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The input resistance of a silicon transistor is 100 $\Omega$. Base current is changed by 40 $\mu$A, which results in a change in collector current by 2 mA. This transistor is used as a common emitter amplifier with a load resistance of 4 k$\Omega$. The voltage gain of the amplifier is
A
1000
B
2000
C
3000
D
4000
Detailed Solution
$R_i = 100\ \Omega$, $\Delta i_B = 40\times10^{-6}$ A, $\Delta i_C = 2\times10^{-3}$ A, $R_o = 4\times10^3\ \Omega$
Current gain $\beta = \frac{\Delta i_C}{\Delta i_B} = \frac{2\times10^{-3}}{40\times10^{-6}} = 50$
$A_V = \beta\frac{R_o}{R_i} = \frac{R_o\times\Delta i_C}{R_i\times\Delta i_B} = \frac{4\times10^3\times2\times10^{-3}}{100\times40\times10^{-6}}$
$A_V = 2000$
Current gain $\beta = \frac{\Delta i_C}{\Delta i_B} = \frac{2\times10^{-3}}{40\times10^{-6}} = 50$
$A_V = \beta\frac{R_o}{R_i} = \frac{R_o\times\Delta i_C}{R_i\times\Delta i_B} = \frac{4\times10^3\times2\times10^{-3}}{100\times40\times10^{-6}}$
$A_V = 2000$