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.



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