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 $
