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Semiconductor Electronics: Materials, Devices and Simple Circuits

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Question
CBSEENPH12039992
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A common emitter amplifier has a voltage gain of 50, an input impedance of 100 Ω and an output impedance of 200 Ω. The power gain the amplifier is 

  • 500

  • 1000

  • 1250

  • 50

Solution

C.

1250

Voltage space gain space equals space straight beta space straight x space impedance space gain rightwards double arrow space 50 space equals space straight beta space straight x space 200 over 100 space rightwards double arrow space straight beta space equals space 25 Also comma space power space gain space equals space straight beta squared space equals space impedance space gain equals space left parenthesis 25 right parenthesis squared space straight x space 200 over 100 space equals space 1250

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Question
CBSEENPH11020955
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A forward biased diode is

Solution

D.

For forward biasing of diode, p-side of diode should be at higher potential than n-side.
The p-n junctioni diode can be shown as:
If p-side of p-n junction diode is given more
positive potential than n-side, then it is forward biased.
In option (d) p-side is at 0V and n-side at -2V, so p is at higher potential. Hence, it is forward biased.
Note: p-n junction diode is mainly used as a rectifier.

Question
CBSEENPH12040018
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A p-n photodiode is fabricated from a semiconductor with a band gap of 2.5 eV. It ca detect a signal of wavelength

  • 6000 Ao

  • 4000 nm

  • 6000 nm

  • 4000 Ao

Solution

D.

4000 Ao

Only signals having wavelength less than threshold wavelength will be detected.
Energy E = hv = hc/λ

λ = hc / E
substituting the values of h, c and E in the above equation

straight lambda space equals space fraction numerator 6.6 space straight x space 10 to the power of negative 34 end exponent space straight x space 3 space straight x space 10 to the power of 8 space over denominator 2.5 space straight x space 1.6 space straight x space 10 to the power of negative 19 end exponent end fraction space equals space 5000 straight A to the power of straight o As space 4000 space straight A to the power of 0 space less than space 5000 space straight A thin space to the power of straight o signal space of space wavelength space 4000 space straight A to the power of straight o space can space be space detected space by space the space photodiode.

Question
CBSEENPH12040046
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A p-n photodiode is made of a material with a band gap of 2.0 eV. The minimum frequency of the radiation that can be absorbed by the material is nearly

  • 10 x 1014 Hz

  • 5 x 1014 Hz

  • 1 x 1014 Hz

  • 20 x 1014 Hz

Solution

B.

5 x 1014 Hz

the p-n photodiode is a semiconductor diode that produces a significant current when illuminated. It is reversed biased but is operated below the breakdown voltage.
Energy of radiation = band gap energy
ie, hv = 2.0 eV
or  v= 2.0 x 1.6 x 10-19 / 6.6 x 10-34 = 5 x 1014 Hz

Question
CBSEENPH12039793
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A p-n-p transistor is connected in a common emitter configuration in a given amplifier. A load resistance of 800 ohm is connected in the collector circuit and the voltage drop across it is 0.8 V. If the current amplification factor is 0.96 and the input resistance of the circuit is 192 ohm, the voltage gain and the power gain of the amplifier will respectively be,

  • 3.69,3.84

  • 4,4

  • 4,3.69

  • 4, 3.84

Solution

D.

4, 3.84

Given,
Resistance across load, RL = 800 ohm
Voltage drop across load, VL = 0.8 V
Input resistance of the circuit, Ri = 192 ohm
Collector current is given by,
straight I subscript straight C space equals space straight V subscript straight L over straight R subscript straight L equals fraction numerator 0.8 over denominator 800 end fraction equals 8 over 8000 equals 1 space mA
Therefore, Current amplification = fraction numerator Output space current over denominator Input space current end fraction
equals straight I subscript straight c over straight I subscript straight B equals 0.96 rightwards double arrow I subscript B space equals space fraction numerator 1 space m A over denominator 0.96 end fraction
Voltage gain, AV is,
straight V subscript straight L over straight V subscript in space equals space fraction numerator V subscript L over denominator I subscript B R subscript i end fraction space space space space space space space equals space fraction numerator 0.8 space x space 0.96 over denominator 10 to the power of negative 3 end exponent space x space 192 end fraction space space space space space space space equals space 4 rightwards double arrow space A subscript V space equals space 4
Power gain,
AP = fraction numerator straight I squared subscript straight C straight R subscript straight L over denominator straight I squared subscript straight B space straight R subscript straight i end fraction space equals space open parentheses I subscript C over I subscript B close parentheses squared. R subscript L over R subscript i space equals space left parenthesis 0.96 right parenthesis squared x 800 over 192
    = (0.96)2 x 800 over 192
AP = 3.84

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