An NPN transistor is used in common emitter configuration as an amplifier with `1 k Omega` load resistance. Signal voltage of 10 mV is applied across the base-emitter. This produces a 3 mA change in the collector current and `15 mu A` change in the base current of the amplifier. The input resistance and voltage gain are:

Options

A. `0.33 kOmega, 1.5 `
B. `0.67 kOmega, 300 `
C. `0.67 kOmega, 200 `
D. `0.33 kOmega, 300 `

Answer

b

Explanation

Given `DeltaV_(i)= 10 xx 10^(-3)V`
`DeltaI_(c)= 3 xx 10^(-3)A`
`DeltaI_(b)= 15 xx 10^(-6)A`
`R_(i)= (DeltaV_(i))/(DeltaI_(b))= (10 xx 10^(-3))/(15 xx 10^(-6))= 0.67kOmega`
`:. "Voltage gain"= (DeltaI_(c))/(DeltaI_(b))xx (R_(0))/(R_(1))`
`= ((3 xx 10^(-3))/(15 xx 10^(-6))) xx (1000)/(670)= 200 xx (1000)/(670)~~ 300`