A transistor is used in an amplifier circuit in common emitter mode. If the base current changes by 100 μA, it brings a change of 10 mA in collector current. If the load resistance is 2 kΩ and input resistance is 1 kΩ, the value of power gain is x × 10⁴. The value of x is _____________.

Solution

Physics sir · 3 steps

Power gain is current gain times voltage gain, so β appears twice: β² × R_L/Rᵢ.

1. Current gain. In common emitter mode the current gain is β = (Δ I_C)/(Δ I_B) = (10 mA)/(100 μA) = (10 × 10⁻³)/(100 × 10⁻⁶) = 100.
2. Voltage gain. The output voltage appears across R_L and the input voltage across Rᵢ. So A_V = β × R_L/Rᵢ = 100 × (2 kΩ)/(1 kΩ) = 200.

   A_P = β × A_V = β² × R_L/Rᵢ = (100)² × 2 = 2 × 10⁴

3. So. The power gain is 2 × 10⁴, so x = 2.

The common slip: 1. Stopping at β² = 10⁴ and forgetting the factor R_L/Rᵢ = 2 gives 1, not 2.

The idea to keep. For a CE amplifier, A_V = β R_L/Rᵢ and A_P = β × A_V = β² R_L/Rᵢ. Find β from the current changes first, then let the resistance ratio do the rest.

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