# Problem Set 3 Solutions

advertisement ```University of California, Berkeley
EE 42/100
Spring 2012
Prof. A. Niknejad
Problem Set 3
Solutions
Please note that these are merely suggested solutions. Many of these problems can be
approached in different ways.
1. The dependent voltage source is floating, so we write one supernode equation around
v1 and v2 :
v2
v1
+
−3=0
−5 +
10 20
In addition, we can relate the voltages across the supernode as follows:
v1
v1 − v2 = 5ix = −5
10
The node voltage solutions are v1 = 45.7 V and v2 = 68.6 V. The power delivered to
2
2) V
2)
the 8-Ω resistance is given by (v1 −v
= (522
= 65.3 W.
R
8Ω
2. First note that node 6 is adjacent to both voltage sources in this circuit. If we choose
this node as our reference, we can knock out two unknowns, nodes 5 and 6. We write
the 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 ) = −is
G8 (v4 − v3 ) + G7 (v4 ) + G9 (v4 − v5 ) = 5i1 = 5G1 Vs
v5 = 10Vx = 10(v4 − v5 )
Writing this in the matrix form Ax = b, we have the following:

  

G2 + G3 + G4
−G3
−G4
0
0
v1
is + G2 Vs

  

−G3
G3 + G5 + G6
−G5
0
0  v2  
0



  

−G4
−G5
G4 + G5 + G8
−G8
0  v3  =  −is 


  

0
0
−G8
G7 + G8 + G9 −G9  v4   5G1 Vs 

0
0
0
10
−11
v5
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-Ω
resistor, and hence no voltage drop.
V2 − 19 V2 V2 − V3
+
+
=0
10
4
20
V3 − V2 V3 − 19 V3
+
+
= 2Ix = 2
20
5
8
19 − V2
10
The solutions are V2 = 6.94 V and V3 = Voc = VT h = 17.49 V. The next step is to find
Isc by shorting the terminals:
5Ω
Ix
V1
+
10 Ω
20 Ω
V2
19 V
2I x
4Ω
2Ω
V3
8Ω
a
I sc
b
V2 − 19 V2 V2 − V3
+
+
=0
10
4
20
19 − V2
V3 − V2 V3 − 19 V3 V3
+
+
+
= 2Ix = 2
20
5
8
2
10
The solutions are V2 = 5.71 V and V3 = 7.71 V. Thus, Isc = IN =
The final step is RT h = RN =
VT h
IN
V3
2
= 3.86 A.
= 4.53 Ω.
4. Since there are four sources, we will consider each one at a time in four different
subcircuits. First we will find the equivalent resistance by zeroing out all the sources:
10Ω
20Ω
30Ω
40Ω
50Ω
Notice that the 50-Ω resistor is shorted out. The equivalent resistance is
Req = RT h = RN = 30||(10 + 20 + 40) = 21 Ω.
The next step is to find the Thevénin voltage. We start with the 10 V source.
2
10Ω
20Ω
30Ω
10 V +
40Ω
+
Voc
−
50Ω
By voltage divider (and shorting out the 50-Ω resistor):
Voc1 =
30
(10 V) = 3 V
10 + 20 + 30 + 40
10Ω
20Ω
7A
30Ω
40Ω
+
Voc
−
50Ω
Voc2 is given by Ohm’s law across the 30-Ω resistor. To find the current across it, we
can use a current divider between it and the other resistors in series (except for the
shorted out 50-Ω):
70
Voc2 = IR = −
(7 A) (30 Ω) = −147 V
100
3
10Ω
20Ω
30Ω
40Ω
+
Voc
−
5A
50Ω
This analysis is similar to the previous, except now the 40-Ω resistor is in parallel
with the others in series (again ignore the 50-Ω).
40
(5 A) (30 Ω) = −60 V
Voc3 = IR = −
100
10Ω
20Ω
30Ω
40Ω
5V
+
+
Voc
−
50Ω
Last step is to deal with the 5 V source. This time the 50-Ω resistor is not shorted
out, but we can ignore it if we consider a voltage divider between the 30-Ω resistor
and the other ones:
Voc4 = −
30
(5 V) = −1.5 V
10 + 20 + 30 + 40
The final step is to sum up all the individual components. Thus, we have
V
Voc = VT h = −205.5 V. Lastly, we can find IN = RVTThh = −205.5
= −9.79 A.
21 Ω
5. This is a problem calling for the usage of the maximum power transfer theorem. To
maximize the power transferred across the load, we need to choose its value such that
it is equal to that of the Thévenin resistance as seen by RL . To find it, we can simply
zero out the current source and find Req between the two terminals.
4
2 kΩ
4 kΩ
R eq
6 kΩ
8 kΩ
This is simply given by Req = RT h = RL = (2k + 6k)||(4k + 8k) = 4800 Ω.
6. We notice that this op amp is in negative feedback, so we can apply the summing
point constraint: v+ = v− and i+ = i− = 0. Using Ohm’s law, v+ = v− = iS RS . At
the output, we again use Ohm’s law and see that iL = Rv+L = iSRRLS . Thus the gain is
Gi =
iL
RS
=
iS
RL
7. We have negative feedback, so we can apply the summing point constraint. So we
simply 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 we
can write a simple nodal equation at v− as follows:
1 − vin (t) 1 − vo (t)
=0
+
10000
R2
Now we want to satisfy some condition for vo (t), so let’s solve for the output voltage:
vo (t) = 1 +
R2
R2
(1 − vin (t)) = 1 +
(1 − 5 − 3 cos(1000t))
10000
10000
The condition we want to satisfy is for the DC component of the output to be 0, so
we want the constant part to equal 0:
1−
4R2
=0
10000
Solving for this gives us R2 = 2500 Ω. Furthermore, the output simplifies to
vo (t) = −0.75 cos(1000t).
9. (a) Again using negative feedback, we can conclude the following nodal voltages by
applying the golden rules at the inputs of both op amps:
5
R1
v1
R2
vx R 3
v2
R4
vo
v1
v2
Now we write two nodal equations at v1 and v2 .
v1
v1 − vx
+
=0
R1
R2
v2 − vx v2 − vo
+
=0
R3
R4
Solving for these equations gives us the following relationship:
vo =
R3 + R4
R4
v2 −
(R1 + R2 )v1
R3
R1 R3
In order for this expression to be in the form vo = K(v2 − v1 ), we must have the
following constraint:
R4
R3 + R4
=
(R1 + R2 )
R3
R1 R3
This simplifies to R1 R3 = R2 R4 .
(b) If the resistors have 1% precision, then any resistor Ri can have a resistance as
high as 1.01Ri or as low as 0.99Ri . Let’s assume that each resistor takes on a
value R = Rnom (1 + ∆) where Rnom is the nominal value, and |∆| &lt; 0.01 is the
tolerance (positive or negative). Now re-write the expression for the gain as
vo = a2 v2 − a1 v1
The a2 coefficient is expanded and then simplified assuming ∆ are small
a2 = 1 +
R4 (1 + ∆4 )
R4
R4
R40
= 1+
≈ 1+
(1 + ∆4 )(1 − ∆3 ) ≈ 1 +
(1 + ∆4 − ∆3 )
0
R3
R3 (1 + ∆3 )
R3
R3
Notice that a2 can be written in terms of the ideal gain G = 1 +
R4
R3
G−1
a2 = G + (G − 1)(∆4 − ∆3 ) = G 1 +
(∆4 − ∆3 ) = G(1 + 2 )
G
Likewise, we can expand the coefficient a1
R0
R0
R4 (1 + ∆4 )
a1 = 40 (1 + 20 ) =
R3
R1
R3 (1 + ∆3 )
6
(1 + ∆2 ) R2
1+
(1 + ∆1 ) R1
Note that R2 /R1 = R3 /R4 so we have
(1 + ∆4 )
a1 =
(1 + ∆3 )
R4 (1 + ∆2 )
+
R3 (1 + ∆1 )
Simplifying assuming ∆ are small
R4
∆2 − ∆1
a1 ≈ (1 + ∆4 − ∆3 )
+ (1 + ∆2 − ∆1 ) = (1 + ∆4 − ∆3 )G 1 +
R3
G
or
∆2 − ∆1
a1 ≈ G 1 + ∆4 − ∆3 +
= G(1 + 1 )
G
For a common-mode input Vc we have
vo = a1 Vc − a2 Vc = G(1 + 1 )Vc − G(1 + 2 ) = GVc (1 − 2 )
The common-mode gain is therefore
Ac =
∆2 − ∆1 G − 1
vo
−
(∆4 − ∆3 ))
= G(∆4 − ∆3 +
Vc
G
G
or
Ac = (∆4 − ∆3 ) + ∆2 − ∆1
The worst case is when the gain is maximized, or the errors in the ∆’s conspire
to maximize the difference
Ac = 4|∆|
7
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