Semiconducting Devices Question 64
Question: A triode whose mutual conductance is 2.5 m A/volt and anode resistance is 20 kilo ohm, is used as an amplifier whose amplification is 10. The resistance connected in plate circuit will be
[MP PET 1989; RPMT 1998]
Options:
A) 1 kW
B) 5 kW
C) 10 kW
D) 20 kW
Show Answer
Answer:
Correct Answer: B
Solution:
$ \mu =r _{P}\times g _{m}=20\times 2.5=50 $
From $ A=\frac{\mu R _{L}}{r _{P}+R _{L}} $
Therefore $ r _{P}+R _{L}=\frac{\mu R _{L}}{A}=\frac{50R _{L}}{10}=5R _{L} $
Therefore $ 4R _{L}=r _{p}\Rightarrow R _{L}=\frac{r _{p}}{4}=\frac{20}{4}=5k\Omega $
