WEBVTT

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<v ->In this problem we're going to design</v>

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the following circuit that has two resistors

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and one NMOS transistor.

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And so our resistor

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connected to the drain of the transistor.

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And then, another resistor connected to the source

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of the transistor.

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The gate is tied to ground.

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And then we have a resistor RD,

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a resistor RS,

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a positive-supply voltage of plus-one volt,

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and a negative-supply of minus-one volt.

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And so the problem,

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we're told that this transistor here

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has the following properties.

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So, it has Mu-n-C-ox

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equal to 400 microamps per volt squared.

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(soft taping noise)

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and VT

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equal to 0.5

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volts.

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We're also told the...

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we're given the dimensions of the transistor.

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So that the width is equal to five microns

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and that the length

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is equal to 0.4 microns.

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(soft taping noise)

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Okay.

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So, the design problem here, essentially, is going to be

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to pick the values of RS

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and RD.

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So, as a design objective here,

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we're actually told that we would like to establish

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a drain current of 0.1 milliamps.

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So, we're going to label that there.

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So, ID

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equal to 0.1

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milliamps

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and that we would like to establish a drain voltage,

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VD

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is equal to 0.3 volts.

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Okay.

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So, now to start tackling the problem, all right?

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So, to pick these two resistor values,

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we're going to start from the top.

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Because if you notice at the top here,

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we know the voltage drop across the resistor

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and we know the current.

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And so, we can use Ohm's law

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to figure that out.

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So, RD is the voltage-drop.

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So one volt minus 0.3 volts,

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divided by the amount of current, right?

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0.1

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milliamps.

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So that gives us RD is equal to seven

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kilo-ohms.

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Now, for RS, it's maybe a little bit more complicated,

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right?

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Because we don't know the source-voltage, yet.

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And then, so we have to look at the transistor

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and basically analyze what's going on

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in the transistor.

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But we have sufficient information to do that.

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So we'd like to start by trying to figure out,

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is this transistor operating in the saturation region

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or in the triode region.

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(soft taping noise)

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Okay.

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So,

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to do that, we need to figure out, you know,

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is VDS

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greater than or less than VGS minus VT.

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Let's figure out what VGS minus VT is equal to.

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So, this is the gate here and that is the source.

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Eventually, we want to figure out that source voltage.

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So, VGS

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minus VT

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that is equal to a gate voltage of zero volts

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minus VS, which we don't know yet,

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minus VT.

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So,

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minus-0.5 volts.

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And we have to figure out what that is

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in relation to VDS.

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So what's VDS.

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VDS, well, it's going to be equal to VD, all right?

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Equal to 0.3

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minus

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VS.

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And so basically,

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if we were going to be in the saturation region...

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So, why don't we hypothesize that we're in saturation.

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(soft taping noise)

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And so, to be in saturation,

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we would need VDS

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greater than VGS

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minus VT.

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So, let's just plug in what we've got for that.

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So VDS we've got 0.3

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minus VS,

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and the question mark, you know, is that greater than

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minus VS,

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minus-0.5?

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And so the VS is canceled out

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and we see .3 is greater than .5;

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in fact that inequality is correct and we are in saturation.

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Okay.

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So that now, essentially, will give us a path to

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towards solving for VS.

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And,

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and so we can write the equation for current

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in the saturation region.

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So, in saturation ID

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is equal to mu-n

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C-ox over two,

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W over L, and times the

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VGS squared.

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So, V-overdrive squared.

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Now, I'd, mu-n-C-ox, W, and L are all known.

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Really, what we're solving for is V-overdrive.

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So, let's turn that around

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so we could rewrite it as the overdrive as the square root.

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All right.

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Of two ID,

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of two times ID

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over mu-n-C-ox, W over L.

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(soft taping noise)

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Okay.

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So now just plugging in numbers for that.

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So equal to square root of two times

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0.1 milliamps,

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over 400 microamps per volts squared,

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(marker squeaking)

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times five microns

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over 0.4 microns.

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And that gives us the overdrive

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of 0.2 volts.

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(clears throat)

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So, VGS

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we know is a V-overdrive

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plus VT.

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Right?

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Because if you overdrive as VGS minus VT.

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So, V-overdrive

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plus VT,

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so equal to

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.2 plus 0.5

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equal to .7 volts.

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And we know the gate voltage, right?

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The gate is a ground.

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So, VS

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must be equal to minus VGS.

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(marker squeaking)

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(soft taping noise)

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Okay.

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Very good.

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So, what are we left with at this point?

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Well, we don't know the value of RS.

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But we now know the voltage-drop

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across the resistor.

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And so, we can solve for RS.

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So, RS will be equal to minus-0.7 volts

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minus the negative-supply of minus-one volts

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all divided by 0.1 milliamps

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and that'll give us three kilo-ohms.

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(marker squeaking)

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And so, just going back to the problem statement,

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we were asked to design the circuit to meet

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certain design constraints.

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And really what we meant by design

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was to find a value for RD and a value for RS.

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So, RD

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seven kilo-ohms.

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RS,

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that is 3 kilo-ohms.

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And there we have it.

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We've solved the problem

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and so, I'd like you to,

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I'd like to invite you

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to take a moment to look a the solution.