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(elec101)[2008](f)quiz~ylmaaa^_10261.pdf
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(13)
12. 2. 4VO
VO
4.
4.
4Vo
4Vo . Vo Vo
= 5A + (6)
2. 4. 6Vo = 20V + Vo Vo = 4V (3)
V1 = -56V (4)
4V
2.
12. 16V
-56V
4A
5A
4.
4.
1A
4Vo
0V
12. In the circuit, find V1. The dependent source (voltage
controlled voltage source) is equal to 4Vo and is in volt. . (13)
12. Vo 2.
4.
(20)
2.
2. 5v
6V . v . 3V 2.
5v . 6V + v
6V . v
2.
2.
2.
5v
5v . 6V + vv 6V . v . 3V
+=
2. 1. 2. 6v . 6Vv 3V . v
+=
2. 1. 2.
6v . 6V + 2v = 3V . v
9v = 9V
v = 1V
1A
0
5V
2.
2.
5V
(20)
9. Find v . The dependent source (current controlled voltage source) is in volt and equal to 5v . (20)
2.
1.5A
5v
(13)
12. 2. 4VO
VO
4.
4.
4Vo
4Vo . Vo Vo
= 5A + (6)
2. 4. 6Vo = 20V + Vo Vo = 4V (3)
V1 = -56V (4)
4V
2.
12. 16V
-56V
4A
5A
4.
4.
1A
4Vo
0V
12. In the circuit, find V1. The dependent source (voltage
controlled voltage source) is equal to 4Vo and is in volt. . (13)
12. Vo 2.
4.
(26)
(a) Find the voltage gain A (=Vo/Vs) and draw the Vo(t) waveform. Assume an ideal op amp. (9)
(a) 8k.
4k.
VS(t) 0V VO(t)
. 2V . 4V
8k.
A =. =.2
(a)
4k. 8V
VO(t)
4V
0V
(b) Show that V1 = Vo/5. Hence find the voltage gain A (=Vo/Vi). Assume an ideal op amp. (17)
R2 = 2k. R3 = 4k.
Vi
.
(b)
VO
V1
R4 = 4k.
R1 = 1k.
QV = V = V
.+ 1
VV . V
1 O 1
=
(b) 1k 4k VO
V1 =
5
Vi . V1 V1 . VO
=
2k 4k 2Vi . 2V1 = V1 . VO
VO
2Vi = 3 . VO
5 2
2V =. V
i 5 O
VO
AV = =.5Vi
(30)
9a
9a. Find Vo. Assume ideal op amp. (18)
Vo
V1 = 1.01V
20k.
2k. 20k. V2 = 1V
2k.
1.01V V1
Vo
R1
I
20k.
V1 . V2
1.01V
I =
R2
1k.
1.01V . 1V
1V
== 10A
20k.
1k. R1
I
1V V2
VO = V1 + I* 20k.= 1.01V +10A* 20k. = 1.01V + 0.2V = 1.21V
(18)
Draw the circuit of an op amp voltage follower and
9b
name two advantages of the circuit. (12)
(6)
voltage follower
1. Very high input resistance
(6)
2. Very low output resistance
(9)
9 (a) Draw the circuit model for a conductance amplifier. What are the ideal value of the output and input resistance?
(9)
(a)
(9)
ideal V to I amplifier, Ri = . Ro =
(21)
12k.
VO
VS =1V
2k. 4k.
In the amplifier, find Vo and I. Assume the op amps are ideal. (21)
VS . 00 . VO 0 . V
= +
1k. 12k. 2k. 2k. V
QV = V * = O (8)
O 4k.+ 2k. 3 12VS =.VO . 6V =.3VO
V
O =.4
VS
VO =.4V (5)
2
I =. mA (8)
3
1
m
312k.
VS =1V
VO
2m 3
04
2k. 4k.
.
3
p.12
In the following ideal op amp circuit, find Vo when Vi = 2V.
R1 = 2kW R3 =2kO
Vo
2kO
+ ideal op amp Vi = 0 (virtual short) Vi -
QV-= V+= v
2k VO
\V = V == v = V
+ O -
2k + 2k 2
Ii
Ii = 0 (virtual open)
Vi