Q.A common-emitter amplifier has a load resistance $R_L = 4\ \text{k}\Omega$ and a current gain $\beta = 60$. If its input (base) resistance is $500\ \Omega$, calculate its voltage gain.

Common Emitter Voltage Gain – From Intuition to Formula

Imagine you have a transistor in the common emitter configuration. The input signal goes into the base, and the output comes from the collector. The transistor is a current amplifier — a small change in base current produces a much larger change in collector current, scaled by $\beta$ (the current gain). But we care about voltage gain, not just current gain.

How does a current change turn into a voltage change? Through resistors. The collector current flows through the collector resistor $R_C$ (or the load resistor $R_L$ connected there). By Ohm's law, a change in collector current $\Delta I_C$ produces a change in voltage across that resistor: $\Delta V_{out} = -\Delta I_C \cdot R_C$. The minus sign appears because an increase in collector current drops more voltage across $R_C$, pulling the collector voltage down — the CE amplifier inverts.

On the input side, the base-emitter junction looks like a diode. A small change in base-emitter voltage $\Delta V_{BE}$ produces a change in base current $\Delta I_B$, and the relationship is governed by the input resistance $r_{in}$ (often called $r_\pi$ or $h_{ie}$). So $\Delta V_{in} = \Delta I_B \cdot r_{in}$.

Now chain it together. The collector current change is $\Delta I_C = \beta \cdot \Delta I_B$. The output voltage change is $-\Delta I_C \cdot R_C = -\beta \cdot \Delta I_B \cdot R_C$. The input voltage change is $\Delta I_B \cdot r_{in}$. The voltage gain $A_V$ is the ratio of output to input voltage:

The negative sign tells you the output is inverted relative to the input. The magnitude of the gain is $\beta$ times the ratio of the output resistor to the input resistance.

Why the ratio matters …