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In an n-p-n transistor 10^10 electron en...

In an n-p-n transistor `10^10` electron enter the emitter in `10^(-6)` s. If 2% of the electrons are lost in the base, the current amplification factor is

A

0.98

B

0.97

C

0.96

D

0.94

Text Solution

Verified by Experts

The correct Answer is:
C
`rArr cot 60^(@) = e cot 30^(@) rArr e = (cos 60^(@))/( cot 30^(@))`
`rArr e = ((1)/(sqrt(3)))/(sqrt(3)) rArr e =(1)/(3)`
`n_(C) = (96)/(100) xx 10^(10) =0.96 xx 10^(10)`
Emitter current `I _(E) = (n_(E) xx e)/( t)`
Collector current `I_(C) = (n_(c) xx e)/(t)`
Current transfer ratio .
`alpha = (I_(C))/(I_(E)) =(n_(C))/(n_(E)) = (0.96 xx 10^(10))/(10^(10)) = 0.96`
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