Semiconductor Physics

Practice Questions:

Question 11: easy

The resistivity of a pure semiconductor is 0.5 Ωm. If the electron and hole mobility be
0.39 m²/V-s and 0.19 m²/V-s respectively then calculate the intrinsic carrier concentration.

1. \[2.16\times 10^{19}/m^{3}\]

2. \[4.32\times 10^{19}/m^{3}\]

3. \[10^{20}/m^{3}\]

4. \[10^{30}/m^{3}\]

View Answer

The resistivity ( $ρ$ ) of a pure (intrinsic) semiconductor is given by the formula:

$\rho = \frac{1}{q (n_i) (\mu_n + \mu_p)}$

$ρ = q ( n ^{i} ) ( μ ^{n} + μ ^{p} ) 1$

Where:

$\rho = 0.5 \, \Omega\text{m}$

$ρ = 0.5 Ω m$ (resistivity),
$q = 1.6 \times 10^{-19} \, \text{C}$

$q = 1.6 \times 1 0^{-19} C$ (charge of an electron),
$n_i$

$n_{i}$ = intrinsic carrier concentration (to be calculated),
$\mu_n = 0.39 \, \text{m}^2/\text{V-s}$

$μ_{n} = 0.39 m^{2} / V-s$ (electron mobility),
$\mu_p = 0.19 \, \text{m}^2/\text{V-s}$

$μ_{p} = 0.19 m^{2} / V-s$ (hole mobility).

Rearranging the formula to solve for

$n_i$

$n_{i}$ :

$n_i = \frac{1}{q \rho (\mu_n + \mu_p)}$

Substitute the values:

$n_i = \frac{1}{(1.6 \times 10^{-19})(0.5)(0.39 + 0.19)}$

$n_i = \frac{1}{(1.6 \times 10^{-19})(0.5)(0.58)}$

$n_{i} = \frac{1}{4.64 \times 1 0^{- 20}} = 2.16 \times 1 0^{19} m^{- 3}$

Thus, the intrinsic carrier concentration

$n_i$

$n_{i}$ is

$2.16 \times 10^{19} \, \text{m}^{-3}$

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Question 12: moderate

In the following circuit of PN junction diodes D1, D2, D3 are ideal then i is :

1. E/R

2. E/2R

3. 2E/3R

4. Zero

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Diode D3 is forward biased so it will short circuit both diodes D1 and D2. So, equivalent resistance of circuit becomes R. Hence,

i= E/R

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Question 13: moderate

If in a p-n junction, a square input signal of 10 V is applied as shown, then the output across RL will be :

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Diode conducts in forward biased case and stops conduction in reverse biased mode.

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Question 14: moderate

The breakdown in a reverse biased p–n junction diode is more likely to occur due to

(a) large velocity of the minority charge carriers if the doping concentration is small
(b) large velocity of the minority charge carriers if the doping concentration is large
(c) strong electric field in a depletion region if the doping concentration is small
(d) strong electric field in the depletion region if the doping concentration is large.

1. a, d

2. c only

3. b only

4. b, c

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The breakdown in a reverse biased p–n junction diode is more likely to occur due to

a. Avalanche Breakdown - large velocity of the minority charge carriers if the doping concentration is small

b. Zener Breakdown - strong electric field in the depletion region if the doping concentration is large

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Question 15: easy

In the following figure, the diodes which are forward biased, are :

1. (a), (b) and (d)

2. (c) only

3. (a) and (c)

4. (b) and (d)

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For Forward biasing of PN Junction diode potential of P side should be greater than potential of N-side semiconductor

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Question 16: moderate

If A and B are input wave forms for given combination of logic gates then output waveform will be :

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Question 17: moderate

Find output Y :

1. P + Q

2. \[\overline{P}+P, Q\]

3. \[\overline{P}+\overline{Q}\]

4. None of these

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Question 18: moderate

If the ratio of the concentration of electrons to that of holes in a semiconductor is 7/5 and the ratio of currents is 7/4 then what is the ratio of their drift velocities :

1. 5/8

2. 4/5

3. 5/4

4. 4/7

View Answer

The ratio of electron to hole currents is given by:

$\frac{I_n}{I_p} = \frac{7}{4}$

The ratio of electron to hole concentrations is:

$\frac{n}{p} = \frac{7}{5}$

Using the current formula

$I = q n v A$

$I = q n v A$ , the ratio of drift velocities (

$\frac{v_n}{v_p}$

$v ^{p} v ^{n}$ ) can be found as:

$\frac{I_{n}}{I_{p}} = \frac{n v_{n}}{p v_{p}}$

Substitute the given ratios:

$\frac{7}{4} = \frac{\frac{7}{5} \cdot v_n}{v_p}$

Simplifying:

$\frac{v_{n}}{v_{p}} = \frac{5}{4}$

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Question 19: moderate

Which of the following circuits provides full-wave rectification of an ac input :

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Theoretical question based on bridge rectifier.

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Question 20: moderate

A sinusoidal voltage of peak value 200 volt is connected to a diode and resistor R in the
circuit shown so that half-wave rectification occurs. If the forward resistance of the diode
is negligible compared to R, the r.m.s. voltage across R is approximately :

1. 200

2. 100

3. 100/√2

4. 283

View Answer

RMS voltage for half wave rectifier is Vmax/2 = 100 Volt

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