Chapter 5, Solution 1.
(a)
(b)
(c)
R in = 1.5 M
R out = 60
A = 8x104
Therefore A dB = 20 log 8x104 = 98.06 dB
Chapter 5, Solution 2.
v 0 = Av d = A(v 2 - v 1 )
= 105 (20-10) x 10-6 = 1V
Chapter 5, Solution 3.
v 0 = Av d = A(v 2 - v 1 )
= 2 x 105 (30 + 20) x 10-6 = 10V
Chapter 5, Solution 4.
v 0 = Av d = A(v 2 - v 1 )
v
4
v2 - v 1 = 0
A 2 x10 6
2 V
v 2 - v 1 = -2 µV = –0.002 mV
1 mV - v 1 = -0.002 mV
v 1 = 1.002 mV
Chapter 5, Solution 5.
I
R0
R in
vd
+
vi
+
-
+
-
-v i + Av d + (R i + R 0 ) I = 0
But
Av d
+
v0
-
(1)
v d = R i I,
-v i + (R i + R 0 + R i A) I = 0
I=
vi
(1 A )R i
R0
(2)
-Av d - R 0 I + v 0 = 0
v 0 = Av d + R 0 I = (R 0 + R i A)I =
v0
vi
R0
R0
RiA
(1 A )R i
10 9
10 4
5
1 10
(R 0 R i A) v i
R 0 (1 A )R i
100 10 4 x10 5
10 4
5
100 (1 10 )
100,000
100,001
0.9999990
Chapter 5, Solution 6.
vi
+ -
I
vd
R0
R in
+
-
+
Av d
+
vo
-
(R 0 + R i )R + v i + Av d = 0
But
v d = R i I,
v i + (R 0 + R i + R i A)I = 0
I=
R0
vi
(1)
(1 A )R i
-Av d - R 0 I + v o = 0
v o = Av d + R 0 I = (R 0 + R i A)I
Substituting for I in (1),
v0 =
=
R0
R0
R iA
vi
(1 A )R i
50 2x10 6 x 2 x10 5 10 3
50 1 2x10 5 x 2 x10 6
200,000 x 2 x10 6
mV
200,001x 2x10 6
v 0 = -0.999995 mV
Chapter 5, Solution 7.
100 k
VS
+
–
R out = 100
1
10 k
2
+
Vd
R in
–
+
AV d
–
At node 1,
+
V out
–
(V S – V 1 )/10 k = [V 1 /100 k] + [(V 1 – V 0 )/100 k]
10 V S – 10 V 1 = V 1 + V 1 – V 0
which leads to V 1 = (10V S + V 0 )/12
At node 2,
(V 1 – V 0 )/100 k = (V 0 – (–AV d ))/100
But V d = V 1 and A = 100,000,
V 1 – V 0 = 1000 (V 0 + 100,000V 1 )
0= 1001V 0 + 99,999,999[(10V S + V 0 )/12]
0 = 83,333,332.5 V S + 8,334,334.25 V 0
which gives us (V 0 / V S ) = –10 (for all practical purposes)
If V S = 1 mV, then V 0 = –10 mV
Since V 0 = A V d = 100,000 V d , then V d = (V 0 /105) V = –100 nV
Chapter 5, Solution 8.
(a)
If v a and v b are the voltages at the inverting and noninverting terminals of the op amp.
va = vb = 0
1mA =
0 v0
2k
v 0 = –2 V
(b)
10 k
2V
-
+
ia
va
2V
-
+
vb
1V
+
+
vo
2k
-
-
+
+-
va
-
+
ia
vo
-
(b)
(a)
Since v a = v b = 1V and i a = 0, no current flows through the 10 k
-v a + 2 + v 0 = 0
10 k
v 0 = v a – 2 = 1 – 2 = –1V
resistor. From Fig. (b),
Chapter 5, Solution 9.
(a)
Let v a and v b be respectively the voltages at the inverting and noninverting terminals of
the op amp
v a = v b = 4V
At the inverting terminal,
1mA =
4 v0
2k
v 0 = 2V
(b)
1V
+-
+
+
-
-
vb
Since v a = v b = 3V,
-v b + 1 + v o = 0
v o = v b – 1 = 2V
vo
Chapter 5, Solution 10.
Since no current enters the op amp, the voltage at the input of the op amp is v s . Hence
vs = vo
10
10 10
vo
2
vo
=2
vs
5.11 Using Fig. 5.50, design a problem to help other students to better understand how ideal op
amps work.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Find v o and i o in the circuit in Fig. 5.50.
Figure 5.50 for Prob. 5.11
Solution
8k
2k
3V
vb =
5k
+
10
(3)
10 5
a
b
io
+
10 k
2V
At node a,
3 va
2
va
8
vo
12 = 5v a – v o
+
4k
vo
But v a = v b = 2V,
v o = –2V
12 = 10 – v o
–i o =
va
8
vo
0 vo
4
2 2
8
2
4
1mA
i o = –1mA
Chapter 5, Solution 12.
Step 1.
Label the unknown nodes in the op amp circuit. Next we write the node
equations and then apply the constraint, V a = V b . Finally, solve for V o in terms of V s .
25 k
5k
VS
Step 2.
+
a
b
+
10 k
+
Vo
[(V a -V s )/5k] + [(V a -V o )/25k] + 0 = 0 and
[(V b -0)/10k] + 0 = 0 or V b = 0 = V a ! Thus,
[(-V s )/5k] + [(-V o )/25k] = 0 or,
V o = ( –25/5)V s or V o /V s = –5.
Chapter 5, Solution 13.
10 k
a
+
100 k
b
1V
+
io
i2
10 k
90 k
50 k
By voltage division,
va =
90
(1)
100
0.9V
vb =
50
vo
150
vo
3
But v a = v b
io = i1 + i2 =
v0
3
vo
10k
0.9
vo
150k
i1
v o = 2.7V
0.27mA + 0.018mA = 288 A
+
vo
Chapter 5, Solution 14.
Transform the current source as shown below. At node 1,
10 v1
5
v1
v2
20
10 k
5k
vo
10
vo
10 k
20 k
v1
10V
v1
v2
+
+
vo
But v 2 = 0. Hence 40 – 4v 1 = v 1 + 2v 1 – 2v o
At node 2,
v1
v2
20
+
v2
vo
v2
0 or v 1 = –2v o
From (1) and (2), 40 = –14v o - 2v o
v o = –2.5V
10
,
40 = 7v 1 – 2v o
(1)
(2)
(a) Let v 1 be the voltage at the node where the three resistors meet. Applying
KCL at this node gives
v1 v1 v o
1
v1
R2
R3
R2
At the inverting terminal,
is
vo
R3
1
R3
0 v1
v1
i s R1
R1
Combining (1) and (2) leads to
vo
R
R
is
R
R
R
(1)
(2)
is
vo
is
(b) For this case,
vo
is
20 40
20 x 40
k
25
- 92 k
=
R
R
RR
R
Chapter 5, Solution 16
Using Fig. 5.55, design a problem to help students better understand inverting op amps.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Obtain i x and i y in the op amp circuit in Fig. 5.55.
Figure 5.55
Solution
10k
5k
ix
va
vb
+
0.5V
-
+
2k
8k
iy
vo
Let currents be in mA and resistances be in k
0.5 v a v a vo
1 3v a vo
5
10
But
va
vb
8
vo
vo
. At node a,
10
va
8
8 2
Substituting (2) into (1) gives
10
8
va
va
1 3v a
8
14
Thus,
0.5 v a
ix
1 / 70 mA
14.28 A
5
vo vb vo va
10
iy
0.6(v o v a ) 0.6( v a
2
10
8
= 85.71 µA
(1)
(2)
va )
0.6 8
x mA
4 14
Chapter 5, Solution 17.
(a)
(b)
(c)
G=
vo
vi
vo
vi
vo
vi
(a) –2.4, (b) –16, (c) –400
Rf
12
5
Ri
80
= –16
5
2000
5
–400
–2.4
Chapter 5, Solution 18.
For the circuit, shown in Fig. 5.57, solve for the Thevenin equivalent circuit
looking into terminals A and B.
10 k
10 k
7.5 V
a
b
+
+
c
A
2.5
B
Figure 5.57
For Prob. 5.18.
Write a node equation at a. Since node b is tied to ground, v b = 0. We cannot write a
node equation at c, we need to use the constraint equation, v a = v b . Once, we know v c ,
we then proceed to solve for V open circuit and I short circuit . This will lead to V Thev (t) = V open
circuit and R equivalent = V open circuit /I short circuit .
[(v a – 7.5)/10k] + [(v a – v c )/10k] + 0 = 0
Our constraint equation leads to,
v a = v b = 0 or v c = –7.5 volts
This is also the open circuit voltage (note, the op-amp keeps the output voltage at –5 volts
in spite of any connection between A and B. Since this means that even a short from A to
B would theoretically then produce an infinite current, R equivalent = 0. In real life, the
short circuit current will be limited to whatever the op-amp can put out into a short
circuited output.
V Thev = –7.5 volts; R equivalent = 0-ohms.
Chapter 5, Solution 19.
We convert the current source and back to a voltage source.
24
(4/3) k
(1.5/3)V
4k
0V
+
4
3
10 k
+
vo
2k
vo
io
10k
1.5
–937.5 mV.
4
3
4
k
3
vo vo 0
–562.5 µA.
2k 10k
Chapter 5, Solution 20.
8k
4k
a
4k
2k
b
+
9V
+
vs
+
+
vo
At node a,
9 va
4
va
8
vo
va
4
vb
18 = 5v a – v o – 2v b
(1)
At node b,
va
4
vb
vb
2
vo
v a = 3v b – 2v o
But v b = v s = 2 V; (2) becomes v a = 6 –2v o and (1) becomes
–18 = 30–10v o – v o – 4
v o = –44/(–11) = 4 V.
(2)
Chapter 5, Solution 21.
Let the voltage at the input of the op amp be v a .
va
1 V,
3-v a
4k
va vo
10 k
3-1 1 vo
4
10
v o = –4 V.
Chapter 5, Solution 22.
A v = -R f /R i = -15.
If R i = 10k , then R f = 150 k .
Chapter 5, Solution 23
At the inverting terminal, v=0 so that KCL gives
vs 0
R1
0
R2
0 vo
Rf
vo
Rf
vs
R1
Chapter 5, Solution 24
v1
Rf
R1
R2
- vs +
+
+
R4
R3
vo
-
v2
We notice that v 1 = v 2 . Applying KCL at node 1 gives
v1
R1
(v1
R2
vs )
v1
Rf
vo
1
R1
0
1
R2
1
v1
Rf
vs
R2
vo
Rf
Applying KCL at node 2 gives
v1 v1 v s
v1
0
R3
R4
Substituting (2) into (1) yields
vo
Rf
R3
R1
R3
Rf
R3
R4
R2
R3
k
Rf
R3
R3
R4
R4
(2)
vs
1
vs
R2
i.e.
R3
R1
R3
Rf
R4
R2
R3
R3
R4
1
R2
(1)
Chapter 5, Solution 25.
This is a voltage follower. If v 1 is the output of the op amp,
v 1 = 3.7 V
v o = [20k/(20k+12k)]v 1 = [20/32]3.7 = 2.312 V.
Chapter 5, Solution 26
Using Fig. 5.64, design a problem to help other students better understand noninverting op amps.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Determine i o in the circuit of Fig. 5.64.
Figure 5.64
Solution
+
vb
+
0.4V
-
vb
0.4
Hence,
-
+
2k
8k
8
8 2
vo
vo
-
0.8vo
vo
io
vo
5k
0.4 / 0.8
0.5
5k
0.5 V
0.1 mA
io
5k
Chapter 5, Solution 27.
This is a voltage follower.
v 1 = [24/(24+16)]7.5 = 4.5 V; v 2 = v 1 = 4.5 V; and
v o = [12/(12+8)]4.5 = 2.7 V.
Chapter 5, Solution 28.
50 k
v1
va
10 k
At node 1,
0 v1
10k
+
+
10 V
v1 v o
50k
But v 1 = 10V,
–5v 1 = v 1 – v o , leads to
v o = 6v 1 = 60V
Alternatively, viewed as a noninverting amplifier,
v o = (1 + (50/10)) (10V) = 60V
i o = v o /(20k) = 60/(20k) = 3 mA.
vo
20 k
Chapter 5, Solution 29
R1
va
+
-
vb
+
vi
-
va
R1
But v a
Or
R2
R2
vb
R2
vi ,
vb
R1
R1
R1
R2
R2
R1
R2
R2
vo
vi
R1
R1
R2
R1
R2
vo
-
vo
vi
+
vo
Chapter 5, Solution 30.
The output of the voltage becomes
v o = v i = 1.2 V
(30k 20k ) 12k
By voltage division,
vx
12
(1.2) 0.2V
12 60
ix
vx
20k
p
0.2
20k
v 2x
R
20
2 x10 6
0.04
20k
10 A
2 W.
After converting the current source to a voltage source, the circuit is as shown below:
+
+
At node 1,
At node 2,
12 v1
3
v1
v1
vo
From (1) and (2),
vo
ix
6
vo
48
11
vo
6k
6
vo
0
6
v1
ix
vo
12
48 = 7v 1 - 3v o
v 1 = 2v o
(1)
(2)
Chapter 5, Solution 32.
Let v x = the voltage at the output of the op amp. The given circuit is a non-inverting
amplifier.
vx
1
60 30
50
(4 mV) = 24 mV
10
20k
By voltage division,
vx
20
12mV
vx
2
20 20
vx
24mV
600 A
ix =
20 20 k
40k
vo =
p=
v o2
R
144 x10 6
60x10 3
204 W.
Chapter 5, Solution 33.
After transforming the current source, the current is as shown below:
1k
4k
4V
vi
+
va
+
vo
2k
3k
This is a noninverting amplifier.
vo
1
1
vi
2
3
vi
2
Since the current entering the op amp is 0, the source resistor has a 0 V potential drop.
Hence v i = 4V.
vo
3
(4)
2
6V
Power dissipated by the 3k
resistor is
v o2
R
ix
12mW, –2mA
va
R
36
3k
vo
12mW
4 6
1k
–2mA.
Chapter 5, Solution 34
v1 vin
R1
but
va
v1 vin
R2
0
(1)
R3
vo
R3 R 4
(2)
Combining (1) and (2),
v1 va
va 1
R1
v2
R2
R1
R2
v1
R1
R 3v o
1
R2
R3 R 4
vo
R3
R3 1
R4
R1
R2
R1
va
R2
0
R1
v2
R2
v1
v1
R1
v2
R2
R1
v2
R2
vO =
R3 R4
(v1 R2
R3 ( R1 R2 )
v2 )
Chapter 5, Solution 35.
Av
vo
vi
1
Rf
Ri
7.5
R f = 6.5R i
If R i = 60 k , R f = 390 k .
Chapter 5, Solution 36
VTh
But
VTh
Vab
vs
Vab
R1
R1
R1
R1
R2
R2
Vab . Thus,
vs
(1
R2
)v s
R1
To get R Th , apply a current source I o at terminals a-b as shown below.
v1
+
-
v2
a
R2
R1
+
vo
-
io
b
Since the noninverting terminal is connected to ground, v 1 = v 2 =0, i.e. no current passes
through R 1 and consequently R 2 . Thus, v o =0 and
RTh
vo
io
0
Chapter 5, Solution 37.
vo
Rf
v1
R1
Rf
v2
R2
30
( 2)
10
30
( 2)
20
v o = 1.5 V.
Rf
v3
R3
30
( 4.5)
30
Chapter 5, Solution 38.
Using Fig. 5.75, design a problem to help other students better understand summing amplifiers.
Although there are many ways to work this problem, this is an example based on the same kind
of problem asked in the third edition.
Problem
Calculate the output voltage due to the summing amplifier shown in Fig. 5.75.
Figure 5.75
Solution
vo
Rf
v1
R1
50
(10)
25
= -120mV
Rf
v2
R2
Rf
Rf
v4
v3
R4
R3
50
50
( 20)
(50)
20
10
50
( 100)
50
Chapter 5, Solution 39
This is a summing amplifier.
vo
Thus,
Rf
R1
v1
Rf
R2
v2
vo
Rf
v3
50
( 2)
10
16.5
9 2.5v 2
R3
50
v2
20
50
( 1)
50
v2
9 2.5v 2
3V
Chapter 5, Solution 40
Determine V o in terms of V 1 and V 2 .
100 k
V1
200 k
100 k
+
V2
Va
Vc
10
+
+
Vb
+
40
Vo
Step 1.
Label the reference and node voltages in the circuit, see above.
Note we now can consider nodes a and b, we cannot write a node equation at c
without introducing another unknown. The node equation at a is [(V a –V 1 )/105] +
[(V a –V 2 )/105] + 0 + [(V a –V c )/2x105] = 0. At b it is clear that V b = 0. Since we
have two equations and three unknowns, we need another equation. We do get
that from the constraint equation, V a = V b . After we find V c in terms of V 1 and
V 2 , we then can determine V o which is equal to [(V c –0)/50] times 40.
Step 2.
Letting V a = V b = 0, the first equation can be simplified to,
[–V 1 /105] + [–V 2 /105] + [–V c /2x105] = 0
Taking V c to the other side of the equation and multiplying everything by 2x105,
we get,
V c = –2V 1 – 2V 2
Now we can find V o which is equal to (40/50)V c = 0.8[–2V 1 –2V 2 ]
V o = –1.6V 1 –1.6V 2 .
Chapter 5, Solution 41.
R f /R i = 1/(4)
R i = 4R f = 40k
The averaging amplifier is as shown below:
v1
v2
v3
v4
R 1 = 40 k
10 k
R 2 = 40 k
R 3 = 40 k
R 4 = 40 k
+
vo
Since the average of three numbers is the sum of those numbers divided by three, the
value of the feedback resistor needs to be equal to one-third of the input resistors or,
Rf
1
R1
3
.
Chapter 5, Solution 43.
In order for
vo
to become
vo
Rf
Ri
Rf
v1
R1
1
v1
4
1
4
Rf
v2
R2
Rf
v3
R3
v2
v4
v3
Rf
Ri
4
Rf
v4
R4
80k
4
20 k .
Chapter 5, Solution 44.
R4
R3
a
R1
v1
At node a,
vb
R1
v1
0 va
R3
vb
va
+
vo
R2
v2
At node b,
b
R2
R4
v2
vb
0
vo
va
v1
R1
1
R1
v2
R2
1
R2
vo
1 R4 / R3
But v a = v b . We set (1) and (2) equal.
vo
1 R4 / R3
R 2 v1 R 1v 2
R1 R 2
or
vo =
R3 R4
R2 v 1
R3 R1 R2
R1 v 2
(1)
(2)
Chapter 5, Solution 45.
This can be achieved as follows:
vo
R
R /3
Rf
R1
v1
v1
R
v2
R/2
Rf
v2
R2
i.e. R f = R, R 1 = R/3, and R 2 = R/2
Thus we need an inverter to invert v 1 , and a summer, as shown below (R<100k ).
R
v1
R
R/3
+
-v 1
v2
R
R/2
+
vo
Chapter 5, Solution 46.
Rf
v1 1
Rx
Rf
1
v3
v1
( v2 )
( v2 )
v3
2
R1
3 3
R2
R3
i.e. R 3 = 2R f , R 1 = R 2 = 3R f . To get -v 2 , we need an inverter with R f = R i . If R f =
10k , a solution is given below.
vo
10 k
v2
v1
10 k
30 k
30 k
+
-v 2
v3
10 k
20 k
+
vo
Chapter 5, Solution 47.
Using eq. (5.18), R1
vo
30(1 2 / 30)
v2
2(1 2 / 20)
2 k , R2
30
V1
2
30k , R3
2k , R4
32
(2) 15(1) 14.09 V
2.2
= 14.09 V.
20 k
Chapter 5, Solution 48.
We can break this problem up into parts. The 5 mV source separates the lower
circuit from the upper. In addition, there is no current flowing into the input of
the op amp which means we now have the 40-kohm resistor in series with a
parallel combination of the 60-kohm resistor and the equivalent 100-kohm
resistor.
+10 mV
Thus,
40k + (60x100k)/(160) = 77.5k
which leads to the current flowing through this part of the circuit,
i = 10 m/77.5k = 129.03x10–9 A
The voltage across the 60k and equivalent 100k is equal to,
v = ix37.5k = 4.839 mV
We can now calculate the voltage across the 80-kohm resistor.
v 80 = 0.8x4.839 m = 3.87 mV
which is also the voltage at both inputs of the op amp and the voltage between the
20-kohm and 80-kohm resistors in the upper circuit. Let v 1 be the voltage to the
left of the 20-kohm resistor of the upper circuit and we can write a node equation
at that node.
(v 1 –10m)/(10k) + v 1 /30k + (v 1 –3.87m)/20k = 0
or
6v 1 – 60m + 2v 1 + 3v 1 – 11.61m = 0
or
v 1 = 71.61/11 = 6.51 mV.
The current through the 20k-ohm resistor, left to right, is,
i 20 = (6.51m–3.87m)/20k = 132 x10–9 A
thus,
v o = 3.87m – 132 x10–9x80k = –6.69 mV.
Chapter 5, Solution 49.
R 1 = R 3 = 20k , R 2 /(R 1 ) = 4
i.e.
R 2 = 4R 1 = 80k
Verify:
vo
4
= R4
R 2 1 R1 / R 2
v2
R1 1 R 3 / R 4
(1 0.25)
v2
1 0.25
4 v1
4 v2
R2
v1
R1
v1
Thus, R 1 = R 3 = 20 k , R 2 = R 4 = 80 k .
Chapter 5, Solution 50.
(a) We use a difference amplifier, as shown below:
v1
v2
vo
R1
R2
R1
+
R2
vo
R2
v 2 v1 2.5 v 2 v1 , i.e. R 2 /R 1 = 2.5
R1
If R 1 = 100 k then R 2 = 250k
(b) We may apply the idea in Prob. 5.35.
v0
2.5v1 2.5v 2
R
R/2
Rf
R1
v1
v1
R
v2
R/2
Rf
v2
R2
i.e. R f = R, R 1 = R/2.5 = R 2
We need an inverter to invert v 1 and a summer, as shown below. We may let R =
100 k .
R
v1
R
R/2.5
+
-v 1
v2
R
R/2.5
+
vo
Chapter 5, Solution 51.
We achieve this by cascading an inverting amplifier and two-input inverting summer as
shown below:
R
v1
R
R
+
va
v2
Verify:
But
v o = -v a - v 2
v a = -v 1 . Hence
vo = v1 - v2.
R
R
+
vo
vo
v
v
v
v
Chapter 5, Solution 53.
(a)
R1
v1
R2
va
vb
R1
v2
At node a,
At node b,
v1
va
R1
vb
va
vo
R2
vo
+
R2
va
R 2 v1 R 1 v o
R1 R 2
(1)
R2
v2
R1 R 2
(2)
But v a = v b . Setting (1) and (2) equal gives
R 2 v1 R 1 v o
R2
v2
R1 R 2
R1 R 2
R1
vo vi
v 2 v1
R2
vo R 2
vi
R1
(b)
v1
vi
+ v2
R 1 /2 v
A
R 1 /2
R2
va
Rg
R 1 /2
R 1 /2
vB
vb
+
R2
+
vo
At node A,
v1 v A
R1 / 2
vB vA
Rg
or
v1
vA
R1
vB
2R g
At node B,
v2 vB
R1 / 2
vB vA
R1 / 2
or
v2
R1
(v B
2R g
vB
vA va
R1 / 2
vA
vA
va
(1)
vb
(2)
vB vb
Rg
vA )
vB
Subtracting (1) from (2),
v2
v1
v2
v1
vB
2R 1
vB
2R g
vA
Since, v a = v b ,
or
vB
2
R1
2R g
1
vi
2
vA
vB
vA
1
R1
1
2R g
vA
vB
vA
vb
va
vi
2
(3)
But for the difference amplifier,
vo
or
vB
R2
vB vA
R1 / 2
R1
vo
vA
2R 2
Equating (3) and (4),
(4)
R1
vo
2R 2
vi
2
vo
vi
1
R1
1
2R g
R2
R1
1
R1
1
2R g
(c)
v1
At node a,
At node b,
va
va vA
R2 / 2
2R 1
2R 1
vA
va
R2
R2
2R 1
2R 1
vb
vB
R2
R2
R1
v1
va
v2
vb
(1)
(2)
Since v a = v b , we subtract (1) from (2),
or
v2
v1
vB
vA
2R 1
(v B
R2
R2
vi
2R 1
vi
2
vA )
(3)
At node A,
va vA
R2 /2
va
At node B,
vB
vA
Rg
R2
vB
2R g
vA
vb vB
R/2
vb
vB
Subtracting (5) from (4),
vB vA
2 vB
vB
vA vo
R/2
vA
vA
vA
Rg
R2
vB
2R g
vA
vB
(5)
vA vB vo
R2
2R g
Combining (3) and (6),
R2
R2
vi 1
R1
2R g
(4)
vB 0
R/2
R2
vB vA
Rg
vA 1
vo
(6)
vo
vo
vo
vi
R2
1
R1
R2
2Rg
Chapter 5, Solution 54.
The first stage is a summer (please note that we let the output of the first stage be v 1 ).
v1
R
vs
R
R
vo
R
= –v s – v o
The second stage is a noninverting amplifier
v o = (1 + R/R)v 1 = 2v 1 = 2(–v s – v o ) or 3v o = –2v s
v o /v s = –0.6667.
Chapter 5, Solution 55.
Let A 1 = k, A 2 = k, and A 3 = k/(4)
A = A 1 A 2 A 3 = k3/(4)
20Log10 A 42
Log10 A 2.1
A = 102 1 = 125.89
k3 = 4A = 503.57
k = 3 503.57 7.956
Thus
A 1 = A 2 = 7.956, A 3 = 1.989
Chapter 5, Solution 56.
Using Fig. 5.83, design a problem to help other students better understand cascaded op
amps.
Although there are many ways to work this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
Calculate the gain of the op amp circuit shown in Fig. 5.83.
40 k
10 k
1k
20 k
–
+
+
vi
–
Figure 5.83
For Prob. 5.56.
Solution
Each stage is an inverting amplifier. Hence,
vo
vs
(
10
40
)(
) 20
1
20
–
+
Chapter 5, Solution 57.
Let v 1 be the output of the first op amp and v 2 be the output of the second op amp.
The first stage is an inverting amplifier.
50
v1
vs1
2 vs1
25
The second state is a summer.
v 2 = –(100/50)v s2 – (100/100)v 1 = –2v s2 + 2v s1
The third state is a noninverting amplifier
vo
(1
100
)v2
50
3v2
6vs1 6vs2
Chapter 5, Solution 58.
684.8 µA
Chapter 5, Solution 59.
The first stage is a noninverting amplifier. If v 1 is the output of the first op amp,
v 1 = (1 + 2R/R)v s = 3v s
The second stage is an inverting amplifier
v o = –(4R/R)v 1 = –4v 1 = –4(3v s ) = –12v s
v o /v s = –12.
Chapter 5, Solution 60.
The first stage is a summer. Let V 1 be the output of the first stage.
10
10
vi
vo
5
4
By voltage division,
10
5
v1
vo
vo
10 2
6
Combining (1) and (2),
5
vo
2v1 2.5v0
6
v1
2 vi
v1
2.5vo
(1)
(2)
10
v0
3
vo
vi
2vi
6 /10
0.6
Chapter 5, Solution 61.
The first op amp is an inverter. If v 1 is the output of the first op amp,
V 1 = –(200/100)(0.4) = –0.8 V
The second op amp is a summer
V o = –(40/10)(–0.2) – (40/20)(–0.8) = 0.8 + 1.6
= 2.4 V.
Chapter 5, Solution 62.
Let v 1 = output of the first op amp
v 2 = output of the second op amp
The first stage is a summer
R2
R
vi – 2 vo
R1
Rf
v1
(1)
The second stage is a follower. By voltage division
vo
R4
v1
R3 R4
v2
v1
R3 R4
vo
R4
From (1) and (2),
1
R3
vo
R4
1
R3
R4
vo
vi
R2
vi
R1
R2
vo
Rf
R2
R1
R2
vi
R1
1
1
R2
vo
Rf
R3
R4
R2 R4 R f
R2
Rf
R1 R2 R4
R3 R f
R4 R f
(2)
Chapter 5, Solution 63.
The two op amps are summers. Let v 1 be the output of the first op amp. For the first
stage,
v1
R2
vi
R1
R2
vo
R3
(1)
For the second stage,
vo
R4
v1
R5
R4
vi
R6
(2)
Combining (1) and (2),
vo
vo 1
R4 R2
vi
R 5 R1
R 2R 4
R 3R 5
R4 R2
vo
R5 R3
R 2R 4
R 1R 5
R4
vi
R6
R4
vi
R6
vo
vi
R2 R4 R4
R1 R5 R6
R2 R4
1
R3 R5
Chapter 5, Solution 64
G4
G
G1
vs
+
1
0V
+
G3
G
G2
2
v
0V +
-
-
At node 1, v 1 =0 so that KCL gives
Gv
(1)
G 2 v s G3 v o
Gv
From (1) and (2),
G1v s G 4 v o G 2 v s
or
(2)
G1v s
G4 vo
+
vo
At node 2,
(G1
G3 v o
vo
vs
G 2 )v s
G1 G2
G3 G4
(G3
G 4 )v o
Chapter 5, Solution 65
The output of the first op amp (to the left) is 6 mV. The second op amp is an inverter so
that its output is
30
(6mV) -18 mV
10
The third op amp is a noninverter so that
vo '
vo '
40
vo
40 8
vo
48
vo '
40
21.6 mV
Chapter 5, Solution 66.
We can start by looking at the contributions to v o from each of the sources and the fact
that each of them go through inverting amplifiers.
The 6 V source contributes –[100k/25k]6; the 4 V source contributes
–[40k/20k][–(100k/20k)]4; and the 2 V source contributes –[100k/10k]2 or
vo
40
100
( 6)
20
25
100
100
( 2)
( 4)
10
20
24 40 20
–4V
Chapter 5, Solution 67.
80
40
4.8 2.8
vo =
80
80
(0.7)
(0.3)
20
20
2 V.
Chapter 5, Solution 68.
If R q = , the first stage is an inverter.
Va
15
(15)
5
45 mV
when V a is the output of the first op amp.
The second stage is a noninverting amplifier.
vo
1
6
va
2
(1 3)( 45) –180mV.
Chapter 5, Solution 69.
In this case, the first stage is a summer
va
15
15
vo
(15)
10
5
45 1.5v o
For the second stage,
vo
7vo
1
6
va
2
180
vo
4v a
4
180
7
45 1.5v o
–25.71 mV.
Chapter 5, Solution 70.
The output of amplifier A is
30
30
(1)
(2)
10
10
vA
9
The output of amplifier B is
vB
20
20
(4)
(3)
10
10
14
40 k
vA
vB
20 k
a
60 k
vo
+
b
10 k
vb
10
( 14)
60 10
At node a,
vA
20
va
2V
va
vo
40
But v a = v b = -2V, 2(-9+2) = -2-v o
Therefore,
v o = 12V
Chapter 5, Solution 71
20k
5k
100k
+
+
1.5 V
–
40k
v2
80k
10k
+
20k
+
10k
v1
+
2.25V
–
v1
+
50k
30k
2.25,
v2
20
(1.5)
5
vo
v3
6,
100
v2
40
v3
(1
100
v3
80
50
) v1
30
6
( 15 7.5) 7.5 V.
+
vo
-
Chapter 5, Solution 72.
Since no current flows into the input terminals of ideal op amp, there is no voltage
drop across the 20 k resistor. As a voltage summer, the output of the first op
amp is
v 01 = 1.8 V
The second stage is an inverter
v2
250
v 01
100
2.5(1.8)
–4.5 V.
Chapter 5, Solution 73.
The first stage is a noninverting amplifier. The output is
50
vo1
(1.8) 1.8 10.8V
10
The second stage is another noninverting amplifier whose output is
vL v01 10.8V
Chapter 5, Solution 74.
Let v 1 = output of the first op amp
v 2 = input of the second op amp.
The two sub-circuits are inverting amplifiers
v1
v2
io
100
(0.9)
9V
10
32
(0.6)
12V
1.6
v1 v 2
9 12
20k
20k
150 A.
Chapter 5, Solution 75.
The schematic is shown below. Pseudo-components VIEWPOINT and IPROBE are involved as
shown to measure v o and i respectively. Once the circuit is saved, we click Analysis | Simulate.
The values of v and i are displayed on the pseudo-components as:
i = 200 A
(v o /v s ) = -4/2 = –2
The results are slightly different than those obtained in Example 5.11.
The schematic is shown below. IPROBE is inserted to measure i o . Upon simulation, the value
of i o is displayed on IPROBE as
io =
Chapter 5, Solution 77.
The schematic for the PSpice solution is shown below.
Note that the output voltage, –6.686 mV, agrees with the answer to problem, 5.48.
6.510mV
3.872mV
–6.686mV
3.872mV
0.0100V
4.838mV
Chapter 5, Solution 78.
The circuit is constructed as shown below. We insert a VIEWPOINT to display v o . Upon
simulating the circuit, we obtain,
v o = 667.75 mV
Chapter 5, Solution 79.
The schematic is shown below.
v o = –4.992 V
R3
R4
R5
20k
10k
40k
V5
1Vdc
V3
20Vdc
0
1.000V
1.000V
1
OS1
6
2.000V OUT
5
OS2
7
-
V+
0
V6
5Vdc
0V
3
2
R1
R2
20k
10k
5.000V
uA741
U1
+
OS2
OUT
-
4
OS1
V-
5
3
-20.00V
V4
20Vdc
1.666V
V2
20Vdc
+
U2
uA741
0V
2
0
0V
0
20.00V
R6
100k
6
1
20.00V
V1
20Vdc
-4.992V
1.666V
0
Checking using nodal analysis we get,
For the first op-amp we get v a1 = [5/(20+10)]10 = 1.6667 V = v b1 .
For the second op-amp, [(v b1 – 1)/20] + [(v b1 – v c2 )/10] = 0 or v c2 = 10[1.6667–1)/20] +
1.6667 = 2 V;
[(v a2 – v c2 )/40] + [(v a2 – v c1 )/100] = 0; and v b2 = 0 = v a2 . This leads to v c1 = –2.5v c2 .
Thus,
= –5 V.
Chapter 5, Solution 80.
The schematic is as shown below. After it is saved and simulated, we obtain
v o = 2.4 V.
Chapter 5, Solution 81.
The schematic is shown below. We insert one VIEWPOINT and one IPROBE to measure v o
and i o respectively. Upon saving and simulating the circuit, we obtain,
v o = 343.4 mV
i o = 24.51 A
Chapter 5, Solution 82.
The maximum voltage level corresponds to
11111 = 25 – 1 = 31
Hence, each bit is worth
(7.75/31) = 250 mV
Chapter 5, Solution 83.
The result depends on your design. Hence, let R G = 10 k ohms, R 1 = 10 k ohms, R 2 = 20 k
ohms, R 3 = 40 k ohms, R 4 = 80 k ohms, R 5 = 160 k ohms, R 6 = 320 k ohms, then,
-v o = (R f /R 1 )v 1 + --------- + (R f /R 6 )v 6
= v 1 + 0.5v 2 + 0.25v 3 + 0.125v 4 + 0.0625v 5 + 0.03125v 6
(a)
|v o | = 1.1875 = 1 + 0.125 + 0.0625 = 1 + (1/8) + (1/16) which implies,
[v 1 v 2 v 3 v 4 v 5 v 6 ] = [100110]
(b)
|v o | = 0 + (1/2) + (1/4) + 0 + (1/16) + (1/32) = (27/32) = 843.75 mV
(c)
This corresponds to [1 1 1 1 1 1].
|v o | = 1 + (1/2) + (1/4) + (1/8) + (1/16) + (1/32) = 63/32 = 1.96875 V
Chapter 5, Solution 84.
(a)
The easiest way to solve this problem is to use superposition and to solve for each
term letting all of the corresponding voltages be equal to zero. Also, starting with
each current contribution (i k ) equal to one amp and working backwards is easiest.
2R
v1
+
R
R
R
2R
ik
v2
+
2R
v3
+
2R
v4
2R
+
For the first case, let v 2 = v 3 = v 4 = 0, and i 1 = 1A.
Therefore,
v 1 = 2R volts or i 1 = v 1 /(2R).
Second case, let v 1 = v 3 = v 4 = 0, and i 2 = 1A.
2R
R
R
2R
i2
v2
Simplifying, we get,
+
R
2R
2R
2R
2R
R
1A
2R
v2
+
Therefore,
v 2 = 1xR + (3/2)(2R) = 4R volts or i 2 = v 2 /(4R) or i 2 =
0.25v 2 /R. Clearly this is equal to the desired 1/4th.
Now for the third case, let v 1 = v 2 = v 4 = 0, and i 3 = 1A.
2R
5R/3
1.5
v3
2R
+
The voltage across the 5R/3-ohm resistor is 5R/2 volts. The current through the
2R resistor at the top is equal to (5/4) A and the current through the 2R-ohm
resistor in series with the source is (3/2) + (5/4) = (11/4) A. Thus,
v 3 = (11/2)R + (5/2)R = (16/2)R = 8R volts or i 3 = v 3 /(8R) or 0.125v 3 /R. Again,
we have the desired result.
For the last case, v 1 = v 2 = v 3 and i 4 = 1A. Simplifying the circuit we get,
R
1A
R
2R
R
2R
v4
5R/3
1.5A
2R
2R
+
R
2R
2R
v4
+
2R
21R/11
11/4A
2R
v4
+
2R
Since the current through the equivalent 21R/11-ohm resistor is (11/4) amps, the
voltage across the 2R-ohm resistor on the right is (21/4)R volts. This means the
current going through the 2R-ohm resistor is (21/8) A. Finally, the current going
through the 2R resistor in series with the source is ((11/4)+(21/8)) = (43/8) A.
Now, v 4 = (21/4)R + (86/8)R = (128/8)R = 16R volts or i 4 = v 4 /(16R) or
0.0625v 4 /R. This is just what we wanted.
(b)
If R f = 12 k ohms and R = 10 k ohms,
-v o = (12/20)[v 1 + (v 2 /2) + (v 3 /4) + (v 4 /8)]
= 0.6[v 1 + 0.5v 2 + 0.25v 3 + 0.125v 4 ]
For
[v 1 v 2 v 3 v 4 ] = [1 0 11],
|v o | = 0.6[1 + 0.25 + 0.125] = 825 mV
For
[v 1 v 2 v 3 v 4 ] = [0 1 0 1],
|v o | = 0.6[0.5 + 0.125] = 375 mV
Design a voltage controlled ideal current source (within the operating limits of the op
amp) where the output current is equal to 200v s (t) µA.
The easiest way to solve this problem is to understand that the op amp creates an output
voltage so that the current through the feedback resistor remains equal to the input
current.
In the following circuit, the op amp wants to keep the voltage at a equal to zero. So, the
input current is v s /R = 200v s (t) µA = v s (t)/5k.
Thus, this circuit acts like an ideal voltage controlled current source no matter what
(within the operational parameters of the op amp) is connected between a and b. Note,
you can change the direction of the current between a and b by sending v s (t) through an
inverting op amp circuit.
v s (t)
+
Chapter 5, Solution 87.
The output, v a , of the first op amp is,
Also,
v a = (1 + (R 2 /R 1 ))v 1
(1)
v o = (-R 4 /R 3 )v a + (1 + (R 4 /R 3 ))v 2
(2)
Substituting (1) into (2),
v o = (-R 4 /R 3 ) (1 + (R 2 /R 1 ))v 1 + (1 + (R 4 /R 3 ))v 2
Or,
If
v o = (1 + (R 4 /R 3 ))v 2 – (R 4 /R 3 + (R 2 R 4 /R 1 R 3 ))v 1
R 4 = R 1 and R 3 = R 2 , then,
v o = (1 + (R 4 /R 3 ))(v 2 – v 1 )
which is a subtractor with a gain of (1 + (R 4 /R 3 )).
Chapter 5, Solution 88.
We need to find V Th at terminals a – b, from this,
v o = (R 2 /R 1 )(1 + 2(R 3 /R 4 ))V Th = (500/25)(1 + 2(10/2))V Th
= 220V Th
Now we use Fig. (b) to find V Th in terms of v i .
a
20 k
a
30 k
20 k
vi
vi
30 k
+
40 k
80 k
40 k
80 k
b
b
(b)
(a)
v a = (3/5)v i , v b = (2/3)v i
V Th = v b – v a (1/15)v i
(v o /v i ) = A v = -220/15 = -14.667
Chapter 5, Solution 89.
A summer with v o = –v 1 – (5/3)v 2 where v 2 = 6-V battery and an inverting amplifier
with v 1 = –12v s .
Chapter 5, Solution 90.
The op amp circuit in Fig. 5.107 is a current amplifier. Find the current gain i o /i s of the
amplifier.
Figure 5.107
For Prob. 5.90.
Solution
Transforming the current source to a voltage source produces the circuit below,
At node b,
v b = (2/(2 + 4))v o = v o /3
20 k
5k
5i s
+
a
b
+
4k
io
+
vo
2k
At node a,
(5i s – v a )/5 = (v a – v o )/20
But v a = v b = v o /3.
20i s – (4/3)v o = (1/3)v o – v o , or i s = v o /30
i o = [(2/(2 + 4))/2]v o = v o /6
i o /i s = (v o /6)/(v o /30) = 5
Chapter 5, Solution 91.
vo
+
R2
R1
is
i2
i1
But
io
io = i1 + i2
(1)
i1 = is
(2)
R 1 and R 2 have the same voltage, v o , across them.
R 1 i 1 = R 2 i 2 , which leads to i 2 = (R 1 /R 2 )i 1
Substituting (2) and (3) into (1) gives,
i o = i s (1 + R 1 /R 2 )
i o /i s = 1 + (R 1 /R 2 ) = 1 + 8/1 = 9
(3)
Chapter 5, Solution 92
The top op amp circuit is a non-inverter, while the lower one is an inverter. The output
at the top op amp is
v 1 = (1 + 60/30)v i = 3v i
while the output of the lower op amp is
v 2 = -(50/20)v i = -2.5v i
Hence,
v o = v 1 – v 2 = 3v i + 2.5v i = 5.5v i
v o /v i = 5.5
Chapter 5, Solution 93.
R3
R1 v
a
+
vb
+
R4
vi
+
R2
At node a,
io
iL
vL
+
vo
RL
(v i – v a )/R 1 = (v a – v o )/R 3
v i – v a = (R 1 /R 2 )(v a – v o )
v i + (R 1 /R 3 )v o = (1 + R 1 /R 3 )v a
(1)
But v a = v b = v L . Hence, (1) becomes
v i = (1 + R 1 /R 3 )v L – (R 1 /R 3 )v o
(2)
i o = v o /(R 4 + R 2 ||R L ), i L = (R 2 /(R 2 + R L ))i o = (R 2 /(R 2 + R L ))(v o /( R 4 + R 2 ||R L ))
Or,
v o = i L [(R 2 + R L )( R 4 + R 2 ||R L )/R 2
(3)
But,
vL = iLRL
(4)
Substituting (3) and (4) into (2),
v i = (1 + R 1 /R 3 ) i L R L – R 1 [(R 2 + R L )/(R 2 R 3 )]( R 4 + R 2 ||R L )i L
= [((R 3 + R 1 )/R 3 )R L – R 1 ((R 2 + R L )/(R 2 R 3 )(R 4 + (R 2 R L /(R 2 + R L ))]i L
= (1/A)i L
Thus,
1
A =
1
R1
RL
R3
R1
R2 RL
R2R3
R4
R 2R L
R2 RL
Please note that A has the units of mhos. An easy check is to let every resistor equal 1ohm and v i equal to one amp. Going through the circuit produces i L = 1A. Plugging into
the above equation produces the same answer so the answer does check.