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 *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*is the input resistance. The negative sign just indicates that in the common emitter configuration the output signal is phase shifted by 180º.

_{i}[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

(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*_{b}*R*_{b}) produced across the base resistance by the flow of the base current.

Therefore, the input signal *i*_{b}*R*_{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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