PN Junction Theory Questions and Answers

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.

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.

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.

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.

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.

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.

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.

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.

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.