Multiple Choice & Assertion-Reason

PHYExam Practice

This is a chapter-wide rapid-revision article spanning semiconductors, diodes, transistors and logic gates. The MCQs and Assertion-Reason items test crisp factual recall: the intrinsic equality $n_e = n_h$ , band-gap magnitudes, depletion-region behaviour under bias, ripple frequencies, Zener operation, the $\alpha$ - $\beta$ link, CE phase reversal and gate truth tables.

These carry quick marks, so the goal is to internalise the standard ISC facts and the few formulas that recur across the whole chapter.

Mass-action law

n_e \, n_h = n_i^{2}

n_e

= electron concentration,

n_h

= hole concentration,

n_i

= intrinsic concentration. For intrinsic material

n_e = n_h = n_i

Current gain relation

\beta = \dfrac{\alpha}{1 - \alpha}, \qquad \alpha = \dfrac{\beta}{1 + \beta}

\alpha

= common-base gain (

I_C/I_E

\beta

= common-emitter gain (

I_C/I_B

). With

\alpha = 0.98

\beta = 49

Ripple frequency of rectifier

f_{out} = m \, f_{in}

f_{in}

= mains frequency,

m = 1

for half-wave and

m = 2

for full-wave. For

50\,\text{Hz}

mains:

50\,\text{Hz}

(half-wave),

100\,\text{Hz}

(full-wave).

Photon energy and band gap

E_g = \dfrac{hc}{\lambda}

E_g

= band gap,

h

= Planck constant,

c

= speed of light,

\lambda

= wavelength of absorbed/emitted photon. Conversion:

1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J}

In an intrinsic semiconductor every broken bond gives one electron and one hole, so $n_e = n_h = n_i$ exactly.
Approximate band gaps: conductor $\approx 0$ (overlapping bands), semiconductor $\approx 1\,\text{eV}$ (Si $1.1\,\text{eV}$ , Ge $0.7\,\text{eV}$ ), insulator $> 3\,\text{eV}$ (diamond $\approx 6\,\text{eV}$ ).
Forward bias narrows the depletion layer and lowers the barrier; reverse bias widens it and raises the barrier.
A Zener used as a regulator always works in reverse bias beyond the breakdown voltage, where $V$ stays nearly constant.
Pentavalent (donor) doping $\to$ n-type, electrons majority; trivalent (acceptor) doping $\to$ p-type, holes majority.
A CE amplifier inverts the signal: input and output a.c. voltages are $180^{\circ}$ out of phase.
NAND output is $0$ only when both inputs are $1$ ; NOR output is $1$ only when all inputs are $0$ (NOR = OR followed by NOT, so R correctly explains A).
Full-wave rectification doubles the ripple frequency relative to the input because both half-cycles produce output pulses.

Where the marks go

Confusing $\alpha$ and $\beta$ : $\alpha < 1$ (close to $1$ ) while $\beta$ is large (tens to hundreds); never write $\beta = \alpha/(1+\alpha)$ .
Saying half-wave doubles the ripple frequency — only full-wave does ( $100\,\text{Hz}$ from $50\,\text{Hz}$ ); half-wave keeps it at $50\,\text{Hz}$ .
Stating a Zener regulates in forward bias — it must be reverse-biased beyond breakdown.
Mixing up NAND and NOR truth-table conditions, or forgetting the $180^{\circ}$ phase reversal in CE.

How the board asks it

Assertion–Reasonzener reverse-bias operation in the breakdown region
Assertion (A): A Zener diode used as a voltage regulator is always connected in reverse bias. Reason (R): In the breakdown region the voltage across the Zener stays nearly constant even though the current changes. Choose: (a) both A and R are true and R is the correct explanation of A, (b) both A and R are true but R is not the correct explanation of A, (c) A is true but R is false, (d) A is false but R is true.
Define / stateintrinsic carrier equality $n_e = n_h = n_i$
Choose the correct option: In an intrinsic semiconductor at a given temperature the carrier concentrations satisfy (a) $n_e > n_h$ , (b) $n_e < n_h$ , (c) $n_e = n_h = n_i$ , (d) $n_e n_h = 0$ .
Numericalfull-wave rectifier doubles the ripple frequency
A full-wave rectifier is fed from $50\,\text{Hz}$ a.c. mains. The ripple frequency of its output is (a) $25\,\text{Hz}$ , (b) $50\,\text{Hz}$ , (c) $100\,\text{Hz}$ , (d) $200\,\text{Hz}$ . Choose the correct option.
Applicationnand gate truth table
Select the correct option: The output of a two-input NAND gate is $0$ only when (a) both inputs are $0$ , (b) both inputs are $1$ , (c) any one input is $1$ , (d) any one input is $0$ .
Distinguishthe $\alpha$ - $\beta$ current-gain relation
For a transistor in the common-emitter configuration the current gains $\alpha$ and $\beta$ satisfy (a) $\alpha < 1$ and $\beta$ is large, (b) $\alpha > 1$ and $\beta < 1$ , (c) $\beta = \alpha/(1-\alpha)$ with $\beta < \alpha$ , (d) $\alpha = \beta$ . Choose the correct statement.

PreviousZener Diode as Voltage Regulator

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.