Semiconducting Devices Question 99
Question: The transfer ratio of a transistor is 50. The input resistance of the transistor when used in the common-emitter configuration is 1 KW. The peak value for an A.C input voltage of 0.01 V peak is
[CBSE PMT 1998]
Options:
A) 100 mA
B) 0.01 mA
C) 0.25 mA
D) 500 mA
Show Answer
Answer:
Correct Answer: D
Solution:
b = 50, Ri = 1000 W, Vi = 0.01V $ \beta =\frac{i _{c}}{i _{b}} $
and $ i _{b}=\frac{V _{i}}{R _{i}}=\frac{0.01}{10^{3}}={{10}^{-5}}A $
Hence $ i _{c}=50\times {{10}^{-5}}A=500\mu A $ .
