Electronic Devices Question 6
Question: A common emitter amplifier is designed with NPN transistor (a = 0.99). The input impedance is 1 KW and load is 10 KW. The voltage gain will be
[CPMT 1996]
Options:
A) 9.9
B) 99
C) 990
D) 9900
Show Answer
Answer:
Correct Answer: C
Solution:
Voltage gain = b x Resistance gain
$ \beta =\frac{\alpha }{1-\alpha }=\frac{0.99}{(1-0.99)}=99 $
Resistance gain $ =\frac{10\times 10^{3}}{10^{3}}=10 $
Therefore Voltage gain = 99 x 10 = 990.
BETA
