Electronic Devices (Semiconductors) — Practice

Question 1

Does n-type semiconductor material carry a net negative charge? Justify your answer.

Accepted Answer

No, it is exactly neutral. Each pentavalent donor contributes one mobile electron but is itself left as a fixed positive ion of equal magnitude, so the totals balance. The letter n refers only to the sign of the majority charge carriers.

Question 2

Give the output frequency of half-wave and full-wave rectifiers fed from a 50 Hz supply.

Accepted Answer

Half-wave gives 50 Hz, since it discards the negative half and produces one pulse per cycle. Full-wave gives 100 Hz, since it inverts the negative half and produces two pulses per cycle. The doubling is the quickest way to identify a full-wave circuit from a trace.

Question 3

State how an LED, a photodiode and a solar cell are each biased.

Accepted Answer

An LED is forward biased and emits light on recombination. A photodiode is reverse biased and detects light through an increased reverse current. A solar cell has no external bias at all and generates its own voltage from incident light.

Question 4

Pure silicon has $n_i = 1.5	imes10^{16}$ m$^{-3}$. After doping the electron concentration is $5	imes10^{22}$ m$^{-3}$. Find the hole concentration.

Accepted Answer

By the mass action law $n_h = n_i^{2}/n_e = 2.25	imes10^{32}/5	imes10^{22} = 4.5	imes10^{9}$ m$^{-3}$. Doping raised the electron count by about three million times and cut the hole count by the same factor.

Question 5

An LED emits red light at 650 nm. Estimate its band gap, and explain why silicon cannot make a visible LED.

Accepted Answer

Using $hc = 1240$ eV nm, the band gap is about 1.91 eV. Silicon's gap of 1.1 eV corresponds to 1127 nm, well into the infrared and invisible. Visible LEDs therefore need wider-gap compounds, and the shorter the wavelength the wider the gap required.

Question 6

Why does a semiconductor's resistance fall with temperature while a metal's rises?

Accepted Answer

Both follow $ho = m/ne^{2}	au$. In a metal the carrier density $n$ is fixed and heating increases lattice vibration, shortening $	au$ and raising resistance. In a semiconductor heating promotes exponentially more electrons across the band gap, and that huge increase in $n$ overwhelms the fall in $	au$, giving a negative temperature coefficient.

Question 7

Write the truth table entries that distinguish NAND from NOR for two inputs.

Accepted Answer

NAND outputs 0 only when both inputs are 1, and 1 in all three other cases. NOR outputs 1 only when both inputs are 0, and 0 otherwise. Both are universal gates, so any logic function can be built from copies of either alone.

Question 8

A 10 V Zener regulates a supply varying from 15 V to 20 V, feeding a 20 mA load, and needs at least 5 mA to stay in breakdown. Find a suitable series resistance and check both extremes.

Accepted Answer

The worst case is the 15 V minimum, where the resistor drops 5 V and must pass at least 25 mA, giving $R \leq 200\ \Omega$. At 20 V with 200 $\Omega$ the resistor passes 50 mA, of which the load takes 20 mA and the Zener absorbs the surplus 30 mA. The output holds at 10 V across the whole range, which is what regulation means.

Question 9

Show how to build a NOT gate and an AND gate using NAND gates alone.

Accepted Answer

Tying both inputs of a NAND together gives $\overline{A\cdot A} = \bar{A}$, that is NOT. For AND, feed $A$ and $B$ into a NAND to get $\overline{A\cdot B}$, then invert that with the NOT just constructed, recovering $A\cdot B$. This is exactly what universality means, and it is why real integrated circuits are dominated by NAND structures.

Question 10

Why can no voltage be measured across the terminals of an unbiased p-n junction, despite a real barrier potential existing across the depletion region?

Accepted Answer

Because attaching metal leads creates additional metal-to-semiconductor junctions, each with its own contact potential, and these exactly cancel the junction's barrier potential around the loop. This cancellation is required by thermodynamics rather than being a coincidence: if any net voltage survived, a junction lying on a bench could drive a current through a resistor indefinitely and act as a perpetual source of energy.

Question 11

A 50 Hz supply feeds a rectifier whose transformer secondary gives 30 V peak. State the ripple frequency and the peak inverse voltage for a half-wave rectifier, a centre-tapped full-wave rectifier and a bridge rectifier.

Accepted Answer

Half-wave passes one hump per cycle, so the ripple frequency is 50 Hz; both full-wave circuits pass two, giving 100 Hz, which is why their filter capacitors have half as long to discharge. For peak inverse voltage: half-wave gives $PIV = 30$ V, since the idle diode sees the whole secondary peak; the centre-tapped circuit gives $PIV = 60$ V, because the idle diode sees its own half winding plus the conducting one; and the bridge gives $PIV = 30$ V, since the idle pair shares the reverse voltage.

Question 12

NOR gates are also universal. Show how to build NOT, OR and AND from NOR gates alone, and say how many gates each needs.

Accepted Answer

**NOT**: tie both inputs together, giving $\overline{A+A} = \bar{A}$ — one gate. **OR**: a NOR is an OR followed by an inversion, so undo it with a second NOR wired as a NOT, giving $\overline{\overline{A+B}} = A+B$ — two gates. **AND**: invert both inputs and NOR the results, since De Morgan gives $\overline{\bar{A}+\bar{B}} = A\cdot B$ — three gates. This is the exact dual of the NAND construction, where AND costs two gates and OR costs three; the two constructions swap which operation is cheap.

Practice — Electronic Devices (Semiconductors)

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