Question
Question: In an n – p – n transistor \(10^{10}\) electron enter the emitter in \(10^{- 6}\) s. if 2% of the el...
In an n – p – n transistor 1010 electron enter the emitter in 10−6 s. if 2% of the electrons are lost in the base, the ratio of current transfer ratio and the current amplification factor is.
A
0.02
B
7
C
33
D
4.9
Answer
0.02
Explanation
Solution
: As, IE=tnE×e
And IC=tnC×e=t(98/100)nE×e=10098×IE
Current transfer ratio,
α=IlEIC=10098=0.98
Current amplification factor,
β=1−αα=1−0.980.98=49∴βα=490.98=0.02