuit, and my good old linear analysis techniques apply.

So that was last week. And this week, we are looking at the nonlinear space. And we looked at a couple of techniques in the nonlinear space, analytical techniques and graphical techniques. And then, I showed you an example.

OK, I showed you an example circuit that was something that I would like to build involving the light emitting expo dweeb, my little garage door opener thingamajig, and I wanted to transmit music over that light beam.

I also showed you that it was highly distorted because it was in the nonlinear space. So, today what I'm going to do is introduce a new part of the playground. There's a new part of the playground, and I'll show you a technique whereby by focusing on this part of the playground and disciplining ourselves in the kind of inputs we apply to circuits, I'm going to show you that certain kinds of nonlinear circuits also move over, when used in a particular way, also move into the linear analysis domain.

OK, so let me leave that for now and go back into quickly reviewing the motivating example of music that I had taken last time. OK, so here was a little example. So I have a music source, VI, and I apply that.

This device that I call the, lightheartedly, the Light Emitting Expo Dweeb has a current, VD, across it, or a voltage, VD, across it, and a current ID through it. And the light intensity, I said, was proportional to the current.

And because of that, I was able to get the light to impinge on a receiving device, which produced a current that was proportional to the intensity of light falling on it. And that signal would then be amplified somehow.

We haven't talked about all of this stuff yet. This will happen next week. But let's say we somehow amplify the signal and then played out through a set of speakers. All right, so if I had some sort of a music signal here, then I could then transmit the music signal over to the side on top of this light beam.

But the problem, as I said the last time, was that our device, the Light Emitting Expo Dweeb had an exponential characteristic, so that I had some trouble in getting undistorted music. So, the characteristic of the VI characteristics of my device looked like so.

The ID versus VD curve looked as follows. OK, it was decidedly nonlinear. And because of that, I was getting a lot of distortions in my signal, and I showed you a little trick to plot, given an input waveform at a transfer function such as here to plot the output function.

OK, let me show you another little animation that I have created here for you that should give you even more intuition in terms of how it happens. So, this is a characteristic I showed you up here. It's on both sides, but I guess it points to only one unless I shuttle back and forth really fast.

So on average, I'll be in both places. But anyway, so here's my ID versus VD characteristic. And as I said, there's an exponential ID versus VD curve. And I want to see what the output looks like, for example, a sinusoidal input.

So I said, let's place the input along a little graph, rotate it so, and take a sinusoid, and apply a sinusoid to the input, VI, which would also appear across the Light Emitting Expo Dweeb. And then, what I wanted to see was how the output looked.

OK, so let me tell you that the output is going to look like this. OK, the output is going to look like so. And, a little artifice to discover curves like this is to think about a point here corresponding to the point on the transfer curve here, because this is VD, looking at the Y intercept.

That's a value of ID, and that's a value of ID here. And, time moves along here, and time moves along here. So, I did this little animation. You'd better be impressed. It took me six hours to do it.

So, here it goes. So, let's say I start by focusing on this little point that corresponds to this point on the transfer function, which then, in turn, points to a time, zero, this point on my ID curve.

OK, I hope this works. So, as my point moves down [LAUGHTER], this was fun to do, I promise you. So notice that as this point has the following excursion, this had the following excursions here. OK, all right.

So let me pause that little animation there. At the end of the lecture, I'll put that up again if you like, and you all can come and play with it. So, you can actually do this in PowerPoint. It took me quite a bit of time to figure out how to do it, though, but it's fun.

OK, so let me show you a little demo, and show you a sinusoid, and show you what the output looks like if I apply a sinusoid for VI. So, I'll show you ID as a function of VI when VI is a sinusoid. There you go.

So, I applied my sinusoid VI, and this is the current that I get. And notice, this is the transfer function that I talked about, the ID versus VD curve of my Light Emitting Expo Dweeb. And I get this highly nonlinear transformation of the input as I get to the output.

OK, so that is a problem. And then, I also played some music for you. Let's do that, too. I played some music for you. I applied the music as an input to the circuit, and that's the output. OK, that's the output that I'm observing at the amplifier.

It's highly distorted. OK, we can stop that. There you go. OK, so that was my problem. OK, so we had covered, we had gone this far last Tuesday. I set the problem up for you, motivated what we had to do, and showed you that I was able to transmit music over my garage door opener, but I did not think I could listen to that music for very long.

So, I challenged all of us to think about how a trick that I could use to be able to transmit music and have a linear response. So, did you people get time to think about it? So how many people here think they know the answer? It's OK, don't be modest.

Go ahead. Could you speak louder? Yeah, you find another something, kind of element, that's got the opposite graph so that when you add them together. Oh, this guy wants to cheat. No. He wants a new element.

So, no, no new elements. Pardon? Build an MP3 encoder. Ah-ha, so that will happen much later. Yes? Digitize the signal before you send it to the LED? Digitize the signal before you send it to the LED.

But in some sense, each of these solutions is a huge sledgehammer approach to look at solving it. There's a much simpler technique I can apply here. Yeah? Add a voltage offset. Ah, ah-ha, that might work.

What else? So let's say, here's my signal, right? If I add a voltage offset, that will just bump the signal up here. Then the curve is still nonlinear. But you're getting there. Well, I'll tell you what.

Let's pause here. Let me quit while I'm ahead. OK, so the answer here, folks, is Zen. OK, what I want you to do is, so, in Zen, what you have to do is you have to sit down in a courtyard, and look at a rock, like a small rock on the ground.

And you got a focus on it till the rest of Earth kind of vanishes. Just focus on the rock. OK, now make like you're in a courtyard, and you're looking at this little area here. Just look at this. OK, and I'll give you ten seconds.

Sit down quietly, and no sounds. Just stare at the spot here. OK, make believe this is your little rock, and just stand there and think about it. OK, I'll give you five seconds to do that. Just stare at it.

And very soon, the answer should pop into your heads. OK, what do you see? This guy, if I focus on this really small region of the graph, this small little piece looks more or less linear. OK, hmm, so that should give me some insight.

This whole thing, the macrograph is nonlinear. But I focus on a little rinky dinky piece of that graph like so, that appears more or less linear. If it's small enough, that appears linear. So, I'm staring at this, and that appears linear.

The question is, how do I exploit this little small, little, linear region to get a linear response from my device. OK, so here's the trick that I'm going to use. The little trick that I'm going to use is the following.

Notice that, let me call this voltage at the center of this region capital VD. What I can do, if I take my input signal, and I just pointed out earlier, I bump it up. I boost it. OK, so I apply a DC offset to my input signal, like so.

So I apply some input signal, VI, which is also equal to the VD if I look at a variable across the nonlinear element. If I apply a DC offset, VI, and I superimpose the music on top of that, let me call my music, just to distinguish between the two, capital VI, and the small vi.

OK, that's my music. So here's my capital VD, my DC offset. And I want to superimpose my music on top of that. OK, so I've gotten halfway there. By superimposing my music here instead of having excursions out here, I now have excursions out here.

OK, and so I'm using some portion of the graph here. But that's still way beyond the small little element there. So a second think that I do in addition to boosting up the signal is shrink it. Think of boost and shrink, BS.

So what I want to do is boost up the signal using a DC offset, and shrink the sucker. OK, so I'm going to go with a small signal and bump it up. OK, so now what happens is that small signal in its excursions, only uses that little portion of the graph.

OK, again, remember: bump and shrink, bump and shrink, two things, boost and shrink. So what do you think of that trick? So, by doing that, what happens is that signal that has excursions here will produce a corresponding response in this region, OK? And I argue that since this is more or less like a straight line, I invoke Zen here, and argue that this little signal now gets transformed, and I get a linear response.

OK: boost and shrink. So in terms of my circuit, let me draw it out for you. My Light Emitting Expo Dweeb, and this whole signal was what I used to call V capital I, and that's made up of two components now, a bump offset, and a shrunk voltage VI.

It shrunk, so therefore I've used the small v and small i, like, really, really small. In the same manner, I get a VD ID across the LED, and the corresponding values here will also have a DC offset and a small response.

Let me call that ID plus I small d. I'll do all this mathematically in a second as well, but first let me do it completely intuitively so you get some insight into what's going on. And, VD is simply capital VD plus small vd.

OK, and this is the same as VI, I, and VI. OK, so what have I done? I've done two things. I have said, as an engineer, OK, I care about getting music across my garage door opener. And I'll do what it takes to do that.

OK, so as an engineer, I'll do two things. I'm going to bump my signal up and shrink it. And the bumping and shrinking, and I do it like this. I shrink my signal, the music signal here, and add a DC offset.

OK, and I claim that the music I listened on the other side now, provided I have enough amplification there, is going to be undistorted. OK, so far I've showing this to you completely intuitively using little sketches, no math.

I promise you, I'll give you a bunch of math in a few seconds, but just get the basic idea, and get the intuition behind it. So let's go back to our demo and take a look. So remember, BS, right, bump and shrink.

So what I'm going to do is first of all, let me bump up the signal. So, what I'll do is I want to add an offset to my input, and let me bump it up. Let me shrink it first. It'll make the point a little clearer.

So, the big input, green, is a big input. Let me shrink it. OK, so I've made my input small, and in the middle of that picture out there, you see the region of the transfer curve that's being articulated.

OK, this region of the curve is being articulated by the small signal. It's a much smaller signal. And the output is still distorted because I have to do two things: bump and shrink. I've only shrunk.

OK, let me bump it up now. What's the yellow curve? It's going to get linear. It's going to get proportional to the input. Then I'm bumping it up now. I can make it smaller, make it even smaller, there you go.

Isn't that fantastic? So, I'm making nature do my bidding here, OK? So, this is one of those, when I learned electronics and so on many, many years ago, this was one of those really big ah-ha moments for me, saying, wow, that stuff is cool.

It's something that I couldn't think about myself, and it's not obvious, and by being disciplined and creative in how I use circuits, I can do really, really cool things. OK, remember this as a big ah-ha moment for you.

So, here's my little signal that I've shrunk and bumped up, and my output is a sinusoid, and not this funny, distorted waveform. And notice that this is the region of the curve that is being articulated.

So, I can make the signal even smaller if I like. OK, and what I'd like to do next is play music for you, and if you don't believe your eyes, you can at least believe your ears. Let me go to the distorted signal again, switch to music, and raise it up.

OK, now what we'll do is shrink the music signal and then bump it up. Can I turn the volume down a little bit? That's good. OK, so if I shrunk the volume a little bit, and let me bump it up, now. [MUSIC PLAYS] Just remember this as a big ah-ha moment.

OK, the signal is really, really small. I like that. I like the enthusiasm. OK, so the signal's very small, and I get a more or less linear response. OK. All right, so that's intuition, and the approach that I've taken is called, it's variously called small signal analysis, incremental analysis, small signal method, small signal discipline, whatever you want.

OK, this simply says that by boosting and shrinking my signal, I get a response that's more or less linear even when I have a nonlinear device. And this technique is called the small signal approach.

So, just to focus on that a little bit longer, switch to page five of your notes and let me draw something out for you. OK, so what I have here, this is my offset VD, and from the VD offset I have my little signal V small d, and the total signal is called V capital D.

Offset, small signal, and that's my total signal. OK, notice the offset is all capital. The total signal is small v capital D, and the music or the small signal is small v small d. Similarly, the output is going to look like this, and here I get an offset in the output ID.

I get a corresponding signal, I small d, and I get a total signal, I capital D, OK? The cool thing to notice is that the signal here, the output signal here corresponding to the input signal, the music signal, VD, is small I small D, and that is more or less linear.

OK, and I can even plot the signal like so. This is my input, v capital D. That's T. This is VD, V small d. That is my total input. And similarly, I have an output. And this is my output ID. And, that looks like this, I capital D, small i small d, total signal I capital D.

OK, so that's the small signal method. So, let me summarize that for you. There are three steps to the method. So, first of all, operate at some DC offset. This is also called DC bias, and in that example it's VDID.

OK, so I choose an operating point that bumps up the operation in some region of interest. The second step is to superimpose small signal on top of VD, capital V capital D, to superimpose a small signal, and the third step is observe the response -- -- and the response, small i small d, that's the music part of the response, ID, is approximately linear.

OK, three steps to the method here, and just remember this notation. And, my notation in the small signal model is as follows. My total signal ID is the sum of two signals, I capital D plus small i small d.

This is called the total signal. That's called the DC offset. And this is the superimposed small signal. OK, total signal, DC offset, plus the small signal. And sometimes, especially when doing math, and so on, we may oftentimes represent ID as a delta, I capital D, OK, to show that ID is incremental change in the value of I capital D.

And because of that, this method is also often called the incremental method, incremental analysis. OK, so far what I've done is given you some intuition. I've developed a small, simple method, given you some insight into why we use this method, and also shown you some demonstrations that show that when I bump and shrink, and observe the response, I do get a more or less linear response.

So let me now do this mathematically and show you that mathematically, you can also derive your response to be a linear response. This is page seven. So, I know that ID is some function of the diode voltage.

F was my nonlinear function. OK, so my function F was a nonlinear function. So therefore, ID was nonlinearly related to VD. So, let's do the math. So as a first step, what we did was replace VD by a DC offset, the small signal method, a DC offset, plus a small incremental change.

OK, by doing the math, let me simply use the delta VD notation to show you that I'm dealing with small increments, and also because in the mathematics community, when you learn about some of these techniques, they will use the incremental change notation, which is the delta VD notation.

In electrical engineering, we use a small v, small d notation. So, this is a large DC offset, and this is a small change about that offset. So, you folks have taken math courses before, and been looking at finding out the value of a function, which is a small change for an input value, which is a small change about a big input value or a big DC point is Taylor's expansion.

OK, so let's use Taylor's series expansion, OK, and substitute VD plus delta VD into this, and see what ID looks like. Again, let me tell you where I'm going with this. ID equals F of VD. This is a nonlinear function, OK? I claim that by replacing VD, the input, with the DC offset plus a small value, the resulting response to the small value will be linear, OK? So what I'm going to do next is replace VD with this sum here, and then do the math, and show you that the response corresponding, or the change in ID corresponding to the change in VD is going to be linear.

All right, so let's expand this function using Taylor's series near the DC offset point, capital V capital D. OK, so ID is simply, by Taylor's series, I want to find out a value of the function close to V capital D.

OK, so I take the value of the function at that point, and then I add a few terms in my Taylor's series expansion. The first term is simply the good old Taylor's series stuff. OK, the first term is the first derivative of the function times the change.

And then, the second one is second derivative. OK, and then I get higher order terms. So this is nothing new here. This is good old Taylor series expansion, and again, let me tell you where I'm going.

I want to look at the response for an input that looks like this, and I want to show you at the end of the day that the response in ID, the effect on ID of using an input like this is as if that effect, the incremental change is linearly related to the small input, delta VD.

So here's my Taylor's series expansion for delta V. Now remember, I told you that delta VD is much, much smaller than V capital D. OK, it's a very, very small quantity. But that quantity is really very small.

Then what I'm going to get is that my output is, I can begin to ignore my second order terms. OK, delta VD is very, very, very small. Then, what I'm going to do is that ignore higher order terms. So I'll go and ignore higher order terms.

They'll all go to zero. Remember, I can do this because by design I've chosen delta VD to be very, very, very small. OK, remember, we are engineers. I've chosen it in a way that this is very small.

OK, so I'm telling you that's the case, and under those conditions, I can ignore second higher order terms, in which case I am left with this expression here. So let me rewrite this. Let me rewrite this down here.

OK, I've just copied this turnout, I've ignored all these terms here, and so I have a more or less equal to sign that remains. So what I'm going to do is when I apply a small input of this form to a large DC offset, my output is also going to look like some output offset with a change in the output offset.

And let me call the output offset I capital D, and some small change in the output delta ID. OK, we'll make sure we can convince ourselves that this is indeed the case. Notice that this guy here, F of capital V capital D is a constant.

That's a constant with respect to the incremental change, delta VD. Similarly, this part here is a constant. Notice that this term here is the first derivative of the function evaluated at the DC bias point, capital V capital D.

OK, so this term is also a constant with respect to delta VD. So notice, then, I have a constant term plus a constant term multiplying a small change, delta VD. So what I can do next is, in this case, given that I have a constant term on both sides, and on this side it's a time varying term, what I can do is equate the two constant terms.

I can go ahead and equate these two terms. Remember, I have a constant plus a time varying term, OK, if I'm assuming here that delta VD, my little music signal is a time varying term. So, this constant will equal this, so ID must equal F of VD.

And I know that's the case because the function evaluated at the DC offset gives me the DC current ID. And similarly, ID is equal to that component. Delta ID is equal to D, F of -- OK, so my incremental change in the output is the first derivative multiplied by the small change in the current.

OK, so I'm pretty much done. So, therefore, notice that delta ID is proportional to delta VD. OK, and that's what I had set out to show. Remember, I had set out to show that provided my input is a small excursion around a large DC offset, then my output could also be a large DC offset with a small excursion on top of it where the two excursions, the input excursion and the output excursion would be linearly related like so.

OK, and the method is very simple. I simply expanded the function about that point, that DC point, neglected higher order terms, and notice that my incremental term was simply the derivative plus the incremental change, a derivative times the incremental change in the input.

Move onto page nine, and I'd like to give you a quick graphical interpretation of this. So I gave an intuitive explanation earlier. This is a mathematical explanation that shows you that the input could be linearly related to the output, provided, the outputs would be linearly related to the input, provided the input has a DC offset, and small excursions about that DC offset.

So, let me give you some intuition in what you've really done here, using a little graph here. So, I'm going to plot ID versus VD, and notice that I have some point here, V capital D, I capital D. That's my DC bias.

So, I have some DC bias point here. OK, what is this? That is simply the slope of the curve at that point. OK, it's the slope of this curve evaluated at this point. So this guy here is simply the slope of this curve evaluated at ID VD.

OK, now, what I care about is this point here, and this point here. So let's say that this is delta VD, all right, and that corresponds to this point here. So what I've done is taken the slope and multiplied that by delta VD.

So I've taken the slope, and multiplied it by delta VD, OK, and that gives me this component here. OK, and so, this is the point that I'm going to get. So in other words, what I've done is approximated point A using the Taylor trick by the point B.

OK, so this is a point, A, which is what I really want, and I've approximated that by taking the slope of the function at V capital D, and multiplying that by the change in the input to get the corresponding Y offset, and that's the point that I get.

And notice that if I make this delta VD small enough, then the error between these two points becomes smaller and smaller. So back to our example, so ID was a e to the BVD. This was the relation for our Expo Dweeb, and let me just plug in the values.

So, ID plus small id. Notice, I'm just shuttling back and forth between the notation delta VD, and small v small d. OK, and so that is given by a e to the BVD, oops, plus, I'm just writing that equation up there.

Let me call this equation X. And so, I get the second term is the derivative, ab times e to the BVD times delta VD, small VD, and equating this term that the DC offset. Notice that this is the DC offset in the output, and the small signal, ID is, further notice that in this particular example, what's that? a e to the BVD.

That's simply ID again. It just happens to be that way in this example. So, I get ID times BVD. So, for my input, small id, my incremental change in the output is some ID times B times VD. And notice that this is a constant.

And because that is a constant, my small signal behavior ID is going to be linearly related to the signal, VD, the input signal VD. OK, in the last three minutes, I'd like to give you one additional insight.

So what we've shown so far is if I have an offset and a small change above it, then my output ID will be linearly related to my input. Now let's stare at this thing again. Let me rewrite it. It's some constant IDB times VD.

So, where have we seen such an expression before? OK, where ID was some constant times VD. OK, remember, I equals V divided by R: Ohm's law. What I want to show you now is how we constantly keep simplifying our lives.

The moment we hit some complication and things get too painful to analyze, as engineers, we come up with some clever tricks to make an analysis and use of circuits simple again. And so, notice that this is similar to some, one by RD VD, where RD is simply one over IDB.

I'm just defining this to be RD. And what that means is that I can take a nonlinear circuit that looks like this. OK, and what I can do is replace this by its incremental equivalent, and build what is called a small signal circuit.

And I'll just introduce it here. And we will revisit the circuit in much more gory detail a couple of weeks from now. So, what I can do is build a small signal circuit where I have all the small signal variables, and replace a nonlinear device by a simple little resistor whose value is given by IDB.

OK, so therefore, what I can do is take my nonlinear circuit, and for small, incremental changes, replace that circuit with this equivalent small signal circuit, and go back to doing simple stuff again.

Thank you.

Transcript - Lecture 8
6.002 Circuits and Electronics, Spring 2007

All right, good morning. So today, we are going to talk about what is both a basic device in itself, the amplifier, and it also serves as a real key example of both nonlinear analysis and small signal analysis.

So, today, dependent sources and amplifiers. So, let me first spend a few seconds just pointing out to you some of the key points from our previous lectures. I also want to point out that each chapter in the course notes has a summary at the end of it.

And if you take a quick scan of the summary at the end of each chapter, it highlights the major takeaway points from each chapter. It stresses what's important, and if you have to remember a few things, what are those things to remember? So, to quickly review, we talked about a few primitive elements: resistors, voltage sources, and so on.

And by now, you should have the facility to play around with these device elements. And then we talked about the Node method, and this is kind of the workhorse of 6.002. When in doubt, use the Node method.

OK, and this will work both for linear circuits and nonlinear circuits. OK, so if you see a problem, or if you see a situation in real life that requires analysis, then as a first step, you should try to think of whether you could apply some of the key intuitive shortcut methods, superposition.

One of my favorites, the Thevenin method, the Norton method, or the method that involves composition, that is very quickly analyzing circuits that have resistors in series and parallel. OK, so if you can apply one of these quick, intuitive, shortcut methods, go do so.

If you can't, then usually you can resort to the Node method irrespective of whether the circuit is linear or nonlinear. So the last week was focused on the nonlinear method or nonlinear circuits, and we spent the first lecture talking about a straightforward application of the Node method, which gave us a bunch of nonlinear equations that we had to solve.

In the last lecture, we talked about the small signal trick. What we said is if you look at the whole space of nonlinear circuits, then within that space, if we focus on small variations, small perturbations about an operating point, then even the behavior of nonlinear circuits in that small regime would be linear.

So small signal method. And as an example, I showed you how I could take a highly nonlinear device like the garage door opener LED, and using that, build a pretty nice transmitter that would transmit music.

And as long as we kept the signal small, and operated the device in a region where its transfer curve was relatively smooth, and I biased, or set the operating point appropriately, I would get a linear, small signal response.

OK. So today, we're going to do a couple things. We're going to look at dependent sources. And the reading for this is section 2.6 of your course notes. And, the dependent source will be a new element in your tool chest.

We will also do amplifiers, and amplifiers are in section 7.1 and section 7.2 of your course notes. So, before I begin with dependent sources, I'm just a huge believer in motivating things with real world examples.

OK, so let me start by motivating: why we need an amplifier? Why do we need to do things like this? Or why do we even bother? And, spend a few minutes really getting you to appreciate that amplification is fundamental.

OK, it's as foundational to life as high fat potato chips and stuff like that. So, let's do some basic examples here. So first, let me talk about, why do we need to amplify signals. Why amplify? Why do we care about building an amplifier? So, an amplifier, think of a little box, and apply some sort of small input.

And I get a larger output. In this example, this may be a voltage with a swing of 10 mV, and in this case, the output might be another voltage with a swing of, say, 100 mV. And commonly, the amplifier, in addition to an input and an output, input port and output port, may also contain the power port, OK, so that I can apply a power supply to the amplifier because commonly as an amplifier signal, I'm looking for a power gain as well, an increase in the power provided by the output.

So, that's an abstract definition of an amplifier, and let's take a look at an example of why we may need this. So let's say I have a small, useful signal, and let's say the signal has 1 mV peak to peak.

And, I'm looking to transmit the signal over a wire to some other point. But let's say that in this environment, I get a bunch of noise that is in a noisy environment. And in this environment, let's assume that some noise may get superimposed.

And if I have a 1 mV signal, and 10 mV of noise, then what I end up with at the output is something that looks like this. And it's really hard to distinguish my 1 mV signal from that large amount of noise.

On the other hand, if I do the following, if I took the signal and passed the signal to an amplifier, and I amplified the signal to be a much larger version of the same signal, let's say in this particular situation 100 mV peak to peak signal.

OK, so I magnified the signal by a factor of 100. OK, let's say it's a linear amplifier, I linearly amplified signal to be 100 mV, then in that case, if I had a noise on top of this, it's going to be less discernible.

The signal will look like this. OK, my 10 mV noise would add on to it. But, this is still pretty decent. I can still recognize the input. And so, this is one application of amplification. If I need to send something from point A to point B as an analog signal, then an amplified signal is less prone to noise attacks than a small signal.

Not surprisingly, a large number of devices that are used in everyday life have amplifiers built into them. So, get a little cell phone, and virtually every single cell phone contains an amplifier. By the way, this is an all digital cell phone.

It's a Kyocera, I forget the number now. It's completely digital. OK, although they say it's completely digital, it turns out that a significant fraction of the circuitry is analog, in particular, so digital is sort of a marketing term to say that there's something special about this.

But remember, there's a bunch of analog stuff. So, here's my little antenna from the cell phone. OK, and typically the first thing that happens to a signal as it comes out of the antenna in your cell phone is, look at cell phone circuits, or cell phone systems would be something that looks like this, OK, this, and may have a label LNA.

If someone were to take a guess at what LNA might stand for? What's that? Linear amplifier. That's pretty good. So that's LNA. Close enough. A is correct. It's amplifier. What does L and N stand for? Low noise.

OK, so this stands for low noise amplifier. So, I get a really rinky dinky small signal here, and then the low noise amplifier amplifies a signal. And in real cell phones, and for that matter, in your 802.11b, or 802.11a, or 802.11g wireless cards, same thing.

Antenna, low noise amplifier, and then you may have a bunch of processing. And commonly, you have a bunch of analog processing. And then, you convert the analog to a digital signal. OK, I recall last week I asked somebody in class here, how would we transmit the signal from point A to point B without it being impacted way too much by noise, and he said, oh, go digital.

Good point. OK, so if I go digital, I can transfer the signal without noise being a real factor. But the analog to digital converters need the signal strengths to be a given value before it can chop it up into digital levels.

OK, so an amplifier is very fundamental. OK, and so in this case, what may be a signal of a few tens of microvolts to be amplified to some large enough value that it can be further processed. So, that's application of amplification in the analog domain.

Let me talk about amplification in the digital domain. So, that's in the analog domain. This amplification is in the domain that I have both analog and digital. OK, and now let me talk about amplification in the digital domain, OK? I'm going to argue that amplification is absolutely foundational to the digital domain.

OK, the digital abstraction would not occur if I did not have basic amplification. OK, and the next minute and 37 seconds I will prove that to you, OK? So, let's do so. So, let's suppose I have a very simple digital system, and the system simply contains a pair of inverters.

So, if I send a one here, it's a zero here and a one here, which is a very simple, trivial, digital system. And here's the input. Here's the output. And we said that for digital systems of this sort to work, they have to follow a static discipline.

OK, our signals and our circuits must follow a discipline for them all to work together. And, the discipline we described comprised of signals adhering to certain voltage thresholds so that all the components in the system could agree on what comprised a zero, and what comprised a one, OK? So the way we did that was we said that you would have a threshold called VIH, V input high, and another threshold called VIL, V input low.

OK, and we said that this circuit must recognize signals that are higher than VIH, 3 V for example as a one, and simultaneously, any signal that has a voltage level less than VIL, say, two volts, should be recognized as a zero.

That was the input constraint. On the output, it had a similar set of constraints, where we had tougher constraints on devices, where we said that the output had to satisfy a output low constraint, output high constraint.

What this said is that for this circuit to be called a good digital circuit that satisfies the static discipline, signals that were ones here should be recognized as such. And if I am producing a one as an output, then the signal level should be higher than VOH.

Similarly, if the signal's a zero, then it should be less than VOL. So as an example, this may be 2 V, this may be 3 V, and this may be 4 V, and this may be 1 V. OK, so input, I should recognize 2 V and less as a zero, but at the output I have to produce a very, very low value, 1 V.

So, I have some noise margin. So as an example, say if I made a plot of the input/output, so I get my VIL here and VIH here. This is time. This would comprise a valid digital signal: zero, one, zero, one, and so on.

OK, now, I had a tougher set of constraints at the output. I would have VOL, VOH. So, at the output, OK, I'm required to stretch the ones and zeros to be further apart from each other so that I get noise margin, and the corresponding signal for our little circuit there would look like so.

Right, if this is a valid input, then this would be the corresponding, valid output. OK, and need I say more? OK, you can see that, intuitively, look, there's amplification happening here, and the reason is that VOL is chosen to be less than VIL, and VOH is higher than VIH.

So therefore, the signal has to be stretched. The signal has to be amplified. OK, and what's the minimum amplification needed for the system to work? The minimum amplification is if I had a signal that looked like this.

OK, that barely skimmed the VIL, VIH level. OK, so if signal were this high peak to peak, VIH minus VIL, and what's the absolute minimum signal at the output? It would look something like this. OK, barely skimming VOL and VOH, OK, so the corresponding output level would be VOH minus VOL.

OK, so this is the absolute minimum amplification that my digital circuit has to provide. OK, and notice, VOH is larger than VIH. VOL is smaller than VIL. Therefore, this quantity needs to be greater than one.

OK, so I've shown you both a simple, graphical, intuitive explanation, and this is a slightly more formal proof that even the digital circuit really requires to have amplification built into it, if it is to satisfy valid static disciplines.

Yes? Yes. The question is, is that the same as gain? Good question. Yes, the term amplification has many, many variants. You could say gain. You could say amplification. You could say increase in signal strength, and so on and so forth.

And in fact, when talking about low noise amplifiers, people sometimes talk about having the low noise, high gain amplifier at the input stage. OK, so let me pause there in terms of motivation. So, I believe I've motivated every which way: pure analog, analog/digital, and digital.

OK, so I've covered every single base here. And so, we need amplification. OK, so let's look at how to build a fundamental, primitive device called the amplifier. Before we do that, however, let me take a quick detour.

It will be convenient for me, as I show you how to build an amplifier, to introduce a new device, a new element, called the dependent source. OK, let me introduce a new device for your arsenal of devices, along with resistors, You learned about a MOSFET, a switch, voltage source, current source, and now a dependent source.

So, a dependent source looks like this, OK, has an output port, and has a control port. So, a dependent source in its simplest form has two ports: an input port and an output port. Remember, a port is a convenient pairing of terminals, and I apply signals to such terminal pairs.

But this is a abstract diagram for a dependent source, and to get a little bit more specific, let me show you an example of a dependent source. So, let's say, here's my input, and I label the terminal variables for the input.

VC is the voltage applied to the input, and IC is the current into this terminal here. And, here is the symbol for the dependent source. Much like a current source or a voltage source has a circle around it, the corresponding symbol for a dependent source is like so.

So this example, for instance, is a dependent, current source. I can apply the corresponding output variables, I0, OK, and I can say that the current, I, is some function. In this example, I've designed the example that the current through the current source, I, is some function of the input voltage or the control voltage, VC.

OK, so notice that the current through a current source, the current through this current source, I, is some function of another variable. OK, in this example, it's the voltage across its control port.

Not surprisingly, this device is called a voltage controlled current source -- -- or a VCCS. So, in like manner I can also devise other forms of sources. You can think of this is a device where a voltage controls an output current.

You can think of all other combinations, current controlling current, voltage controlling voltage, current controlling voltage, and so on. So, another example, I give you another dependent source, and in this situation, my output current is controlled by an input current, VC.

IC rather. And I claim that I for this one is some function of a current, IC. OK, it's another dependent source where the output current for its output port is related to the current, IC. And, this is a current controlled current source.

OK, it's a current controlled current source. And, if I had lots of time on my hands, and I was wanting to kill time, I'd sit around drawing for you, other types of dependent sources. I would draw for you a current controlled voltage sourced, and I could also draw for you a voltage controlled voltage source.

OK, so that's an abstract diagram for such a source. And so, let's do a few examples involving elements like this. To begin, just so you can build up your intuition, let me start by doing a very simple circuit, involving an independent current source, OK, just so we can relate back to what we've been doing so far.

So, let's say I have some resistor, and I have a standard current source with current I nought. This is an independent current source. Remember the circle? And, some resistor, R, and let's say I care about the voltage across the resistor.

OK, so I have a current I nought flowing through it. So, I can very quickly write down VR as, simply, I0 R. OK, it's the drop across the resistor when a current I nought flows through it. OK, so this is what you've been used to doing.

Correspondingly, I can do an example with a dependent current source. And, as an example, I'll use a voltage controlled current source. OK, a voltage controlled current source is a dependent current source whose output current depends on the voltage applied at the control port of the current source.

So let me build a little circuit. OK, so here's my current. And let's say it's VC IC for the control port, and similarly, let's say my current I here is some function of the control port voltage. And let's say, to be specific, there is some K over VC, some function.

OK, there are a variety of dependent sources that can be built, and here's a hypothetical device where the output current is mathematically related to the input in the following manner. So, let me build a circuit of the following form.

So, let's add the resistor, R, and here's my circuit, OK? And, as before, let me look to figuring out what VR is. So, notice that I have to supply some voltage at the input so that the output can depend on the input because right now I don't know what the input here.

So what I'll do is let me apply VR over here. OK, so let me make this connection. OK, let me make the connection from here to here. What I've done is I've applied VR at the control port of the dependent current source.

OK, and I often draw a circuit like this. This looks pretty messy. I will often draw the circuit like so: R, VR. OK, short form circuit drawing would look like this. This is a complete drawing that I show you the explicit connections of the control port, but oftentimes, when the control port does not have any other impact in the circuit, you can eliminate, don't explicitly show the control port.

Rather, you can simply show the dependence of the output current on whatever circuit variable you have in mind. So, you can draw the diamond like this, and see its current is some function of VR. VR in this is case is K divided by VR, OK? OK, so let's go ahead and analyze this little circuit here, and look at what this might give us.

Our goal, as before, is to find out the value, VR. So, in this case, let's apply the Node method to this node, and sum the currents into that node to be zero. OK, so sum the currents going into that node to be zero.

The current going down is simply VR divided by R. OK, and that is equal to the current that is going out of the node. And so that is equal to F of VR. And I know that F of VR is given by K divided by VR.

OK, a simple application of the Node method. So then, I collect VR's on the left hand side, and I get VR squared is K times R, OK, and VR is simply the square root of KR. There you go: I'm done. OK, I've gone ahead an applied the Node method to this, and when have to figure out the current here, I simply reflect the fact that it depends on VR like so, and I just go ahead and solve the circuit.

Remember, the workhorse of the circuit industry, the Node method, when in doubt, apply it. It simply works. And notice, this is a nonlinear circuit. OK, the dependence is nonlinear, and I get the response like so.

So, to plug in some numbers, supposing K was 10 to the minus 3 amperes per volt, and R was one kilo ohm, then I can plug the numbers in and the kilo here cancels with the 10 to the minus 3, and I get VR equals 1 V.

OK, this simply says, if I build a circuit like this, then this voltage here will be 1 V. So, again, as long as you remember that the dependent source is simply another little circuit element, OK, and you usually draw just the output port for dependent sources, and reflect the way that the control affects the current, that'll suffice, and you get, through the application of the Node method, the variable you're interested in.

Let's do another example, OK, of another fun current source, a voltage controlled current source, and look at it this way. So, let's say I have a resistor, and I have a current source, a resistor, RL, and this goes to some, I apply a VS here.

Remember this short form notation; that's simply applying a supply VS between that node and the ground. OK, and let us say the current IV through the device is some function of the current at its control port.

OK, so I'm not going to show you that. But remember that the device already looks like this, that there is a control port here. I'm not showing that to you. And let us say that I apply some voltage, VI, to the input port.

The reason we often don't show the input port is for many practical dependent sources, the input has no other effect on the circuit. So, for example, in this case, the input has infinite resistance looking in.

So therefore, if I apply a VI here, it doesn't draw any current from VI. I simply apply the voltage, VI. It doesn't affect the circuit in any other way except in terms of how it controls the current ID.

So let's say the current ID is some function of VI because VI is applied at the control port. OK, and as I pointed out before, I oftentimes, just for clarity, just to show this dependent source explicitly.

OK, so let's work the example. So as I said, I'm going to choose ID to be F of VI, and let's pick some specific parameters here. Let's say it's K by two VI minus one, both squared. OK, and let's say this is true for VI less than equal to one volt.

And let us also say that ID equals zero for VI less than one volt. OK, it's a dependent source, and it can have various forms of dependences on the input. And, I just picked an example of some hypothetical, or as yet, hypothetical dependent source, the current through which is related to the input using a square law relation, VI minus one all squared as long as VI is greater than one.

And if VI is less than one, then the current is simply zero, it shuts off. So, I can go ahead and apply. So, let's say I want to find out V0 versus VI. So, I care about finding out V0. V0 is the voltage of this node with respect to ground.

OK, so it's a slightly more complicated circuit than you saw up here, than you saw up there. So, let's go ahead and do this example. Start by applying the workhorse of the circuits business, the Node method, and let's start with doing this for VI.

Let's first do it for VI greater than one, notice the behavior of this is different for different ranges of VI. So let's first do it for VI greater than or equal to one and apply the Node method. Node method says sum the currents going into this node; we know the voltage at this node.

It's VI. We know the voltage at this node. It's VS. OK, the only unknown is V nought. And so, let's go ahead and write the node equations for that node. So, the current going up, let me simply equate the current going up to the current that has been supplied by this particular node here.

And, that should equate that the two of them should sum to zero, the current going up plus the current going down should sum to zero. So, I get V0 minus VS divided by R. That's the current going up.

Plus, the current going down must sum to zero, plus ID must sum to zero. And ID is going to be K divided by two VI minus one all squared. That must equal zero. Straightforward application of Node method, current going up plus the current going down at this node should equal zero because the total current leaving the node must be zero, OK? So I can go ahead and simplify this, multiply it throughout by, I call this RL here.

So, multiply it throughout by RL, and move all of this to the other side, so I get VS divided by RL, multiply it throughout by RL. I get VS at this side. I take this term to the other side. This becomes a minus.

RL multiplies here, so I get KRL. That's the expression I get. V nought is VS minus KRL divided by two times VI minus one all squared. Let me put a box around this because I will be referring to this more times in 6.002 for a variety of reasons than probably any other equation on Earth.

OK, this is the first time you saw it. You saw it here. OK, mark it down. You'll smile every other time you look at it in quizzes, and you will find out why this comes up very often in 6.002. So, I'll just give you a few seconds to savor this big moment in your 6.002 life.

All right, OK, so it's pretty simple actually. I mean, there's really not much. A lot of this stuff is just a plain old, simple application of the Node method, and things just fall out. It's just so simple.

So, the V nought, I apply the Node method, I get V nought for this nonlinear circuit. I can also it for VI less than one. For VI less than one, when VI is less than one, what happens? ID is zero. OK, since ID is zero, think of this as an open circuit.

OK, so there's no voltage drop across RL. And, this voltage V nought is equal to VS. So, I like to see things in pictures. I'm not an equations kind of person. I'm much more of a graphical person.

So, let me draw a little graph to show how V nought, to see the form of V nought, and then let's study that little system a little bit more carefully. So, this is page seven, and we plot V nought versus VI for you.

And let's take a look at how this really simple circuit looks. This has got nothing. It's got an RL resistor connected to a supply, and a dependent current source, and I apply some voltage VI at the input.

It's a very, very simple circuit. So, let's see. So as long as VI is less than one, the output stays at VS. OK, that makes intuitive sense, right? As long as the current here is zero, this is like an open circuit here.

If this is an open circuit, then effectively, V nought is simply the voltage VS. V nought simply appears here. If you want to grunge through KVL and KCL, go ahead. VS minus RL times the current is V nought, and the current is zero so it's, yes.

So, this is simply VS. When VI goes above one volt, fun stuff begins to happen. OK, when V nought goes above one volt, then this equation applies because VI is greater than one. This equation applies.

And, when VI is a one, one minus one is zero. This term cancels out, so this is VS. OK, phew! So, I start off here. As VI increases, what happens now? As VI increases, this term here becomes increasingly negative, OK, subtracting from VS.

OK, so I get some behavior like this. V nought begins to drop. And it makes intuitive sense, right? As ID begins to increase, the voltage here will begin to drop because I'm drawing more and more current through RL.

I'm dropping more and more across RL. So more and more drops across RL, so V nought begins to drop too. So, it looks something like this. I'll show you a little demo, but my claim is that you have just seen an amplifier.

Whoa. You just saw an amplifier. So, I snuck an amplifier by you, OK? So, I just snuck an amplifier past you. I'll show you why in a second. So, let's take a look at this waveform here. Let's not worry about what happens way down here.

We'll talk about that a little later. But, look at this curve here. I claim there is amplification in the following sense. Focus on some change in the input voltage, delta VI, OK, and for that change in input voltage, I get some change in the output voltage.

OK, for some change in the input voltage, delta VI, I get some change in the output voltage. And guess what? In this, at least the way I have drawn it, delta V nought divided by delta VI, if I can find regions of the curve where this is greater than one, then I have amplification.

OK, so what's that saying? What that's saying is that if I apply some voltage here, OK, and I change that voltage by a small amount from, let's say, 2 V to 2.1. OK, I am going to find the output voltage.

Let's say I go from 2 V to 2.1 here. OK, abstractly out there, I might have an output that goes from three to, let's say, two V perhaps. OK, so for a 0.1 change here, I'm going to get a bigger drop here, so from 3 V to 2 V, giving me an amplification in this little circuit.

OK, so we'll see this again and again, and you'll really understand it. So, I have a small change in the input, and I have a corresponding larger change in the output. So, I've shown you an amplifier.

I haven't shown you a linear amplifier. There's an extra charge for that. OK, that'll happen later. OK, all I've shown you so far is an amplifier, and this happens to be a crummy amplifier. It's a nonlinear amplifier because, notice, this is not linear.

It's a nice little curve, and so it's not linear. But, I promised you an amplifier, and I'm cheap, and that's all you get for now. OK, we'll see linear stuff later, but for now, I have a little amplifier.

So, let's do some real numbers, and plot some numbers down, and also look at a demo. So, let's do an example. Let's say VS is 10 V, that the K is two milliamps per V squared, and let's say RL is five kilo-ohms, OK? So, let me substitute these values into that equation, and I get V nought is, VS is ten.

So, it's ten minus, KRL divided by two. So, K is two milliamps. Two milliamps times five kilo-ohms is ten divided by two gives me five, and VI minus one squared. That's what I have. I just plug in a bunch of numbers, and that's what I get.

So, what I'll do is let me just do a little table for you, and plot using real numbers, simply plot those values for you.

Transcript - Lecture 9a
6.002 Circuits and Electronics, Spring 2007

All right. Let's get started. I guess this watch is a couple minutes fast. First a quick announcement. In case you have forgotten, your lab notebooks are due tomorrow with the post-lab exercises for the first lab.

OK, so I am going to continue with amplifiers today. And to just give you a sense of where we headed, we have this five lecture sequence covering different aspects of amplifiers with dependent sources and showed how we could build an amplifier with it on Tuesday.

Today I am going to show you a real device that implements a dependent source. And then next Tuesday we will talk about analysis of an amplifier. Wednesday is our quiz. Thursday and the Tuesday after that we then talk about small signal analysis and small signal use of the amplifier.

Today we will talk about the MOSFET amplifier. So let's start with a quick review. And in the last lecture, I showed you that I could build a amplifier using a dependent source. And a dependent source worked as follows.

Let's say I had a circuit and I connected a dependent source into the circuit. Let's say in this example I have a current source. So this is some circuit. And the current i is a function of some parameter in the circuit.

That's why this is a dependent source. This is a dependent current source. So it could be that I have some element inside. And I measure, I sample the voltage across the element or between any two points in the circuit.

And, in this little example here, this current could be dependent on that voltage. So notice that although I showed you the two terminals of the dependent source that carried a current, there is another implicit port, another implicit terminal there.

And that terminal there is called the "control port" of the dependent source at which I apply a voltage or current that will control the value of the current source. As a quick aside. There is a small glitch with the tools in your tool chest.

We talked about the superposition technique where you were taught to turn on one source at a time, for a linear circuit one source at a time, and then sum up the responses to all the sources acting one at a time.

Well, what do you do about dependent sources? A dependent source is a source. And we have to modify the superposition statement just a little bit. And for details you can look at Section 3.5.1 of your course notes on the details and some examples on how to do this.

So the approach is very simple, actually. The approach is, for the purpose of superposition, to not treat your dependent source as sources that you turn on and turn off. So what you do is when you do superposition with dependent sources simply leave all your dependent sources in the circuit.

Just leave them in there and turn on and off only your independent sources. So look at the response of the circuit by turning on your independent sources one at a time and summing up the responses.

And your dependent sources stay within the circuit and simply analyze them as you do anything else. So essentially what it says is that just be a little cautious when you have dependent sources, but the basic method applies almost without any change.

The readings for today's lecture are Section 7.3 to 7.6. So since we are going to build up on the dependent source amplifier, let me start with a quick review of that amplifier. We built our amplifier as follows.

We connected our dependent source in the following manner. And the current through the dependent source in the example we took was related to an input voltage vI. So some voltage vI. And so these two were the control port of the dependent source and a vI was applied there.

And I showed you a simple amplifier built with a dependent source that behaved in this manner. And again I will keep reminding you, just remember that the dependent source is actually this box here, the control port and the output port.

And commonly we don't explicitly show the control port for those dependent sources for which the control port does not have any other affect on the circuit, like it doesn't draw any current or things like that.

So in this particular example we said that this behaved in the following manner for vI greater than or equal to 1 volt and iD was zero otherwise. So we can analyze the circuit to figure out what vO is going to look like.

And a simple application of KVL at this loop here, again, you know, when I say this loop here, I am pointing at something here. That is the VS source that is implicitly across these two nodes. Again, this is a shorthand notation where this little up arrow here implies that I have a voltage source connected between these two terminals here.

And so there is a loop here that involves VS. So Vo is simply VS minus the drop across this resistor. So it's VS minus the drop across this resistor gives me vO. And the drop across the resistor is simply iD RL.

iD is the current here and that's the drop across the resistor. And I could get the explicit relationship of vO versus vI by substituting for iD as vI minus one all squared. So vO relates to vI in the following manner.

Nothing new so far. I have pretty much reviewed what we did the last time. Here is where we take our next step forward with some new material. Up to now I have talked as a theoretician would where I said just imagine that you had spherical cow or something like that.

Here I just asked you to imagine this ideal dependent source, control port and an output port, and it behaved in this manner. So as a next step what I would like to do is show you a practical dependent source which turns out to be a little bit more complicated than this idealized dependent source that I showed you in many dimensions.

Real life tends to impose a bunch of practical constraints on you, and we will look at those in a second. If I could find a dependent source that looked like this -- We had a control port A prime and output port B prime.

And I looked at some examples where the current through the dependent current source was some function of the input voltage. This is a "voltage controlled current source". What I am going to do is talk about a device that can give me this behavior or some close approximation to it.

It turns out that under certain conditions the MOSFET that you have already looked at behaves in this manner. The MOSFET that you've seen sort of behaves like this. And let me show you under what conditions the MOSFET behaves in that manner.

Let me create some room for myself. Notice that I need a control port, needed an output port. And I am going to view my MOSFET in a slightly different manner than you have seen before. I draw these two terminals here.

And this was a three terminal MOSFET. This was my drain, my gate and my source terminal. It was a three terminal device, but what I do is I view the MOSFET slightly differently. I will just use this terminal to be common across both the gate and the drain.

And so this voltage here is vGS. I am just using the source port, the source terminal along with the gate as a terminal pair. I am using the same source along with the drain as another terminal pair.

So I have a vDS out there and I have some current iDS that flows out here. Notice that when I view the MOSFET in this manner I have accomplished my first step, which is I seem to have a box which has a port here and a port here.

And I also explained to you that a MOSFET behaves in a particular manner. For one, the output port behaved as an open circuit under certain conditions when -- This was vGS, G, drain and source. When vGS was less than a threshold voltage VT this MOSFET had an equivalent circuit that looked like this.

So when vGS was less than some threshold voltage VT then there was an open circuit between the drain and the source. And you saw this before. So far nothing new here. However, when vGS is greater than or equal to VT -- vGS was greater than VT.

The MOSFET behavior we looked at earlier showed that this behaved either like a short circuit in the simplest form or in a slightly more detailed form it behaved like a resistor. We call that the SR model of the MOSFET.

So when vGS was greater than VT we said that a simple way to approximate MOSFET behavior was to view this as a resistor connected between the drain and the source. That was our SR model use of the MOSFET.

It turns out that we kind of lied. We were sort of looking at the MOSFET in a really funny way. And I shone the light on the MOSFET in a really, really clever way. Well, I shouldn't say clever. A really, really tricky way.

And tricked you into believing that it was just a resistor. And we constrained how you use the MOSFET. So that behavior was indeed a resistive behavior. But it turns out that in real life the behavior of the MOSFET between the drain and the source terminals is much more complicated than the limited form in which you saw it.

So today what I am going to do is take the wraps off the complete MOSFET and show you its full behavior in all its gory glory. And I will spend a bit of time on that to clearly emphasize under what conditions the MOSFET behaves like a resistor, as you saw when you did digital circuits, or behaves differently in other domains of use.

Let me pause for a second and leave this space blank here. And let's do some investigations. Let me leave this here. I won't draw in anything yet. You will figure out what it looks like yourselves under certain conditions.

What I will do next is apply some voltages on a MOSFET and observe the current versus vDS behavior and plot that on a scope and take a look at it. What I am going to do -- -- is figure out what iDS looks like for -- Remember iG into the gate for 6.002 is always going to be zero.

In much more detailed analyses of the MOSFET, in future courses you may see slightly more complex behavior. But as far as we are concerned it is an open circuit looking into the gate. So I am going to apply a vGS across the MOSFET, apply a vDS across the MOSFET and plot iDS versus vDS.

First let me show you what you already know. What you already know -- This is vDS. I will just keep doing as much as I can of what you already know. And then when I do some new stuff I will tell you explicitly.

You've seen this before. The MOSFET behaves like an open circuit when vGS less than VT. That is when vG is less than a threshold voltage VT, I have zero current flowing through the MOSFET. And when vGS was greater than VT then the S model of the MOSFET the switch model simply said that look, we can model the D2S as a short circuit.

You saw this in your labs and you saw that it was a very, very small resistance between the drain and the source and it kind of looked like a short circuit. But then we said well, that's not quite it.

There is some resistance. And so we said a slightly more accurate model would have this line droop a little bit to imply that there was some resistance R_on between the drain and the source, so vDS iDS.

So this was when vGS less than VT and vGS greater than or equal to VT. I have some resistance. And that showed me a straight line kind of like behavior. And I showed you that behavior. So far absolutely nothing new.

Now what I have plotted there for you is that behavior. Up here notice that this is the vDS axis, this is the iDS axis. I am plotting iDS versus vDS. And when vGS -- The gate voltage is more than a threshold, notice that I see what looks like something more or less like a straight line.

And this is a straight line with some slope, more or less a straight line implying resistive behavior. And we also had some fun and games here. We said hey, what if I turn vGS off? Boom. That would be my iDS of zero implying that the MOSFET behaved like an open circuit between the drain and the source.

I applied a positive vGS more than VT and it began to look like a resistor. Open circuit, resistor, open circuit, resistor, OK? Up until now nothing new. So you shouldn't have learned anything at all that is new until now in today's lecture.

Now watch. What I am going to do is, as I said, I kind of lied all this time and I just showed you this behavior. And what I have been doing all along is very carefully using a very small value of vDS.

Notice it's a small values of vDS. I haven't told you what it looks like as vDS increases. Well, let's go try it out. We have a scope here. We have the MOSFET here. Now, I am not sure what is going to happen now.

You may see smoke or have an explosion, who knows what? But look up there for a second. I am just going to increase vDS and you can figure out what happens for yourselves. I increase vDS. Whoa, what a liar.

Agarwal is a liar. I have been kind of tricking you. I have been putting -- Covering up all this part here and showing you just this region of the curve for small values of vDS. But as I increase vDS this is nothing that looks even close to that of resistive behavior.

So what's happening here? What's happening is that as I increase my vDS the iDS curve tails off and saturates at some value of current. Notice it saturates at some value of current. And so I am going to look at this region of behavior.

Notice that what we have looked at so far was the behavior for small vDS. It kind of looks resistive. But when I pump up the vDS, really whack this node really hard with a much larger vDS the guy says, oh, I give up.

And the current saturates out and flattens out and holds the value steady at some value. So what's that behavior look like? What is my horizontal line above the X axis in terms of V I elements? What is that behavior like? Current source, exactly.

So this is current source like behavior. And so let me start by drawing you a little model and explaining it in more detail. What happens is that under certain conditions, and the conditions are the following, when vDS, that is my drain to source voltage is greater than or equal to vGS minus VT.

When my drain voltage goes above vGS minus VT, so if vGS is 3 volts and if VT is 1 volt, then if vDS goes above 2 volts, if I am hammering the drain of the MOSFET with a higher voltage then this guy says I give up, can't show you nice restive behavior, and the current saturates out and it doesn't allow you draw any more current than a maximum value.

And that's the current source behavior. This one behaves like a current source. And the current iDS is given by the following expression. The current is given by iDS is equal to a constant K divide by two times (vGS-VT) all squared.

Kind of reminiscent of the carefully chosen dependent source example, just that this one here is VT. This model, which applies when vGS is greater than VT, the MOSFET has to be on and the drain to source voltage in the MOSFET must be larger than some value, and that value is vGS minus VT then this guy begins to behave like a current source.

This model of the MOSFET is called the "switch current source model". So in the region of the MOSFET characteristics where vGS is greater than VT and the drain to source voltage is larger than vGS minus VT, the MOSFET behaved like a current source between its drain and source terminals.

And in that part we model the MOSFET as a current source. And so not surprisingly that part of the model is called the SCS model in contrast with the SR model where we had a resistor. Again, remember, this is not meant to be conflicting.

It is not like gee, how can the MOSFET look like a resistor, and then suddenly what happens it becomes a current source. Well, the two regions are different. It is not that it is behaving as a current source for the same parameters, no.

When vDS is less than this right-hand side it does behave resistive. The SR model applies. But increase vDS beyond a point, the current saturates and the SCS applies like so. So let's draw. The SCS behavior can be drawn here vDS and iDS.

As I mentioned to you, for small values of vDS, let's say I pick some value of vGS, let's say vGS3, some value vGS, it is going to look like a resistor until vDS becomes equal to vGS3 minus VT. And after that it saturates out and begins to look like a current source.

And this point is where vDS becomes equal to vGS minus VT. And this way is when this equal sign becomes a greater than sign, vDS becomes larger then I move into this part of the curve. Similarly, for various other values of vGS it will look like this -- -- and so on.

And it behaved like an open circuit as before when vGS less than VT. When vGS less than VT it is still behaving like an open circuit. And so as I increase my vGS, provided I keep my vDS greater than vGS minus VT, I get current source like behavior.

And notice that this is increasing vGS. I have purposely drawn these curves at greater distances from each other to imply that it is a nonlinear relationship in that if I increase vGS by some amount, the increase in vDS is related to the square of vGS.

It is vGS minus VT all squared. So I get a family of curves of that look like this. And this is in the region of operation where vDS equals vGS minus VT. And this applies in this regime where vDS less than vGS minus VT.

This region of operation is called, as you might expect, the "saturation region". We say the MOSFET has been hammered, the MOSFET has been walloped, the MOSFET is in saturation. So the MOSFET is in saturation.

This region, corresponding to this, is called the triode region. This is really very simple. All we are doing is saying that when vDS is increased beyond a certain limit, given my vGS minus VT, the MOSFET begins to behave like a current source.

It cannot draw any more current. It limits the current to a given value like a current source. But on the left-hand side of this it behaves in a resistive manner. So what I would like to do is -- What I will do is, we've plotted for you, for the MOSFET, all its characteristics in its full glory for a whole bunch of values of vGS and a whole bunch of values of vDS.

And let me stare at those curves with you for a few seconds and walk you through them. So what do I have here? One of these curves corresponds to a given value of vGS. This may be vGS equals 2 volts.

This is vDS, the drain to source voltage, and this is the current. So focus on this curve for now. In the beginning I hid the right-hand side behavior from you and showed you just the resistive behavior out here.

When I increase vDS to be much larger the curve saturated and I got the saturation region operation of the MOSFET. And notice as I increase my value of vGS the saturation current also increases according to a square law behavior.

So these are the entire curves of the MOSFET. Finally the truth comes out. And notice that when vDS is less than vGS minus VT, I have more or less resistive behavior. But when vDS is greater than vGS minus VT I get current source like behavior.

So one question you may ask is when do I use one model or the other? When do I use the SR model and when do I use the SCS model? If you want to do a real detailed analysis then you can use the SR model when vDS is less than vGS minus VT.

And you would use this model when vDS is greater than or equal to vGS minus VT. That is simple enough. In 6.002, to eliminate confusion we constrain how we look at things a little bit more stringently.

And what we do is that for our entire digital analysis, for the entire digital world we focus on the SR model. And I will tell you why in a second. So for all digital circuits, invertors, look at power of invertors, look at delay, a bunch of other things, we will be using the SR model in 6.002.

And I will tell you why in a second. And for analog -- That is for amplifier designs and situations like that, we will be operating the MOSFET in a saturation region. And I will talk about that in a second.

What I am saying here is that in 6.002, when we do analog designs, we are going to discipline ourselves to using the MOSFET only in this region. We are going to constrain ourselves to play in only this region of the playground where vDS is quite large.

Why? Because I am asking you to. I am saying let's play in that part of the playground and keep your vDS high. And so the MOSFET is going to be operating somewhere in here. So we can apply just the SCS model, just the current source behavior in that region.

There is another important reason, which I will get to in a second. And for digital designs we will simply use the SR model. And it turns out that this is realistic because in the digital designs that you have you seen and will be seeing in this course, the pull down MOSFET is on, or when these pull down MOSFETs are on, the output voltage is pulled down close to ground.

So vDS is very, very small. So it does make sense that this model apply. And when we talk about amplifiers, I am asking you to follow this discipline. I will tell you why in a second. I am saying analog designs follow this discipline that I call the saturation discipline.

It says simply operate the MOSFET operating in saturation as a current source. We will look at an amplifier in a second, and I will tell you why. Now let's do a MOSFET amplifier. Remember my amplifier had an input port and an output port.

And in general in our use we are going to have a common ground. And we have a VS and a ground here as well. That is the power port of the amplifier. The input port and the output port. And let me redraw the circuit putting a MOSFET in place of the current source, RL, VS, vO, drain, gate, source, vI.

So my input is vI. Again, the MOSFET output is vO. And I have a resistor RL. Hey, we've seen that before. It turns out this is not surprising. You've seen this before. This was our primitive inverter circuit.

So what's different here? We showed you the circuit as an inverter. What's different here is that when we look at MOSFET behavior as a current source, this behaves like an amplifier. In other words, when vDS is greater than some value then this behaves like a current source.

When vDS is small, in other words, in the digital design when vDS was small here, because when the MOSFET was on it pulled the voltage down to ground, we could view this behavior as a resistor. And exactly the same thing, it is an amplifier.

And with digital designs, I was driving it with 5 volts and 0 volts and that was it, rail to rail. As an amplifier, what I am doing now is looking at a small region of its behavior when vDS is greater than vGS minus VT.

What I am saying is that for amplification let's follow the saturation discipline. And the reason is that when this behaves like a current source, what I have shown you is that if this behaves like a current source I have shown you that this expression up here gives you amplification.

In last lecture we plotted a bunch of values for vO versus vI, and we saw that we were getting amplification. For a small change in vI, I was getting a larger change in vO, and that was when I had the equation for a current source in there.

And so we know for a fact that if I can operate this as a current source, with a reasonable choice of values here, I am going to be able to get amplification. What I haven't told you is if this is operated in the linear region, in fact, you do not get amplification.

I won't cover that, but you can check that out in your course notes as a discussion or you can try it out for yourself. Replace this with the SR model for small vDS and you can show yourselves that you don't get any amplification.

In order to get the amplification we are telling ourselves let's focus on this part of the playground where vDS is greater than or equal to vGS minus VT. And for vGS greater than or equal to VT. So when vGS is greater than VT the MOSFET is on.

Further, when vDS is large, larger than vGS minus VT this behaves like a current source. So we have now created a small playground for ourselves where we can build lots of fun little amplifiers and other circuits.

And provided our circuits follow the saturation discipline where for the MOSFET or MOSFETs in the circuit these expressions are true then the MOSFETs are going to be in saturation, the current source model applies, and I will be indeed getting saturation.

In future courses you may actually see the MOSFET used in other regimes of operation for a variety of reasons. But in 6.002 when we talk about amplifiers and so on we will be adopting the saturation discipline.

And your homework problems and so on will state that. Assume that the MOSFETs are in saturation. What that means is that you can begin to model them as a current source and simply analyze their behavior accordingly.

One minor nit. Note that vDS for the MOSFET is the same as vO. And vGS for the MOSFET is the same as vI. So if you see me jumping back and forth using vOs and vIs or vDSs and vGSs they are the same thing in this circuit.

If you are dealing with circuits with many MOSFETs then you will have vDS1s and vGS1s and so on and so forth. But for this simple circuit, vO and vDS are the same, vI and vGS are the same. So we could go ahead and analyze that circuit.

What I do to analyze the circuit, I am telling you this. I am telling you that the MOSFET is behaving in saturation. I am telling you this. We have disciplined ourselves to say that in that circuit the MOSFET is in saturation.

As soon as we tell you that we can then go ahead and analyze that circuit. And to analyze that circuit what you will do is simply replace the MOSFET with its equivalent model, and that looks like this.

Since you have been told that it is in saturation, we can replace the MOSFET with its current source model. And the current iDS for the MOSFET is given by K/2(vI-VT)^2. And it is always good to write the constraints under which you are implicitly working close by.

So the constraints are one, vGS is greater than or equal to VT, vDS is greater than or equal to vGS minus VT. These constraints immediately follow from a statement of the type we are operating under the saturation discipline or the MOSFET is in saturation.

Let me just mark this equation as A, and we will refer to it again. So with this new little circuit with the MOSFET working as a current source, let's go ahead and analyze our amplifier. Notice that to analyze the circuit I have a current source.

It's a dependent current source where the current depends on the square of the input. So I want to go and analyze it. This is a nonlinear circuit. So I can apply any one of the methods that we talked about last week for nonlinear circuits.

To analyze it I will go ahead and use the analytical method. And my goal will be to obtain vO versus vI. Again, remember where are we here? The MOSFET circuit operating in saturation so I can replace this with a current source.

It is nonlinear. And so I can apply one of the two methods, the analytical method or the graphical method. Let's do both and start with the analytical method. The analytical method simply says go forth, apply the node method and solve.

Simple stuff. Let's go ahead and do that. Node method. I have a single node here that is of interest. I know the voltage vI at this node. I know the voltage VS at this node. So the only unknown is here at vO.

So I will go ahead and do that. Let me go ahead and equate the currents into the node to be zero. So the currents out of the node here are iDS. And that was equal the current into that same node.

So iDS must equal VS minus vO divided by RL. iDS=VS-vO/RL. For later reference, let me call that B. Simplifying, what I can do is, we know that iDS is given by K/2(vI-VT)^2. So I replace iDS with this expression and I multiply that by RL.

So I get K/2(vI-VT)RL. So iDS gets multiplied by RL and I get vO on this side and VS remains out here. All I have done is multiplied both sides by RL. So it is RL iDS, taken RL iDS to this side, that is here, I get the minus sign, and VS stays here, vO comes here.

So that is my final expression. Remember this is true under certain conditions. I will keep hammering that home because some of the most common errors made by people is in forgetting the constraints under which this was obtained.

And the constraint under which this was obtained is the saturation discipline. And that was true when vGS for a MOSFET was greater than or equal to VT and vDS for a MOSFET was greater than or equal to vGS minus VT.

I also know that for vGS less than VT, vO=VS. So when vGS is less than VT then this one turns off. That's why it is the SCS model, switch current source model. When vGS is less than zero it turns off and VS directly appears at vO.

I would like to stare at this constraint with you for a second, vDS greater than or equal to vGS minus VT here. And vDS is simply vO. I want to rewrite this constraint in terms of iDS. It will come in handy.

So iDS is K/2(vI-VT)^2. This is vI-VT. So vI-VT is simply square root of 2iDS/K. In other words, I can write iDS less than or equal to K/2vO^2. So this constraint expressed in terms of iDS is simply iDS less than or equal to K/2vO^2.

So all I've done here is analyzed this nonlinear circuit. I can also analyze it using the graphical method. And in order to do that, for my nonlinear circuit, in order to do that, all I have to do is plot.

Let's have iDS here and vDS here. And as we did with a nonlinear expo dweeb, what I do is I plot the device characteristics iDS versus vDS. The device characteristics under saturation look like this, so vGS increasing.

iDS versus vDS has a bunch of curves that look like current sources of increasing values. That simply reflects equation A. And then I superimpose on top of that the expression that comes up due to equation B which is iDS equals, let me write that down here, iDS equals VS/RL - vO/RL.

That's B. And let me plot that. That is a straight line relationship between iDS and vO. And so when vO is zero iDS is VS/RL. And when iDS is zero vO equals VS. Remember, vO and vDS are the same.

So this is what I get. This is the straight line corresponding to equation B here. And, as before, we just find the point where the two intersect. Let's say I am given some value of vGS. And let's say I am given some known value of vDS.

So for that I can go ahead and find out the corresponding value of iDS from this graph. Just as I told you when we did the expo dweeb stuff, this line here is called a load line. You will be seeing that again and again and again where we have the equation corresponding to the one shown here, the equation written for the output loop superimposed on the device characteristics.

That's called a load line. So I can get this point corresponding to the operating point of the MOSFET for this iDS, vDS and vGS by using the graphical method. In the next lecture we are going to look at, given a device of this sort, how do we figure out the boundaries of valid operation so that the MOSFET stays in saturation?

Transcript - Lecture 9b
6.002 Circuits and Electronics, Spring 2007

Good morning, all. Let's get going. So I guess you had your quiz review yesterday. I hope you guys didn't beat up on (name of TA) and who else was it? (Name of TA) too much. As you know, the quiz is tomorrow.

And unfortunately MIT couldn't give us one big room so we are broken up into three rooms. And you will go to your room based on the first letter of your last name. OK, so today we shall cover a topic called "Large Signal Analysis".

So in the last couple lectures we looked at one dependent sources abstractly, and then we looked at an amplifier built using a practical dependent source called the MOSFET. Now, the MOSFET had to be operated in a given region of its operation in order to behave like a current source.

And while it behaved like a current source you would get amplification or a FET amplifier. So that was in the past two lectures. What you are going to do today is called large signal analysis, and this is a loaded term.

So large signal analysis means something very specific in our business, and I will describe to what that is. This analysis involves looking at a circuit containing, for example, a MOSFET and figuring out how to get that device to operate in a way that the MOSFET was always in saturation.

So you had to figure out, based on parameters that you could control, how to establish those parameters so that the circuit operated in a way that the MOSFET was always in saturation. So large signal analysis involves that.

And although the examples we use, use the MOSFET, the same kind of analysis can apply to any other device. Remember, your MOSFET is a primitive element that we use as an example in this course. There are other primitive elements that you can use.

The course notes, for example, discusses a couple other devices. One is the "bipolar junction transistor", the BJT, and works through a complete example from start to finish involving a circuit containing a bipolar junction transistor.

And you can do a large signal analysis of that device as well. It turns out that you need to operate that device in an interesting region of its operating space, and so you can conduct a large signal analysis of a circuit containing that device and figure out how best to operate that circuit.

So that is large signal analysis, and we will do an example and explain how this is done using an example today. So to quickly review where we have been so far, we looked at this little structure here, our MOSFET amplifier.

Notice that when I write a voltage at a node, that's a short form for saying I am looking at the voltage between the ground node and the node at which the voltage is written down. So VO here and VI applied here.

This is a very, very common circuit that we use. To emphasize one more point, in general, in the kind of circuits we look at both in this course and in real life, there are a few patterns that we use very commonly that keep repeating all the time.

Very often you don't have to look at every possible permutation and combination of how things could be connected. This sort of connecting thing is very, very, very common. And you will see a lot of this pattern.

And we do the equivalent circuit for this. In the equivalent circuit we replace the MOSFET with a dependent source provided this operated in the saturation region. So I will just say while operating under saturation the equivalent circuit would look like this, VO, VI.

And IDS for the dependent source was given by K/2 (VI-VT)^2. So this was an amplifier. Here was the equivalent circuit while this device was in saturation. And to operate in saturation, I said that certain properties need to be true for the MOSFET.

And there are two properties that need to be true for this to be operating in saturation. One is that its gate to source voltage needs to be greater than VT, so VGS for the MOSFET should be greater than VT.

And the second one was that the output voltage needed to be greater than the input voltage minus one threshold drop. And this was the same as VDS for the MOSFET, this was the same as VGS for the MOSFET.

So what are we really saying here? What we are saying is that look, we built this circuit using a MOSFET, and it is up to us as engineers to choose its operating points in a way that these two properties hold.

For example, to make the first condition true, I can discipline myself to operate such that VI is always greater than VT. Similarly, I can choose VS, RL and VI in a way that this condition is true, which says that the drain to source voltage across my MOSFET drain and source should be greater than VI minus VT.

As an example, if VI was 2 volts and VT was, say, 1 volt, then what I am saying is that VO should be greater than or equal to 2 minus 1 or 1 volt. So I need to keep this high, 2, 3, 4, 5, whatever, a high voltage so that this guy stays in saturation.

The relevant readings for the material that we are going to cover in the course notes are in 7.5.1 and 7.6. So that is pretty much a review of where we were. We said we could build an amplifier. Its equivalent circuit was shown on the right.

And, provided that, I discipline myself to operate in the saturation region or to have the MOSFET operating in the saturation region, then this would work like an amplifier and all would be good with the world.

So today -- -- we look at large signal analysis of a circuit. And an example would be this circuit up here containing a MOSFET. And, again, as I mentioned earlier, a large signal analysis is a loaded term in 6.002, or for that matter in circuits.

And large signal analysis involves two steps. The first step involves writing down the transfer function of your little circuit. In our case, VO is the output, VI is the input, so involves writing down VO versus VI.

Simply write down the transfer function. In other words, the relationship between the output and the input for that circuit. And, in our case, again, we've disciplined ourselves to adhere to the "saturation discipline".

And the second part of large signal analysis is to find out the valid input operating range. Find out for the given circuit parameters, let's say I apply a VS and I use some value of RL and I use a given MOSFET, which has a given value of VT.

The question then is that what is a valid set of input voltages that would operate the circuit in a way that I would be in saturation. And so find out the valid input range, and this would give me a corresponding output range -- -- for saturation operation of the MOSFET.

That is what we will dwell on in the lecture today. So what we are saying here is that if I am careful with how I apply VI for a given value of RL and VS and for a given choice of my MOS transistor then I can stay within saturation provided I select my input voltages carefully.

And the analysis that we will go through today will figure out what that range of input voltages is. And, again, I will use this as a motivating example, the MOSFET amplifier. But in general large signal analysis would apply to any other circuit as well.

For example, in recitation you may learn about other circuits containing a MOSFET. And you can do a large signal analysis of other circuits containing a MOSFET or you might learn about some other devices like the bipolar junction transistor, and you could do the same kind of analysis for that device.

So remember that the MOSFET amplifier here is an example. I will be using that as a driving example to explain large signal analysis. So the first step, as I mentioned earlier, is to get the VO versus VI.

And in general for some circuit that you build the output will not even be a voltage. There are certain circuits where the output might be some kind of a current. Let's say I am building some kind of a circuit where I would like the output current or the current through some edge of the circuit to depend on some input.

In that case the transfer function would be the output current versus VI. And if I had an input current here it would be output current versus input current, you know, whatever the given problem tells you.

So this is under the saturation discipline. And I will not rederive this for you. You can apply a good old technique like the analytical method. Or you can use the graphical method to get the appropriate answer here.

I wanted to point out in a quick aside that why do we care about graphical analysis? Once you have the analytical method, why do you care about the graphical method? And a student asked me a question after lecture last Thursday, and it occurred to me that it's not obvious why you need the graphical method.

So it turns out that often times you do not have an equation describing the device. So let's say, for example, I am a manufacturer. Let's say I am AMD. As AMD I sit down and my semiconductor division builds a MOSFET.

And when you build a MOSFET your experiments and your fabrication division often times doesn't give you an equation with the MOSFET. They build something and then you look at it and you experiment with it.

You apply various input voltages and you measure currents and output voltages and so on. And so what you end up getting is a graph that describes the behavior of the MOSFET. And you have seen this in your lab as well, your 2N7000 or was it 2000? 7000.

So your 2N7000, the MOSFET you use in the lab also gives you a data sheet. And in that data sheet you see a bunch of curves. So very often devices come with data sheets. And when you have a data sheet but no equation then you can apply the graphical method and solve your circuits.

In this example, assuming I can apply the analytical method, here was the expression that I had derived for you in the last lecture. So VO was related to VI using the square law relationship. And we can plot and do other fun stuff with this equation.

So here is the input voltage VI. That is my VT. So notice that VO is VS. This is true when VI greater than or equal to VT and VO greater than or equal to VI minus VT. So these are the constraints of the saturation discipline.

And in our particular situation when VI was less than VT output would simply be VS. If VI is less than VT the MOSFET would turn off, switch off, and I would have no current flowing through RL, and VS would appear at the output.

So until VT, I have VS, and then following that I get the square law behavior articulated by this equation. And this was simply VS-K/2 (VI-VT)(RL^2). So that's the first part. You have seen this before.

The transfer function shows that I have a square law dependence between VI and VO. So now I can embark on the second step of my large signal analysis, and my goal is to find the valid input operating range.

So what does that mean? What I am looking to do here is, for this little circuit, is drain, source, gate, VI, VO, RL and VS. What I am looking to do is that given the value of the supply VS, RL and a MOSFET, in our case given a MOSFET implies that it is a given value of K and a given value of VT for that MOSFET.

So what I am going to do is find out, let's assume VI is my free variable here. So my goal will be to find out the range of VI for which this device stays in saturation. And I will use a couple of methods to do that, and I will use both a combination of a graphical method to give you intuition and then apply analytical analysis to get down to specific answers.

So let's start with the intuitive part. So here is VI, VO. I will use the transfer curve VO versus VI to help build intuition here. So that is what it looks like. So the first step, looking at this graph, we know that this point here, that VI needs to be greater than VT to satisfy the first equation.

Let me just write down both equations here. So VI greater than or equal to VT is one of them, and VO is greater than VI minus VT is a second equation. And just remember that this is the same as VDS and this is the same as VGS.

So VI must be greater than VT for the MOSFET to turn on. And so therefore the valid operating range starts at this point and is somewhere up here. So the first part is pretty easy. Somewhere here -- Somewhere at that point, my output voltage VO.

I'm not quite sure what that point is. My output voltage VO, as this keeps falling down, my output voltage VO goes lower than one threshold below VI. And at that point it goes into its triode region, and I need to find out what that point is.

So somewhere here I go into my triode region and begin to show a different behavior than the amplifying square law relationship there and go into my triode behavior. So I need to find out what this point is.

Once I find out what that point is then this will be my valid operating range. So let's figure out what that point is. At that point the following is true. Certainly VI is greater than VT. And at that point the output goes below one threshold, the input minus one threshold.

So at this point the following is true, VO is equal to VI minus VT. At that point the output voltage is equal to the input minus VT. And if the output goes lower then it will violate this equation here.

It is no longer greater than that number. So how do we find out what this point is? The principle intuition. Let's draw some lines here. Let's assume that VI and VT use the same scale, say, volts.

So if I draw a straight line at 45 degrees then that is a curve representing VI equals VO. We all know that. No big shakes. So the line at 45 degrees here is the line at which VI equal VO. And if I take that line now, the VI equals VO line, and I begin translating it to the right.

So let's take a line here. Let's take a line there. That line will be simply equal to VO equals VI minus VT. I have translated that to the right. And so this line is simply VO equals VI minus VT.

So this line is a locus of points at which VO is equal to this value. This minus VT shows up as a translation to the right. So I take my VO equals VI line, translate that to the right and it becomes VO equals VI minus VT.

Elementary geometry 101 or whatever. So what do we have here? Above this we have the condition VO greater than or equal to VI minus VT, and below that we have VO less than VI minus VT. If we look at this graph here, this is the valid input operating range.

Starting at this point greater than VT, and at this point my output equals VO equals VI minus VT. This must be the valid operating range for the input here to here. And correspondingly the outputs are from here to this point to here like so.

So this is my valid input operating range and this is my valid output operating range or the corresponding valid output operating range. So what does this say? What this is saying is that if I, as the designer of the circuit, am disciplined enough to apply inputs that are in this range, VT to some value here, graphically shown here.

Then my MOSFET will remain in saturation. And correspondingly my outputs will go between VS and some value here. So hopefully that gives you some of the intuition behind how we get it. And let's continue.

Let me label this point X. So continuing with two to get the valid operating range. I have shown you intuitively where that point is, but what I will do next is actually compute that for you. It is a pretty simple computation.

Note that point X is the intersection of two curves VO equals VI minus VT. And the second curve is VO equals VS minus K divide by 2, VI minus V2 all squared RL. So the point X iss at the intersection of these curves, and I can very easily get that as follows.

What I will do is I will simply substitute for VI minus VT from this equation here and then solve for it, so I get VO equals VS minus K divide by 2 VO squared RL. And so this gives me a quadratic in VO.

And I can solve for it pretty easily. And I get for a quadratic AX squared plus BX plus C equals 0. The solution is given by VO is minus B plus/minus square root of B squared minus 4AC divided by 2A.

And so I am just going to get those numbers here. So the coefficient of VO, that is B, is minus 1. Take the positive root because we are up in the positive voltages here. And square root of B squared, that is 1, minus 4AC.

So I get a plus 4 times K divide by 2 RL. And 2A is simply 2 times K divided by 2 times RL. So that is what I get. That gives me VO. So it tells me that VO, at the point where the output just equals one threshold drop below VI is given by the other circuit perimeter such as VS, RL and so on.

Oh, I am missing a VS here. I just forgot the VS up here. That is my VO. So what is VI equal to? Remember that at this point VO equals VI minus VT, so VI is simply VT plus -- I have not taught you anything earth shattering here.

I have just done some grubby math here to solve these two equations. So this is a straight line at 45 degrees from VT and this is the transfer function. And I need to find the intersection. And the intersection is given by this point.

So that point, VI being VT plus something, is simply the second dot on the X axis. So therefore I am pretty much done. My valid input range for VI goes from VT. So it starts at VT. That is where the transistor just turns on.

And then goes all the way to this point, VT plus minus 1 plus square root of 1 plus 2 K RL VS K RL. So this is my valid operating range. And again remember I won't dwell on this equation because, in some sense, you will get a different set of limits for other devices, for other circuits containing a MOSFET.

Or, for that matter, for other outputs that one may be focusing on. So what is more important here is not so much the results that you see but the process that I have gone through. So what is more important here is how did I get here? And the way I got here was looked at the graph and said look, the MOSFET is in saturation in that regime.

And I am finding the bounding points of the regime of saturation operation. So now, as an engineer, I can say that hey, look, if you build a MOSFET circuit like so, with a given value of RL, a given MOSFET and a given VS, then if you limit yourself you are operating with input voltages in this range thou shalt be happy.

If you go beyond that range then you will be violating the saturation discipline. So the corresponding output range -- I can write the corresponding output range, and that goes from VS, when the input is at VT the output is at VS and goes from VS down to the input minus VT.

Which is simply minus 1 plus -- Let me go back and quickly show you a little MOSFET circuit, amplified circuit so you can stare at a real transfer curve yourselves. And indeed convince yourselves that roughly at the point where proportionately shown in the curve up there the MOSFET indeed goes into its triode region and begins heading out of its saturation region.

Notice that here that is the same curve, the transfer function. And the amplified output is at VS until input reaches a threshold voltage VT. And once input goes beyond VT the output begins to drop precipitously.

And at some point here this begins to go into its triode region. And what I am going to do is simply increase the input voltage VI, and you will see that the output them begins to go into its triode region.

It keeps dropping. And, as you can see, the output begins to go into a space where the gain is no longer more than 1. And this is a triode region of MOSFET operation. So the MOSFET is in saturation, things are going great.

As I increase my VI notice at some point I begin to go out of my saturation region of the MOSFET. And somewhere here I go from the saturation region and transition into the triode region. And this value shown here gives you the corresponding input voltage and the output voltage.

Other practical devices like bipolar junction transistors or MOSFETs and other circuits and so on can be subjected to a similar analysis. And you can find out the valid operating regions for that device as well, or for that circuit.

So as a next step what I would like to do -- Out here I began by looking at the transfer function, the VO versus VI curve, and used that to drive the intuition behind how we calculated the bounding regions.

You can do the same kind of analysis intuitively looking at yet another curve, another set of graphs that you are familiar with, and that is a load line characteristic. And it is interesting to get two interpretations.

And you can use whichever one you feel comfortable with. So I will do two alternatively and show you another set of curves that you can use to get that. Here I am going to plot IDS versus VDS, which is the same as VO.

This was the load line graph that we had seen earlier. And, just for our reference, remember that VI must be greater than VT for saturation operation. Similarly VO should be greater than or equal to VI minus VT for saturation operation.

Those are my limits. The way we got the load line graph was we superimposed the load line equation over the device characteristics. And so let me plot the device characteristics in the saturation region.

Remember that this constraint could be related to the current as I derived for you in the last lecture as follows. IDS being less than or equal to K divided by 2 VO squared. So in terms of my IDS versus VDS relation, this lateral constraint is equivalent to IDS being less than K by 2 VO squared.

So this is that equation. So this line is IDS equals K by 2 VO squared. And in this region I have the valid operating region where IDS is less than that quality. So here are all my other curves for various values VGS.

So here are my devices curves, IDS versus VDS. Remember that these curves come down like this, for the MOSFET, right? Just that we focus on the right-hand side because that is where the MOSFET is in saturation.

And on this side the MOSFET is in its triode region, and we discipline ourselves not to operate the MOSFET such that it is in its triode region. So those were the device characteristics. And then I could plot my load line equation.

My load line equation, if you recall, was IDS = VS/RL - VO/RL. This was simply obtained by writing KVL at the loop containing the output node and the supply VS. Notice there that VO is equal to VS minus IDS times RL.

And that is simply by dividing by RL on both sides and moving IDS to the left-hand side we get this equation. And this equation gives rise to a curve that looks like this. And what is this point here? This point is where VO is 0.

So when VO is 0 my IDS is simply VS divided by RL. And this point is obtained when IDS is 0. And under those conditions VS and VO are equal so this is VS. This is my saturation region and this is the triode region.

This was another interesting graph. We often times fondly call it the load line graph. So here is a load line superimposed on the MOSFET device IDS versus VDS curves for a variety of values of VGS.

So by looking at this curve, we can also intuitively determine the valid operating range. So what are the two points here? I will let you stare at it for a couple of seconds yourselves to figure out what two points here bound the valid operating range of the MOSFET, the valid operating range of the circuit.

I will start. One is this point, because at this point the output is VS and VGS has just begun to equal VT. So think about where the second point is for valid operation. It is here, and, somewhere along that load line.

Remember the load line is a constraint that must be met by the output VO. It is the constraint imposed by KVL on the output. So the output is constrained to operate in this regime for various values of VGS.

So as the output keeps going from here all the way here, at some point I exit my saturation region. And that other point is given by this one. So notice that this is the curve that bounds. On the left-hand side of this the MOSFET is no longer in saturation.

It is on the right-hand side, and so therefore this is the valid operating region. Here to here. This is good. This is VS. That is good to know. And for this point I know that VI, which is VGS, equals VT.

I know VO is equal to VS. And IDS, at this point, is 0. So VO and IDS being VS and 0 correspondingly are the output operating perimeters when VI equals VD. So that is one point. And let's find out what this point is.

At that point I get my output just entering the range of the MOSFET triode region operation. Notice that this point is the intersection of two curves, this line and this curve. So this curve here is given by IDS equals K divided by 2 VO squared.

And this is my load line equation. So that is VS divided by RL minus VO divided by RL. That's it. So I won't go ahead and solve that for you. You can go and check it out and convince yourselves that if you solve these two equations and find out the VO for this, it should be the same VO that you obtained using the other graph.

What I have done here, obtaining the valid regions of operation is no different from what I did here. The two are alternative approaches to getting to the same place. Just that over the years what I have discovered is that there are one class of people that are output transfer function people, this graph, and another set of people that are load line people that like to think that way.

I have always been a transfer function person myself, but some of you may be load line people and so you can use that to drive your intuition. It is pretty amazing. As we get into this business and keep going down the path, it is amazing how some people really kind of get the load line thing and others feel much more comfortable with the transfer function.

So pick whatever you want. So what we have so far is we have conducted a large signal analysis of a MOSFET amplifier. It is an analysis of a circuit, and we found two things. One is the transfer function under saturation operation, and we found the valid input operating ranges and the corresponding valid output operating ranges for the circuit.

In the last five or six minutes let me talk about a couple of other issues. And the first issue is what we have done so far is intuitively and mathematically shown you what the valid regions are. Now you are thinking that's fine, but how do I get there? This region is good, VT through that other point, that's good, but how do I get there? How do I make my amplifier operate in that region? The answer is pretty simple, and let me drive the intuition again using a graph.

So this is a graph. And I showed you that -- That was my valid region here. Take a 45 degree line, find out where it intersects the transfer function, then this is the valid region here, VT through that coordinating function that we developed out there.

If I have an input that looks like so, some input whose gyrations fall within this range, will constantly keep the MOSFET in saturation. And the corresponding output will look like this. If my input is in this range, my output will be within this range.

And how do I get my input to be here? Let's say I have a sinusoid that is 1 volt peak to peak or whatever. How do I get my sinusoid up there? Well, you have learned the trick on how to boost things.

Remember boost? All you have to do is boost up your signal by some value capital VI. And the way you do that is as follows. VS, RL, VO. What you do is you apply a DC offset to your input. You take your sinusoid and boost it up so that all the gyrations of the input are in the valid range.

This is my input, some VA. Then I apply some DC offset capital VI given by this value here. And boost up the interesting input? My interesting input is the VA. And I boost it up by capital VI so that this guy is always in saturation.

I would like to show you a little demo now. I am going to show you an input that is a triangular wave. And what we will do is I'll play with a wide variety of offset voltages. This guy is a triangular wave.

And what I am going to do is apply a triangular wave and we'll look at the output and convince ourselves that I get amplification when VI is big enough that the input goes into a valid operating range.

And we will look at a variety of ranges here. You can put it a little larger. OK. So the triangular wave is my input. And this is my output. This looks nothing like a triangular wave. And the reason is that I do not have the right offset.

So what I will do is gradually increase the offset on the MOSFET. So at this point the offset is very low, a very small near zero offset. And so therefore my output is a disaster. My MOSFET is not in saturation all the time.

So what I will do here is apply some sort of offset. Is this the one? We want to switch. This is the input. You can see I am applying an offset by bumping and boosting up the input. I don't have clipping happening at both ends, but I get something.

And I get amplification. Now let me apply way too much of an offset. With this offset I am kind of operating here. What I will do now is apply an even higher offset so that this triangular wave begins to move here.

If I apply a very high offset what I am doing is overdriving the amplifier, boosting it so high that the MOSFET is going to go into its triode region and you are going to see that I won't have any gain.

My output is going to shrink noticeably if I overdrive the input. You will notice the input going higher and higher. Pull the trigger point down. There you go. Notice that as I boost up my input even higher notice that the output is a really small image of what the right input should be.

The right answer here, of course, is that I apply some right amount of offset to boost up the input into the right regime so that the output is seen to be some amplified version of this input. So I showed you three things.

One is very little offset. That was like so, as the thing comes down. A very high offset, it gets killed again. And the right amount of offset. But notice that we still have a problem, even with the right offset.

The output is not linearly related to the input. It is nonlinear. And the answer to get a linear response is good old small signal stuff. And we will be looking at the small signal part in the next lecture.

Transcript - Lecture 10
6.002 Circuits and Electronics, Spring 2007

Good morning, all. Good morning. I hope you guys did not spend all of last night celebrating the Red Sox victory, but there is one more tonight. OK. Let's see. I trust the quiz went OK. What I will do today is take off from where we left off on Tuesday.

And continue our discussion of the large signal and small signal analysis of our amplifier. Today the focus will be on "Small Signal Analysis". So let me start by reviewing some of the material. And, as you know, our MOSFET amplifier looks like this.

One of the things you will notice in circuits, as I have been mentioning all along in this course, is that certain kinds of patterns keep repeating time and time again. And this is one such pattern.

A three terminal device like the MOSFET with an input and the drain to source port connected to RL and VS in series in the following manner, this is a very common pattern. There are several other common patterns.

The voltage divider is a common pattern. We keep running into that again and again and again. The Thevenin form, a voltage source in series with the resistor is another very common form. The Norton equivalent form, which is a current source in parallel with a resistor is also very common.

And it behooves all of us to be very familiar with the analyses of these things. Voltage dividers in particular are just so common that you need to be able to look at it and boom, be able to write down the expression for voltage dividers.

I would also encourage you to go and look at current dividers. When you have two resistors in parallel and you have some current flowing into the resistors to find out the current in one branch versus the other very quickly.

The expression is very analogous to the voltage divider expression. And some of these very common patterns are highlighted in the summary pages in the course notes, so it is good to keep track of those and be extremely familiar with those patterns to the point where if you see it you should be able to jump up and shout out the answer just by looking at it without having to do any math.

So here was an amplifier. And then we noticed that when the MOSFET was in saturation it behaved like a current source. And this circuit would give us amplification while the MOSFET was in saturation.

So we agreed to adhere to the saturation discipline which simply said that I was going to use my circuit in a way that the MOSFET would always remain in saturation in building things like amplifiers and so on.

And by doing that throughout the analysis I could make the assumption that the MOSFET was in saturation. I didn't have to go through -- Analysis became easier. I didn't have to figure out now, what region is the MOSFET in? Well, because of my discipline it is always going to be in saturation.

But in turn what we had to do was conduct a large signal analysis. Again, in follow on courses you will be given circuits like this. In fact, this very circuit with a very high likelihood. And you will be looking at more complicated models of the MOSFET.

Or you will be given the MOSFET like this and, let's say in that course the designers do not adhere to the saturation discipline, in which case you have to first figure out is my MOSFET in its triode region or in the saturation region? And depending on the region it is in you have to apply different equations.

So it is one step more complicated than in 002. In 002 we simplified our lives by following a discipline. And let me tell you that following a discipline is quite OK. When it simplifies our lives and we can do good things with it, it is quite OK to do that.

We are not wimps or anything like that. It is quite OK to have a discipline and agree that we are going to play in this region of the playground and build circuits in that manner. By doing so, we could assume the MOSFET was in saturation all the time.

And analysis simply used a current source model. By the same token, what becomes important is to figure out what are the boundaries of valid operation of the MOSFET in saturation? To do that we conducted a large signal analysis.

And it had two components to it. One of course was to figure out the output versus input response. And what this usually does is that it does a nonlinear analysis of this circuit. If it is a linear circuit it is a linear analysis.

And figures out what the values of the various voltages and currents are in the circuit as a function of the applied inputs and chosen parameters. And the second step we said was to figure out valid operating ranges -- -- for input and corresponding ranges for the other dependent parameters such as VO.

You could also find out the corresponding operating range for the current IDS and so on. So by doing this you could first analyze the circuit, find out the "bias" parameters, find out the values of VI and VO and so on.

And then you could say all right, provided, as long as VI stays within these bounds my assumption that this is in saturation will hold and everything will be fine. The reading for this is Chapter 8.

And today we will take the next step and revisit small signal analysis. In the demo that I showed you at the end of last lecture, I showed you an input triangular wave. And the input triangular wave gave rise to an output.

And we noticed that we did have amplification, I had a small input and a much bigger output. I did have amplification when the MOSFET was in saturation but it was highly nonlinear. The input was a triangular wave and the output was some funny, it kind of looked like a sinusoid whose extremities had been whacked down and kind of flattened.

And its upward going peak had been shrunk. So it was a kind of weird nonlinear behavior. I will show that to you again later on. And so it amplified but it was nonlinear. And remember our goal of two weeks ago? We set out to build a linear amplifier.

So today we will walk down that path and talk about building a linear amplifier. So to very quickly revisit the input versus output characteristic, VI versus VO, this is VT and this is VS, this is what things looked like.

Also to quickly review the valid ranges, until some point here the amplifier was in saturation, the MOSFET was in saturation and somewhere here I had VO being equal to VI minus a threshold drop. At that point the MOSFET went into its triode region and I no longer was following the saturation discipline.

So therefore this is my valid region of operation. We also know that the output was given by VS minus K (VI-VT) all squared RL over 2. Again assuming the MOSFET is in saturation. It is very important to keep stating this because this is true only when the MOSFET is in saturation, when I am following the discipline.

Notice that this is a nonlinear relationship. So VO depends on some funny square law dependence on VI. The key here is how do we go about building our amplifier? Take a look at this point here. At this point here let's say I have a VI input.

Corresponding output is VO. Focus is this point. And left to itself this was a nonlinear curve. Remember the trick that we used in our nonlinear Expo Dweeb example? We used the Zen Method. Remember the Zen Method? We said look, this is nonlinear, but if you can focus your mind on this little piece of the curve here this looks more or less linear.

If I look at a small itty-bitty portion of the curve and I do the Zen thing, and kind of zoom in on here. This looked more or less linear. This means that if I could work with very small signals and apply the signal in a way that I also had a DC offset of some sort.

Then I would be in a region of the curve, I would be delineating a small region of the curve which would be more or less linear. This was a small signal trick. And what we will do here is simply revisit the small signal model.

Most of what I am going to do from here on will be more or less a repeat of what you saw for the light emitting expo dweeb. Just that here I have a three terminal device, with a little bit more complication.

The equation is different. I don't have to resort to a Taylor series expansion. I will just do a complete expansion of this expression and develop the small signal values for you. Recall the small signal model.

It had the following steps. The first step will operate at some bias point, VI, VO, and of course some corresponding point IDS. This is Page 3. And then superimpose a small signal VI on top of the big fat bias.

Remember the "boost"? So VI is the boost. Boom. And above VI, I have small signal VI that I apply. And our claim is that response of the amplifier to VI is approximately linear. The key trick with this is that for my small signal model here, this is Page 3 here, and Page 2.

The key trick here is that with the small signal model, I operate my amplifier at some operating point, VO, VI. I superimpose a small signal VI on top of small VI on top of big VI. And then I claim that the response to VI is approximately linear.

And let me just embellish that curve a little bit more. Notice that in this situation this was my VI, which is my bias voltage, this is VO, which is the output bias, and of course not shown on this graph is the output operating current which is IDS.

One nice way of thinking about this is to redraw this and think that your coordinate axes have kind of shifted in the following manner. This is VI. This is also on your Page 3. This is VT. Remember this was the operating point, VO and VI.

And notice that we were operating in this small regime of our transfer curve here. And in effect what we are saying is that I am going to apply small variations about VI and call those variations delta VI or small VI.

And the resulting variations are going to look like delta VO. Also referred to as small V, small O. So I will have small variations here. And they give rise to corresponding small variations there.

One way to view this is as if we are working with a new coordinate system. Another way to view this is that so the capital VI and capital VO correspond to my VI and VO as the total voltages in my circuit, but at this bias point I can think of another coordinate system here with small VI and VO out there.

And for small changes to VI, I can figure out the corresponding small changes to VO. Just that all the analysis I perform here is going to be linear. And I will prove it to you in a couple of different ways in the next few seconds.

When I am doing small signal analysis I am operating here in this regime at some bias point. You have also seen this before. How do I get a bias? This is my amplifier RL and VS. This is Page 4. VO.

The way I get a bias is I apply some DC voltage VI and superimpose on top of that my small signal small VI. This is my DC bias that has boosted up the signal to an interesting value. And because of that what I can get is by varying VI as a small signal with a very small amplitude, I am going to get a linear response here.

And I can draw that for you as well. This is my bias point here. And if I vary my signal like so then my output should look like this. This is point VI, this is point VO, and this is my small signal VI and this is my small signal VO and this is capital VO.

So this small thing here is VI. I would like to show you a little demo. I will start with the same demo I showed you the last time. I showed you the amplifier. In the demo I am going to apply a triangular wave.

And initially I start with a large signal. And you will see that the output looks really corny, is going to look something like this. That's large signal response. And then I will begin playing with the input making it smaller, and you can see how it looks yourselves.

There you go. So this is where I stopped the last time. The last lecture I applied this input, time is going to the right, and the purple curve in the background is the output. It looks much more like a sinusoid with some flattening of its tips.

Nothing like an interesting triangular wave. What I will do next is that let me make sure I have enough of a boost here, enough of a DC voltage so that I am operating at some point here. I believe I already have that.

Notice that I can shift up the triangular wave input, or I can shift it down. So let me bias it here. I have chosen a VI that's about, I forget how many volts per division it is, but I have chosen some VI here.

And I biased it such that this is the input. You get a nonlinear response. It is amplified. It is much bigger. What I will do next is make VI that I apply smaller and smaller. I have already done the boosting.

Boom, that's a boost. So I have boosted up your VI already. Next is I am going to shrink it, and hopefully you will see that if all that I am saying is truthful here you will see a triangular response.

Let's go try it out. Watch the yellow. I am going to shrink the yellow and make it smaller and smaller. There you go. It is great when nature works like you expect it to. I have never seen a triangular wave looks so pretty in my life.

It is awesome. Look at this. Here is a tiny triangular wave. And the output is also a triangular wave but it is much more linear. Yes. Question? What's that? The question is that the output here is only as big as the input used to be before.

That's a good question. What I have done here is I am showing you a laboratory experiment. And let's assume that this input is the input I am getting from some sensor in the field. Assume that this is my input, not what I had before.

Assume that this is my input to begin with and this is the amplified output. What I can also do is I can also change the bias. And we will see this at the end of the lecture, in the last ten minutes of lecture.

How do you select a bias point? By changing your bias point you can change the properties of an amplifier to give you a preview of upcoming attractions. Let me ask you, what do you think should happen if I change the bias point? I have not shown you the math yet, so intuitively what do you think should happen? If I increase the bias what do you think is going to happen? Yes.

Good insight. Higher bias will be more amplification. Let's see if our friend is correct. Let me set a higher bias. Not necessarily, I guess. You're actually right, by the way. I am playing a trick on everybody here.

As I change my input bias. Notice that under certain conditions my output becomes smaller and gets more distorted. Under other conditions what is going to happen to my output is that it is becoming smaller and is going to get distorted again.

So there are a bunch of funny effects happening that reflect on the bias point, but for an appropriate choice of bias point as I increase the bias the amplification should increase. And I will show you that in a few minutes.

But it is a complicated relationship. Yes. This is finally getting fun. Here is the question. Professor Agarwal, we love your song and dance, but if you really want to get a high signal at the output and you want to amplify your big input signal how do you do it? So the question is let's say I have an input that is this big here, if it is this big, I have shown you how I can get things that are this big, but what if my input was this big? How do I get an output that is this big? Well, I will use one of those learned by questioning methods and have you tell me the answer.

Someone tell me the answer. How do I do that? Yes. Use another amplifier. So the answer is I will use one amplifier to go from here to here. And the suggestion is use another amplifier to go from here to here.

And, in fact, I believe that you may have a problem in your problem set where you will do that. And so you have only yourselves to blame. So how do you make this work? What you have to do is this VI has to be much smaller than the bias point VI on this one.

I have to build a different amplifier, choose a different set of parameters such that VI prime, which is the VI for this guy, is much less than V capital I prime for this guy. It's a design question.

You need to design it in a way that the signals of interest need to be much smaller than the bias voltage of this amplifier. So you may have to use much higher supply voltages. My amplifier, I believe, has a 4 volt supply or 5 volt supply.

You might have to use an amplifier with a much bigger supply, different values of RL and so on. And I know that the course notes also have some exercises and problem sets that discuss that in more detail.

Yes. This is even more fun. The question is, good question. The question is why do you need this guy here? Just use this guy, right? Why do you need this guy? Big guys rule, right? Who needs the little guys? Well, let me use the Socratic method again.

Why don't you give me the answer? You guys are smart. Why do you need little guys? Why do you need the small guy here? Anybody with the answer? Yeah. The big guy may not be as sensitive. I like that.

You know what? He is almost correct. I will show you why in a second. Anything else? Any other reason? Yes. Bingo. That is another good answer. So let me address both the answers. The answer given was that look, this amplifier is amplifying the signal by a certain amount, by a factor of 7.

And I have designed this such that this amplifies a signal by a factor of maybe 10. So in all I am getting an amplification of 70. This would be a great design question for lab next year. I give you a bunch of components and ask you to design an amplifier given the constraints with the highest amount of amplification.

It turns out that when you design your amplifier, in order to meet the saturation discipline and so on, you have to choose values of RL and VS and stuff like that and be within power constraints so the amplifier doesn't blow up and stuff.

And by the end of it all you are going to get a measly 7X gain out of it. The same way here, to be able to deal with a very small signal here and get some amplification, another set of values and you get 10X.

So they multiply. It is much harder to build one amplifier with a much larger gain. You know what? I just realized that we will be looking at this in the last five or seven minutes of lecture. I am going to show you what the amplification depends upon.

It depends upon K. It depends upon RL. It depends upon VI. Now the question is I have had all this time to think about how to stitch in sensitive into this, and I believe I can. It turns out that when you have large voltages and so on and you have practical devices, it turns out that the more current you pump through devices they tend to produce noise of various kinds.

So very powerful amplifiers are not very good at dealing with really tiny signals because they have some inherent noise capabilities. And so I guess that is sensitive. It is sensitive to noise. Another question? Yes.

Ask me the question again. I didn't follow. Let me just explain it. It turns out that I will not be able to pass this through the big amplifier to begin with because it is just going to give me a gain of just a factor of 7.

However, if I have a signal that is this big to begin with then I may just need this amplifier. I don't need the smaller guy. If my signal was this big to begin with, if I had a strong sensor that produced a strong signal to begin with, yeah, I can deal with just a single stage.

I don't need to two stages. It is all a matter of design. And it is actually a fun design exercise. Given a budget, dollars, right? You go to your supply room and look at the parts that you have and you go to build what you have to build with the parts that you have.

And so sometimes you need to build two amplifiers to get the gain or build a signal amplifier. It's all a design thing. All right. Moving on to Page 7. That brings us to the small signal model. Page 5.

What I showed you up on the little demo was that provided the signal input in this example VI was much smaller than capital VI out there as I shrank my input, I was able to get a more or less linear response at the output.

And so to repeat my notation at the input, the total input is a sum of the operating point input plus a small signal input. This is called the total variable. This is called the DC bias. It is also called the operating point voltage.

And this is called my small signal input. It is also variously called incremental input. This is more a mathematical term relating to incremental analysis or perturbation analysis. So VI, call it small signal, call it small perturbation, call it increment, whatever you want.

Similarly, at the output I have my total variable at the output a sum of the output operating voltage and the small signal voltage. I do not like using Os in symbols because big O and small O is simply a function of how big you write them.

It is not super clear. And in terms of a graph, let me plot the input and output for you. Let's say this is the total input and that is the total output. I may have some bias VI. And corresponding to that I may have some bias VO.

Hold that thought for a second while I give you a preview of something that we will be covering in about three or four weeks. Notice that as I couple amplifiers together, the output operating point voltage of this amplifier in this connection becomes the input operating point voltage of this amplifier, right? So when they connect this output to this input, the output operating point voltage becomes coupled to the input here so it becomes the input operating point voltage here.

Now I have a nightmare on my hands. As I adjust the bias of this guy, the bias of this guy changes, too. The two are dependent. It is a pain in the neck. And we being engineers find ways to simplify our lives.

And you will learn another trick in about three or four weeks. And that trick will let you decouple these two stages in a way that you can design this stage in isolation, go have a cup of coffee and then come back to this stage and design this stage in isolation.

For those of you who want to run ahead and think about how to do it, think about it. What trick can you use to get them in isolation? Moving on. What I would like to do next is address this from a mathematical point of view.

And much as I did for the light emitting expo dweeb analyze this mathematically and show you that if VI is much smaller than capital VI, I indeed get a linear response. This time around I won't use Taylor series because it turns out that this expression can be expanded fully.

So you don't have to buy into Taylor series and so on. I am going to list everything down for you. We know, to begin with, that VO for the amplifier is VS-RLK/2 (VI-VT)^2. What I am going to do for this, much as I did for the LED, what I'm going to do is derive for you the output as a function of the input when the input VI is very small.

In other words, when I substitute for VI, V capital I squared plus small VI. Much as I did for the expo dweeb, I want to substitute for VI a big DC VI. So VI is much smaller than VI. And show you for yourselves that the output response, V small O is going to be linearly connected to VI.

Notice that, let me write another equation here. This is a total variable. This simply says that if the input is VI then the output is going to be VO, which means that the operating point input voltage should satisfy this equation, correct? In other words, the operating point output voltage V capital O should equal VS-RLK/2 (VI-VT)^2.

This is at VI equals capital VI. This is very simple but may seem confusing. All this is saying is that look, this equation gives me the relationship between VI and VO. Therefore, if I apply capital VI as the input, I'm given that my corresponding output is capital VO, so they must satisfy this equation, right? Those are bias point values and that must satisfy this equation.

Simple. I know that. So hold that thought. Stash it away in the back of your minds. Now let me go through a bunch of grubby math and substitute for VI in this expression here. Let me go ahead and do that.

VS-RLK/2((VI+vi)-VT)^2. When I do something that is other than math I will wake you up. I will just keep doing a bunch of steps that are pure math. No cheating. No nothing. Watch my fingers. When I do anything that is not obvious math I will wake you up.

Next I am going to simply move VT over and rewrite this as follows, RLK/2((VI-VT)+vi)^2. Again, I haven't done anything interesting so far. I have just substituted this. I am just juggling things around just to pass away some time, I guess.

All right. Next what I am going to do is simply expand this out and write it this way RLK/2, expand that out and treat this as one unit VS - RLK/2((VI-VT)^2+2(VI-VT)vi+vi^2). Nothing fancy here. This is like the honest board.

Nothing fancy here. Standard stuff. Only math. I will move to this blackboard here where I do some fun EE stuff. Yes. Good. At least one person isn't asleep here. Thank you. So just math here.

Nothing fancy. Plain old simple math. I have not done any trickery. I still have all my ten fingers. Now what I am going to do, now watch me. I am not using Taylor series here because this expression lends itself to this analysis.

Notice VI squared here. I made the assumption that VI is much smaller than capital VI, so what I can do is assuming that VT is small enough that VI minus VT is still a big number compared to small VI, what I can do is ignore this in comparison to the capital VI terms.

So I have a capital VI term here. I am going to ignore VI squared. So, for example, if capital VI was 5 volts and small VI was 100 millivolts 0.1, so 0.1 squared is 0.01. So it is comparing 0.01 to 5.

So I am off by a factor of 500. So now watch me. Now I begin playing some fun and games here. I eliminate this, and because I eliminate that it now becomes approximately equal. What I do in addition is let me write down the output.

The total variable is the sum of the DC bias and some variation of the output. And let me simply expand that term and write it down again. VS-RLK/2(VI-VT)^2-RLK/2. I get a two here. And I get VI-VT.

I won't forget the VI this time. Again, from here to there nothing fancy. This is the one step where I have used a trick. I have said small VI is much smaller than capital VI, and so I have simply expanded this out and written it here.

So do you see the obvious next trick here? From star look at this guy. I can cancel this out from star because I know that at the operating point these two expressions are equal, and so therefore I can cancel out the operating point voltage and this.

What I am left with is small VO is simply minus RLK(VI-VT) times vi. Only one place where I did something funny. Other than that it is purely math. So this is what I get. Notice that this whole thing is a constant, minus RLK(VI-VT).

This whole thing is a constant. And so VO is equal to some constant times VI. Let me just define some terms for you that you will use again and again. For reasons that will be obvious next lecture, I am going to call this term here GM.

I am going to call this term a constant, K(VI - VT). It is a constant for a given bias point voltage. So I am going to call that GM. And then I am going to call this whole thing A. And of course this is VI.

There you go. I have my linear amplifier. A is the gain times small VI. And the gain has these terms in it. I just call this GM. You will see why later. But notice that the gain relates to RL. The size of the load resistor RL, how big it is, 1K, 10K, whatever.

K, this is a MOSFET parameter, and VI minus VT. That is a constant for a given bias point voltage and small VI. So VO equals small VI. I won't give you a graphical interpretation, but I encourage you to go and look at Figure 8.9 in the course notes.

And it gives you a graphical interpretation of that expression. Move to Page 7. Another way of looking at this, another way of mathematically analyzing it, here I went through a full blown expansion and pretty much deriving the small signal response.

What I can also do is take a shortcut here. So let me just give you the shortcut. You might find this handy. VO=VS-KRL/2(VI-VT)^2. And my shortcut is as follows. My small signal response is simply this relationship.

I find the slope at the point capital VI and multiply by the increment. Slope times the increment gives me the incremental change in VO as follows. d/dI (VS-KRL/2(VI-VT)^2) evaluated at vI=VI times vi.

This is math again. I want to find out the change in VO for a small change in VI, and I do that by taking the first derivative of this with respect to VI substituting V capital I and multiplying by the small change delta VI or small VI.

So this is simply the slope of the VO versus VI curve at VI. And so therefore taking the derivative here of this. This is a constant so it vanishes. But twice 2 to cancel out, so I get KRL(VI-VT) times small vi evaluated at capital VI.

So I get twice KRL, VI evaluated at capital VI, so it is VI minus VT times small VI. Same thing. Oh, and I have a minus sign here. I get the same expression that I derived for you up there, and this is just taking the slope and going with it.

And this, as I mentioned before, this is A. The last few minutes let me kind of pull everything together and also hit upon something that many of your questions are touched upon. And that all relates to how to choose the bias point.

So here I have taken an analysis approach. When teaching we often teach you are given something, you analyze it, but as you begin to master it you can begin to design things where you can ask a lot of questions and so on.

And here what we have is an analysis given a value of RLK, VI and so on. How to choose the bias point becomes more of a design issue. If you are designing an amplifier, you asked me the question, how do I choose two small amplifiers versus one big amplifier, that sort of stuff? It boils down to how do you choose the bias point? How do you choose VI? How do you choose RL and so on? What I would like to do is touch upon some of these things.

First of all, gain or the amplification. One of the most important design perimeters for an amplifier is what is the gain? Let's say you get a job at Maxim Integrated Technologies, and they say we would like you to build a linear power amplifier for cell phones.

You can say I know how to do that. And then they say the next stage needs a 100 millivolt input. While this thing coming from the antenna is only a few tens or a few hundreds of a microvolt. So you sit down and say oh, my gosh, I need an amplification of so much, and you go design an amplifier.

So gain tends to be a key parameter. And notice that gain is proportional to RL. It relates to VI minus VT, so proportional to VI. It is also related to RL. The second point is the gain point determines where I bias something.

If I choose my bias too high I get distortion, or if I choose my bias too low I get distortion. So depending on how I choose my bias point, as a signal goes up it may begin clipping or begin distorting.

And I will show you a demo the next time on that particular example. So bias point will determine how big of a signal you can send without getting too much distortion. And the other thing is that, relates to how big of an input, what is a valid input range? So let's say you have a signal.

And you want that signal to have both positive and negative excursions of the same value. Then, depending on where you choose a bias point, your input range may become smaller or larger. And we will go through these in the context of and amplifier and look at some design issues in the next lecture.