Chapter 14: Semiconductor Electronics: Materials, Devices and Simple Circuits

An n-type silicon crystal is produced by doping pure silicon with a pentavalent impurity (such as As, P or Sb). Four of the five valence electrons of each impurity atom form covalent bonds with neighbouring Si atoms, and the fifth electron is loosely bound and easily becomes a free electron.

So in n-type silicon the donated free electrons greatly outnumber the holes; electrons are the majority carriers.

This statement says electrons are majority carriers — which is correct — but it claims the dopants are trivalent. Trivalent atoms produce p-type, not n-type, material. Hence the statement is false.

In n-type silicon the pentavalent dopant donates a large number of free electrons. These electrons far outnumber the thermally generated holes, so electrons are the majority carriers and holes are the minority carriers.

This statement correctly identifies the dopants as pentavalent, but it wrongly calls electrons the minority carriers. Hence the statement is false.

n-type silicon is obtained by doping with a pentavalent impurity, which donates free electrons. The donated electrons are the majority carriers, while the few thermally generated holes are the minority carriers.

Both parts of this statement are therefore correct: holes are indeed the minority carriers, and the dopants are indeed pentavalent atoms. Hence the statement is true — this is the correct option.

This statement describes a p-type semiconductor, not an n-type one. In n-type silicon the dopants are pentavalent (not trivalent) and the majority carriers are electrons (not holes).

Both parts of the statement are therefore wrong for n-type silicon, so the statement is false.

The measured band gaps of these Group-14 elements are approximately

$$(E_g)_\mathrm{C} \approx 5.4 \ \mathrm{eV}, \qquad (E_g)_\mathrm{Si} \approx 1.1 \ \mathrm{eV}, \qquad (E_g)_\mathrm{Ge} \approx 0.7 \ \mathrm{eV}$$

As we move down the group, the valence electrons occupy higher orbits and are bound less tightly, so the band gap decreases steadily: $$(E_g)_\mathrm{C} > (E_g)_\mathrm{Si} > (E_g)_\mathrm{Ge}$$.

This option claims $$(E_g)_\mathrm{Si} < (E_g)_\mathrm{Ge}$$, i.e. silicon has a smaller gap than germanium. That is wrong, since $$1.1 \ \mathrm{eV} > 0.7 \ \mathrm{eV}$$. Hence the statement is false.

Using the standard values $$(E_g)_\mathrm{C} \approx 5.4 \ \mathrm{eV}$$, $$(E_g)_\mathrm{Si} \approx 1.1 \ \mathrm{eV}$$ and $$(E_g)_\mathrm{Ge} \approx 0.7 \ \mathrm{eV}$$:

This option asserts $$(E_g)_\mathrm{C} < (E_g)_\mathrm{Ge}$$ — that carbon has a smaller gap than germanium. In fact carbon has the largest band gap of the three ($$5.4 \ \mathrm{eV} \gg 0.7 \ \mathrm{eV}$$). It also asserts $$(E_g)_\mathrm{Ge} > (E_g)_\mathrm{Si}$$, which is wrong as well.

Hence the statement is false.

The energy band gaps of carbon, silicon and germanium are approximately

$$(E_g)_\mathrm{C} \approx 5.4 \ \mathrm{eV}, \qquad (E_g)_\mathrm{Si} \approx 1.1 \ \mathrm{eV}, \qquad (E_g)_\mathrm{Ge} \approx 0.7 \ \mathrm{eV}$$

All three elements have four valence electrons and the same lattice structure, but the valence electrons lie in successively higher orbits — the 2nd for C, the 3rd for Si and the 4th for Ge. The farther the valence electrons are from the nucleus, the more weakly they are bound, so the band gap decreases on going down the group.

This gives the ordering

$$(E_g)_\mathrm{C} > (E_g)_\mathrm{Si} > (E_g)_\mathrm{Ge}$$

which is exactly what this option states. Hence the statement is true — this is the correct option.

This option claims all three elements have equal band gaps. They do not. Even though C, Si and Ge all have four valence electrons and the same crystal structure, their band gaps are markedly different:

$$(E_g)_\mathrm{C} \approx 5.4 \ \mathrm{eV} \neq (E_g)_\mathrm{Si} \approx 1.1 \ \mathrm{eV} \neq (E_g)_\mathrm{Ge} \approx 0.7 \ \mathrm{eV}$$

It is precisely this difference in band gaps that makes carbon an insulator while silicon and germanium are semiconductors. Hence the statement is false.

Diffusion of carriers across a junction is driven by a difference in concentration, not by electrostatic attraction. Holes move from the p-side to the n-side simply because there are far more holes on the p-side — they spread out from a region of high concentration to one of low concentration, exactly as a gas diffuses.

The electrons of the n-region do not "pull" the holes across. (In fact, once the depletion layer forms, the built-in field opposes further hole diffusion.) Hence this statement is not the correct reason.

Motion of carriers caused by a potential difference (an electric field) is called drift, not diffusion. In an unbiased p-n junction the built-in potential difference across the depletion layer actually opposes the diffusion of holes from p to n — it does not cause it.

The diffusion of holes from the p-region to the n-region is caused purely by the concentration gradient. Hence this statement is not the correct reason.

In the p-region holes are the majority carriers, so their concentration is very high there. In the n-region holes are only the minority carriers, so their concentration is very low. This creates a large concentration gradient of holes across the junction.

Whenever there is a concentration gradient, particles diffuse from the region of higher concentration to the region of lower concentration. Therefore holes diffuse from the p-region (high hole concentration) to the n-region (low hole concentration).

This is exactly the correct reason, so this statement is true — option (c) is the correct choice.

Since options (a) and (b) are both incorrect explanations of hole diffusion — (a) attributes it to attraction by n-region electrons and (b) attributes it to the potential difference (which is actually drift, and in an unbiased junction opposes diffusion) — the option "All the above" cannot be correct.

Only option (c) gives the true reason. Hence statement (d) is false.

Let the equilibrium (unbiased) barrier be $$V_0$$. When an external forward-bias voltage $$V_F$$ is connected with the p-side positive and the n-side negative, the external field is opposite to the junction’s internal field, so the effective barrier becomes

$$V_b = V_0 - V_F$$

Because $$V_F$$ is a positive quantity, the difference $$V_0 - V_F$$ is smaller than $$V_0$$. Hence the potential barrier is not raised; it is reduced. Therefore the statement “raises the potential barrier” is false.

Under forward bias the barrier height is lowered (see part (a)), so majority carriers can more easily diffuse across the junction. The majority-carrier current therefore increases sharply rather than dropping to zero. Hence the statement is incorrect.

Using the same reasoning as in part (a), the effective barrier under forward bias is $$V_b = V_0 - V_F$$, which is smaller than the original barrier $$V_0$$. Thus forward bias lowers the potential barrier. The statement is correct.

The four alternatives offered for what a forward bias does to a p-n junction are:

• (a) raises the potential barrier,
• (b) reduces the majority carrier current to zero,
• (c) lowers the potential barrier,
• (d) none of the above.

Let the equilibrium (unbiased) barrier height be $$V_0$$. When a forward bias $$V_F$$ is applied (p-side made positive with respect to the n-side), the external field opposes the built-in field across the depletion layer, so the effective barrier becomes

$$V_b = V_0 - V_F$$

Since $$V_F$$ is positive, $$V_b$$ is smaller than $$V_0$$. That is, forward biasing lowers the potential barrier — exactly what option (c) states. The lowered barrier allows majority carriers (electrons in the n-region and holes in the p-region) to diffuse across the junction in large numbers, so the majority-carrier current actually increases sharply rather than being raised to a higher barrier (ruling out (a)) or being reduced to zero (ruling out (b)).

Since option (c) is a correct description of what forward bias does, the choice "None of the above" cannot be right. Hence statement (d) is false.