NCERT Exemplar Solutions Class 12 Physics Semiconductor Electronics — Very Short and Short Answer Questions Q17–30

NCERT Exemplar Problems, Class 12 Physics — Semiconductor Electronics: Materials, Devices and Simple Circuits, Very Short Answer Questions 17 to 22 and Short Answer Questions 23 to 30. All 14 questions solved. The book numbers them 14.17 to 14.30.

Three ideas carry this page. Doping and band gap decide what a material does: a dopant of similar size with one valence electron more or less supplies carriers, and the gap $E_g$ decides whether light of a given wavelength is absorbed or emitted. A diode is a one-way switch: it conducts when its anode is at the higher potential, and every rectifier, clipper and logic gate here follows from deciding which diodes conduct. A transistor controls power, it does not create it: the base current sets the collector current, and the extra output power comes from the d.c. supply.

Key insight. For every circuit, first mark which diodes are forward biased for each input, then read off the output. The OR and AND gates of Questions 26 and 29 differ only in which side of the diodes the inputs are connected, and the NOT gate of Question 27 is a transistor driven between cut-off and saturation.

Very Short Answer Questions

Question 17

Why are elemental dopants for Silicon or Germanium usually chosen from group XIII or group XV?

Solution. A dopant must fit into the Si or Ge lattice without distorting it, so its atomic size should be close to that of the host atom. Group XIII and group XV elements (B, Al, Ga, In; P, As, Sb) are the neighbours of group XIV in the periodic table and so are of comparable size. They also have one valence electron less (group XIII) or one more (group XV) than Si or Ge. On forming the four covalent bonds, a group XV atom has one electron left over, which is easily freed (n-type), and a group XIII atom is one electron short, which creates a hole (p-type).

Question 18

Sn, C, and Si, Ge are all group XIV elements. Yet, Sn is a conductor, C is an insulator while Si and Ge are semiconductors. Why?

Solution. The behaviour depends on the energy gap $E_g$ between the valence and conduction bands. Going down the group the atoms get larger, the valence electrons are held less tightly and the gap shrinks:

Carbon’s gap is far too large for thermal energy to lift electrons across at room temperature. Si and Ge have small gaps, so a few electrons are excited, and tin’s bands overlap, so it has free electrons at all temperatures.

Question 19

Can the potential barrier across a p–n junction be measured by simply connecting a voltmeter across the junction?

Solution. No. The depletion region has almost no free charge carriers, so its resistance is very large. A voltmeter reads correctly only if its resistance is much larger than that of the part it measures, and no ordinary voltmeter satisfies this for a depletion region. In addition, when the metal leads are joined to the p and n sides, contact potentials arise at the two metal–semiconductor junctions. Round the closed loop they add up to cancel the barrier voltage, so no net voltage, and hence no current, can be obtained from the junction at equilibrium.

Question 20

Draw the output waveform across the resistor (Fig. 14.8).

Solution. The figure shows the input at A, a square wave switching between $+1$ V and $-1$ V. A is joined to the resistor, and the other end of the resistor goes to a diode whose anode is towards the resistor (the triangle’s flat side) and whose cathode leads back to the earth side. The output $V_0$ is taken across the resistor.

• When the input is $+1$ V, the anode is positive, the diode conducts, and with an ideal diode all of the $1$ V appears across the resistor.
• When the input is $-1$ V, the diode is reverse biased, no current flows and the voltage across the resistor is zero.

The output is a train of $+1$ V pulses, one for each positive half of the input, with zero in between. This is a half-wave rectified square wave, which is the same circuit as a half-wave rectifier.

Question 21

The amplifiers X, Y and Z are connected in series. If the voltage gains of X, Y and Z are 10, 20 and 30, respectively and the input signal is 1 mV peak value, then what is the output signal voltage (peak value) (i) if dc supply voltage is 10V? (ii) if dc supply voltage is 5V?

Solution. In series the gains multiply: $$A_V = 10\times20\times30 = 6000.$$ The ideal output is $6000\times1\ \text{mV} = 6$ V peak.

(i) With a $10$ V supply, $6$ V peak is within the supply, so the output is $6$ V.

(ii) With a $5$ V supply, the output cannot exceed the supply voltage. The amplifier saturates and the peaks are clipped, so the output is about $5$ V.

Question 22

In a CE transistor amplifier there is a current and voltage gain associated with the circuit. In other words there is a power gain. Considering power a measure of energy, does the circuit violate conservation of energy?

Solution. No. The a.c. output power is larger than the a.c. input power, but the extra energy does not come from the transistor. It is drawn from the d.c. supply battery. The small input signal only controls how much of the supply’s power is passed to the load, so energy is conserved overall: the supply provides the d.c. power, and part of it is converted to a.c. signal power and the rest is dissipated as heat.

Short Answer Questions

Question 23

(Fig. 14.9 shows two current–voltage characteristics. In (a) the current is zero for negative voltage until a point P on the negative voltage axis, where it drops steeply downwards, and rises with the usual forward knee for positive voltage. In (b) the curve lies in the fourth quadrant, running from a point Q on the negative current axis to a point P on the positive voltage axis.)

(i) Name the type of a diode whose characteristics are shown in Fig. 14.9 (A) and Fig. 14.9 (B). (ii) What does the point P in Fig. (A) represent? (iii) What does the points P and Q in Fig. (B) represent?

Solution. (i) Fig. 14.9(a) is the characteristic of a Zener diode: in reverse bias the current stays tiny and then rises abruptly at the breakdown voltage. Fig. 14.9(b) lies in the fourth quadrant, where the device delivers power, so it is a solar cell (photodiode in the photovoltaic mode) under illumination.

(ii) In (a) the point P is the Zener breakdown voltage $V_Z$, where the reverse current suddenly increases while the voltage stays almost fixed.

(iii) In (b) the point P on the voltage axis, where the current is zero, is the open-circuit voltage $V_{oc}$. The point Q on the current axis, where the voltage is zero, is the short-circuit current $I_{sc}$.

Question 24

Three photodiodes D1, D2 and D3 are made of semiconductors having band gaps of 2.5 eV, 2 eV and 3 eV, respectively. Which ones will be able to detect light of wavelength 6000 Å?

Solution. A photodiode responds only if the photon energy is at least the band gap. For $\lambda = 6000\ \text{Å} = 6\times10^{-7}$ m, $$E = \frac{hc}{\lambda} = \frac{1240\ \text{eV nm}}{600\ \text{nm}} \approx 2.07\ \text{eV}.$$ Compare with the gaps: D1 needs $2.5$ eV, D2 needs $2$ eV and D3 needs $3$ eV. Only $E_g = 2$ eV is below $2.07$ eV, so only D2 can be excited.

Question 25

If the resistance $R_1$ is increased (Fig. 14.10), how will the readings of the ammeter and voltmeter change?

(The circuit is a transistor in common-emitter connection. $R_1$ and a battery are in the base–emitter loop. In the collector circuit the ammeter, the resistor $R_2$ and a second battery are in series, and the voltmeter is connected across $R_2$.)

Solution. The base current is $$I_B = \frac{V_{BB} – V_{BE}}{R_1}.$$ Increasing $R_1$ lowers $I_B$. The collector current is $I_C = \beta I_B$, so it falls too. The ammeter is in the collector circuit, so its reading decreases. The voltmeter measures the drop across $R_2$, $V = I_C R_2$, and that falls as $I_C$ falls.

Question 26

Two car garages have a common gate which needs to open automatically when a car enters either of the garages or cars enter both. Devise a circuit that resembles this situation using diodes for this situation.

Solution. The gate must open for inputs $A$ (garage 1), $B$ (garage 2) or both, which is the OR function. Use two diodes with their anodes connected to the two garage sensors, so that they serve as inputs $A$ and $B$. Join the two cathodes together and connect them to earth through a resistor $R$. The output $Y$ is taken across $R$, and it operates the gate.

If either input is high, that diode conducts and pulls the output high (the other diode is then reverse biased, so inputs do not interfere). If both are low, no diode conducts and the output is zero.

Question 27

How would you set up a circuit to obtain NOT gate using a transistor?

Solution. Use an npn transistor in the common-emitter connection as a switch. The input $V_i$ is applied to the base through a resistor $R_B$, the emitter is earthed, and the collector is connected to the supply $V_{CC}$ through a resistor $R_C$. The output is taken from the collector.

• Input $0$ (low): the base–emitter junction is not forward biased, so the transistor is in cut-off, $I_C = 0$, and there is no drop across $R_C$. The output is high, $V_{CC}$, which is $1$.
• Input $1$ (high): the base current is large enough to drive the transistor into saturation, so $V_{CE}\approx0$ and the output is low, which is $0$.

Question 28

Explain why elemental semiconductor cannot be used to make visible LEDs.

Solution. A LED emits photons of energy about equal to the band gap. Visible light has photon energies from about $1.8$ eV (red) to $3.1$ eV (violet), so a visible LED needs a gap of at least $1.8$ eV. The elemental semiconductors have much smaller gaps, so the emitted wavelength $\lambda = 1240/E_g$ nm lies in the infrared: $$\text{Si } (1.1\ \text{eV}):\ \lambda \approx 1.13\ \mu\text{m},\qquad \text{Ge } (0.7\ \text{eV}):\ \lambda \approx 1.77\ \mu\text{m}.$$ There is also a second reason: Si and Ge are indirect-gap semiconductors, in which electron–hole recombination mostly gives heat (lattice vibrations) rather than light, so they are poor emitters even at the right energy. Visible LEDs use compound semiconductors with direct and wider gaps, such as GaAsP or GaP.

Question 29

Write the truth table for the circuit shown in Fig. 14.11. Name the gate that the circuit resembles.

(Two diodes D$_1$ and D$_2$ have their cathodes towards the inputs $A$ and $B$, and their anodes joined at the output $V_0$. The common point is also connected to $+5$ V through a resistor.)

Solution. The output node is pulled up to $+5$ V through the resistor. A diode conducts only when its anode is more positive than its cathode.

• If either input is at $0$, that diode is forward biased and conducts. It pulls the output down to about $0$ V.
• Only if both inputs are at $1$ ($+5$ V) is there no drop across either diode, so neither conducts and $V_0$ stays at $+5$ V.

The output is high only when both inputs are high, which is the AND function.

Question 30

A Zener of power rating 1 W is to be used as a voltage regulator. If zener has a breakdown of 5 V and it has to regulate voltage which fluctuated between 3 V and 7 V, what should be the value of $R_s$ for safe operation (Fig. 14.12)?

Solution. The series resistor $R_s$ takes up the difference between the input and the Zener voltage. The largest Zener current allowed by the power rating is $$I_{Z,\max} = \frac{P}{V_Z} = \frac{1\ \text{W}}{5\ \text{V}} = 0.2\ \text{A}.$$ The Zener current is largest when the input is highest, $7$ V, with no load drawing current. For this to stay at $0.2$ A, $$R_s = \frac{V_{in,\max} – V_Z}{I_{Z,\max}} = \frac{7-5}{0.2} = 10\ \Omega.$$ A larger $R_s$ would also be safe for the Zener but the minimum value is fixed by the $7$ V end, so $10\ \Omega$ is the value needed. Note also that when the input is below $5$ V (the $3$ V end of the range) the Zener does not break down, so no regulation is possible there; the $7$ V end alone fixes $R_s$.

Common mistakes

• Treating the depletion region as a source in Question 19. The barrier is internal; the contact potentials at the leads cancel it, so a voltmeter reads nothing.
• Reading the output of Question 20 as a full wave. Only the positive level passes; the negative level is blocked, not inverted.
• Using the Zener value in Question 21 instead of the supply limit. An amplifier cannot give more than its supply, so $6$ V becomes $5$ V.
• Saying the amplifier creates power in Question 22. The energy comes from the d.c. battery.
• Comparing wavelength to gap directly in Question 24. Compare photon energy $hc/\lambda$ with $E_g$; any gap above the photon energy fails to detect it.
• Confusing the OR and AND diode gates in Questions 26 and 29. With anodes at the inputs the output follows the highest input (OR); with cathodes at the inputs and a pull-up the output follows the lowest (AND).
• Choosing the $3$ V end for $R_s$ in Question 30. The Zener is in danger at the highest input, $7$ V, and the $3$ V end gives no regulation at all.

Practise next

• A photodiode is made of a semiconductor with a band gap of $1.4$ eV. What longest wavelength can it detect?

• A Zener of breakdown $6$ V and power rating $0.9$ W regulates an input that varies between $8$ V and $10$ V. Find the smallest safe $R_s$.

• Exemplar Semiconductor Electronics: Multiple Choice Questions — the band, junction and transistor ideas behind these answers.
• Exemplar Semiconductor Electronics: Long Answer Questions — the gates, transistor biasing and diode networks at full length.