NCERT Exemplar Solutions Class 12 Physics Semiconductor Electronics — Multiple Choice Questions Q1–16

NCERT Exemplar Problems, Class 12 Physics — Semiconductor Electronics: Materials, Devices and Simple Circuits, MCQ I and MCQ II. All 16 questions solved.

The book numbers these questions 14.1 to 14.16; here they are Questions 1 to 16. Questions 1 to 8 are MCQ I (exactly one option is correct) and Questions 9 to 16 are MCQ II (one or more options may be correct). Five ideas carry the page:

• Carriers and temperature. Conductivity $\sigma = ne^2\tau/m$; in a semiconductor $n$ rises exponentially with temperature while $\tau$ falls only slightly.
• Bias and the barrier. Forward bias lowers the potential barrier of a p-n junction, reverse bias raises it.
• Diode circuits. Decide first which diodes conduct, then replace each by a wire or an open circuit.
• Transistors. Emitter current splits into a large collector part and a small base part; the transfer curve has cut-off, active and saturation regions.
• Breakdown and filtering. Avalanche and Zener breakdown depend on doping; ripple falls as $f$, $C$ and $R_L$ rise.

Key insight. Nearly every question is answered by one decision: which way is each junction biased? Once that is settled, the circuit reduces to wires and gaps. Question 13 is the one place where the printed key and the physics part company.

Choose the correct answer from the given options. (MCQ I: one correct. MCQ II: one or more correct.)

Question 1

The conductivity of a semiconductor increases with increase in temperature because

Solution. Conductivity is $\sigma = ne^2\tau/m$. Heating a semiconductor breaks many more covalent bonds, so the number density $n$ of electron-hole pairs rises very steeply, roughly exponentially. At the same time the lattice vibrates more strongly, electrons collide more often and the relaxation time $\tau$ falls. But $\tau$ falls only slowly (a power law in $T$), whereas $n$ grows exponentially, so the product $n\tau$ increases. (a) leaves out the fall in $\tau$ and (b), (c) wrongly say $\tau$ increases; only (d) states the whole picture.

Question 2

In Fig. 14.1, $V_0$ is the potential barrier across a p-n junction, when no battery is connected across the junction

The figure shows three curves rising from the same base level and levelling off at different heights: curve 1 is the highest, above the marked barrier $V_0$; curve 2 levels off at $V_0$; curve 3 is the lowest.

Solution. Curve 2 is the unbiased barrier $V_0$. A reverse bias adds to the barrier, because the applied field points the same way as the junction field, so the barrier becomes higher than $V_0$: that is curve 1. A forward bias opposes the junction field and lowers the barrier below $V_0$: that is curve 3. The two curves therefore lie on opposite sides of curve 2, which rules out (a) and (d), and the order in (b) is the correct one.

Question 3

In Fig. 14.2, assuming the diodes to be ideal,

The figure shows a series string: terminal A at $-10$ V, a resistor $R$, diode $D_1$ (drawn conducting to the right, towards the corner), then diode $D_2$ pointing upward from the earthed terminal B.

Solution. A diode conducts only when its anode (the arrow end) is at the higher potential. Point A is at $-10$ V and B is at $0$ V, so B is the higher end of the string.

• $D_1$: its anode is on the side of A, at about $-10$ V, and its cathode is on the side of B, which is higher. The anode is lower than the cathode, so $D_1$ is reverse biased.
• $D_2$: its anode is at B, at $0$ V, and its cathode is towards A, which is lower. So $D_2$ is forward biased.

The diodes are in series. One of them blocks, so no current flows in either direction. The statement that $D_2$ conducts and $D_1$ blocks, with no current, is (b).

Question 4

A $220$ V A.C. supply is connected between points A and B (Fig. 14.3). What will be the potential difference $V$ across the capacitor?

The figure shows the supply between A and B, a diode in the top wire, and the capacitor $C$ connected across the output, with $V$ marked across it.

Solution. The diode with the capacitor is a half-wave rectifier with a smoothing capacitor and no load. On the half-cycle when A is positive the diode conducts and the capacitor charges up to the peak voltage of the supply. When the supply voltage falls the diode becomes reverse biased and the capacitor has nowhere to discharge, so it holds that peak. The quoted $220$ V is the r.m.s. value, so the peak is

$$V_0 = 220\sqrt{2}\ \text{V} \approx 311\ \text{V}.$$

Question 5

Hole is

Solution. When a bond breaks, the electron that leaves it becomes free and a vacancy is left behind in the bond. Neighbouring bond electrons can jump into it, so the vacancy moves and acts like a positive charge. A hole is not an anti-particle (the positron is) and not a real particle at all, which excludes (a) and (d). (c) is too loose: the mere absence of free electrons describes an insulator, which has no holes in this sense.

Question 6

The output of the given circuit in Fig. 14.4.

The figure shows an a.c. source $v_m\sin\omega t$ in series with a resistor in the top wire; across the output terminals, from the top rail to the bottom rail, is a diode with its anode (arrow) at the top.

Solution. This is a diode in parallel with the output (a shunt), not in series with it. During the positive half-cycle the top rail is positive, the diode is forward biased and, being ideal, acts as a short circuit across the output terminals, so the output is zero. During the negative half-cycle the diode is reverse biased and open, so the full source voltage appears across the output (no current flows, hence no drop across the series resistor). The output therefore consists of the negative half-cycles only, while the positive ones are removed.

Question 7

In the circuit shown in Fig. 14.5, if the diode forward voltage drop is $0.3$ V, the voltage difference between A and B is

The figure shows a current of $0.2$ mA entering at A, passing through a $5\ \text{k}\Omega$ resistor, then to a junction from which a capacitor branch leaves sideways; the main path continues through a diode (forward drop $0.3$ V) and a second $5\ \text{k}\Omega$ resistor to B.

Solution. The capacitor branch carries no steady (d.c.) current, so the whole $0.2$ mA flows down the main path through both resistors and the diode:

$$V_{AB} = (0.2\ \text{mA})(5\ \text{k}\Omega) + 0.3\ \text{V} + (0.2\ \text{mA})(5\ \text{k}\Omega) = 1 + 0.3 + 1 = 2.3\ \text{V}.$$

Question 8

Truth table for the given circuit (Fig. 14.6) is

The figure shows inputs A and B. A passes to the upper gate directly and also, through a NOT gate, to the lower gate; B goes to the lower gate and, by a wire that crosses up the left side, to the upper gate. The upper gate gives output C and the lower gate gives output D; C and D are the two inputs of the last gate, whose output is E.

The four tables are given as $(A,B)\to E$ in the order $00, 01, 10, 11$:

Solution. The upper gate receives A and B, so $C = A\cdot B$. The lower gate receives $\bar A$ (through the NOT gate) and B, so $D = \bar A\cdot B$. The last gate combines them:

$$E = C + D = AB + \bar A B = B(A + \bar A) = B.$$

So $E$ simply copies $B$: $0, 1, 0, 1$ for $AB = 00, 01, 10, 11$.

Note on the figure. The last gate is drawn with an AND outline. If it really were an AND gate, $E = AB\cdot\bar AB = 0$ for every input, which matches none of the four tables. The intended gate is OR, which is how the book’s own key draws the same gate in a later question, and it gives (c). This is a slip in the drawing, not an error in the key.

Question 9

When an electric field is applied across a semiconductor

Solution. The field does work on the carriers. A conduction electron gains kinetic energy from the field, so it moves to a higher energy level within the conduction band: (a) is true and (b) is false. A hole is the absence of an electron. As the field pushes the valence electrons one way, the vacancy shifts the other way, which on the electron-energy diagram means that the vacancy moves to a lower energy level. So (c) is true and (d) is false.

Question 10

Consider an npn transistor with its base-emitter junction forward biased and collector base junction reverse biased. Which of the following statements are true?

Solution. The forward-biased emitter-base junction injects electrons from the heavily doped emitter into the thin base: (c) is true. The base is very thin and lightly doped, so very few of these electrons recombine; the field of the reverse-biased collector-base junction sweeps almost all of them across into the collector. So electrons do cross over from emitter to collector, and (a) is true. Holes are the majority carriers in the p-type base, and the reverse-biased collector junction stops them crossing, so (b) is false. (d) is false because the majority of the injected electrons do reach the collector, not leave through the base.

Question 11

Figure 14.7 shows the transfer characteristics of a base biased CE transistor. Which of the following statements are true?

The figure plots output voltage $V_0$ against input voltage $V_i$. The output stays at a high constant value from $0$ to $0.6$ V, falls steadily between $0.6$ V and $2$ V, and is close to zero beyond $2$ V.

Solution. The curve has three regions.

At $0.4$ V the transistor is in cut-off, not in the active state, so (a) is false. At $1$ V it is active, where a small change in input gives a large change in output, so (b) is true. At $0.5$ V it is in cut-off, a switch that is off, so (c) is true. At $2.5$ V it is saturated, a switch that is on, so (d) is true.

Question 12

In a npn transistor circuit, the collector current is $10$ mA. If $95$ per cent of the electrons emitted reach the collector, which of the following statements are true?

Solution. The collector receives $95\%$ of the emitter current, so $I_C = 0.95\,I_E$:

$$I_E = \frac{10}{0.95} = 10.53\ \text{mA}, \qquad I_B = I_E – I_C = 10.53 – 10 = 0.53\ \text{mA}.$$

(a) wrongly takes $I_E$ below $I_C$, and (d) comes from taking $5\%$ of the wrong quantity. Always remember that $I_E = I_B + I_C$ and that $I_E$ is the largest of the three.

Question 13

In the depletion region of a diode

Solution. The depletion region is the layer on either side of the junction from which the mobile carriers have diffused away and recombined. What remains are the fixed donor ions (positive) on the n side and the fixed acceptor ions (negative) on the p side. So there are no mobile charges, (a), and immobile ions do exist, (d). Option (c) names the process that built the region, not a property of what is in it now, so it is not offered as correct. Option (b) is false: it says mobile holes and electrons are present, which contradicts (a), and the region is not neutral point by point, since its two sides carry equal and opposite ion charge that produces the barrier field.

Printed key error. The key prints (a), (b), (d). Option (b) cannot be included with (a): (a) says there are no mobile charges, (b) says holes and electrons exist in the region. They contradict each other. The region holds equal and opposite immobile ion charge, not mobile carriers. The correct answer is (a), (d).

Question 14

What happens during regulation action of a Zener diode?

Solution. In breakdown the voltage across the Zener stays almost constant but the current through it varies widely. That variation is what absorbs changes in input voltage or load: when the input rises, the Zener takes more current, the current through $R_s$ increases and the extra drop across $R_s$ keeps the output steady. So the current through $R_s$ changes: (b) is true. The current in the Zener is not fixed, so (a) is false. Because the same voltage goes with different currents, the Zener’s dynamic resistance $\Delta V/\Delta I$ changes, so (d) is true and (c) is false.

Question 15

To reduce the ripples in a rectifier circuit with capacitor filter

Solution. Between charging peaks the capacitor discharges through the load with time constant $R_LC$, and the time available for discharge is about $1/f$ (half that for a full-wave circuit). The ripple voltage is therefore roughly

$$\Delta V \propto \frac{1}{f\,C\,R_L}.$$

To reduce ripple, raise $R_L$ (slower discharge), raise $f$ (shorter gaps between recharges) or raise $C$ (more stored charge). So (a), (c), (d) are true and (b) is false.

Question 16

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

Solution. Two mechanisms exist. In avalanche breakdown the doping is light, so the depletion region is wide; minority carriers are accelerated over a long path and gain enough velocity to knock out more carriers by collision: (a). In Zener breakdown the doping is heavy, so the depletion region is very thin and even a modest reverse voltage gives an extremely strong field that pulls electrons straight out of their bonds: (d). The other pairings (fast carriers with heavy doping, strong field with light doping) do not occur together.

Common mistakes

• Mixing up curves 1 and 3 in Question 2. Reverse bias raises the barrier above $V_0$; forward bias lowers it.
• Reading a diode’s direction from the wire, not the arrow, in Question 3. The anode is the arrow end and must be the higher potential to conduct.
• Using $220$ V instead of the peak in Question 4. An unloaded capacitor holds the peak, $220\sqrt2$ V.
• Treating the shunt diode in Question 6 as a series diode. It shorts the output on the positive half, so the negative halves survive.
• Forgetting that the capacitor blocks d.c. in Question 7. All $0.2$ mA goes through both resistors.
• Taking the last gate in Question 8 literally as AND. Then $E$ is always $0$; the intended gate is OR and $E = B$.
• Calling $0.4$ V “active” in Question 11. The active region starts at $0.6$ V.
• Ticking (b) in Question 13. Mobile holes and electrons do not remain in the depletion region.

Practise next

• Emitter and base current. In an npn transistor $98\%$ of the emitter electrons reach the collector and the collector current is $19.6$ mA. Find $I_E$ and $I_B$.

• Peak on a capacitor. A $110$ V (r.m.s.) supply feeds an unloaded half-wave rectifier with a capacitor. What voltage does the capacitor hold?

• Exemplar Semiconductor Electronics: Very Short and Short Answer Questions — Questions 17 to 30, with the waveform, gain and logic-gate problems.
• Exemplar Semiconductor Electronics: Long Answer Questions — Questions 31 to 40, with transistor biasing and gate networks.
• NCERT textbook Chapter 14 — Semiconductor Electronics for the theory of p-n junctions, rectifiers, transistors and logic gates.