The manifestation of band structure in solids is due to

Bohr’s correspondence principle
Pauli’s exclusion principle
Heisenberg’s uncertainty principle
Boltzmann’s law

Solution

Pauli’s exclusion principle.

The transfer ratio β of transistor is 50. The input resistance of a transistor when used in C.E. (Common Emitter)configuration is 1kΩ. The peak value of the collector A.C current for an A.C input voltage of 0.01V peak is

100 μA
.01 mA
.25 mA
500 μA

Solution

In a npn transistor 10¹⁰ electrons enter the emitter in 10^–6s. 4% of the electrons are lost in the base. The current transfer ratio will be

0.98
0.97
0.96
0.94

Solution

In a p-type semiconductor the acceptor level is situated 60meV above the valence band. The maximum wavelength of light required to produce a hole will be

0.207 × 10^–5 m
2.07 × 10^–5 m
20.7 × 10^–5 m
2075 × 10^–5 m

Solution

What is the conductivity of a semiconductor if electron density = 5 × 10¹²/cm³ and hole density = 8 × 10¹³/cm³(µ^e = 2.3 m² V–1 s–1, µ_h = 0.01 m²V^–1s^–1)

5.634
1.968
3.421
8.964

Solution

The gate for which output is high if atleast one input is low?

NAND
NOR
AND
OR

The current gain of a transistor in common base mode is 0.995. The current gain of the same transistor in common emitter mode is

When the forward bias voltage of a diode is changed from 0.6 V to 0.7 V, the current changes from 5 mA to 15 mA.Then its forward bias resistance is

0.01 Ω
0.1 Ω
10 Ω
100 Ω

Solution

In a reverse biased diode when the applied voltage changes by 1 V, the current is found to change by 0.5 µA. The reverse bias resistance of the diode is

2 × 10⁵ Ω
2 × 10⁶ Ω
200 Ω
2 Ω

Solution

The intrinsic conductivity of germanium at 27° is 2.13 mho m^–1 and mobilities of electrons and holes are 0.38 and 0.18 m²V^–1s^–1 respectively. The density of charge carriers is

2.37 × 10¹⁹ m^–3
3.28 × 10¹⁹ m^–3
7.83 × 10¹⁹ m^–3
8.47 × 10¹⁹ m^–3

Solution

The cause of the potential barrier in a p-n diode is

depletion of positive charges near the junction
concentration of positive charges near the junction
depletion of negative charges near the junction
concentration of positive and negative charges near the junction

Solution

During the formation of a junction diode, holes from p-region diffuse into n-region and electrons from n-region diffuse into p-region. In both cases, when an electron smeets a hole, they cancel the effect at each other and as a result, a thin layer at the junction becomes free from any of charges carriers. This is called depletion layer.There is a potential gradient in the depletion layer,negative on the p-side,and positive on the n-side. The potential difference thus developed across the junction is called potential barrier.

The truth-table given below is for which gate?

XOR
OR
AND
NAND

Solution

Relation between A, B and C shows that C=\(\overline{AB}\) So NAND Gate

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