The typical transfer characteristics of a transistor in CE configuration is shown in figure. A load resistor of 2 k Ω is connected in the collector branch of the circuit used. The input resistance of the transistor is 0.50 k Ω. The voltage gain of the transistor is ______________.

Solution

Physics sir · 4 steps

Read the current gain off the slope of the graph, then scale it by the resistance ratio.

1. Current gain. The transfer characteristic is a straight line through the origin, so its slope is the current gain β = Δ I_C / Δ I_B. From the graph, I_C rises by 5 mA for every 100 μA rise in I_B.
2. Match the units. Convert both currents to the same unit: β = (5 × 10⁻³ A)/(100 × 10⁻⁶ A) = 50.

   A_V = β × R_L/Rᵢ

3. Put in numbers. With R_L = 2 kΩ and Rᵢ = 0.5 kΩ, the ratio is 2/0.5 = 4. So A_V = 50 × 4.
4. So. The voltage gain of the transistor is 200.

The common slip: 0.2. Dividing 5 by 100 without converting mA and μA to the same unit gives β = 0.05 and a gain of 0.2. The correct value is β = 50.

The idea to keep. In CE mode, voltage gain = current gain × (load resistance / input resistance). Get β from the slope of the I_C against I_B graph, and always put both currents in the same unit first.

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