Semiconducting Devices Question 66
Question: For a triode m = 64 and gm =1600 m mho. It is used as an amplifier and an input signal of 1V (rms) is applied. The signal power in the load of 40 kW will be
Options:
A) 23.5 mW
B) 48.7 mW
C) 25.6 mW
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ r _{p}=\frac{\mu }{g _{m}}=\frac{64}{1600\times {{10}^{-6}}}=4\times 10^{4}\Omega $
Voltage gain $ A _{v}=\frac{\mu }{1+\frac{r _{p}}{R _{L}}}=\frac{64}{1+\frac{4\times 10^{4}}{40\times 10^{3}}}=32 $
Output signal voltage $ V _{0}=A _{v}\times V _{i}=32\times 1=32V(r.m.s.) $
Signal power in load $ =\frac{V _{0}^{2}}{R _{L}}=\frac{{{(32)}^{2}}}{40\times 10^{3}}=25.6mW $
