Semiconducting Devices Question 265

Question: In a triode, $ g _{m}=2\times {{10}^{-3}}oh{{m}^{-1}};\mu =42 $ , resistance load, $ R=50 $ kilo ohm. The voltage amplification obtained from this triode will be

[MNR 1999]

Options:

A) 30.42

B) 29.57

C) 28.18

D) 27.15

Show Answer

Answer:

Correct Answer: B

Solution:

Voltage gain $ A _{v}=\frac{\mu }{1+\frac{r _{p}}{R _{L}}} $ and $ \mu =r _{p}\times g _{m} $

Therefore $ r _{p}=\frac{42}{2\times {{10}^{-3}}}=21000\Omega $

Therefore $ A _{v}=\frac{42}{1+\frac{21000}{50\times 10^{3}}}=29.57 $



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