Free PN Junction MCQs with Answers

1. A semiconductor doped with a pentavalent impurity such as phosphorus becomes

• A. n type, with electrons as the majority carriers
• B. p type, with holes as the majority carriers
• C. an insulator
• D. a conductor with no charge carriers

Explanation: A pentavalent atom has five valence electrons but only four are needed for bonding, so the fifth is loosely bound and becomes a free electron, making electrons the majority carriers. Doping with a trivalent impurity such as boron leaves a vacancy, a hole, and gives p type material. The crystal as a whole stays electrically neutral in both cases.

2. The majority charge carriers in a p type semiconductor are

• A. electrons
• B. holes
• C. protons
• D. neutrons

Explanation: Trivalent dopants create vacancies in the bonding structure, and these holes behave as mobile positive charges as neighbouring electrons shift to fill them. Electrons are still present as minority carriers, generated thermally. Protons and neutrons are locked in the nuclei and never move through the lattice.

3. The depletion region of a pn junction is a layer that

• A. contains a very high concentration of free carriers
• B. has been swept clear of mobile charge carriers, leaving fixed ions and a potential barrier
• C. conducts better than the rest of the crystal
• D. consists only of electrons

Explanation: Electrons diffusing across the junction recombine with holes, leaving exposed fixed ions on both sides and a region with almost no mobile carriers, so it acts as an insulator. The resulting internal field forms a potential barrier of about 0.7 V for silicon and 0.3 V for germanium. Forward biasing narrows this region and reverse biasing widens it.

4. A pn junction diode is forward biased when

• A. the p side is connected to the positive terminal and the n side to the negative
• B. the n side is connected to the positive terminal
• C. both sides are connected to the same terminal
• D. no voltage is applied

Explanation: Connecting positive to p repels holes and negative to n repels electrons, driving both towards the junction, so the depletion layer narrows and current flows once the applied voltage exceeds the barrier. Reverse bias does the opposite, widening the layer so that only a tiny leakage current passes. This one way behaviour is what makes the diode useful.

5. In reverse bias, the current through an ideal diode is

• A. large
• B. the same as in forward bias
• C. almost zero, apart from a small leakage current
• D. infinite

Explanation: The widened depletion region blocks the majority carriers, leaving only the thermally generated minority carriers to cross, which gives a leakage current of the order of microamperes. If the reverse voltage rises far enough the junction breaks down and conducts heavily, which is destructive for an ordinary diode but is the intended operating mode of a Zener diode. Ideal diode analysis simply treats the reverse current as zero.

6. The barrier potential across an unbiased silicon pn junction is about

• A. 0.3 V
• B. 0.7 V
• C. 5 V
• D. 12 V

Explanation: Silicon has a barrier of roughly 0.7 V, while germanium's is about 0.3 V, which is why a silicon diode needs about 0.7 V across it before it conducts appreciably. This forward voltage drop must be allowed for in circuit calculations. Silicon is preferred for most applications because it tolerates higher temperatures and has lower leakage.

7. A Zener diode is normally operated

• A. in forward bias as a rectifier
• B. in reverse bias at its breakdown voltage, as a voltage regulator
• C. with no bias at all
• D. only at very high frequency

Explanation: Beyond the breakdown voltage a Zener conducts heavily while the voltage across it stays almost constant, which makes it a simple and effective voltage reference. Unlike an ordinary diode this breakdown is non destructive provided a series resistor limits the current. In forward bias it behaves like any other silicon diode.

8. A light emitting diode produces light when

• A. it is reverse biased and breaks down
• B. it is forward biased and electrons recombine with holes, releasing energy as photons
• C. it is heated
• D. current passes through a filament inside it

Explanation: Each recombination across the junction releases an energy equal to the band gap of the semiconductor, and in materials such as gallium arsenide phosphide that energy falls in the visible range. The colour therefore depends on the band gap of the material rather than on any coating. Because no filament is heated, an LED wastes far less energy than an incandescent lamp.

9. A photodiode is used to detect light because incident photons

• A. heat the junction
• B. create extra electron hole pairs, increasing the reverse current
• C. reduce the leakage current to zero
• D. change the doping of the material

Explanation: Photons with enough energy free electrons across the band gap in the depletion region, so the reverse current rises in proportion to the light intensity and gives a measurable signal. The device is therefore operated in reverse bias, the opposite of an LED. Solar cells work on the same principle but are designed to deliver power rather than a signal.

10. Compared with a conductor, a pure semiconductor at room temperature has

• A. far fewer free charge carriers, and a conductivity that rises with temperature
• B. more free carriers
• C. the same number of carriers
• D. no carriers at any temperature

Explanation: A pure semiconductor has a small energy gap, so only a few electrons have enough thermal energy to cross it at room temperature, but heating promotes many more and conductivity climbs sharply. In a metal the carrier number is fixed and heating only increases scattering, so resistance rises. Doping raises the carrier concentration enormously without needing heat.