MOSFET Large Signal and Small Signal Models: Analysis and Biasing

Greetings. Welcome to electronic circuits one. I am Behzad Razavi and this is lecture number 33.

Today, we will look at the MOS device again and this time we will delve into what we call large signal operation and

small signal operation and based on these two types of operation develop models for the for circuit

representation of transistors. But before we go there, let's take a look and see what we did last time.

In the last lecture, we went into one second order of second order effect in MOS devices, namely

channel length modulation. And we saw that as the drain voltage increases,

assuming that we are already in saturation, so we are over this hump here,

then what we see is that the pinch-off point actually gradually moves to the left towards the source.

And that means that the actual length of the channel decreases, which according to our equation

indicates that the current should increase. But rather than make L a voltage

dependent quantity and making things complicated, we simply multiply the original equation by this factor of 1

plus plus lambda VDS. So that when VDS goes up, the drain current also increases.

And that's a very simple representation that we use in many types of analysis and design.

All right. We also looked at the concept of biasing. We saw that if we have a MOS device and we just apply a signal to it

without anything else, the device is dead. There's no current in it. We are hoping that it will convert a

voltage to a current, but it doesn't do that if we just apply a little signal to, for example, a gate source

voltage, but nothing else. So we what we realized was that we have to apply a certain voltage between the

gate and the source even before a signal is applied to keep the transistor alive, so to

speak. We saw that if we apply a certain voltage, we have a certain current and that allows the drain current

to be a sharper function of the voltage than if you are over over here or if you are over here.

So in fact, this brought us to the concept of how much current change we get for a given voltage change and

that's what we might consider the strength of the device as a voltage to current converter.

So we actually defined the slope to be the transconductance of the device as a measure of its strength.

So what we expect is that if we bias the device at a higher voltage, higher VGS in the absence of the signal,

which means we also have a higher current, perhaps, then the transconductance is higher because this

is a non-linear equation, right? It's a non-linear equation. So because of this non-linearity, the

slope, because of the second order effect, this second the power of two, this slope keeps increasing as VGS

increases. And that tells us that a higher bias voltage for gate source gives us a higher current and a higher

slope. All right. So today, we will delve into GM a little more and try to understand

its properties. As I mentioned last time, we have three different expressions for GM. We simply take the

derivative of the current with respect to VGS. We take the derivative of this and the

first equation that we find is this. So GM

expressions. So we saw that if we take the derivative,

this two cancels this one half and then we just have mu n Cox W over L times VGS minus VTH. All right, so that's one

equation that we can use. mu n Cox W over L VGS minus VTH.

But there are other manipulations that we can perform to obtain other types of expressions. For example, I know here

that VGS if I divide both sides by VGS minus VTH

and multiply both sides by two, on this side I have two times ID over VGS minus VTH

and on this side I have mu n Cox W over L VGS minus VTH, which happens to be GM. So GM is also equal to two ID

over VGS minus VTH. Okay. And lastly, what I can do is

I can just remove this VGS substitute for this VGS minus VTH from this equation. So VGS minus VTH is equal to

this divided by this square root of that, right? So I replace this VGS minus VTH with that and that gives you

the third equation, which is two mu n Cox W over L times the bias current, the drain current.

Okay. Now, if you look at these carefully, something looks to be a little strange.

For example, here, GM is proportional to VGS minus VTH, whereas here, it is inversely

proportional to VGS minus VTH. Which one is correct? Well, both are correct, but it just depends on what we

are doing. So let's go back to this equation and understand that there are three

parameters under the designer's control that are related by this equation. W over L is one of parameters. VGS minus

VTH is the second one and ID is the third one. These three are related by this equation. Mu n Cox is a constant of

the technology. We don't have much control over that. So when we are looking at these

dependencies and we want to know how GM varies with this or this, we have to remember that

in addition to the overdrive voltage and in addition to W over L, there's a third parameter, ID.

Or in addition to ID and in addition to the overdrive voltage, there's a third parameter, W over L and so on.

So let me elaborate on this and see how what we mean exactly. Okay, so here, we can say that GM, for

example, is proportional to W over L if we keep VGS minus VTH constant.

So VGS minus VTH constant. All right, that's fine, no problem.

Or we can say GM is proportional to the overdrive voltage proportional to this if we keep W over L

constant. That's fine, too. Nothing exciting yet, right?

Okay. Now, let's go to this equation and see what it means. So again, we keep one of them constant. Let's say we keep ID

constant. So, if ID is constant, it means that I have biased the device

at a current of 1 mA, right? There's a 1 mA current in it before any signal comes in because I need a certain amount of

transconductance. Right? I need to be somewhere on the on the higher parts of the ID versus VGS

curve, right? I have to be somewhere around here. So, I need a certain bias current, ID0.

So, that's the amount that we're assuming here. Okay. So, if ID is constant, GM is

inversely proportional to the overdrive. So, GM is inversely proportional to the overdrive.

All right. So, here's the tricky part. This says that if the overdrive voltage increases,

GM decreases. Okay, provided that ID is constant.

So, now we have to ask ourselves, we are increasing the overdrive voltage, but ID is constant. Is that possible?

Is it possible to increase the overdrive voltage of a MOSFET, but keep its drain current constant?

Well, we go back here. Remember, there are three parameters here. So, is it possible to increase VGS

- VTH while ID is constant? Yes. We have to reduce W over L.

Is it possible? Yes, we can do that. So, this condition is possible. You just have to be careful

in terms of what it means. It means that if you are changing VGS - VTH in this experimentation,

and you'd like ID to be constant because we're trying to see the dependence of GM upon overdrive, then implicitly we're

assuming that W over L is decreased. All right. Okay. Now, let's consider the other case. Let's assume that the

overdrive is constant, but ID is changing. So, VGS - VTH

is constant, but and then that means GM is proportional to ID.

All right. What does this mean? So,

I would like to increase ID so that GM increases. That happens if I keep VGS - VTH

constant. But how can I keep VGS - VTH constant while ID increases?

All right. So, we just go back to that same equation. How can I keep VGS - VTH constant

while ID increases? Well, that implicitly assumes that W over L increases.

All right. So, this scenario assumes that the overdrive voltage is constant, meaning you connect the battery from

here to here. That's it. The constant battery, we don't touch it anymore. But then I'm trying to get a higher

transconductance by increasing the current. How is that possible? The only way

that's possible is by making the device wider. W has to increase or L has to decrease. So, W over L has to increase.

All right. And similar observations apply to this one. I just won't go there, but I want you to know

that these three are all valid. It just depends on what we're assuming constant and what's variable for each of these

scenarios. Okay. So, to drive this home, let's look at an example.

Uh we take a MOS device like this uh with a certain W over L.

All right. It has some mu and Cox, and it has some threshold. Now, we copy this device over here. So,

exactly the same type of device. So, here's another type of device. Another copy of the same device.

So, it has the same mu and Cox, the same threshold, and the same W over L. Doesn't have to, but we chose it to have

the same W over L. Now, I place these devices in parallel. And by that, I mean I connect the gates

together, I connect the sources together, and I connect the drains together.

All right. So, my question is, if

VGS is constant, which means VGS - VTH is constant, how does the GM

of the composite device compare with the GM of one device? So, let's think about that for a second.

We had one device with a dimension of W over L, and it had a certain battery between the

gate and the source. So, it had some current and it had some GM. We can calculate it

using any of these, right? Now, I came along and I added another device in parallel to this device. The

drains are shorted, the gates are shorted, the sources are shorted. In a sense, now

these two devices are equivalent to one device with

twice W. Okay? W over L here, W over L here, the result is 2W over L.

And you can actually see that if you look at the MOS structure, right? So, here's the MOS structure that we

have uh developed before in three dimensions,

approximately. So, this is W. I'm sorry, this is L, and this is W.

And I take another one, exactly the same, and I place these two in parallel.

So, now what happens is when these two gates are connected to each other, and this is also W,

we see that it is as if the two devices are connected like so, as if the W the width of the device has

doubled. Okay? Imagine you just take this and take this and place them next to each other. If you look at the top,

it might be more it might be clearer. So, from the top, this is what we have. Uh we have

these are the source and drains, or source and drain. This is the gate, and this is W, right?

From here to here is W. And then we have another one down here. Source and drain,

and this is also W. And now we just short these together, right? So, we short the sources

together, the drains together, the gates together. So, we just bring this closer and closer

to each other, these two merge, and we see that the total W is 2W. All right. So, what this tells us is

that we have two devices uh whose equivalent width is 2W. Now, if you don't like this visualization, no

problem. You can just write equations and see what happens, right? So, we have one current for this guy,

ID1. We have another current for this guy, ID2. And because the drains are connected and

the sources are connected, we have ID1 + ID2 for the overall composite device. And they have the same VGS because the

gates are shorted and the sources are shorted, right? So, I write ID1 is equal to ID2

is equal to 1/2 of mu n Cox W over L VGS - VTH squared. So, that means that the sum

ID1 + ID2 it's just two of these, right? So, it's

equal to 1/2 mu n Cox two times W over L VGS - VTH squared.

Realize that you realize that this two has to go here. It has nowhere else to go.

It cannot go to this voltage. We cannot take this two inside a square root because they have the same VGS.

This VGS is the same as before when I didn't have one device. And mu and Cox again is a constant of the technology.

It's the same for both of these devices. So, equivalently, if I want to put these two devices in a black box, the only

equivalency I can create is to say the new composite device is twice as wide twice as wide as each of these devices.

All right. Okay. So, now what happens to the transconductance of the device?

Well, we write GM. We can try to use any of these. The perhaps the simplest one is this one

because we kept VGS - VTH constant. Right? They have the same voltage. We just added one more device. So, GM,

which was 2 ID 1 over VGS - VTH.

This is the older GM. Old GM meaning one device. The new GM meaning both of these are in

a box. So, the new GM is

two times Now, we said that the current has doubled. So, 2 ID 1

divided by VGS - VTH. So, we see that the transconductance is doubled.

All right, so you you really have to be careful when we are considering various cases as to what is

kept constant and what is varied because then you get different results. So, here because VGS is the same for one device

or two devices, the overall GM is twice that of one of them. Uh but if you did something else, you

could get something else, right? For example, this equation says, if W goes up by a factor of two,

GM goes up by square root of two. All right, so you have resolved to have resolved all of these in your mind to

make sure there's no confusion. Very well. Let's uh go on

to the uh one other

uh question that I raised last time. I just want to make sure that uh you have the correct picture in mind.

So, let me go back to that example where we said, let's build an amplifier last time and I raised the question.

So, here's where we are. So, let's build

an amplifier. So, this is what we saw. We eventually uh had a MOSFET,

which is a voltage-dependent current source. We had a microphone. We had a battery.

And uh the battery gave us some bias voltage to make sure we have a certain

uh current and transconductance and all of that. And then we had a resistor because we

wanted the current coming out of the MOS device to be converted to a voltage. We usually like voltages as outputs and

maybe even inputs. So, we pass it through a resistor. And we said, this is our output voltage.

So, if the microphone signal uh is superimposed on a voltage, when the microphone signal changes, it

wiggles this current. This current changes. This current the changes current flows through this resistor.

This voltage changes and that's the amplified microphone signal. But then I mentioned that uh

we do need to make sure that the device is in saturation. In saturation, you have to have this

condition, right? So, the drain-source voltage cannot be arbitrarily small. It has to be greater than VGS - VTH. This

has to hold before the signal comes in and it has to hold even after the signal comes in. So, at least let's make sure

that is true before the signal comes in. All right, so when we have no signal, VGS is equal to V0.

Okay. So, that's some amount. Uh so, let's just give numbers here. So, let's say V0 is uh

I don't know, 0.9 V. And uh let's say VTH is 0.5 V.

Okay. So, what is the minimum amount of VDS that we can tolerate? This equation says it has to be 0.9 -

0.5. So, for this device to be in saturation, we require that VDS be greater than or equal to 0.4

V. All right, if VDS is less than that, we have a problem. We are in triode region,

not in saturation. Okay. Now, does this circuit allow or enable this device to be in saturation?

In other words, can we have this much VDS here? Well, so if we need this much VDS here,

um and there's a current flowing this way, where is that current coming from?

Uh this part is open. So, all of this current has to come from this resistor. So, we really are saying that a current

is circulating here. A DC current before the signal comes in without a source of energy, without a

battery. How could that be? It's not possible, right? If you think of this MOSFET just for a second as a resistor

between source and drain, you have a resistor here, we have a resistor here. That's it. We have nothing else. How

could we have a current? So, this circuit does not provide VDS of this much. In fact, there won't be any

current here at all. VDS will be zero. VDS will be zero. The device will be in triode region. Everything is dead.

Nothing Nothing good comes out of it. Okay, so we have a problem, right? And we need to figure out what to do.

All right, so I really want the VDS to be, for example, more than 0.4 V, right? With the numbers I have chosen.

Okay, so can I do this? If I want that, can I go ahead and uh do this? I will come along and I just

put a battery maybe right here. Put a battery between the source and drain and choose

that to be 0.4 V. So, now the battery guarantees that this voltage minus this voltage is 0.4 V.

So, it guarantees that the device is in saturation. So, is that a good idea?

I'll give you 90 seconds to think about it. Okay. So, what did you decide?

Can we use a battery like so to guarantee VDS equals 0.4 V? Well, yes, we could. The only problem is

that now we don't have an output signal anymore, even though we I am talking to the microphone.

Why? Well, you can see that if there's a battery here, by definition, the voltage

across the battery doesn't change. Right? That's the definition of a constant constant voltage source.

Which means the voltage across this resistor doesn't change with time. Which means this value is always 0.4 V.

If I speak into the microphone, if I speak loudly, if I speak softly, nothing comes out, right? This is always 0.4 V.

So, this battery in a sense has shorted out this resistor. Uh the resistor has no role in the

circuit anymore because uh this voltage is dominated by the voltage source.

So, that won't do. We have to think of something else. Okay, so we have to fix this situation

and uh let's draw the circuit again. Uh with all the stuff, we have a microphone,

a battery for VGS. And we still need this resistor.

The resistor is fundamentally necessary because the current coming out of the MOSFET has to go through a resistor to

create a voltage. Right? So, we need that resistor. All right, so uh

that resistor, let's draw it like this. Except that I will not connect the bottom of the resistor to here, to this

common rail. I will put a battery here. Another battery and we call this V1 for now.

So, now something interesting happens. Okay. All right, so let's hope that with the

addition of this battery, now we have a current flowing through the device, the transistor, and we have a reasonable

VDS, right? Let's hope something like this is happening. Okay, so VDS is from here to here.

And uh the question is, can I guarantee that this VDS is 0.4 V

while this resistor is still there and the current from the uh

MOS device is flowing through RL to generate some sort of voltage, which represents the amplified version of the

microphone signal? Okay, well, let's just give some numbers here just to

have something to work with. So, let's say this resistor is 1 kΩ. And let's say we are shooting for a bias

current of 1 mA here. So, we'll call this ID0 to indicate that this is the current

before the microphone signal comes in, when the microphone signal is 0 V. Okay.

All right. Now, how much is VDS in this case?

Well, 1 mA flows upward through RL, right? That's to go this way through the uh transistor.

So, it generates a voltage. 1 mA * 1 kΩ is 1 V, but that 1 V is like this, plus here,

minus here, cuz the current going is going upward. So, that's 1 V.

So, that would be RL * ID0, positive here, negative here. Okay, so let's write a quick KVL around

this loop to see what we get. All right, so we have VDS here.

Then we have, if I start from this terminal, I say minus VDS, minus V 1 V, which is RL * ID0, plus V1

= 0, right? That's my KVL. So, minus VDS, then I'm going this way, minus 1 V,

uh or this would be really minus ID0 RL, and then plus V1

= 0. I'm interested in VDS, so we see that VDS

is equal to V1 minus I will write ID0 RL, knowing that it's equal to 1 V right now, right? Minus 1 V. Uh

Okay, so this says that if we have, for example, 1 V of drop across the resistor,

and we need 0.4 V for VDS at least, that is not that hard. We just pick V1 to be how much? So, this should be 0.4 V,

and and this should be that we are subtracting 1 V,

right? So, that means that V1 has to be 1.4 V. Uh So, indeed, if I come in and add a

battery from here to here with a value of 1.4 V, assuming that we have 1 mA of current

and 1 kΩ resistance here, then the voltage that remains for this VDS will be 0.4 V,

and the transistor will be happy. All right, so this allows us to provide this proper biasing.

So, proper biasing of a MOSFET requires some sort of voltage between gate and source when there's no signal, and some

sort of voltage between drain and source, so that this condition is satisfied, right? For saturation.

That means that we need a battery here, and we need a battery here. So, as we will see later, we can try to

get rid of this battery, but this is eventually our power supply, right? It's an amplifier.

It's a it requires to draw some energy from a battery or something to amplify, and

that is the source of energy for this amplifier. All right, this answers that question,

and now we are ready to move on to the next topic. Okay, the next topic is a large signal

and small signal operation. And I want to talk about that. If you watched the bipolar videos, we talked

about these before, but I'm assuming you haven't, so we'll start uh looking at those concepts again.

All right, so large signal

operation. Well, uh we'll start with an example

just to get warmed up. So, here's an example. Uh suppose we have something very simple

like this. I have two batteries connected to the MOSFET, V1 here,

V2 here, or sorry, I I guess I'll call this V0 and V1. So, V0 and V1 just to be

consistent with the previous example. Okay, and I have this resistor

tied between the source and this point. Why? Uh just just for the heck of it. We're we're curious to see what happens.

All right, suppose that I have everything I need to know. I have W over L,

I have μnCox, uh threshold, everything else, and I would like to find the current that flows

through this MOS device. All right. Okay. So, here's the what we are trying to

find, ID. So, we have μnCox, W over L, μnCox, W over L, we have VTH, the

threshold of the device. These are properties of the device itself. Do we have VGS? No, we don't know this

gate-source voltage. Why? Well, because I know V0, but then there's a drop from here to here,

right? So, I don't know VGS yet. And I don't know ID yet. Okay, so I have to find ID

directly, or I have to find VGS and then find ID from this equation, assuming that the device is in saturation.

Okay, is that difficult? No, not really. I write a a KVL in this loop here. Uh if I have a current of ID in the

drain, that also ends up in the source. So, we have a current of ID here, right? So, that current is ID.

Which means the voltage drop across RS is equal to ID * RS. So, now let's let's write a KVL in this

loop here. We have this voltage, V0, is equal to

this VGS, VGS, gate-source voltage,

plus this voltage, which is ID * RS. Okay, very simple.

Now, we have two unknowns here, VGS and ID. Fortunately, there is a relationship

between VGS and ID elsewhere. Let's look. This one also gives us a relationship. So, we have two equations

and two unknowns, and we have to solve them. Right? So, let's move forward and see

what we can do. I will try to find VGS from here and plug it in here, because I'm trying to find ID.

So, let's try to find VGS from here. In fact, this is something good to know. VGS is equal to what?

We multiply by two, divide by μnCoxW over L, and then take the square root,

and then bring the threshold to the other side. So, we have 2 ID

over μnCoxW over L, and then plus VTH.

Okay, that's a good equation to remember. Previously, we always found ID in terms

of the overdrive. Now, we find VGS in terms of everything else. Okay. So, let's go ahead and plug that

in here, see what happens. So, that's equal to square root of 2 ID

divided by μnCoxW over L, plus VTH,

plus ID * RS. So, what we do know in this equation are

V0, the voltage we have applied here, and VTH, and μnCox, and W over L, and

RS. What we don't know is ID. So, we can find ID.

We can even find a closed-form equation for ID if we really want to. Uh what we would do is we take V0 to

that side, we take the square root to this side. Then we square both sides and simplify

and so on, we end up with a simple uh second-order equation that we can solve. Okay.

All right. The purpose of this exercise was to show you that because of this nonlinear relationship,

uh the calculations do get a little complex as we go along. Now, here it's not a big deal, because

we can find even a closed-form equation for ID. But

it's not always that easy. Uh let's suppose that I give you something like this. I have

one device like so.

And then I have another device like so. I connect the battery from here to here.

And stuff from here to here, another battery from here to here. Uh this is some voltage V0, this some other voltage

V0 prime. And then a battery from here to here, which is our usual

V1, right? Now this becomes substantially more complicated if you want to analyze it.

Because these two devices now interact here. We have some current from this guy, some current from this guy flowing

through this resistor. So it's a little messy. Okay, so things can become a little more

complex. All right. So far so good. I just calculated the bias conditions here,

right? I just found the current device. There was no signal yet. I used just how much current I might have. Could be a

milliamp, half a milliamp, 2 milliamps, some amount. All right. Now let's go ahead and apply

a signal on top of this bias that we have. After all, we're trying to build an amplifier. So a signal should be

coming in and we're curious to see what the signal does. So let's go ahead and add a signal

and see what happens. So let's add

a signal. So here's the situation. I add microphone signal here. I have the

original mic original battery of V0. This goes to the gate source of the MOSFET.

And then for now we just have a battery here. Don't worry about anything else. So we call this V1,

V0. This generates a signal Vmic

uh from here to here. And then we have a certain current here ID.

So we're just interested in the drain current for now. Remember that eventually this current has to flow

through a resistor, all sorts of things. But for now we're considering a very simple case.

Okay. So let's imagine that Vmic is just a simple sinusoid. So Vmic

is equal to VM sine of omega t.

That's all. It's a little sinusoid with zero DC. It just goes up and down around zero, right?

Okay. Uh but it's added to V0. So this VGS now has V0 plus this in it, right?

Okay. So can I find ID? Well, I have the same equation as before except that V0 is replaced with V0 plus Vmic.

So I can say V0 plus Vmic or actually I'll just write the equation for it.

V0 plus VM sine of omega t is equal to

uh what did we have there? We had square root of 2 ID over mu n c ox w over l plus the

threshold voltage and then plus ID times RS. Okay, so we simply replace V0 with V0

plus the signal from the microphone. Okay, I'm still trying to find ID. Can I find ID here? Yes, we can find ID.

Uh again, we have to take all of this to this side and this square root to this side. Then we have to

square both sides. We end up with a quadratic equation. We solve the quadratic. We end up with some complex

square roots and things in there for ID in terms of all of this. So this will go under square root and all sorts of

things. It'll be a little messy. All right. So this is what we call large signal

operation. Why? Because I never mentioned whether this VM small or not. So in the general

case it might be large. So we say large signal

operation if VM

is not small. Now we don't know what small means.

Small with respect to what? But that's okay. For now just assume it's not small.

So by large signal operation we mean the signal that's coming to the circuit has a certain amplitude

and that amplitude is not assumed to be small enough. Small enough for what? We'll have to

see. But it's not assumed to be small. So it can be anything you want, right? It can become very large. It can be this

VM can be 10 millivolts, 15 millivolts, 100 millivolts. We haven't specified anything.

If you haven't specified anything, we continue to use the complete model of the MOS device.

Which is ID is equal to 1/2 of mu n c ox w over l times VGS minus VTH squared. Now if you want to include channel

length modulation, then well, times 1 plus lambda VDS. All right. So uh by large signal

operation, that's what we mean. The signal is coming in. We don't have any particular limit on it.

Uh a common mistake by students is that they say large signal operation means DC calculations.

That is not correct. They say bias calculations. That is not correct. It happens that for bias calculations we

use the large signal model, but it doesn't necessarily mean all the time that large signal means just bias.

So if we have large signal operation, we use the complete model, which we call the large signal model.

So we say use large signal

model of the transistors.

So what was the model of the transistor? Do you remember from last time? Well, we said that we have a gate.

We have a source. Not much going on between gate and source, so that's sort of open.

We have VGS. And then we have the current source here.

After all, it's a voltage dependent current source. And then here we had 1/2 of mu n c ox w over l VGS

minus VTH squared. This is the drain. And if you want to include channel length

modulation, then 1 plus lambda VDS. So if we don't have any restriction on the signal amplitude coming in, then we

have to use this model, which we call the large signal model. All right?

Okay, very good. Now as I said, once we have a few transistors or more, things can get pretty complicated.

Right? Because there each of these has this type of model and then we have various devices, various currents,

various voltages. We end up with some number of equations, a number of unknowns, and then many of these

equations are non-linear. So the calculations can get messy. All right. Now under some conditions

we can simplify the situation. Okay, so we have to see what those what those conditions are. And those are what

we call small signal operation. So let's uh talk about small signal operation and see what that means.

Small signal operation.

So as the name implies, we are hoping that in some cases the signal that's coming in, the microphone signal,

whatever it is, has a small amplitude. Again, how small is small? We have to see. But it's small.

So let's see. If the signal amplitude is small, is there any simplification that we can apply to our models and circuits,

etc., so that we can quickly find the properties of the circuit? Okay. So again, we start with an

example. So here's what we have. We have a microphone as usual

and the battery as usual. And again, I've just simplified everything, so we just have another

battery here. V1, V0, Vmic.

And in fact, this time I don't even have uh I'm sorry, there was a resistor here I

forgot to draw RS. But this time I don't have a resistor, so no RS here.

Okay. So um again, I know that this drain current

is related to this VGS. And it just happens that now VGS is equal to the microphone signal

plus V0. If you remember that was not the case here, right? VGS was not equal to this

because of the drop across the resistor, this term here. But here, it's very simple.

VGS is just equal to this. So, I can write ID directly. I can write ID

is equal to 1/2 of μn Cox

W over L VGS - VTH

squared. Um so, by the way, lambda is zero as usual, and VGS has a known value. VGS is

the microphone signal plus V0. And the microphone signal again we assume to be Vm sin of omega t, right?

Okay, so that's equal to 1/2 of μn Cox W over L I will group these a little.

So, I will write this as VGS - VTH plus uh Vm sin of omega t

squared. Now, if Vm is not small, this is it. That's how we calculate ID. No question

about it. But if I if Vm is small, maybe we can make some simplifications.

All right. So, if Vm is small, what's our first reaction here? Well, maybe I can factor out VGS - VTH

to get one plus a small number, and then expand that using a simple approximation.

Okay, so we will write this. So, we say, "Assume Vm

is much less than VGS - VTH." And that tells us now what we mean by small.

Okay, so then ID is equal to uh let me actually

write this down here because I need more room. So, ID

is equal to 1/2 of μn Cox

W over L All right. So, I will factor VGS - VTH out of here.

So, what I get is VGS - VTH squared. So, VGS - VTH

squared. And then what remains inside here is one plus Vm sin of omega mt omega t divided

by VGS - VTH. One plus Vm sin of omega t divided by VGS - VTH.

Now, because this term is much less than one I can write I can exploit this. One plus

epsilon to the power of two is approximately equal to one plus two epsilon, right?

So, let's write it like that. So, we have one plus two times Vm

sin of omega t divided by VGS - VTH. So, the key is that because this term

over this over this is much less than one, we managed to go away from this non-linear behavior and represent the

result by this simple linear expression. Okay, so let's examine this closely and see whether we can make make any sense

out of this. All right. Okay, so what is this? Do we know what

this means, what this is here? Uh given that uh VGS Okay, sorry. Let me go back here and

make a small correction. I'll do it in a different color so that you remember what happened here.

So, here uh what we really have for VGS is V0 plus V microphone, right? So, this

should be V0. And this should be V0. And then this should be V0.

And then this should be V0. Okay, the battery, the bias value that we have for the gate source before the

microphone signal comes in. Okay, so do we recognize this whole thing here?

Is the is a current quantity, right? And it's a current quantity given by this battery.

So, that would be the bias current of the device, which we usually call ID0. So, this is ID0.

Just the bias current. So, let's uh write this and see what we get. So, we have ID

is equal to ID0 plus two

ID0 over V0 - VTH times Vm sin of

omega t. So, now that's not surprising. What we got is the following.

The total current that flows through the transistor, ID, is equal to the bias current. Of course, we have bias

current. Plus the signal current. The signal current is the perturbation created in the current when the signal

comes in and go as applied to the gate source, right? So, we might call this the bias current.

And we might call this the signal current. You can see that if there's no signal

if Vm zero, there's no signal current, right? That's the perturbation that we have created inside this device here

when the microphone signal goes up and down. Okay, that's very interesting. So, that

uh suggests something. Maybe when I'm considering this MOSFET here with all the things that are going

on I can in my mind separate it into two separate circuits, two independent

circuits. All right. So, here's what we will do. We take that circuit. So, let me draw it

quickly again here. Uh sorry. Let me do that with the microphone first.

So, we have a microphone. Then we have a battery. And then we have this.

And what we see is that when I'm talking to the microphone uh V0 V1

then I get this total ID here, right? So, I'm thinking maybe I can decompose this into two.

Here's one. I have a MOS device with no signal coming in, just the bias quantities.

So, here's the MOS device with just the bias quantities. So, we have V0 V1, a certain

current ID0. This scenario takes care of this current here.

All right. And then on top of this superimposed on this, I have another circuit which is like this.

It's taking the microphone signal and generating a certain amount of current, which is all of that stuff

there. Two ID0 over VGS uh sorry, V0 - VTH times Vm sin of omega t.

So, in other words, what I'm really trying to do is this. I'm saying that this device

consists of really two currents. One current is constant and equal to ID0.

That's the bias current. And then superimposed on that, we have another current that's flowing, and that current

is this guy here. So, two ID0 over VGS - VTH times Vm

sin of omega t. Okay. And uh what we what's going on here is

we have a microphone signal. And then we have a battery. Uh this represents what we have so far

here. All right. Now,

uh it's a little strange, uh but I guess I would need to draw this battery as well just to have a complete

situation. And that's okay. We'll draw that, too, V1.

And that's the overall picture, right? Between the gate and the source, we don't have any impedance because the

current drawn by the gate is very small. So, but we still have a voltage here, right? That's the VGS that we have

that's total VGS that we have. Okay. All right.

So, uh one interesting observation What is this this uh factor here? Is it

familiar? This factor is the following. It's a bias quantity because there's no signal

in it, right? And it's equal to twice the bias current divided by

the bias overdrive voltage. Because V0 is here.

And when there's no signal, the gate source is equal to V0. So, it's like V VGS under bias

conditions minus a threshold. So, that's the overdrive voltage under bias conditions.

So, this is entirely bias dependent. It's not single signal dependent, hopefully. So, and it looks familiar.

Twice the drain current divided by the overdrive. What is that? Well, if we rewind

and look what we obtained before, we see that it is the transconductance of the device.

That's not surprising. The transconductance device says, "If you give me a voltage change of

delta V between the gate and source, I will give you a current change equal to GM times delta V."

Right? That is the definition of transconductance. So, if we go back here, you can see

if you give me a change in VGS, then I give you a change in the current equal to GM times the change in VGS.

So, the microphone signal is creating a change in VGS. And as a result, we get a change in the

drain current, and these are related by GM. So, it's not surprising that what we have here is just a transconduct-

transconductance of the device times the voltage that we applied. So, this is GM

times VM sin of omega t. Okay, so the pen is sick again. All right, so sin of

Let's see. sin of omega t. All right, so let's take a step back and

see what happened here. We have a bias voltage between gate and source. That gave us a certain bias

current. Because device is presum- is presumably in saturation.

All right, then on top of that we came in with a signal. We had a signal voltage that was applied

to the gate source on top of that bias voltage, right? So, signal voltage on top of the bias voltage.

So, this signal voltage changed this VGS. And that change got multiplied by GM to

give us a change in the current. And that's what happened. We had the voltage multiplied by GM, and that gave

us a change in the current. This current represents the change in the bias current. This is the bias current. This

is the change in the current. This bias current is just from bias conditions. This is from signal conditions. So, we

can look at this as the change in the bias current. All right, so maybe what we can do is we

decompose this according to this picture, and we say, "Here's one picture where we have

V0 and we have a current. And we have a voltage here, V1.

And this current is given by 1/2 of μn Cox W over L uh V0 - VTH squared.

All right. Okay. And then so, this took care of this voltage, this current, and this

voltage because they are all bias quantities, and they do not change with time.

Now, next to this, we draw another circuit for this guy. What is that? Okay, that has a voltage

coming from the microphone. So, we draw the voltage for the microphone,

V mic. And then we go to another current source,

which is equal to GM times the input voltage, the microphone

voltage, right? So, GM V mic. Okay. Uh this is the same as V mic. And then

what happens to this point here? Well, this point comes back here because I am using superposition.

I'm saying that uh this circuit plus this circuit should give me this circuit.

Right? So, uh if I have a V voltage of V1 between drain and source, I have to have it only

in one of these two models. I cannot have it in both models. So, this is what we call the large

signal model, if you will, which also used for bias calculations. And this is what we call the small signal model.

Small signal model of the MOS device.

So, it's very important to understand that small signal operation means the signal that's coming in

creates a very small change in everything. In particular, it's creates a small change in the overdrive voltage

because that was our condition. Right? The overdrive voltage is changed negligibly as a result of this. You can

see here that uh this term is much less than this term, so the overdrive voltage is not affected much by the voltage

that's coming by the signal voltage. All right. Alternatively, we can say that the signal creates small changes in

the drain current. So, we have a bias current of, let's say, 1 mA.

And then we have some small changes around that. These could be maybe 0.1 mA or something 0.1 mA or something like

that. So, that's called small signal operation.

And then in that situation, and only in that situation, can we simplify the circuit to a bias model and a signal

model. And the signal model, the small signal model, consists of something that represents only changes

in the current or voltage. We look at anything that is changing with time.

The microphone signal changing with time has to be there. The There's a component in the current

of the device that's changing with time. That has to be there, too. But everything that doesn't change with

time, everything whose change with time is zero, is zero here. Is zeroed out.

So, this voltage doesn't change with time. Its change with time is zero. Because we

are looking for only time varying components, it's not here anymore. You see, it's not in series with this.

This voltage represents a constant value. Its change with time is zero. Because its change with time is zero, it

has zero value here in the in this small circuit. So, this is what we call the small

signal model. You can see that small signal model now easily lends itself to linear

calculations. We don't have any uh in terms of the signal itself, it's just a linear model. We don't have any square

or anything else, so it's much easier to handle. All right. Our time is up. I will see

you next time.