Semiconductor Electronics: Materials, Devices and Simple Circuits: common doubts, answered
The questions students ask most often about Semiconductor Electronics: Materials, Devices and Simple Circuits, each with a short answer. For the full chapter, read the Semiconductor Electronics: Materials, Devices and Simple Circuits notes.
About the chapter
What is the difference between intrinsic and extrinsic semiconductors?
An intrinsic semiconductor is pure, while an extrinsic one has been doped with a tiny, controlled amount of impurity. In pure silicon or germanium, free electrons and holes are made only by heat breaking bonds, so their numbers are equal. Pentavalent dopants give n-type material, with electrons as majority carriers; trivalent dopants give p-type, with holes in the majority. Doping raises the conductivity enormously, yet the crystal stays neutral.
From valves to semiconductors
Read this section in the notes →Why did semiconductor devices replace vacuum tubes?
Because they are smaller, cheaper, more reliable and use far less power. A vacuum tube needs a heated cathode, an evacuated glass envelope and fairly high voltages, so it is bulky, wasteful and wears out. A semiconductor device controls the flow of charge inside a solid, with no heater and no vacuum, so it works at low voltage and can be made tiny.
Classifying solids by conductivity
Read this section in the notes →How are metals, semiconductors and insulators told apart by resistivity?
Metals have very low resistivity, about 10⁻² to 10⁻⁸ Ω m; insulators very high, about 10¹¹ to 10¹⁹ Ω m; and semiconductors lie between, about 10⁻⁵ to 10⁶ Ω m. These ranges sort materials quickly, but the deeper explanation comes from energy bands: the size of the gap between the valence and conduction bands decides how easily electrons can conduct.
Energy bands
Read this section in the notes →What is the difference between the valence band and the conduction band?
The valence band holds the electrons tied up in bonds between atoms, while the conduction band is the higher band in which electrons can move freely through the crystal. A solid conducts only if there are electrons in the conduction band, or vacancies in the valence band. The energy gap between the top of the valence band and the bottom of the conduction band is Eg = EC − EV.
Why do insulators not conduct while semiconductors conduct a little?
Because an insulator's band gap is large, above about 3 eV, so at ordinary temperatures almost no electrons can jump into the conduction band. A semiconductor's gap is smaller, about 1.1 eV for silicon and 0.7 eV for germanium, so some electrons gain enough thermal energy to cross it. In a metal the bands overlap or the conduction band is partly filled, so conduction is easy.
Intrinsic semiconductors
Read this section in the notes →What is a hole in a semiconductor?
A hole is the vacancy left in a covalent bond when an electron breaks free, and it behaves as a positive charge of size e. A bound electron from a neighbouring bond can jump into the vacancy, leaving a new hole behind, so the hole effectively moves through the crystal opposite to the electrons. Its motion adds to the current, so the total is I = Iₑ + Iₕ.
Why are the numbers of free electrons and holes equal in an intrinsic semiconductor?
Because every free electron is produced by breaking a bond, and each broken bond leaves exactly one hole behind. A pure semiconductor has no other source of carriers, so nₑ = nₕ = nᵢ. Raising the temperature breaks more bonds and increases both numbers equally. Both kinds of carrier move under an applied field and contribute to the current.
Why does a pure semiconductor behave like an insulator at absolute zero?
Because at 0 K no electron has the thermal energy needed to cross the band gap, so the valence band is completely full and the conduction band completely empty. With no free electrons and no holes there are no charge carriers at all. Only as the temperature rises above 0 K do some bonds break, and conduction begins.
n-type and p-type semiconductors
Read this section in the notes →Is an n-type semiconductor negatively charged?
No, it is electrically neutral. Every extra free electron in n-type material came from a donor atom, which is left behind as a fixed positive ion. These ions balance the extra electrons exactly, so the crystal as a whole has no net charge. The name n-type describes only its majority carriers, which are electrons, not its overall charge.
Which impurities make n-type silicon and which make p-type silicon?
n-type uses pentavalent donors with five valence electrons, such as phosphorus, arsenic or antimony; p-type uses trivalent acceptors with three, such as boron, aluminium or indium. A pentavalent atom forms four bonds with its silicon neighbours and has one electron spare to donate. A trivalent atom is one electron short of four bonds, leaving a hole that can accept an electron.
Carrier concentration and band levels
Read this section in the notes →Why does doping reduce the number of minority carriers?
Because the product of electron and hole concentrations stays fixed in thermal equilibrium: nₑnₕ = nᵢ². Adding donors pushes nₑ far above nᵢ, so nₕ must fall below nᵢ; with so many electrons around, holes are filled more often. For example, silicon doped so that nₑ = 5 × 10²² m⁻³ has only about 4.5 × 10⁹ holes per cubic metre.
Where do the donor and acceptor energy levels lie in the band diagram?
A donor level lies just below the bottom of the conduction band in n-type material, and an acceptor level lies just above the top of the valence band in p-type material. So very little energy lifts an electron from a donor level into the conduction band, or from the valence band into an acceptor level. That is why most impurity atoms are already ionised at room temperature.
Formation of a p-n junction
Read this section in the notes →How is the depletion layer formed at a p-n junction?
Electrons diffuse from the n-side into the p-side and holes from the p-side into the n-side, because each is more concentrated on its own side. They leave behind fixed ions: positive donor ions on the n-side and negative acceptor ions on the p-side. This thin region, emptied of mobile carriers, is the depletion layer, and its ions set up a field and a barrier potential across the junction.
What is the difference between diffusion current and drift current in a p-n junction?
Diffusion current comes from the difference in carrier concentration, as majority carriers spread across the junction; drift current comes from the junction's electric field, which sweeps minority carriers across the other way. At first diffusion dominates, but the field it builds up increases the drift. In equilibrium the two are equal and opposite, so no net current flows.
Diode under forward and reverse bias
Read this section in the notes →Why does forward bias narrow the depletion layer?
Because the applied voltage opposes the built-in barrier, lowering it from V₀ to V₀ − V. With the positive terminal on the p-side, majority carriers on both sides are pushed towards the junction, refilling part of the depletion region. Once the barrier is low enough, large numbers of carriers diffuse across, and the forward current rises steeply.
Why does reverse bias widen the depletion layer?
Because the applied voltage adds to the built-in barrier, raising it to V₀ + V. With the positive terminal on the n-side, majority carriers on both sides are drawn away from the junction, uncovering more fixed ions and widening the depletion region. Majority carriers can hardly cross the higher barrier, so only a tiny current, carried by minority carriers, flows.
Diode V-I characteristics
Read this section in the notes →Why does the reverse current of a diode stay almost constant as the reverse voltage increases?
Because it is limited by the supply of minority carriers, not by the voltage. Even a small reverse voltage sweeps across every minority carrier that reaches the junction, so raising the voltage further cannot increase their number. The reverse current therefore saturates at a very small value, of the order of microamperes or less, and stays there until the voltage becomes large enough to cause breakdown.
What is the threshold or cut-in voltage of a diode?
It is the forward voltage beyond which the current starts to rise sharply, about 0.7 V for a silicon diode and about 0.2 V for a germanium diode. Below it, the barrier still holds back most carriers and the current is very small. Above it, even a small increase in voltage produces a large increase in current, so the diode conducts well.
What is the dynamic resistance of a diode?
It is the ratio of a small change in voltage to the small change in current it causes, at a particular point on the V-I curve: rd = ΔV/ΔI. Because the diode's curve is not a straight line, this resistance differs from point to point. It is small in forward bias above the threshold, where the curve is steep, and very large in reverse bias.
Half-wave and full-wave rectifiers
Read this section in the notes →Why is the output frequency of a full-wave rectifier twice the input frequency?
Because it uses both halves of every input cycle, giving two output pulses per cycle instead of one. One diode conducts during the positive half cycle and the other during the negative half, and both drive current through the load in the same direction. So 50 Hz mains gives a 100 Hz output from a full-wave rectifier, but only 50 Hz from a half-wave one.
Why does a full-wave rectifier use a centre-tapped transformer?
So that the two diodes receive voltages that are opposite in phase, each equal to half the secondary voltage. Measured from the centre tap, one end of the secondary is positive while the other is negative. In one half cycle the first diode conducts and the second blocks; in the next the roles swap. Both currents flow through the load the same way.
Filtering with a capacitor
Read this section in the notes →How does a capacitor smooth the output of a rectifier?
It charges to the peak voltage as each pulse rises, then feeds current to the load while the rectified voltage falls. The capacitor is connected in parallel with the load, not in series. Between pulses it discharges slowly through the load, so the voltage sags only a little before the next pulse tops it up. A larger product C × RL means a smaller sag and steadier output.
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