Multiple Choice Questions (including KEAM 2009) on Transistor Amplifiers

Questions from electronics are generally simple and interesting at the level expected from you. Here are a few questions:

(1) A transistor amplifier circuit is operated with an emitter current of 2 mA. The collector current is 1.98 mA. The common emitter current gain (β_dc) of the transistor used in the circuit is

(a) 45

(b) 50

(c) 100

(d) 125

(e) 200

Current gain (β_dc) in the common emitter configuration is given by

β_dc = I_C /I_B = 2 mA/(2 – 1.98) mA = 100

(2) A common emitter low frequency amplifier has an effective input resistance of 1 kΩ. The collector load resistance is 5 kΩ. On applying a signal, the base current changes by 10 μA and the collector current changes by 1 mA. What is the power gain of this amplifier?

(a) 100

(b) 500

(c) 1000

(d) 10000

(e) 50000

The power gain or power amplification (A_P) is the product of the current gain β_ac and the voltage gain A_v:

A_P = β_ac× A_v

Here β_ac = ∆I_C /∆I_B = (1×10^–3)/(10×10^–6) = 100 and

A_v = – β_ac×R_L/R_i where R_L is the load resistance and R_i is the input resistance. The negative sign just indicates that in the common emitter configuration the output signal is phase shifted by 180º.

[We use the load resistance R_L instead of the output resistance R_o (of the amplifier stage) in the above expression since the load resistance is small compared to the transistor output resistance. (The output resistance of the amplifier stage is the parallel combined value of R_L and the transistor output resistance)].

Ignoring the negative sign, A_v = 100×5/1 = 500

Therefore, power gain A_P = β_ac× A_v = 100×500 = 50000.

(3) For a common emitter amplifier, the audio signal voltage across the collector resistance 2 kΩ is 2V. If the current amplification factor of the transistor is 200 and the base resistance is 1.5 kΩ, the input signal voltage and base current are

(a) 0.1 V and 1 μA

(b) 0.15 V and 10 μA

(c) 0.015 V and 1 A

(d) 0.0015 V and 1 mA

(e) 0.0075 V and 5 μA

This MCQ appeared in KEAM (Engineering) 2009 question paper.

The collector signal current i_c is given by

i_c = v_c/R_c = 2 V/ 2 kΩ = 1 mA.

Since the current amplification (β_ac) of the transistor is 200, the base signal current i_b is given by

i_b = i_c /β_ac = 1 mA /200 = 0.005 mA = 5 μA

The input signal is the voltage drop (i_bR_b) produced across the base resistance by the flow of the base current.

Therefore, the input signal voltage = i_bR_b = 5 μA×1.5 kΩ = 7.5×10^–3 V = 0.0075 V.

(4) In an NPN transistor 108 electrons enter the emitter in 10^–8 s. If 1% electrons are lost in the base, the fraction of current that enters the collector and current amplification are respectively

(a) 0.8 and 49

(b) 0.9 and 90

(c) 0.7 and 50

(d) 0.99 and 99

(e) 0.88 and 88

This MCQ appeared in KEAM (Medical) 2009 question paper.

The number of electrons (108) entering the base and the time (10^–8 s) given in this question are not required to solve the problem. They can serve as distraction while attempting this simple question!

Since 1% electrons are lost in the base, 99% reach the collector so that the fraction of current that enters the collector is 99/100 = 0.99.

The current amplification is the ratio of the collector current to the base current and is equal to 99/1 = 99.

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