NCERT Class 12 Physics — Semiconductor Electronics: Materials, Devices and Simple Circuits, Chapter 14 Exercises. All 6 questions solved.

The six exercises at the end of Chapter 14 are short. Five are multiple choice — two on doped semiconductors, one on the band gaps of carbon, silicon and germanium, and two on the p-n junction — and the last asks for the output frequency of a rectifier. They test whether the ideas are clear rather than any arithmetic, and three facts from the chapter settle all of them:

• Doping — a pentavalent impurity such as arsenic, antimony or phosphorus donates an extra electron and makes an n-type semiconductor, with electrons as majority carriers. A trivalent impurity such as boron, aluminium or indium leaves a hole and makes a p-type semiconductor, with holes as majority carriers. Since $n_en_h = n_i^2$, the more there are of one kind of carrier, the fewer of the other.
• Band gaps — for C (diamond), Si and Ge the energy gaps are $5.4\ \text{eV}$, $1.1\ \text{eV}$ and $0.7\ \text{eV}$; a material with $E_g > 3\ \text{eV}$ is an insulator.
• The p-n junction — diffusion of majority carriers across the junction leaves a depletion layer and a barrier potential $V_0$. A forward bias $V$ lowers the barrier to $V_0 – V$; a reverse bias raises it to $V_0 + V$.

Key insight. Keep two kinds of carrier motion apart. Diffusion is driven by a difference in concentration and needs no force at all; drift is driven by an electric field. Holes cross from p to n by diffusion (14.4), the field of the depletion layer then pushes them back, and forward bias works by weakening that field (14.5). The doping questions (14.1, 14.2) come down to counting valence electrons: a dopant with five supplies spare electrons, a dopant with three leaves holes.

Question 14.1

In an n-type silicon, which of the following statement is true:

Solution. Silicon has four valence electrons, all used in its four covalent bonds. An n-type semiconductor is made by adding a pentavalent dopant such as phosphorus or arsenic: four of its five valence electrons join the bonds, and the fifth is so loosely held that it is freed at room temperature. These donated electrons far outnumber the electron-hole pairs made thermally, so electrons are the majority carriers. The holes are the minority carriers — fewer, in fact, than in pure silicon, because the many free electrons fill them in ($n_en_h = n_i^2$).

That rules out (a), which has the wrong dopant, and (b), which has the electrons as the minority. Option (d) describes a p-type semiconductor. Only (c) is right on both counts.

Question 14.2

Which of the statements given in Exercise 14.1 is true for p-type semiconductors?

Solution. A p-type semiconductor is made with a trivalent dopant such as boron, aluminium or indium. It has only three valence electrons for silicon’s four bonds, so one bond is left an electron short — a hole. Each dopant atom supplies a hole, so holes are the majority carriers and electrons the minority. Option (b) is tempting because “electrons are minority carriers” is true of p-type, but its dopant is wrong; only (d) has both the carriers and the dopant right.

Question 14.3

Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to $(E_g)_{\text{C}}$, $(E_g)_{\text{Si}}$ and $(E_g)_{\text{Ge}}$. Which of the following statements is true?

Solution. The chapter gives the gaps as $5.4\ \text{eV}$ for carbon (diamond), $1.1\ \text{eV}$ for silicon and $0.7\ \text{eV}$ for germanium, so $(E_g)_{\text{C}} > (E_g)_{\text{Si}} > (E_g)_{\text{Ge}}$.

The trend has a reason. Going down the group from carbon to silicon to germanium, the atoms get larger and the valence electrons sit farther from the nucleus, so they are less tightly bound and less energy is needed to free one into the conduction band. That is why diamond, with a gap well above $3\ \text{eV}$, is an insulator, while silicon and germanium are semiconductors. Having the same number of valence electrons does not make the gaps equal, which rules out (d).

Question 14.4

In an unbiased p-n junction, holes diffuse from the p-region to n-region because

Solution. Diffusion is the movement of particles from where they are crowded to where they are sparse, and it needs no force — it is the random thermal motion of a great many particles that makes more of them wander from the crowded side than back. The p-region is full of holes and the n-region has very few, so holes spread from p to n. That is (c).

The other options describe forces that are either absent or pull the wrong way. A free electron does not attract a hole from a distance across the crystal, so (a) is wrong. The potential difference across the junction is the barrier set up by the diffusion: the negative ions it leaves on the p-side and the positive ions on the n-side make a field pointing from n to p, which pushes holes back towards p. It opposes the diffusion of holes rather than causing it, so (b) is wrong, and with it (d).

Question 14.5

When a forward bias is applied to a p-n junction, it

Solution. In forward bias the p-side is connected to the positive terminal of the battery and the n-side to the negative. The applied voltage $V$ then acts opposite to the barrier potential $V_0$, so the effective barrier falls to $V_0 – V$ and the depletion layer narrows. With a lower barrier, many more majority carriers — holes from p, electrons from n — can cross the junction, so the majority carrier current increases sharply rather than falling to zero. That rules out (a) and (b), and since (c) is true, (d) is wrong too.

Question 14.6

In half-wave rectification, what is the output frequency if the input frequency is $50\ \text{Hz}$? What is the output frequency of a full-wave rectifier for the same input frequency?

Solution. The output frequency is the number of times per second the output pattern repeats. A $50\ \text{Hz}$ input completes a cycle every $1/50\ \text{s} = 20\ \text{ms}$.

A half-wave rectifier lets through only one half of each cycle — the half in which the diode is forward biased — and blocks the other. So the output is one pulse per input cycle, followed by a gap, and the pattern repeats every $20\ \text{ms}$: the output frequency is $50\ \text{Hz}$, the same as the input.

A full-wave rectifier uses both halves, turning the negative half the other way up, so there are two identical pulses in every input cycle. The pattern now repeats every $10\ \text{ms}$, and the output frequency is $2\times50 = 100\ \text{Hz}$.

Common mistakes

• Question 14.1: thinking an n-type crystal is negatively charged. The “n” names the majority carrier, not the charge of the crystal. Every donated electron leaves behind a positive donor ion, so the material stays electrically neutral.
• Question 14.1: taking the dopant from the carrier’s name. Students sometimes reason that n-type needs “negative” trivalent atoms. It is the pentavalent dopant, with one electron more than silicon needs, that supplies the free electrons.
• Question 14.2: stopping at the first half-true option. “Electrons are minority carriers” in (b) is true for p-type, but its dopant is pentavalent. Each option makes two claims, and both have to be right.
• Question 14.3: choosing (a) because silicon is the common semiconductor. How widely a material is used says nothing about the size of its gap. Carbon (diamond) has by far the largest gap, which is why it is an insulator.
• Question 14.4: choosing “All the above”. The potential difference across the junction does not drive the holes across; it is the barrier that builds up against their diffusion. Only the concentration difference causes diffusion.
• Question 14.5: choosing (b). Forward bias makes it easier, not harder, for majority carriers to cross the junction, so the majority carrier current rises steeply. It is under reverse bias that the majority current is cut off.
• Question 14.6: giving $25\ \text{Hz}$ or $50\ \text{Hz}$ for full-wave. Half-wave rectification throws half of each cycle away but does not slow the repetition, so it stays at $50\ \text{Hz}$. Full-wave rectification produces two pulses per cycle, which doubles the frequency to $100\ \text{Hz}$.

Practise next

• Chapter 12, Atoms — the discrete energy levels of a single atom, which spread into the valence and conduction bands of a solid.
• Chapter 7, Alternating Current — the alternating voltage that a rectifier converts, and the frequency and period used in Question 14.6.
• Chapter 3, Current Electricity — resistivity and conductivity, whose enormous range across metals, semiconductors and insulators the band gap explains.