Semiconductor Electronics: Materials, Devices And Simple Circuits
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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
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500
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1000
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1250
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50
C.
1250
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A forward biased diode is
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.
A p-n photodiode is fabricated from a semiconductor with a band gap of 2.5 eV. It ca detect a signal of wavelength
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6000 Ao
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4000 nm
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6000 nm
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4000 Ao
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
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
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10 x 1014 Hz
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5 x 1014 Hz
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1 x 1014 Hz
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20 x 1014 Hz
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
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,
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3.69,3.84
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4,4
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4,3.69
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4, 3.84
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,
Therefore, Current amplification = 
Voltage gain, AV is,
Power gain,
AP = 
= (0.96)2 x 
AP = 3.84
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