WEBVTT
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So in this problem, we're going to
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look at the following circuit
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as two MOS type transistors and one resistor.
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So here's the circuit.
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And you'll notice that both transistors
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are diode connected, right?
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So that the gate and the drain are tied together.
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And a positive supply voltage of 2.5 volts,
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and then we're going to have a resistor, R.
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And let's label the transistors,
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so we'll call this transistor here Q2,
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and that transistor Q1.
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And so we're given various parameters for the transistors.
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So we're told that mu-m-C-ox
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is equal to 250 microamps per volt squared
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and that the threshold voltage is 0.5 volts.
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We're also told to ignore channel length modulation.
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So lambda is equal to zero.
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And we're told that both transistors here
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are required to have lengths of 0.25 microns.
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L1 and L2 is equal to 0.25 microns.
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And this is a design problem, and what we mean by that is
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that we're going to be given one more constraint here
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but given that constraint, we're going to need
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to pick a value of R, and we're going to need
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to finish designing those transistors.
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And what that means is, well, we know the transistor length,
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but we don't know the width.
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So we have to find the width for both transistors
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and the resistance.
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So the one last constraint is that we're told
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that we need to establish a drain current of 0.5 milliamps.
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And so we have two more constraints.
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So the drain voltage for this transistor, here,
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that should be 1.8 volts and we're told
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that the drain voltage for this transistor, here
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should be one volt.
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Okay, so now we can start solving the problem.
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So we'll start tackling what's easiest.
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If you look at this resistor, here,
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we know the voltage drop across the resistor
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and we know the drain current.
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So we could, using Ohm's law, say
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that the drop is two and a half volts, minus 1.8 volts,
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divided by 0.5 milliamps.
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So that gives us a resistance, R, of 1.4 kilo-ohms.
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Okay, now moving on to the two transistors,
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we know the transistor length but don't know the width.
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And usually when we try to figure this out,
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we need to figure out what is
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the transistor in the saturation region
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or the triode region, or cut off.
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But it can't be cut off because clearly current is flowing.
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But also, you'll notice that both transistors
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are diode-connected, and a diode-connected transistor
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is naturally in the saturation region.
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Okay, so they're diode-connected, so they are
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known to be in the saturation region.
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So because they're in the saturation region,
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we can write an equation that relates
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the voltages to the current.
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So we can write ID is equal to mu-n-C-ox over two,
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W over L, times VGS minus VT, all squared.
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And so if we look at this,
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in each case, for both transistors,
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we know the current, we know mu-n-C-ox, we know L
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and if we look at this carefully,
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for both transistors, we know the gate voltage
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and we know the source voltage.
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So we know VGS and we're also given VT.
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But what we're trying to solve for is the width.
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So we're just going to take
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this equation here and rearrange it
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so that we can figure out width.
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And so we have width is equal to, let's see,
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bring the two up, so two times ID,
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divided by, we'll have the mu-n-C-ox,
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VGS minus VT squared on the bottom.
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Okay, mu-n-C-ox, VGS minus VT,
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all squared on the bottom,
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and then L has to come out on top.
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Okay, so now all of these parameters
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on the right are known for both transistors,
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so we can figure out both transistor widths.
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So why don't we do transistor Q1 first.
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And so for Q1, so the width of transistor one
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is going to be equal to two times ID,
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so two times 0.5 milliamps,
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divided by mu-n-C-ox, so 250
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microamps per volt squared, times VGS minus VT.
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So we've got Q1, so what is VGS for Q1?
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Well it's one volt minus zero volt
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minus VT, so minus .5, all squared, times L.
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So times .25 microns.
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And that's equal to four microns.
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Okay, so that's the first transistor, Q1.
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Q2, we have W2
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is equal to two times 0.5 milliamps,
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divided by 250 microamps per volt squared, times,
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well now we have a different VGS.
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We have 1.8 volts minus one volt,
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minus VT, .5, all squared,
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times .25 microns.
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And that's equal to 11.1 microns.
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So there we have it, the problem asked us
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to design the circuit and design
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and to define the values of R
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and then the widths of Q1 and Q2.
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So we have R of 1.4 kilo-ohms and we have,
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for the bottom transistor we have
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four microns, a width of four microns.
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And then for the top transistor,
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we have a width of 11.1 microns.
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So there it is, I invite you to take a look at the solution.