Special Purpose Diodes — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Special Purpose Diodes MCQs with step-by-step solutions (21 questions). Part of Semiconductor Devices. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Special Purpose Diodes · easy · theory
A Zener diode is designed to operate in the:
A. Forward bias region before the knee
B. Unbiased state
C. Forward breakdown region
D. Reverse breakdown region ✓ Correct
Solution: Its breakdown voltage is sharply defined and stable, which is what makes voltage regulation possible.
Q2 — Special Purpose Diodes · hard · theory
A Zener diode is heavily doped compared with an ordinary diode because heavy doping produces a:
A. Narrow depletion layer with an intense internal field ✓ Correct
B. Wide depletion layer with zero field
C. Wide depletion layer with a weak internal field
D. Complete absence of a depletion layer
Solution: The thin layer gives a field above $10^6\text{ V/m}$ at only a few volts, triggering breakdown.
Q3 — Special Purpose Diodes · easy · theory
The chief application of a Zener diode is as a:
A. Light source
B. Current amplifier
C. Voltage regulator ✓ Correct
D. Rectifier
Solution: It holds the voltage across a load constant despite changes in supply voltage or load current.
Q4 — Special Purpose Diodes · easy · theory
A light-emitting diode emits light when it is:
A. In the breakdown region
B. Unbiased
C. Reverse biased
D. Forward biased ✓ Correct
Solution: Electrons and holes injected across the junction recombine, releasing photons.
Q5 — Special Purpose Diodes · hard · theory
Light-emitting diodes are made from materials such as gallium arsenide rather than silicon because silicon has:
A. An indirect band gap, so recombination rarely produces photons ✓ Correct
B. Too large a band gap
C. Too high a melting point
D. No valence electrons
Solution: In an indirect-gap material the recombination energy is mostly given up as lattice vibrations, not light.
Q6 — Special Purpose Diodes · medium · theory
A photodiode is normally operated in:
A. Forward bias
B. The breakdown region
C. Reverse bias ✓ Correct
D. The unbiased state
Solution: The fractional change in the small reverse current on illumination is easy to detect.
Q7 — Special Purpose Diodes · medium · theory
A solar cell generates electrical energy from light by means of the:
A. Photoelectric emission into vacuum
B. Thermionic emission of electrons
C. Photovoltaic effect across a p-n junction ✓ Correct
D. Peltier effect at a semiconductor contact
Solution: Absorbed photons create electron-hole pairs which the junction field separates, producing a voltage.
Q8 — Special Purpose Diodes · medium · theory
Unlike a photodiode, a solar cell:
A. Must be reverse biased to work
B. Must be forward biased to work
C. Requires no external bias to produce a voltage ✓ Correct
D. Cannot produce any current
Solution: It is a source of energy rather than a detector, so it generates rather than modulates a current.
Q9 — Special Purpose Diodes · easy · theory
A light-emitting diode converts:
A. Light energy into electrical energy
B. Heat into light
C. Light into heat
D. Electrical energy into light energy ✓ Correct
Solution: The reverse conversion is performed by a photodiode or a solar cell.
Q10 — Special Purpose Diodes · hard · numerical
A $6\text{ V}$ Zener diode is connected through a $400\,\Omega$ series resistor to a $10\text{ V}$ supply with no load. The current through the Zener is:
A. $1\text{ mA}$
B. $15\text{ mA}$
C. $25\text{ mA}$
D. $10\text{ mA}$ ✓ Correct
Solution: The resistor drops $10 - 6 = 4\text{ V}$, so $I = \dfrac{4}{400} = 10\text{ mA}$.
Q11 — Special Purpose Diodes · hard · numerical
A $5\text{ V}$ Zener diode is fed through a $700\,\Omega$ resistor from a $12\text{ V}$ supply with no load. The Zener current is:
A. $7\text{ mA}$
B. $24\text{ mA}$
C. $17\text{ mA}$
D. $10\text{ mA}$ ✓ Correct
Solution: $I = \dfrac{12 - 5}{700} = \dfrac{7}{700} = 10\text{ mA}$.
Q12 — Special Purpose Diodes · hard · numerical
A $9\text{ V}$ Zener diode is fed through a $300\,\Omega$ resistor from a $15\text{ V}$ supply with no load. The Zener current is:
A. $5\text{ mA}$
B. $30\text{ mA}$
C. $20\text{ mA}$ ✓ Correct
D. $50\text{ mA}$
Solution: $I = \dfrac{15 - 9}{300} = \dfrac{6}{300} = 20\text{ mA}$.
Q13 — Special Purpose Diodes · medium · numerical
A Zener diode of breakdown voltage $5\text{ V}$ carries $20\text{ mA}$. The power it dissipates is:
A. $0.1\text{ W}$ ✓ Correct
B. $0.01\text{ W}$
C. $1\text{ W}$
D. $100\text{ W}$
Solution: $P = V_ZI_Z = 5 \times 0.02 = 0.1\text{ W}$.
Q14 — Special Purpose Diodes · hard · numerical
A red LED has a band gap of about $1.9\text{ eV}$. The wavelength of the light it emits is approximately:
A. $653\text{ nm}$ ✓ Correct
B. $496\text{ nm}$
C. $380\text{ nm}$
D. $1240\text{ nm}$
Solution: $\lambda = \dfrac{1240}{1.9} \approx 653\text{ nm}$, in the red part of the spectrum.
Q15 — Special Purpose Diodes · medium · numerical
An LED emits light of wavelength $550\text{ nm}$. Its band gap is approximately:
A. $0.55\text{ eV}$
B. $1.1\text{ eV}$
C. $4.5\text{ eV}$
D. $2.25\text{ eV}$ ✓ Correct
Solution: $E_g = \dfrac{1240}{550} \approx 2.25\text{ eV}$.
Q16 — Special Purpose Diodes · medium · numerical
An LED has a band gap of $2.0\text{ eV}$. The wavelength it emits is:
A. $620\text{ nm}$ ✓ Correct
B. $1240\text{ nm}$
C. $496\text{ nm}$
D. $310\text{ nm}$
Solution: $\lambda = \dfrac{1240}{2.0} = 620\text{ nm}$.
Q17 — Special Purpose Diodes · hard · numerical
Gallium arsenide has a band gap of $1.43\text{ eV}$. An LED made from it emits radiation of wavelength approximately:
A. $434\text{ nm}$, in the violet
B. $867\text{ nm}$, in the infrared ✓ Correct
C. $1240\text{ nm}$, in the visible
D. $620\text{ nm}$, in the red
Solution: $\lambda = \dfrac{1240}{1.43} \approx 867\text{ nm}$, just beyond the red end of the visible spectrum.
Q18 — Special Purpose Diodes · medium · numerical
The current through a reverse biased photodiode is:
A. Independent of the incident light
B. Proportional to the intensity of the incident light ✓ Correct
C. Zero under all conditions
D. Inversely proportional to the intensity
Solution: Each absorbed photon creates an electron-hole pair, so the photocurrent counts the photons arriving.
Q19 — Special Purpose Diodes · medium · numerical
A Zener diode in a regulator circuit holds the load voltage at:
A. Its breakdown voltage, regardless of moderate supply changes ✓ Correct
B. Half the supply voltage
C. The full supply voltage
D. Zero volt
Solution: Once in breakdown, large changes in current produce almost no change in the Zener voltage.
Q20 — Special Purpose Diodes · hard · numerical
A $6\text{ V}$ Zener regulator draws a total current of $25\text{ mA}$ from the series resistor while the load takes $15\text{ mA}$. The Zener current is:
A. $25\text{ mA}$
B. $15\text{ mA}$
C. $40\text{ mA}$
D. $10\text{ mA}$ ✓ Correct
Solution: The Zener carries whatever the load does not: $25 - 15 = 10\text{ mA}$.
Q21 — Special Purpose Diodes · medium · numerical
A solar cell is illuminated but left on open circuit. The quantity it develops is:
A. A steady alternating voltage
B. Neither voltage nor current
C. An open circuit voltage with no current flowing ✓ Correct
D. A short circuit current with no voltage
Solution: With no external path the separated charges simply build up a potential difference across the junction.