COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 27.
Known quantities:
For the circuit shown in Figure P8.27, let:
R = 2 MΩ AV (OL) = 200,000 R0 = 50Ω
RS = 1kΩ R1 = 1kΩ R2 = 100kΩ RLOAD = 10kΩ
Find:
The gain AV =
v0
vi
Analysis:
The op-amp has a very large input resistance, a very large "open loop gain" [ΑV(OL)], and a very small output
resistance. Therefore, it can be modeled with small error as an ideal op-amp.
The amplifier shown in Figure P8.10 is a noninverting amplifier, so we have
AV = 1 +
R2
100 ⋅ 10 3
= 101
= 1+
R1
1 ⋅ 10 3
Principles and Applications of Electrical Engineering, 5/e, Giorgio Rizzoni
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 30.
Known quantities:
For the circuit shown in Figure P8.30:
vS (t ) = 0.02 + 10−3 cos(ωt ) V , RF = 220 kΩ , R1 = 47kΩ , R2 = 1.8kΩ
.
Find:
a) expression for the output voltage.
b) The corresponding value of vo(t).
Analysis:
a) the circuit is a noninverting amplifier, then
 R 
v0 = 1 + F vS
R2 

 220 
b) v0 = 1 +
vS = 2.464 + 0.1232 cos(ωt ) V
 1.8 
Principles and Applications of Electrical Engineering, 5/e, Giorgio Rizzoni
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 84.
Known quantities:
Figure P8.84.
Find:
a)
If the op-amp shown in Figure P8.84 has an open-loop gain of 45 X 105, find the closed-loop gain for
RS = RF = 7.5 kΩ .
Rf
b) Repeat part a if RF = 5RS = 37.5 kΩ .
RS
Analysis:
v-
a)
vin = v +
v+
and
(
v0 = AOL vin − v −
Writing KCL at v-:
)
vin
 v

⇒ v − = − 0 − vin  .
 AOL

+
+
+ A (v -v- )
OL
v out
+
-
v − − 0 v − − v0
+
=0.
RS
RF
− v0
−v0
+ vin
+ vin
A
A
v
= 0
Substituting, OL
+ OL
RS
RF
RF

 1
1
1
1 
1 
 = −vin 

⇒ v0  −
−
−
+



 AOL RS AOL RF RF 
 RS RF 
 1
1 

− 
+
R
R
 1
v0
1 
S
F 


= AOL RS RF 
+
=

1
1
vin − 1 −
 K1 K 2 
−
AOL RS AOL RF RF
where:
K1 = RF RS + RS 2 + RS AOL RF
K 2 = RF RS + RF 2 + RS AOL RF
!
 


 


v0 RF 
1
1



=
+
vin RS  RS + RF + 1   RS + RF + 1 
 


  AOL RS

 AOL RS
For the conditions of part a we obtain:
1
1
ACL =
+
= 1.999
2 ×105 + 1 2 ×105 + 1
45
45
1
1
b) ACL = 5
+
= 5.999 .
6 ×105 + 1 6 × 105 + 1
45
45
Principles and Applications of Electrical Engineering, 5/e, Giorgio Rizzoni
© 2007 The McGraw-Hill Companies.
-
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 15.
Note: The +/- terminals of the op amp are inverted in Figure P8.15.
Known quantities:
The circuit of Figure P8.15. open-loop gain A.
1
Find:
O
Show that I RL is independent on RL.
Analysis:
Applying KVL at node 1, we have
Vs − V + V + − Vo
(1)
=
Rs
RL
Because V − = 0 ⇒ A(V + − V − ) = AV + = Vo
With (1) and (2), we have
V+ =
(2)
Vs
V − V + Vs
≈ 0 , So I RL = s
=
, it’s independent on RL, and is constant.
(1 − A) Rs RL + 1
Rs
RS
Principles and Applications of Electrical Engineering, 5/e, Giorgio Rizzoni
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 25.
Known quantities:
The values of the two 10 percent tolerance resistors used in an inverting amplifier:
R F = 33 kΩ ; RS = 1.2 kΩ .
Find:
a.
b.
The nominal gain of the amplifier.
The maximum value of Av .
c.
The minimum value of Av .
Analysis:
a.
b.
c.
− 33
= −27.5 .
Rs
1 .2
First we note that the gain of the amplifier is proportional to Rf and inversely proportional to RS. This
tells us that to find the maximum gain of the amplifier we consider the maximum Rf and the minimum
RS .
R f max
33 + 0.1(33)
=
Av max =
= 33.6 .
RS min 1.2 − 0.1(1.2)
R f min
33 − 0.1(33)
=
To find Av min we consider the opposite case: Av min =
= 22.5 .
RS max 1.2 + 0.1(1.2 )
The gain of the inverting amplifier is: Av = −
Rf
=
Principles and Applications of Electrical Engineering, 5/e, Giorgio Rizzoni
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 7.
Note: The +/- terminals of the op amp are inverted in Figure P8.7.
Known quantities:
The circuit of Figure P8.7.
Find:
Voltage Vo .
Analysis:
Looking at the figure, we can immediately see that
V2 = 0 V . Using nodal analysis:
1
1
1
1
+
+
)V1 −
11 = 0
(
2000
4000 2000 6000
1
1
−(
)V1 − (
)Vo = 0
6000
12000
Solving for V1 and Vo ,
V1 = 6 V and Vo = −12 V
Principles and Applications of Electrical Engineering, 5/e, Giorgio Rizzoni
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 8.
Note: The +/- terminals of the op amp are inverted in Figure P8.8.
Known quantities:
The circuit of Figure P8.8.
Find:
Show this circuit is a noninverting summing amplifier.
Analysis:
Applying KCL at the inverting terminal:
Rf −
)v
v3 = (1 +
R
Applying KCL at the noninverting terminal:
 1
R1
R2
1  +
1
1

v −
+
v2 − v1 = 0 or
v+ =
v2 +
v1
R1 + R2
R1 + R2
R2
R1
 R2 R1 
therefore, v3 = (1 +
Rf
R
)(
R1
R2
v2 +
v1 )
R1 + R2
R1 + R2
and the circuit does indeed compute the weighted sum of the inputs.
Principles and Applications of Electrical Engineering, 5/e, Giorgio Rizzoni
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 26.
Known quantities:
For the circuit shown in Figure P8.26, let v1 (t ) = 10 + 10 −3 sin (ωt )V , RF = 10 kΩ , Vbatt = 20 V .
Find:
a.
b.
The value of RS such that no DC voltage appears at the output.
The corresponding value of vout(t).
Analysis:
a.
The circuit may be modeled as shown:
Applying the principle of superposition:
For the 20-V source:
−10kΩ
vo 20 =
(20)
RS
 10kΩ 
+ 1(10)
For the 10-V source: vo 10 = 
 RS

The total DC output is:
Solving for RS
b.
vo
DC
= vo 20 + vo 10 = −
 10,000 
10,000
(20) + 
+ 1(10) = 0
RS

 RS
10,000
(10 - 20) = -10 ⇒ RS = 10 kΩ .
RS
Since we have already determined RS such that the DC component of the output will be zero, we can
simply treat the amplifier as if the AC source were the only source present. Therefore,

 Rf
+ 1 = 0.001sin (ωt ) ⋅ (1 + 1) = 2 ⋅ 10 − 3 sin (ωt ) V .
vo (t ) = 0.001sin (ωt ) ⋅ 

 RS
Principles and Applications of Electrical Engineering, 5/e, Giorgio Rizzoni
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 41.
Known quantities:
The circuit shown in Figure P8.41.
Find:
Show that the current Iout through the light-emitting
diode is proportional to the source voltage VS as long
as :
VS > 0.
Analysis:
Assume the op amp is ideal:
V − ≈ V + = VS 
 ⇒ if

I− ≈ 0
VS > 0
I out =
V − VS
=
.
R2 R2
Principles and Applications of Electrical Engineering, 5/e, Giorgio Rizzoni
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 23.
Known quantities:
The circuit of Figure P8.23.
Find:
The gain of the circuit.
Analysis:
V1
Using nodal analysis at the two nodes shown in the figure,
jω
1
1 1 jω
Vout = 0
)V1 − Vin −
( + +
3
6
2 3 6
1
 1 jω 
− V1 +  +
Vout = 0
2
6 
2
Therefore,
Vout
6
=− 2
Vin
ω − j 5ω − 6
Principles and Applications of Electrical Engineering, 5/e, Giorgio Rizzoni
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 62.
Known quantities:
For the circuit shown in Figure P8.62, let
C = C1 = C2 = 220 µF
R = R1 = R2 = RS = 2kΩ .
Find:
Determine the frequency response.
Analysis:
With reference to the Figure shown below, we have
I R 2 ( jω ) =
1
1
1
VO ( jω ) ⇒ VC 2 ( jω ) =
I R 2 ( jω ) =
VO ( jω ),
R2
jωC 2
jωC 2 R2
I R1 ( jω ) = −
1
1
VC 2 ( jω ) = −
VO ( jω ),
R1
jωC 2R2 R1

1  1

 R2 +
VO ( jω )
jωC 2  R2


1

 jωC1 VO ( jω ),
I C1 ( jω ) =
= 1 +
1
jωC 2 R2 

jωC1


C
1
1
VO ( jω ) =
I in ( jω ) = I C1 ( jω ) + I R 2 ( jω ) − I R1 ( jω ) =  jωC1+ 1 +
+
C 2 R2 R2 jωC 2 R2 R1 

2
1 
VO ( jω ),
=  + jωC +
R
jωCR 2 

Vin ( jω ) = −VC 2 ( jω ) − RS I in ( jω ) = −

2 
VO ( jω ) ⇒
= − 2 + jωCR +
jωCR 

1
V ( jω )
=−
Hυ ( jω ) = O
Vin ( jω )
2 + jωCR +

1
1 
VO ( jω ) =
VO ( jω ) −  2 + jωCR +
jωCR
jωCR 

2
jωCR
=−
jωCR
( jω ) (CR )2 + 2 jωCR + 2
2
Principles and Applications of Electrical Engineering, 5/e, Giorgio Rizzoni
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 73.
Known quantities:
Find:
Derive the differential equation corresponding to the analog computer simulation circuit of Figure P8.73.
Analysis:
x(t ) = −200∫ z dt or z = −
Therefore, y =
1 dx
. Also, z = −20 y .
200 dt
1 dx
. Also, y = ∫ (4 f (t ) + x(t )) dt or
4000 dt
dy
= 4 f (t ) + x(t ) .
dt
1 d 2x
d 2x
=
4
f
(
t
)
+
x
(
t
)
or
− 4000 x(t ) − 16000 f (t ) = 0 .
4000 dt 2
dt 2
______________________________________________________________________________
Therefore
Principles and Applications of Electrical Engineering, 5/e, Giorgio Rizzoni
© 2007 The McGraw-Hill Companies.