Questions: 1. For the circuit of Figure (1), the voltage gain from base to colector is approximately: (a)l (b)12 (c)112 (d)224 2. The output signal of the common-emitter amplifier is out-of-phase with the input by: (a)0° (b)45° (c)90°: (d) 180°. 3. If the emitter bypass capacitor in Figure (1) is removed, the Amplifier voltage gain will:

EBK ELECTRICAL WIRING RESIDENTIAL
19th Edition
ISBN:9781337516549
Author:Simmons
Publisher:Simmons
Chapter25: Television, Telephone, And Low-voltage Signal Systems
Section25.1: Television Circuit
Problem 5R: From a cost standpoint, which system is more economical to install: a master amplifier distribution...
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Questions: -
1. For the circuit of Figure (1), the voltage gain from base to collector
is approximately:
(a)l (b)12 (c)112 (d)224
2 The output signal of the common-emitter amplifier is out-of-phase
with the input by:
(a)0° (b)45° (c)90°: (d) 180°.
3. If the emitter bypass capacitor in Figure (1) is removed, the
mplifier voltage gain will:
(a) Increase.
(b) Decrease.
(c) Remain essentially the same.
Transcribed Image Text:Questions: - 1. For the circuit of Figure (1), the voltage gain from base to collector is approximately: (a)l (b)12 (c)112 (d)224 2 The output signal of the common-emitter amplifier is out-of-phase with the input by: (a)0° (b)45° (c)90°: (d) 180°. 3. If the emitter bypass capacitor in Figure (1) is removed, the mplifier voltage gain will: (a) Increase. (b) Decrease. (c) Remain essentially the same.
main
R,
RL
to polnt 8
Uoc = 10 V
4.7 kA
10 ks2
C1
I pF
Y2
Yi
Rparl 1
• A
0.22 1F
1 kn
Uoul
1k n
Rparl 2
1=1 kHz
100 (2
Uin= 11 x Un
R2
R2
47 2
30 mV
10 n
Vé hps Typo 9118 2
Fig.1
Transcribed Image Text:main R, RL to polnt 8 Uoc = 10 V 4.7 kA 10 ks2 C1 I pF Y2 Yi Rparl 1 • A 0.22 1F 1 kn Uoul 1k n Rparl 2 1=1 kHz 100 (2 Uin= 11 x Un R2 R2 47 2 30 mV 10 n Vé hps Typo 9118 2 Fig.1
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Simmons
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