Carrier Transport and Mobility Questions and Answers

Answer: C

Electron mobility falls with increasing field in GaAs.

In gallium arsenide, the conduction band has a low-energy valley where electrons are light and highly mobile, and a higher satellite valley about 0.3 eV above it where they are heavy and slow. As the field rises past about 3 kV/cm, electrons gain enough energy to transfer to the upper valley, so their average drift velocity actually falls as the field rises. This negative differential mobility is the transferred electron (Gunn) effect used in Gunn diodes.

In Si and Ge the drift velocity simply saturates at high fields; it does not fall. In metals, mobility is essentially independent of the field.

Answer: A

The transit time is 7 × 10^-11 s.

The transit time is the thickness of the active region divided by the drift velocity:

\[ \begin{aligned} t &= \frac{L}{v_{d}} = \frac{7 \times 10^{-6}\ \text{m}}{1 \times 10^{5}\ \text{m/s}} \\ &= 7 \times 10^{-11}\ \text{s} \end{aligned} \]

That is 70 ps, corresponding to an oscillation frequency of about \(1/t \approx 14\) GHz, typical of a Gunn oscillator. The value \(7 \times 10^{-6}\) s forgets to divide by the velocity (it is the thickness in metres), and the other options are many orders of magnitude off.

Answer: A

The electron mobility in intrinsic silicon at 300 K is 1350 cm\(^{2}\)/s as printed, that is 1350 cm\(^{2}\)/V·s.

The standard textbook values at room temperature are about 1350 (often 1400 or 1450) cm\(^{2}\)/V·s for electrons and about 480 cm\(^{2}\)/V·s for holes in silicon. The paper drops the per volt from the unit, but the number identifies the answer.

3900 cm\(^{2}\)/V·s is the electron mobility of germanium, and about 8500 cm\(^{2}\)/V·s is that of gallium arsenide, whose light electrons make it a high-speed material. 5000 is not the value for any of these common semiconductors.

Answer: A

The drift velocity is \(\mu\).E, mobility times field.

Under an electric field, carriers accelerate between collisions and acquire an average velocity proportional to the field. Mobility \(\mu\) is defined as the constant of proportionality:

\[ v_{d} = \mu E \]

Its units are cm\(^{2}\)/V·s, so that cm\(^{2}\)/V·s times V/cm gives cm/s. E/μ and μ/E have the wrong dimensions and would make the velocity fall as the field rises, and 2.E has no mobility at all. At very high fields the velocity saturates, but the linear relation holds at normal fields.

Answer: B

The donor concentration is about \(9.25 \times 10^{14}\ \mathrm{cm^{-3}}\). In n-type silicon at 300 K the conductivity is almost entirely due to electrons, because the donors are fully ionised and the hole concentration is negligible. So the conductivity is \(\sigma = q N_D \mu_n\), and the resistivity is its reciprocal \(\rho = 1/\sigma\).

Solving for the donor concentration with \(q = 1.6 \times 10^{-19}\) C:

The options with negative exponents, \(2.86 \times 10^{-14}\) and \(1.1 \times 10^{-15}\), cannot be concentrations, since a doping level of a fraction of an atom per cubic centimetre is meaningless. The value \(11.46 \times 10^{15}\) is about twelve times too large. The answer is \(9.25 \times 10^{14}\ \mathrm{cm^{-3}}\).

Answer: C

With rising temperature the conductivity of a metal decreases and that of a semiconductor increases. In a metal the number of free electrons is essentially fixed, while lattice vibrations grow and scatter them more, so mobility and conductivity fall. In a semiconductor, heat frees more electrons and holes across the band gap, and the rise in carrier concentration outweighs the fall in mobility, so conductivity rises. Options A, B and D give the wrong direction for one or both materials, so option C is correct.

Answer: A

Mobility is the drift velocity per unit electric field, \(\mu = v/E\), so its unit is (m/s) divided by (V/m), that is \(m^{2} V^{-1} s^{-1}\). The option \(m V^{-1} s^{-1}\) lacks one power of metre, while \(V s m^{-1}\) and \(V m s^{-1}\) are not dimensions of a velocity-per-field ratio at all. Typical values at room temperature are about 0.14 m^2/Vs for electrons in silicon and 0.05 m^2/Vs for holes, and the drift velocity is the product of this mobility and the field, which is why the unit is the one given. The answer is \(m^{2} V^{-1} s^{-1}\).

Answer: B

In an n-type bar under an electric field, electrons drift against the field and holes along it, and both motions give a current in the same direction, because the charges are opposite. So the total current is the sum of the electron and hole currents, with electrons as the majority carriers and holes as the minority carriers. Statements 1 and 2 are right, while the difference of the two currents, statement 3, is wrong. Holes move along the field and electrons against it, so the two currents have the same direction and add, and in the n-type bar the electron part is far the larger because they are the majority carriers. The answer is 1 and 2.

Answer: C

A semiconductor has a negative temperature coefficient of resistance, because heating frees more charge carriers across the band gap and so the resistance falls as the temperature rises. Metals have a positive coefficient since the lattice scattering rises with temperature. A zero or unit coefficient does not describe a semiconductor. So the negative coefficient is the answer.

Answer: C

Semiconductors have a negative temperature coefficient of resistance. As the temperature rises, more electrons are excited across the band gap, so the carrier concentration increases rapidly and outweighs the small reduction in mobility. The resistance therefore falls with temperature.

A positive coefficient is the behaviour of metals, where more lattice scattering at higher temperature raises the resistance. Zero or no change would describe special alloys such as constantan, not semiconductors.