From the given transfer characteristic of a transistor in CE configuration, the value of power gain of this configuration is 10ˣ, for R_B=10 k Ω, and R_C=1 k Ω. The value of x is __________.

Solution

Physics sir · 3 steps

Read β off the slope of the line, then power gain is β² times the resistance ratio.

1. Current gain. The slope of the I_C against I_B line is the current gain. Going from 100 μA to 200 μA, I_C rises from 10 mA to 20 mA, so β = (10 mA)/(100 μA) = (10 × 10⁻³)/(100 × 10⁻⁶) = 100.
2. Voltage gain. In CE mode the voltage gain is A_V = β R_C/R_B = 100 × (1 kΩ)/(10 kΩ) = 10.

   A_P = β × A_V = β² R_C/R_B = 100² × 1/10 = 10³

3. So. The power gain is 10³, so x = 3.

The common slip: 4. Taking power gain as just β² = 10⁴ and forgetting the factor R_C/R_B = 1/10 gives 4, not 3.

The idea to keep. Power gain is current gain times voltage gain. In CE mode that is β² R_C/R_B, and β always comes from the slope of the transfer characteristic.

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