Question

A transistor is used in common-emitter mode in an amplifier circuit. When a signal of $10\text{mV}$ is added to the base-emitter voltage, the base current changes by $10\mu A$ and the collector current changes by $1.5\text{mA}$. The load resistance is $5k\Omega$. The voltage gain of the transistor will be

Answer 1

$R_B=\frac{10\times {10}^{-3}}{10\times {10}^{-6}}$
$={10}^3\Omega$
Voltage gain of transistor can be written as:
$\therefore A_V=\left(\frac{\Delta I_C}{\Delta I_B}\right)\times \left(\frac{R_C}{R_B}\right)$
$=\frac{1.5\times {10}^{-3}}{10\times {10}^{-6}}\times \frac{5\times {10}^3}{1\times {10}^3}$
$=\frac{1.5\times 5}{10}\times \left(1000\right)$
$=750$