Carrier Concentration and Fermi Level Questions and Answers

Answer: D

The correct statements are 1 and 3.

A semiconductor becomes degenerate when it is doped so heavily that it behaves almost like a metal. For P-type material that means a very high acceptor concentration (statement 1), so high that the Fermi level moves below the top of the valence band, into the valence band itself (statement 3).

A very high donor concentration (statement 2) and a Fermi level in the conduction band (statement 4) describe a degenerate N-type semiconductor instead, so options containing 2 or 4 are wrong.

Answer: B

In a p-type material the Fermi level lies near the valence band.

Acceptor doping makes holes the majority carriers. The hole concentration is \(p = N_{V} e^{-(E_{F} - E_{V})/kT}\), so a large hole density requires E_F to sit close to E_V. The heavier the doping, the closer the Fermi level moves to the valence band edge.

The Fermi level lies at the middle of the gap only in an intrinsic semiconductor, near the conduction band in an n-type material, and inside the conduction band only in a degenerately doped n-type material.

Answer: D

The correct code is (a)-(2), (b)-(3), (c)-(4), (d)-(1).

Donor levels lie just below the conduction band, so (a) is (2); the Fermi level of an intrinsic semiconductor lies at the middle of the forbidden gap, so (d) is (1). Those two pairings alone leave only this option, since the others put the donor band near the valence band or the intrinsic Fermi level elsewhere.

The key then places both the acceptor band and the p-type Fermi level near the valence band: acceptor levels sit a few hundredths of an eV above E_V, and in a p-type material at room temperature the Fermi level is also pulled down close to the valence band edge.

Answer: A

The Fermi-Dirac distribution is \(f(E) = \dfrac{1}{e^{(E - E_f)/KT} + 1}\).

It gives the probability that an energy state E is occupied by an electron at temperature T. At \(E = E_f\) the exponential is 1, so \(f = 1/2\); far above \(E_f\) the probability falls towards 0, and far below it rises towards 1.

The form with \(E_f - E\) in the exponent describes the probability that a state is empty (occupied by a hole), not by an electron. The form with -1 in the denominator is the Bose-Einstein distribution, which applies to bosons such as photons. The last option drops the +1 and misplaces KT, so it is not a distribution at all.

Answer: B

Doping introduces impurities to modify the electrical properties of the material.

Doping is the deliberate addition of small, controlled amounts of impurity atoms to a pure (intrinsic) semiconductor. Pentavalent donors such as phosphorus give extra electrons and make n-type material; trivalent acceptors such as boron create holes and make p-type material. This is what lets engineers set the carrier type and concentration in each region of a device.

Doping does not increase the bandgap, which is a property of the host crystal, and improving thermal conductivity is not its purpose. It usually lowers resistivity, but that is a consequence of adding carriers rather than the role itself, and compensated doping can leave the resistivity high; the general statement of the role is the modification of electrical properties.

Answer: B

The resistivity of the intrinsic germanium is 44.6 ohm-cm, option (B). For an intrinsic semiconductor the conductivity is \(\sigma = n_{i}\,q\,(\mu_{n} + \mu_{p})\), and the resistivity is its reciprocal.

\[ \sigma = n_{i}\,q\,(\mu_{n}+\mu_{p}) \]

\[ = 2.5\times10^{13}\times1.6\times10^{-19} \times5600 \]

\[ \sigma = 2.24\times10^{-2}\ \text{S/cm} \]

\[ \rho = \frac{1}{\sigma} = 44.6\ \Omega\text{-cm} \]

Option (C), 0.022, is the conductivity in S/cm and not the resistivity. Option (A), 0.446, is out by a factor of one hundred. Option (D), 22.2, comes from using twice the sum of the mobilities.

Answer: D

Statements I and IV are correct, so option (D) is right. Statement I is true: the acceptor energy level in a p-type semiconductor lies just above the valence band, so that electrons can jump to it with little energy and leave holes. Statement II is false, because the donor level lies close to the conduction band, not the valence band. Statement III is false: at 0 K all the donors in an n-type semiconductor hold their electrons, no free carriers exist, and the material behaves as an insulator. Statement IV is true for the same reason, since at 0 K the acceptors are empty and the p-type material is an insulator. Options (A), (B) and (C) include a false statement.

Answer: C

The electron concentration is \(10^{5}\)/cm\(^{3}\). By the mass action law \(np = n_{i}^{2}\), so the minority electron concentration is

\[ \begin{aligned} n &= \frac{n_{i}^{2}}{p} \\ &= \frac{(1.5 \times 10^{10})^{2}}{2.25 \times 10^{15}} \end{aligned} \]

\[ \begin{aligned} n &= \frac{2.25 \times 10^{20}}{2.25 \times 10^{15}} \\ &= 10^{5}\ \text{cm}^{-3} \end{aligned} \]

Option (A), zero, is wrong because a doped semiconductor always keeps some thermally generated minority carriers. Option (B), \(10^{10}\), is close to \(n_{i}\) itself and would hold only for intrinsic material. Option (D) multiplies the numbers instead of applying \(np = n_{i}^{2}\).

Answer: C

Statements 1 and 4 are correct, the third option. The acceptor level lies close to the valence band, so that holes are created easily, and at zero kelvin the acceptor levels are unionised and a p-type semiconductor has no free carriers, so it behaves as an insulator.

The donor level lies close to the conduction band and not the valence band, so statement 2 is wrong, and an n-type semiconductor at zero kelvin has no free electrons and also behaves as an insulator, so statement 3 is wrong. So the other options are wrong.

Answer: D

In an intrinsic semiconductor the electron density is the same as the hole density.

Every electron that is excited from the valence band to the conduction band leaves exactly one hole behind, so electrons and holes are created in pairs and n = p = n_i. The equal numbers are what makes the material intrinsic, that is, undoped.

Twice, thrice or square-root relations would need extra carriers of one sign, which only come from donor or acceptor impurities. Those give n-type or p-type material, and not an intrinsic one.