Daniel Townsend

Exercise 5.7

(a) Using the inequality on the formula sheet, we know that the cutoff region is vSG < |Vtp|, so therefore it would conduct whenvSG ≥ |Vtp|. Vtp is given to use as -1V. By definition vSG = vS − vG. It’s easy to tell that vS = 5V from the circuit diagramin figure E5.7.

Knowing all this, let’s plug in the values to the original inequality.

5− vG ≥ | − 1|

5 ≥ 1 + vG

4 ≥ vG

∴ vG ≤ 4

(b) The triode region operation is defined as either vDG > |Vtp| or equivalently as vSD < |VOV |. In this case it is easier to workwith the former. By definition vDG = vD − vG. We can substitute that into our inequality.

vD − vG > | − 1|

vD > vG + 1

Note that ≥ versus > is not really all that important, since the drain current is a continuous function.

(c) Saturation is literally just the opposite of part (b).

vD ≤ vG + 1

(d) Given ID = 7.5 · 10−5, and knowing what we were told in the problem description we can solve for the value of VOV

7.5 · 10−5 =1

2(6 · 10−5)(10)(vOV )2

|vOV | = 0.5V

|vOV | = VSG − |Vtp| ∴ VSG = 1.5V

VSG = VS − VG ∴ VG = 5− 1.5 = 3.5V

Finding the range of VD is just plugging our VG into part (c) answer

vD ≤ 3.5 + 1 ∴ VD ≤ 4.5V

(e) The equation for ro =|VA|ID

=1

|λ|ID

ro =1

|λ|ID=

1

| − 0.02|7.5 · 10−5

∴ ro = 666666Ω

(f) The equation for saturation when a λ is specified is iD =1

2kp(vOV )2 · (1 + |λ|vSD)

Plugginging in all we know into this equation, it’s easy to find the value of ID. The book asks to find the current at twodifferent values of VD

For VD = 3

iD =1

2(6 · 10−5)(10)(0.5)2 · (1 + 0.02(5− 3)) = 7.8 · 10−5A

For VD = 0

iD =1

2(6 · 10−5)(10)(0.5)2 · (1 + 0.02(5− 0)) = 8.25 · 10−5A

1

Daniel Townsend

Exercise 5.8

Before we begin it would be useful to draw the circuit diagram of interest.

RS

−2.5V

+2.5V

RD

0.3mA+0.4V

For RD a simple Ohm’s law calculation will suffice.

0.3mA =2.5− 0.4

RD

∴ RD =2.5− 0.4

0.0003= 7000Ω

Since VG − VD = 0− 0.4 = −0.4 ≤ 1V, this circuit operates in saturation mode.

We can set up the saturation equation to solve for VS . Note that VGS = −VS since VG = 0.

0.3mA =1

2(6 · 10−5)

(120

3

)(−VS − 1)

(−VS − 1)2 =0.3mA

1

2(6 · 10−5)

(120

3

)

VS = −√√√√√ 0.3mA

1

2(6 · 10−5)

(120

3

) − 1

VS = −1.5V

We can now solve for RS .

RS =−1.5− (−2.5)

0.0003

RS = 3333Ω

2

Daniel Townsend

Exercise 5.9

To start, it is useful to note that VD = VG ∴ VGD = 0. Since 0 ≤ 0.5V this transistor is operating in saturation mode.

We have two equations governing ID

ID =1

2kn(vOV )2 ID =

1.8− VDR

Setting them equal to each other and plugging in known values, we can solve for R.

1

2(0.0004)

(72

18

)(0.8− 0.5)2 =

1.8− .8R

Note that VGS = VD since VD = VG and VS = 0

R =1

1

2(0.0004)

(72

18

)(0.8− 0.5)2

R = 13888Ω

3

Daniel Townsend

Exercise 5.10

The right side of this circuit is exactly the same as Excercise 5.9, meaning VD = 0.8V and R = 13888Ω.

The left side has the same rail voltage of 1.8V. The problem statement says to operate this new transistor at the edge ofsaturation, or when VGD = Vtn.

Since Vtn = 0.5V this would mean that VGD would have to equal 0.5V as well.

Since we know VG = 0.8V, the value of VD = 0.8− 0.5 = 0.3V

We now have an Ohm’s law equation governing the resistor.

ID =1.8− .3R2

We also know that Q2 is identical to Q1 so we can write out the saturation equation for ID.

ID =1

2(0.0004)

(72

18

)(0.8− 0.5)2 = 72µA

Now we can solve for R2

72µA =1.8− .3R2

R2 =1.8− .372µA

∴ R2 = 20833Ω

4

Daniel Townsend

Exercise 5.11

Example 5.5 is already worked out, giving a resistor value RD = 12400Ω, so the resistor value we need to use is twice that,or RD = 24800Ω

Excercise 5.11 asks us to double the resistor value and then find the values of ID and VD. Note that we are no longer tryingto make VD = 0.1V, that was just what they were asking in the example.

VOV = VGS − Vtn = 5− 1 = 4V

First, let’s guess that this circuit is operating in saturation mode (just to illustrate what happens when you guess wrong)

We can solve for ID using the equation for saturation mode of an NMOSFET

ID =1

2kn(vOV )2 =

1

2(0.001)(4)2 = 8 · 10−3A

From this we can find VD by using Ohm’s law

5− VD24800

= 8 · 10−3 ∴ VD = −193.4V

You don’t really need to use the inequalities to realize that this is really wrong, but let’s do so anyway.

VGD = 5− (−193.4) = 198.4 1

This is a silly step, but it is necessary to illustrate our point.

After you are done with solving a MOSFET circuit at DC, ALWAYS CHECK THE CONDITIONS. If we had gotten 3V forVD it would have still been wrong, but harder to notice.

Since our guess of saturation mode was wrong, we need to completely re-do this problem. Start with the triode equation.

ID = kn

(vOV −

1

2vDS

)vDS = (0.001)

(4− 1

2VDS

)VDS

We don’t know VDS yet, but we know another relation to it, Ohm’s law.

ID =5− VD24800

Both equations are equal to ID, therefore can both be set equal to each other. Also note that since VS = 0, VDS = VD.

(0.001)

(4− 1

2VD

)VD =

5− VD24800

Solving this and setting it equal to zero gives us

12.4V 2D − 100.2VD + 5 = 0 ∴ VD =

100.2±√

100.22 − 4(12.4)(5)

2(12.4)

∴ VD = 8.03V OR VD = 0.0502V

Since we get two answers, we get to pick whichever one makes more sense. The second answer makes more sense.

∴ VD = 0.0502V

We can find ID with this equation by plugging this value into the Ohm’s law equation.

ID =5− 0.0502

24800= .200mA

5

Daniel Townsend

Exercise 5.14

This is a design problem where they ask you to get the value of a resistor based on a specified overvoltage and thresholdvoltage, and they give you all of the necessary parameters to solve.

First of all, we need to figure out what mode that the circuit is operating in.

Since the Gate and the Drain are grounded, VDG = 0V. Since 0 < | − 0.4|, this is operating in the saturation region.

We must use the equation for drain current in the saturation mode.

ID =1

2kp(vOV )2 =

1

2(0.0001)

(10

.18

)(.6)2 = 1mA

The resistor has its own Ohm’s law equation

1mA =1.8− VS

R

But we don’t know VS yet!

We can find it with the overvoltage and the threshold voltage, both given to us.

|VOV | = VSG − |Vtp| ∴ VSG = .6 + .4 = 1V

Since VG = 0, VSG = VS

Now let’s solve for the resistance.

1mA =1.8− 1

R

R = 800Ω

Boom.

6

Daniel Townsend

Exercise 5.27

This is an amplifier problem. All MOSFET amplifiers operate in the saturation region. If they didn’t they wouldn’t be ableto find a linear region of the I-V curve to create amplification.

Therefore when you see an amplifier problem, always assume it’s in saturation region when you do DC analysis on it.

In this problem there is no need to perform DC analysis at all, since they tell you that it’s biased at ID = .25mA and has aVOV = .25V. Therefore we can just jump straight into the small signal analysis.

Small signal analysis parameters:

ro =VAID

=50

.00025= 200kΩ

gm = knVOV

Note that in both the textbook and the solutions on Sakai they made an error. They said that gm = 2IDVOV

, which would only betrue if there was no early voltage VA. However the problem statement explicitly states an early voltage.

Ignore this, I emailed the TA and he said that you can ignore channel length modulation when performing DC analysis on aMOSFET amplifier.

To solve for gm use the equation gm =2IDVOV

gm =2 · 0.00025

0.25= 0.002

With the small signal parameters, it’s now time to draw the small signal equivalent circuit.+−vsig

Rsig

RLRD

Replace the MOSFET with its equivalent circuit (don’t forget the ro)

+−vsig

Rsig

G

S

gmvgs ro RLRD

Plugging in all the known values we get the following circuit

+−vsig

100kΩG

S

0.002 · vgs 200kΩ 20kΩ20kΩ

7

Daniel Townsend

The circuit can be rearranged to look prettier.

+−vsig

100kΩ

G

S

0.002 · vgs 200kΩ 20kΩ20kΩ

Some simple observation of the equivalent circuit can allow us to see the value of many variables.

Rin =∞

Ro =1

1

200kΩ+

1

20kΩ

≈ 18.182kΩ

Note that Output resistance does NOT include the load resistance.

The gain without load Avo is found by multiplying the current source with the output resistance. Since the current is goingaround the opposite way our voltage output is being measured it will be negative.

Avo =voutvin

∣∣∣∣RL=∞

= −(18.182kΩ)(0.002) ≈ −36.4

It’s pretty clear that there is an open circuit between the gate and the source. This means that Rsig really has no effect.

Therefore vin = vsig

Because of this, Av = Gv.

Therefore Av = Gv = −(0.002)(Ro‖RL) ≈ −19.05

8

Daniel Townsend

Exercise 5.29

The wording on this problem is a little weird. It tells us that a CG amplifier is required to match a signal source withRsig = 100Ω. I still don’t really understand what it is trying to say, but from the solutions manual it basically means that

the resistance1

gm= Rsig = 100Ω.

The circuit is shown below, with its equivalent circuit shown next to it.

+−vsig

Rsig

RD

+−vsig

Rsig

1

gm−

+

vgs

gmvgs

RD

We can fill in the resistor values for this circuit, and change it up a little bit, to obtain the following.

+−vsig

100Ω

100Ω

+

−vgs gmvgs 2000Ω

Knowing that1

gm= 100 ∴ gm = 0.01 we can solve for kn

kn =0.01

0.20= 0.05

A

V2

Knowing this, and that the circuit operates in saturation mode, we can now solve for the value of ID.

ID =1

2kn(vOV )2 =

1

2(0.05)(0.20)2 = 1mA

Calculating the voltage gain is simple. First you need to find vgs in terms of vsig.

vgs = − 100

100 + 100vsig = −1

2vsig

vout = −gm(vgs)RD = −(0.01)

(−1

2vsig

)(2000)

Gv =voutvsig

= 10V

V

9

Daniel Townsend

Problem 5.79

This is more likely the type of problem you’d see on an exam.

The first step in small signal analysis is to find the DC values and draw the DC equivalent circuit.

5MΩ

10MΩ

+15V

3kΩ

+15V

7.5kΩ

+5V

3kΩ

+15V

7.5kΩ

Let’s determine the equivalent bias on the left side. Note that no current goes through the gate of the MOSFET, so all youhave to do is voltage divider.

VG = 15

(5

5 + 10

)= 5V

We can redraw the equivalent circuit as shown above on the right.

Since this is an amplifier circuit, we automatically assume it’s operating in saturation mode.

(a) For part (a) They tell us values of the bias and ask us to verify these values. To do this, we will need to use the saturationequation.

ID?=

1

2kn(VOV )2

0.001?=

1

2(0.002)(1)2

0.001 = 0.001 This checks out.

The more important values we need to check are our biasing. Since the problem states that VGS = 2V, this means thatVS = VG − VGS = 5− 2 = 3V

The problem also directly states that VD = 7.5V.

We can perform the Ohm’s law calculation on each resistor and check that it equals 1mA.

0.001?=

15− 7.5

7500

0.001 = 0.001

0.001?=

3− 0

3000

0.001 = 0.001

This means that all the values are consistent and we completed part (a).

(b) Part (b) is just plug and chug.

gm = knVOV = 0.002

ro =VAID

=100

0.001= 100kΩ

10

Daniel Townsend

(c) Part (c) asks us to draw the small signal equivalent circuit.

Remember that when drawing a small signal equivalent you short all voltage sources and open circuit all current sources.This means that you have to ground all the rail voltages too.

In addition all capacitors become shorts.

5MΩ

10MΩ

3kΩ

7.5kΩ

100kΩ

+−vsig10kΩ

−

+

vout

Now we need to replace the MOSFET with it’s equivalent circuit. While doing this, we can also get rid of the 3kΩ resistorsince it is being shorted out by ground, and we can also rearrange the circuit to better understand it.

5MΩ 10MΩ100kΩ+−vsig

G

S

gmvgs 100kΩ 7.5kΩ 10kΩ

(d) From the circuit diagram many values can be determined.

Rin =1

1

5+

1

10

= 3.33MΩ

Ro =1

1

100+

1

7.5

= 6.98kΩ

vgsvsig

=3333

100 + 3333≈ 0.971

vovgs

= −(gm)(Ro‖RL) = −(0.002)(6.98kΩ‖10kΩ) ≈ −8.22V

V

vovsig

= −(gm)(0.971)(Ro‖RL) ≈ −8V

V

11