Mid-band voltage gain of a common-emitter amplifier
BlinkNBuild practice problem · GATE standard, authored and verified in-house
A common-emitter BJT amplifier is biased at a collector current $I_C = 1\ \text{mA}$ and has a collector resistor $R_C = 2\ \text{k}\Omega$. The emitter is fully bypassed, the thermal voltage is $V_T = 25\ \text{mV}$, and the Early effect and load are negligible.
The magnitude of the mid-band voltage gain $|A_v|$ is ________.
Show the step-by-step derivation
Answer
80 unitless (V/V) unitless (V/V)
Step-by-step derivation
- Find the transconductance at the bias point. For a BJT, $$g_m = \frac{I_C}{V_T} = \frac{1 \times 10^{-3}}{25 \times 10^{-3}} = 0.04\ \text{A/V} = 40\ \text{mA/V}.$$
- Draw the small-signal picture. With the emitter fully bypassed, the emitter is at AC ground, so the whole input signal appears across $v_{be}$ and the controlled source $g_m v_{be}$ drives $R_C$.
- The output voltage is the collector current flowing through $R_C$, with a sign inversion because increasing $v_{be}$ pulls the collector down: $$v_{out} = -g_m v_{be} R_C.$$
- Divide by $v_{in} = v_{be}$: $$A_v = \frac{v_{out}}{v_{in}} = -g_m R_C.$$
- Substitute: $$A_v = -(40 \times 10^{-3})(2 \times 10^{3}) = -80,$$ so $|A_v| = \mathbf{80}$.
- Useful shortcut. $g_m R_C = I_C R_C / V_T$, and $I_C R_C$ is just the DC drop across the collector resistor. So $|A_v| = V_{RC}/V_T = 2\ \text{V} / 25\ \text{mV} = 80$ - gain is set by how many thermal voltages you drop across $R_C$, which is why supply headroom directly limits achievable gain.
The trap this question is built around
Including $\beta$ in the gain. Current gain does not appear in $-g_m R_C$ at all: $\beta$ sets the input resistance $r_\pi = \beta/g_m$, and so affects loading from a real source, but the intrinsic voltage gain is independent of it.