Your Answer : (Not Answered)

Correct Answer : A

The answer is both (A) and (R) are true and (R) is the correct explanation of (A).

A tunnel diode has a region of negative differential resistance in its forward characteristic, caused by quantum mechanical tunnelling of carriers through its very thin, heavily doped junction. With a suitable load line this gives two stable operating points, one on each positive-resistance branch, so the diode can store a binary state and switch between states in picoseconds. That bistability is why it was used as a fast memory and switching element.

Since the memory use follows directly from the negative resistance, R explains A; the options calling either statement false, or denying the link, do not fit.

Your Answer : (Not Answered)

Correct Answer : C

X-rays are used because their wavelengths are comparable to inter-atomic distances.

Diffraction is strong only when the wavelength is of the same order as the spacing of the scattering objects. X-rays have wavelengths of roughly 0.1 to 10 Å, the same order as the spacing of atomic planes in crystals, so a crystal acts as a three-dimensional diffraction grating and Bragg's law, 2d sin θ = nλ, gives the plane spacings.

Penetrating power alone does not produce a diffraction pattern. The claim that X-rays do not interact with matter is false, since diffraction is itself an interaction. Speed of analysis is not the physical reason.

Your Answer : (Not Answered)

Correct Answer : C

X-ray crystallography cannot give the vibration frequency of a carbonyl group.

X-ray diffraction measures the electron density in a crystal, from which atomic positions are found. That gives the packing arrangement of molecules, the spacing between parallel planes or rings of atoms (from Bragg's law), and the geometry of hydrogen bonds between them.

A vibration frequency, such as the C=O stretch near 1700 cm^-1, is a dynamic property measured by infrared or Raman spectroscopy, not by diffraction from a static average structure.

Your Answer : (Not Answered)

Correct Answer : C

The electron is surrounded by a large number of unoccupied energy states.

The conduction band of a semiconductor contains a very large number of allowed states, and at ordinary temperatures only a tiny fraction of them hold electrons. An electron excited into the band therefore finds plenty of empty states next to it in energy, which is why it can gain a little energy from an applied field, move into a neighbouring state, and conduct.

The options describing a very small number of states, or a band full of occupied states, describe the valence band of an insulator-like case, where a full band leaves no room for the electron to move.

Your Answer : (Not Answered)

Correct Answer : B

The material that is not a semiconductor is indium.

Indium is a metal of group III (group 13). It conducts like a metal, with no band gap, and in semiconductor work it is used as a dopant (an acceptor in germanium and silicon) or as part of compounds such as indium phosphide, not as a semiconductor on its own.

The other three are semiconductors. Germanium and silicon are the classic elemental semiconductors of group IV, with band gaps of about 0.67 eV and 1.1 eV. Gallium arsenide (spelt 'Galium' in the paper) is a III-V compound semiconductor with a direct band gap of about 1.43 eV, used in LEDs and high-speed devices.

Your Answer : (Not Answered)

Correct Answer : D

The band gap of SiO₂ is about 8 eV.

Silicon dioxide is an excellent insulator, with a band gap usually quoted as about 8 to 9 eV. This wide gap is why thermally grown SiO₂ serves as the gate dielectric and as an insulating and masking layer in silicon technology.

1.12 eV is the band gap of silicon and 1.42 eV that of gallium arsenide, both semiconductors, and 5 eV is far too small for SiO₂; it is closer to the gap of diamond.

Your Answer : (Not Answered)

Correct Answer : D

The typical doping of a tunnel diode is about 1 part in \(10^{3}\).

A tunnel diode is doped so heavily, roughly a thousand times more than an ordinary diode, that both sides are degenerate and the depletion layer is only about 10 nm thick. Electrons can then tunnel straight through the barrier, which produces the negative-resistance region of its characteristic.

An ordinary p-n junction diode is doped at about 1 part in \(10^{8}\). Concentrations of 1 part in \(10^{6}\) or \(10^{10}\) are far too light to make the barrier thin enough for tunnelling.

Your Answer : (Not Answered)

Correct Answer : A

Reverse biasing lowers the carrier transit period through the depletion layer and reduces the depletion capacitance.

A reverse bias widens the depletion layer, and since the junction behaves like a parallel-plate capacitor, \(C = \epsilon A/W\), a larger width W gives a smaller capacitance. The reverse bias also sets up a strong electric field across the layer, which sweeps photo-generated electrons and holes across at high drift velocity, so the transit is fast. Lower capacitance and fast collection together give the photodiode a fast response.

Option B states both effects in reverse. A bias voltage does not change the doping concentrations of either side, so options C and D, which speak of changing impurity concentrations, are wrong.

Your Answer : (Not Answered)

Correct Answer : C

The diode will conduct.

A diode is forward biased when its P-side is more positive than its N-side. Here the P-side is at +0.5 V and the N-side at −1.0 V, so the forward voltage across the junction is 0.5 − (−1.0) = 1.5 V.

That is well above the cut-in voltage of a silicon diode (about 0.7 V) or a germanium diode (about 0.3 V), so the diode conducts fully. 'Not conduct' and 'breakdown' would need reverse bias, and breakdown also needs a large reverse voltage. 'Conduct partially' fits a forward voltage below the cut-in level, not 1.5 V.

Your Answer : (Not Answered)

Correct Answer : B

The correct matching is a-2, b-3, c-4, d-1.

• (a) Drift current flows when there is an electric field that pushes carriers (2).
• (b) Diffusion current flows when there is a concentration difference of electrons, holes or both (3).
• (c) \(I_{D} \approx I_{S} e^{qV_{D}/\eta kT}\) is the ideal diode current when V_D is positive and much larger than \(\eta kT/q\) (4).
• (d) \(I_{D} \approx -I_{S}\) is the ideal current when V_D is negative: the exponential term vanishes and only the reverse saturation current remains (1).

The last option has a and b right but swaps the forward and reverse expressions, and the others misplace the two kinds of current.

Your Answer : (Not Answered)

Correct Answer : B

Under reverse bias drift current flow is favourable.

When the applied voltage is less than zero, the PN junction is reverse biased. The external field adds to the built-in field, so the potential barrier rises and the depletion region widens. The higher barrier stops majority carriers from diffusing across, so diffusion current is suppressed. The stronger field, however, sweeps minority carriers across the junction, and this drift of minority carriers is the small reverse saturation current.

The barrier becomes larger, not smaller, and it is diffusion current, not drift current, that is no longer favoured, so the other three options describe forward bias or invert the effect.

Your Answer : (Not Answered)

Correct Answer : C

The compound not suitable for LEDs is SiCa.

LED materials are compound semiconductors with a suitable band gap, ideally a direct one so that electron-hole recombination releases its energy as light. Gallium arsenide (GaAs) is used for infrared LEDs, gallium phosphide (GaP), doped with nitrogen or zinc-oxygen centres, for red, yellow and green LEDs, and gallium nitride (GaN) for blue, green and white LEDs.

'SiCa', as printed, is not an LED semiconductor at all. Even read as silicon carbide (SiC), it is an indirect-gap material that gave only very dim blue LEDs and was abandoned once GaN arrived, so on either reading it is the unsuitable one.

Your Answer : (Not Answered)

Correct Answer : C

The colour depends on the energy band gap.

In an LED, light is produced when electrons recombine with holes across the band gap, and each emitted photon carries energy roughly equal to the gap. Since \(\lambda = hc/E_{g}\), or \(\lambda(\mu m) \approx 1.24/E_{g}(eV)\), the band gap fixes the wavelength and so the colour. That is why different semiconductor compounds, such as GaAs (infrared), GaAsP (red) and GaN-based materials (blue), are used for different colours.

The forward bias and forward current change the brightness, not the colour, and the reflector only directs the light.

Your Answer : (Not Answered)

Correct 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.

Your Answer : (Not Answered)

Correct 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.

Your Answer : (Not Answered)

Correct Answer : A

The correct matching is a-iii, b-iv, c-i, d-ii.

• A Gunn diode works by the transferred electron effect in GaAs, where electrons move into a higher, low-mobility valley (iii).
• A GaAs FET (MESFET) is the classic low noise microwave amplifier device (iv).
• A conventional transistor suffers from stray capacitances and lead inductances at high frequencies, which limit its microwave use (i).
• A PIN diode has a very small capacitance when reverse biased, because the wide intrinsic layer separates the charges, which suits it to microwave switching (ii).

The codes that swap the last two items give the PIN diode the parasitic problem that it is designed to avoid.

Your Answer : (Not Answered)

Correct Answer : C

The leakage current increases.

The reverse leakage (saturation) current of a diode is carried by minority carriers, which are generated thermally. Their number rises steeply with temperature: as a rule of thumb the reverse saturation current roughly doubles for every 10 °C rise in a silicon or germanium diode.

So it cannot remain the same or decrease, and it never becomes zero, because thermal generation of electron-hole pairs never stops at any practical temperature. This temperature sensitivity is one reason transistor bias circuits need stabilisation against thermal runaway.

Your Answer : (Not Answered)

Correct Answer : C

At equilibrium, drift current = diffusion current.

When a pn junction forms, holes and electrons diffuse across it because of the concentration difference, leaving behind uncovered ions that make up the depletion region. Those ions set up an electric field, which drives a drift current in the opposite direction. Diffusion continues only until the field, and so the potential barrier, has grown just enough for the drift current to cancel the diffusion current exactly. That balance is what 'stabilized' means, and it is why no net current flows through an unbiased junction.

If drift exceeded diffusion, or vice versa, a net current would flow with no external source, which cannot happen in equilibrium.