A transistor is used in common-emitter mode in an amplifier circuit. When a signal of 10 mV is added to the base-emitter voltage, the base current changes by 10 μA and the collector current changes by 1.5 mA. The load resistance is 5 kΩ. The voltage gain of the transistor will be _________.

Solution

Physics sir · 4 steps

Voltage gain is output voltage over input voltage: get the output from the collector current through the load.

1. Output voltage. The change in collector current flows through the load, so the output voltage changes by Δ Vₒ = Δ I_C × R_L = 1.5 × 10⁻³ A × 5000 Ω = 7.5 V.
2. Input voltage. The input signal is the change in base-emitter voltage, Δ Vᵢ = 10 mV = 0.01 V.

   A_V = (Δ Vₒ)/(Δ Vᵢ) = (Δ I_C R_L)/(Δ V_(BE)) = (7.5 V)/(0.01 V) = 750

3. Cross-check. The current gain is β = (1.5 mA)/(10 μA) = 150 and the input resistance is rᵢ = (10 mV)/(10 μA) = 1 kΩ. Then A_V = β R_L/rᵢ = 150 × 5 = 750, the same result.
4. So. The voltage gain of the amplifier is 750.

The common slip: 150. Stopping at the current gain β = Δ I_C / Δ I_B gives 150. That is the current gain, not the voltage gain; you still need to multiply by R_L / rᵢ = 5.

The idea to keep. In a CE amplifier, voltage gain is current gain times the resistance ratio: A_V = β R_L/rᵢ. If you are given Δ V_(BE) directly, just divide Δ I_C R_L by it.

Answer from the NTA final answer key. Solution by PrepWiser.