Electronic Devices Chapter-Wise Test 4

Correct answer Carries: 4.

Wrong Answer Carries: -1.

The ionization energy of donor electrons in an n-type semiconductor is:

The fifth electron of a pentavalent dopant (e.g., As) in Si or Ge is weakly bound, requiring very small energy (~0.01 eV for Ge, ~0.05 eV for Si) to become free, much less than the intrinsic energy gap.

Very small
Equal to the energy gap
Very large
Zero
1

The drift current in a p-n junction is caused by:

Drift current results from the electric field in the depletion region, moving electrons from p-side to n-side and holes from n-side to p-side, opposing diffusion at equilibrium.

Electric field
Concentration gradient
Thermal energy
External voltage
1

The recombination process in a semiconductor:

Recombination occurs when a free electron collides with a hole, annihilating both and releasing energy, balancing the generation rate at equilibrium.

Generates new electron-hole pairs
Increases minority carriers
Reduces free carriers
Enhances conductivity
3

In an n-type semiconductor, the minority carriers are:

In an n-type semiconductor, electrons are majority carriers due to pentavalent doping, while holes, generated intrinsically or reduced by recombination, are minority carriers.

Electrons
Holes
Positive ions
Negative ions
2

In forward bias, the resistance of a p-n junction diode is:

In forward bias, the barrier height decreases, allowing significant current (mA) to flow, resulting in low resistance compared to reverse bias, where resistance is high due to minimal current (\( \mu \text{A} \)).

High
Low
Infinite
Zero
2

In an intrinsic semiconductor at \( T > 0 \, \text{K} \), the conduction is due to:

At \( T > 0 \, \text{K} \), thermal energy excites electrons from the valence band to the conduction band, creating electron-hole pairs that contribute to conduction.

Electrons only
Electrons and holes
Holes only
Ionic movement
2

A Ge crystal with \( 4 \times 10^{28} \, \text{atoms} \, \text{m}^{-3} \) is doped with 1.5 ppm of trivalent impurity. The number of holes per cubic meter is approximately:

1.5 ppm = \( 1.5 \times 10^{-6} \). Number of acceptor atoms = \( 1.5 \times 10^{-6} \times 4 \times 10^{28} = 6 \times 10^{22} \, \text{m}^{-3} \), each creating one hole, assuming full ionization.

\( 2 \times 10^{22} \, \text{m}^{-3} \)
\( 4 \times 10^{22} \, \text{m}^{-3} \)
\( 8 \times 10^{22} \, \text{m}^{-3} \)
\( 6 \times 10^{22} \, \text{m}^{-3} \)
4

The process of adding impurities to a pure semiconductor is called:

Doping is the deliberate addition of impurities (e.g., pentavalent or trivalent atoms) to a pure semiconductor to increase its conductivity by providing additional charge carriers.

Diffusion
Doping
Drift
Recombination
2

The output frequency of a full-wave rectifier with a 60 Hz input is:

A full-wave rectifier uses both half-cycles of the AC input, doubling the output frequency. For a 60 Hz input, the output frequency is \( 60 \times 2 = 120 \, \text{Hz} \).

60 Hz
120 Hz
30 Hz
180 Hz
2

The purpose of a centre-tap transformer in a full-wave rectifier is to:

The centre-tap transformer splits the secondary voltage into two equal halves, allowing two diodes to rectify alternate half-cycles, producing a full-wave output.

Increase voltage
Reduce current
Split the AC input
Filter the output
3

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