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?

omg…dese videos r so valuable..almost cried a little(out of joy) ðŸ™‚

lool ðŸ™‚

@savita1926: I am also under the impression that saturation region and cutof region are for digital logic.. but if you operate mosfet in the linear region (triode) and expect it to amplify the signal, then we can use just a resistor amplification ðŸ˜‰

Only in saturation region, we have I(DS) squarely dependent on V(GS) making it a dependent current source and hence amplifiying the signal.. not sure if i understood it correctly..

For a MOSFET, the saturation region is the region where the FET acts like a current source. It is the opposite name used for a BJT transistor. In a BJT this same region is called the linear region.

i think you're lost.

at 23:09 how does he know that the dependent current source is by [k/2][(vgs-VT)^2] and not anything else?

He defines the current source as iDS=[blah]. If you are asking how he arrived at a MOSFET's parameters of iDS=(k/2)(vgs-vt)^2, that is the characteristic of a MOSFET in saturation, where the k/2 parameter is dependent on oxide thickness, channel size, and other properties, and is indicated in the manufacturer's datasheet. See wikipedia Mosfet entry, Modes of operation, for more information.

here what is the value of vt

If you didn't study MOSFETs in a solid state devices course, you'll be lost in this lecture.

as VDS=VGS-VT is the turning point of current source and voltage source. it seems like the extra voltage other than vt, the part of voltage that is released over the pn knot between G and S, the extra voltage would work against the electric conduction of pn knot between D and S

Dense lecture content wise, but good stuff.

WHY DID HE CUT HIS HAIR? BA

I am confused about something here. If I am not mistaken, the 'saturation' behavior of the mosfet( Id saturate when Vds gets large enough so that Vds> Vgs-VT) is because its resistance converge to a constant value. And the saturated current is supplied by the power source on the right(the circuit at 16:18). Now back to our primitive inverter, the inverter circuit isn't exactly the same as the circuit at 16:18, it has got a resistor above the mosfet, so i think what control the ids should be the combination of Vs and R (AND Vg, but since it's saturated, it shouldn't matter anymore?)?

quality: trve black metal kvlt