In a silicon transistor, the base current is changed by `20 muA`. This results in a change of `0.02 V` in base to emitter voltage and a change of

Options

A. `0.5 k Omega`
B. `1 k Omega`
C. `1.5 k Omega`
D. `2 k Omega`

Answer

B

Explanation

Input resistance
`=0.02/(20xx10^(-6))=(2xx10^(4))/20 Omega=10^(3) Omega`
Again, current gain
`=(2xx10^(-3))/(20xx10^(-3))=(2xx10^(-3))/(2xx10^(-5))=100`
Now, trans-conductance
`=(beta_(aC))/(R_(in))=100/(1xx10^(3))=0.1 omega^(-1)`
Voltage gain
`=g_(m)g_(L)=0.1xx5xx103=500`
Change in output across load `=500xx0.22 V=10 V`