Communication System Question 18
Question: The transfer ratio of a transistor is 50. The input resistant of the transistor when used in the common-emittej configuration is 1 k $ \Omega $ . The peak value for an AC input voltage of 0.01 V peak is
Options:
A) $ 100\mu A $
B) $ 0.01mA $
C) $ 0.25mA $
D) $ 500\mu A $
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Answer:
Correct Answer: D
Solution:
[d] $ \beta =50, $
$ R _{i}=1000\Omega , $
$ V _{i}=0.01V $
$ \beta =\frac{i _{c}}{i _{b}} $ and
$ ib=\frac{V _{i}}{R _{i}}=\frac{0.01}{10^{3}}={{10}^{-5}}A $
Hence $ i _{c}=50\times {{10}^{-5}}A=500\mu A. $
BETA
