University of California, Berkeley Spring 2012EE 42/100 Prof. A. Niknejad

Problem Set 3Solutions

Please note that these are merely suggested solutions. Many of these problems can beapproached in different ways.

1. The dependent voltage source is floating, so we write one supernode equation aroundv1 and v2:

−5 +v1

10+v2

20− 3 = 0

In addition, we can relate the voltages across the supernode as follows:

v1 − v2 = 5ix = −5

(v1

10

)The node voltage solutions are v1 = 45.7 V and v2 = 68.6 V. The power delivered to

the 8-Ω resistance is given by (v1−v2)2

R= (5222) V

8 Ω= 65.3 W.

2. First note that node 6 is adjacent to both voltage sources in this circuit. If we choosethis node as our reference, we can knock out two unknowns, nodes 5 and 6. We writethe rest of the nodal equations in order, using conductances:

G2(v1 − Vs) +G3(v1 − v2) +G4(v1 − v3) = is

G3(v2 − v1) +G5(v2 − v3) +G6(v2) = 0

G4(v3 − v1) +G5(v3 − v2) +G8(v3 − v4) = −isG8(v4 − v3) +G7(v4) +G9(v4 − v5) = 5i1 = 5G1Vs

v5 = 10Vx = 10(v4 − v5)

Writing this in the matrix form Ax = b, we have the following:G2 +G3 +G4 −G3 −G4 0 0−G3 G3 +G5 +G6 −G5 0 0−G4 −G5 G4 +G5 +G8 −G8 0

0 0 −G8 G7 +G8 +G9 −G9

0 0 0 10 −11

v1

v2

v3

v4

v5

=

is +G2Vs

0−is

5G1Vs

0

3. The first and most straightforward step is to find the open-circuit voltage using nodal

analysis. Notice that V3 = Voc. This is because there is no current across the 2-Ωresistor, and hence no voltage drop.

V2 − 19

10+V2

4+V2 − V3

20= 0

V3 − V2

20+V3 − 19

5+V3

8= 2Ix = 2

(19− V2

10

)The solutions are V2 = 6.94 V and V3 = Voc = VTh = 17.49 V. The next step is to findIsc by shorting the terminals:

+ 19 V

V1

10 Ω

5 Ω

20 Ω

4 Ω 2I x

V2 V3

8 Ω

2 Ω a

b

I x

I sc

V2 − 19

10+V2

4+V2 − V3

20= 0

V3 − V2

20+V3 − 19

5+V3

8+V3

2= 2Ix = 2

(19− V2

10

)The solutions are V2 = 5.71 V and V3 = 7.71 V. Thus, Isc = IN = V3

2= 3.86 A.

The final step is RTh = RN = VTh

IN= 4.53 Ω.

4. Since there are four sources, we will consider each one at a time in four differentsubcircuits. First we will find the equivalent resistance by zeroing out all the sources:

30Ω

40Ω

50Ω

20Ω

10Ω

Notice that the 50-Ω resistor is shorted out. The equivalent resistance isReq = RTh = RN = 30||(10 + 20 + 40) = 21 Ω.

The next step is to find the Thevenin voltage. We start with the 10 V source.

2

+

30Ω

40Ω10 V

50Ω

20Ω

10Ω

Voc

+

−

By voltage divider (and shorting out the 50-Ω resistor):

Voc1 =30

10 + 20 + 30 + 40(10 V) = 3 V

30Ω

40Ω

7 A

50Ω

20Ω

10Ω

Voc

+

−

Voc2 is given by Ohm’s law across the 30-Ω resistor. To find the current across it, wecan use a current divider between it and the other resistors in series (except for theshorted out 50-Ω):

Voc2 = IR =

[− 70

100(7 A)

](30 Ω) = −147 V

3

30Ω

40Ω

50Ω

20Ω

10Ω

5 A

Voc

+

−

This analysis is similar to the previous, except now the 40-Ω resistor is in parallelwith the others in series (again ignore the 50-Ω).

Voc3 = IR =

[− 40

100(5 A)

](30 Ω) = −60 V

+

30Ω

40Ω

5 V 50Ω

20Ω

10Ω

Voc

+

−

Last step is to deal with the 5 V source. This time the 50-Ω resistor is not shortedout, but we can ignore it if we consider a voltage divider between the 30-Ω resistorand the other ones:

Voc4 = − 30

10 + 20 + 30 + 40(5 V) = −1.5 V

The final step is to sum up all the individual components. Thus, we haveVoc = VTh = −205.5 V. Lastly, we can find IN = VTh

RTh= −205.5 V

21 Ω= −9.79 A.

5. This is a problem calling for the usage of the maximum power transfer theorem. Tomaximize the power transferred across the load, we need to choose its value such thatit is equal to that of the Thevenin resistance as seen by RL. To find it, we can simplyzero out the current source and find Req between the two terminals.

4

2 kΩ 4 kΩ

6 kΩ 8 kΩ

R eq

This is simply given by Req = RTh = RL = (2k + 6k)||(4k + 8k) = 4800 Ω.

6. We notice that this op amp is in negative feedback, so we can apply the summingpoint constraint: v+ = v− and i+ = i− = 0. Using Ohm’s law, v+ = v− = iSRS. Atthe output, we again use Ohm’s law and see that iL = v+

RL= iSRS

RL. Thus the gain is

Gi =iLiS

=RS

RL

7. We have negative feedback, so we can apply the summing point constraint. So wesimply have vO = vS, and the gain is just Gv = 1.

8. The best way to go about this problem is to first find a relationship between vo(t)and vin(t). Notice that because of negative feedback, we have v− = v+ = 1 V. So wecan write a simple nodal equation at v− as follows:

1− vin(t)

10000+

1− vo(t)

R2

= 0

Now we want to satisfy some condition for vo(t), so let’s solve for the output voltage:

vo(t) = 1 +R2

10000(1− vin(t)) = 1 +

R2

10000(1− 5− 3 cos(1000t))

The condition we want to satisfy is for the DC component of the output to be 0, sowe want the constant part to equal 0:

1− 4R2

10000= 0

Solving for this gives us R2 = 2500 Ω. Furthermore, the output simplifies tovo(t) = −0.75 cos(1000t).

9. (a) Again using negative feedback, we can conclude the following nodal voltages byapplying the golden rules at the inputs of both op amps:

5

v1

v2

R 1 R 2 R 3 R 4vo

v1 v2vx

Now we write two nodal equations at v1 and v2.

v1

R1

+v1 − vx

R2

= 0

v2 − vx

R3

+v2 − vo

R4

= 0

Solving for these equations gives us the following relationship:

vo =R3 +R4

R3

v2 −R4

R1R3

(R1 +R2)v1

In order for this expression to be in the form vo = K(v2 − v1), we must have thefollowing constraint:

R3 +R4

R3

=R4

R1R3

(R1 +R2)

This simplifies to R1R3 = R2R4.

(b) If the resistors have 1% precision, then any resistor Ri can have a resistance ashigh as 1.01Ri or as low as 0.99Ri. Let’s assume that each resistor takes on avalue R = Rnom(1 + ∆) where Rnom is the nominal value, and |∆| < 0.01 is thetolerance (positive or negative). Now re-write the expression for the gain as

vo = a2v2 − a1v1

The a2 coefficient is expanded and then simplified assuming ∆ are small

a2 = 1 +R′4R′3

= 1 +R4(1 + ∆4)

R3(1 + ∆3)≈ 1 +

R4

R3

(1 + ∆4)(1−∆3) ≈ 1 +R4

R3

(1 + ∆4−∆3)

Notice that a2 can be written in terms of the ideal gain G = 1 + R4

R3

a2 = G+ (G− 1)(∆4 −∆3) = G

(1 +

G− 1

G(∆4 −∆3)

)= G(1 + ε2)

Likewise, we can expand the coefficient a1

a1 =R′4R′3

(1 +R′2R′1

) =R4

R3

(1 + ∆4)

(1 + ∆3)

(1 +

(1 + ∆2)

(1 + ∆1)

R2

R1

)

6

Note that R2/R1 = R3/R4 so we have

a1 =(1 + ∆4)

(1 + ∆3)

(R4

R3

+(1 + ∆2)

(1 + ∆1)

)Simplifying assuming ∆ are small

a1 ≈ (1 + ∆4 −∆3)

(R4

R3

+ (1 + ∆2 −∆1)

)= (1 + ∆4 −∆3)G

(1 +

∆2 −∆1

G

)or

a1 ≈ G

(1 + ∆4 −∆3 +

∆2 −∆1

G

)= G(1 + ε1)

For a common-mode input Vc we have

vo = a1Vc − a2Vc = G(1 + ε1)Vc −G(1 + ε2) = GVc(ε1 − ε2)

The common-mode gain is therefore

Ac =vo

Vc

= G(∆4 −∆3 +∆2 −∆1

G− G− 1

G(∆4 −∆3))

orAc = (∆4 −∆3) + ∆2 −∆1

The worst case is when the gain is maximized, or the errors in the ∆’s conspireto maximize the difference

Ac = 4|∆|

7