Junction Transistors
ICSE · Class 12 · Physics
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Quick Quiz: Junction Transistors
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A common-emitter amplifier has input resistance 1.0 kΩ and AC current gain beta(AC) = 40. What is the transconductance g_m?
A common-emitter amplifier has beta(AC) = 50, input resistance 2 kΩ, and load resistance 5 kΩ. What is the AC voltage gain?
A common-emitter amplifier has beta(AC) = 30 and voltage gain A_V = 600. What is the resistance gain R_out/R_in?
In a common-emitter circuit, V_CC = 12 V, I_C = 2 mA, and R_L = 3 kΩ. What is V_CE?
Sample Questions
A common-emitter transistor has I_B = 10 microamperes and I_C = 0.50 mA. What are beta(DC) and alpha(DC) respectively?
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50 and 0.98
First, beta(DC) = I_C / I_B = 0.50 mA / 10 microamperes = 0.50 / 0.010 = 50. Then I_E = I_B + I_C = 10 microamperes + 0.50 mA = 0.510 mA. So alpha = I_C / I_E = 0.50 / 0.510 ≈ 0.98. The correct option is 50 and 0.98.
A p-n-p transistor carries 2% of emitter charge in the base. If the emitter current is 1.6 mA, what is the collector current?
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1.568 mA
If 2% is lost in the base, then 98% reaches the collector. So I_C = 0.98 × 1.6 mA = 1.568 mA. The correct option is 1.568 mA.
A common-emitter amplifier has input resistance 665 Ω and transconductance 0.1 S. What is beta(AC)?
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66.5
Use g_m = beta(AC) / R_in, so beta(AC) = g_m × R_in. Substituting values gives beta = 0.1 × 665 = 66.5. The correct option is 66.5.
A common-emitter transistor has I_B = 105 microamperes and I_C = 2.05 mA. What is beta(DC)?
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19.5
Use beta(DC) = I_C / I_B. Convert 105 microamperes to 0.105 mA. Then beta = 2.05 / 0.105 ≈ 19.5. The correct option is 19.5.
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