A transistor is used in common-emitter mode in an amplifier circuit. When a signal of 20 mV is added to the base-emitter voltage, the base current changes by 20 $\mu$A and collector current changes by 2 mA. The load resistance is 5 k$\Omega$. The transconductance of the amplifier and voltage gain is respectively

Solution:

The transconductance ($g_m$) is defined as the ratio of the change in collector current ($\Delta I_c$) to the change in base-emitter voltage ($\Delta V_{be}$) for a constant collector-emitter voltage. In this case, the signal voltage added to the base-emitter voltage is $\Delta V_{be} = 20 \text{ mV} = 20 \times 10^{-3} \text{ V}$. The change in collector current is $\Delta I_c = 2 \text{ mA} = 2 \times 10^{-3} \text{ A}$.

$g_m = \frac{\Delta I_c}{\Delta V_{be}} = \frac{2 \times 10^{-3} \text{ A}}{20 \times 10^{-3} \text{ V}} = \frac{2}{20} \text{ A/V} = 0.1 \text{ A/V} = 0.1 \text{ mho}$.

The voltage gain ($A_v$) of a common-emitter amplifier is given by $A_v = -g_m R_L$, where $R_L$ is the load resistance. The magnitude of the voltage gain is $|A_v| = g_m R_L$. The load resistance is $R_L = 5 \text{ k}\Omega = 5 \times 10^3 \text{ }\Omega$.

$|A_v| = (0.1 \text{ mho}) \times (5 \times 10^3 \text{ }\Omega) = (0.1 \text{ A/V}) \times (5000 \text{ }\Omega) = 500$.

Thus, the transconductance is 0.1 mho and the voltage gain is 500.