The collector resistance and the input resistance of a CE amplifier are respectively 10 kΩ and 2 kΩ. If β of the transistor is 49, the voltage gain of the amplifier is | Semiconductor Electronics: Materials, Devices and Simple Circuits

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 Multiple Choice QuestionsMultiple Choice Questions

41.

In a semiconductor, 2/3rd of the total current is carried by electrons and remaining 1/3rd by the holes. If at this temperature, the drift velocity of electrons is 3 times that of holes, the ratio of number density of electrons to that of holes is

  • 32

  • 23

  • 53

  • 33


42.

In an p-n-p transistor, 1010 holes enter the emitter in 10-6 s. If 29% of holes is lost in the base, then the current amplification factor is

  • 49

  • 19

  • 29

  • 39


43.

The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than 600 nm is incident on it. The energy band gap (in eV) for the semiconductor is

  • 1.50

  • 0.75

  • 2.06

  • 1.35


44.

The inputs A, B and C to be given in order to get an output Y = 1 from the following circuit are

         

  • 0, 1, 0

  • 1, 0, 0

  • 1, 0, 1

  • 1, 1, 0


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45.

The collector resistance and the input resistance of a CE amplifier are respectively 10 kΩ and 2 kΩ. If β of the transistor is 49, the voltage gain of the amplifier is

  • 125

  • 150

  • 245

  • 200


C.

245

The voltage gain of the amplifier (AV)

           = β . RoutRin = 49 × 102= 49 × 5 = 245 V


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46.

The light emitting diode (LED) is

  • a heavily doped p-n junction with no external bias

  • a heavily doped p-n junction with reverse bias

  • a heavily doped p-n junction with forward bias

  • a lightly doped p-n junction with no external bias


47.

Identify the mismatched pair from the following

  • Zener diode : Voltage regulator

  • Germanium doped with phosphorus : n-type semiconductor

  • semiconductor : band gap > 3 eV

  • p-n junction diode : rectifier


48.

In a common-emitter configuration, a transistor has β = 50 and input resistance 1 kΩ. If the peak value of AC input is 0.01 V, then the peak value of collector current is

  • 0.01 µA

  • 500 µA

  • 100 µA

  • 0.5 µA


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49.

The waveforms A and B given below are given as input to a NAND gate. Then, its logic output y is

  • for t1 to t2 ; y = 0

  • for t2 to t3 ; y = 1

  • for t3 to t4 ; y = 1

  • for t4 to t5 ; y = 0


50.

The logic gates giving output '1' for the inputs of '1' and 'O' are

  • AND and OR

  • OR and NOR

  • NAND and NOR

  • NAND and OR


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