Semiconductors & Semiconductor Devices
From energy bands to logic gates.
Energy Bands in Solids
Why silicon sits between copper and quartz.
Copper conducts, quartz doesn’t, and silicon sits in between, about 10¹¹ times worse than copper and 10¹³ times better than quartz. Odder still, silicon conducts better when it is warmed. Energy bands explain all three.
Three kinds of solid
In a metal, a field E drives a current density j = σE, with conductivity σ. Heating a metal lowers σ: its free electrons collide more often and gain less drift speed between collisions, so its resistivity ρ = 1/σ rises. An insulator carries almost no current unless the field is enormous. A semiconductor, such as silicon or germanium, conducts weakly, and its conductivity rises with temperature.
From levels to bands
An isolated atom has sharp energy levels (1s, 2s, 2p, 3s…), each holding a fixed number of electrons by the Pauli principle. Take sodium: 11 electrons as 1s² 2s² 2p⁶ 3s¹. The 3s level has room for two electrons but holds one.
Now gather N sodium atoms, far apart, as in a vapour. There are 2N identical 1s states (all filled), 2N 2s states, 6N 2p states, and 2N 3s states of which only N are filled. Bring the atoms close, as in a solid, and the outer 3s electrons of neighbours start to interact. The 2N states no longer share one energy: some move up, some down, and the sharp level spreads into a band of very closely spaced levels. Inner levels spread far less, because inner electrons barely feel the neighbours.
Who can take energy?
At 0 K electrons fill the bands from the bottom up. At temperature T, collisions trade energies of order kT (about 0.026 eV at 300 K). An electron can only accept a small amount of energy, from heat or from a battery’s field, if there is an empty state just above it. In a completely full band there isn’t, so those electrons can neither absorb heat nor carry current. Only electrons in a partly empty band conduct.
The four band pictures
| Type | Band structure | Example | Result |
|---|---|---|---|
| Metal (sodium-type) | Highest occupied band only partly full | Na: the 3s band is half full | Good conductor |
| Metal (zinc-type) | Top band full, but the next band overlaps it | Zn: 4s full, 4p overlaps it | Good conductor |
| Insulator | Top band full, big gap (several eV) to an empty band | Diamond, gap ≈ 6 eV | Practically no current |
| Semiconductor | Top band full, small gap (about 1 eV) to an empty band | Si 1.1 eV, Ge 0.68 eV | Weak conduction, rising with T |
The full band is called the valence band, the empty one above it the conduction band. In a semiconductor, thermal collisions lift a few electrons across the small gap. Those electrons have the whole empty conduction band to move in, and the vacancies they leave let valence electrons shuffle too. There are few such electrons, so the conductivity is small, but it grows quickly as the temperature rises.
In a metal, conduction electrons travel on average 3 × 10⁻⁸ m between collisions. What field would give an electron 1.5 eV of energy over one such free path?
- 1Atomic levels spread into bands when atoms are packed together; band gaps separate them.
- 2Only electrons with empty states just above them can gain energy and conduct.
- 3Metals: top band part-full or overlapping. Insulators: full band, big gap. Semiconductors: full band, small gap (Si 1.1 eV, Ge 0.68 eV).
- 4Semiconductors conduct more as T rises; metals conduct less.
Check yourself
Holes, Doping & Conductivity
Two kinds of carrier, and how to choose one.
A semiconductor carries current two ways: electrons in the conduction band, and holes in the valence band. Add one impurity atom in a million and you can multiply its conductivity a million times, and choose which carrier does most of the work.
Bonds, broken bonds and holes
Silicon (Z = 14, outer shell 3s² 3p²) has four outer electrons. In the crystal, each atom shares one electron with each of four neighbours, forming four covalent bonds of two electrons each. All these bonding electrons are in the valence band.
A violent enough collision can tear one electron out of its bond. It moves away in a large orbit, hopping from atom to atom: it is now in the conduction band. The broken bond is left one electron short. A neighbouring bond electron can slip into that gap with very little energy (all bond electrons have about the same energy), which breaks its bond instead. The vacancy has moved one step, in the opposite direction to the electron.
In pure silicon every conduction electron leaves one hole, so their numbers are equal. Both drift in a field (electrons against it, holes along it) and both carry current. A pure semiconductor like this is called intrinsic. Germanium has the same crystal structure and behaves the same way.
Doping: n-type and p-type
Replace a few silicon atoms with phosphorus (five outer electrons). Four go into bonds; the fifth circles the phosphorus ion in a wide orbit (about 3 nm). Its energy lies only about 0.045 eV below the conduction band, so heat frees nearly all of them. Electrons become the majority carriers.
Replace a few with aluminium (three outer electrons). One bond is left incomplete. A valence electron needs only about 0.057 eV to move in and complete it, which leaves a hole behind in the valence band. Holes become the majority carriers.
Adding an impurity on purpose is called doping, and a doped semiconductor is extrinsic. The effect is dramatic: phosphorus at one part in a million raises silicon’s conductivity by about a million. On a band diagram, donors add impurity levels just below the conduction band, and acceptors add them just above the valence band.
How many carriers?
Pairs are created by heat and destroyed when an electron falls back into a hole; the two rates balance at a steady density. For pure silicon at 300 K that is about 7 × 10¹⁵ pairs per m³, against about 9 × 10²⁸ free electrons per m³ in copper. When a semiconductor is doped, the product of the electron and hole densities stays roughly constant, so adding many electrons suppresses the holes.
| Material | Electrons per m³ | Holes per m³ |
|---|---|---|
| Copper (metal) | ≈ 9 × 10²⁸ | — |
| Pure silicon | ≈ 7 × 10¹⁵ | ≈ 7 × 10¹⁵ |
| Silicon + phosphorus (1 in 10⁶) | ≈ 5 × 10²² | ≈ 1 × 10⁹ |
| Silicon + aluminium (1 in 10⁶) | ≈ 1 × 10⁹ | ≈ 5 × 10²² |
Conductivity
For a metal, j = nevd; writing the mobility μ = vd/E (drift speed per unit field) gives σ = neμ. A semiconductor has two kinds of carrier with different mobilities, so their contributions add:
An n-type sample has 6 × 10¹⁹ conduction electrons and 2 × 10¹⁸ holes per m³. The mobilities are 0.39 m² V⁻¹ s⁻¹ for electrons and 0.19 m² V⁻¹ s⁻¹ for holes. Find its resistivity.
Temperature
The number of thermally created pairs grows roughly as T3/2e−ΔE/2kT, where ΔE is the gap. Warmer also means more collisions and lower mobility, but the growth in carriers wins by far. So a semiconductor’s resistivity falls as it heats: its temperature coefficient is negative (silicon averages about −0.07 K⁻¹), the opposite of a metal.
A doped crystal has impurity levels 45 meV below the conduction band. (a) Is it n-type or p-type? (b) At what temperature does kT equal 45 meV?
A semiconductor’s band gap equals the energy of a 620 nm photon. Find (a) the minimum energy to make an electron–hole pair, and (b) how many times larger it is than kT at 300 K.
In a p-type crystal the acceptor levels lie 45 meV above the valence band. What is the longest wavelength of light that can create a hole?
Silicon’s gap is 1.1 eV. Taking the number of carriers proportional to e−ΔE/2kT, by what percentage does it grow from 300 K to 310 K?
Pure silicon at 300 K has 7 × 10¹⁵ electron–hole pairs per m³ and 5 × 10²⁸ atoms per m³. Arsenic is added at 1 atom per 10⁸ silicon atoms, and every arsenic atom donates an electron. By what factor does the number of carriers rise?
- 1A broken bond gives a conduction electron and a hole; holes behave as +e carriers.
- 2Intrinsic: electrons = holes. Doping with pentavalent donors gives n-type; with trivalent acceptors, p-type.
- 3Impurity levels sit just below the conduction band (donors) or just above the valence band (acceptors).
- 4σ = e(nₑμₑ + nₕμₕ); nₑnₕ stays about constant under doping.
- 5Carriers ∝ T^(3/2) e^(−ΔE/2kT): semiconductor resistivity falls with temperature.
Check yourself
The p-n Junction & Diodes
A one-way valve for current.
Join p-type and n-type silicon in one crystal and you get a one-way valve for current. That simple junction is a diode, and with light or a carefully chosen breakdown it becomes a light sensor, an LED or a voltage regulator.
Making the junction
A p-n junction is a single crystal doped with donors on one side and acceptors on the other. One way to make it is to dope a whole wafer n-type, then diffuse a stronger dose of acceptors in from one face to turn that layer p-type. Metal contacts are added to both sides.
The depletion layer
Imagine the junction just formed. The p side is full of holes, the n side full of conduction electrons. Each kind spreads from where it is crowded to where it is scarce: holes diffuse into the n side, electrons into the p side, and near the junction they meet and cancel. The heavy ions cannot move, so they are left behind uncovered: negative acceptor ions on the p side, positive donor ions on the n side.
These fixed charges create a field pointing from the n side to the p side. It pushes any hole near the junction back to the p side and any electron back to the n side, so a thin region around the junction is left with no mobile carriers at all: the depletion layer. The n side ends up at a higher potential than the p side: a potential barrier of a few tenths of a volt.
Two currents that balance
Only the few holes and electrons energetic enough to climb the barrier diffuse across. They make a current from the p side to the n side.
Heat keeps creating electron–hole pairs everywhere. A pair born in the depletion layer is pulled apart at once by the field: the electron to the n side, the hole to the p side. That makes a small current from the n side to the p side.
With no battery connected the two are equal, and no net charge crosses the junction.
Forward and reverse bias
Forward bias: the battery’s positive terminal goes to the p side. That raises the p side’s potential, lowering the barrier and narrowing the depletion layer. Many more carriers can diffuse across, while the drift current (set by how fast pairs are made) barely changes. The net current from p to n grows very rapidly with voltage, not in proportion to it: a junction does not obey Ohm’s law.
Reverse bias: the positive terminal goes to the n side. The barrier grows and the depletion layer widens, so diffusion almost stops. What remains is the small drift current, typically microamperes, flowing from n to p. The junction now behaves like a very large resistance.
Dynamic resistance
Because the i–V graph is curved, a single “resistance” doesn’t describe it. Instead we use the response to a small change:
From a diode’s characteristic: the current is 1.0 mA at 0.50 V and 2.0 mA at 0.55 V; further up, 20 mA at 0.70 V and 30 mA at 0.72 V. Find the dynamic resistance in each region.
A junction has a barrier of 0.70 V across a depletion layer 1.0 μm wide. (a) Estimate the field in the layer. (b) An electron approaches from the n side at 6.0 × 10⁵ m/s. How fast is it moving when it reaches the p side?
Points A and B are joined by two parallel branches. Branch 1 has an ideal diode D₁ pointing from A to B in series with 15 Ω. Branch 2 has an ideal diode D₂ pointing from B to A in series with 30 Ω. A 3 V battery is connected between A and B. Find the current when the positive terminal is at (a) A, (b) B.
Diodes that use light
Photodiode: if light with photon energy hc/λ at least equal to the gap falls on the junction, it breaks bonds and creates extra electron–hole pairs. In reverse bias, where the current is just the drift of such pairs, the current rises in proportion to the light intensity: a light sensor.
Light-emitting diode (LED): in forward bias, conduction electrons fall into holes near the junction, and the energy released can come out as a photon. Silicon and germanium give infrared; compounds such as gallium arsenide, gallium phosphide and indium phosphide give visible light (380–780 nm). LEDs are now everywhere, from indicator lamps to room lighting.
A photodiode’s reverse current rises only for light of wavelength 870 nm or shorter. Assuming the light acts by creating electron–hole pairs, find the band gap.
Breakdown and the Zener diode
Increase a reverse bias far enough and, at a particular breakdown voltage, the reverse current suddenly shoots up. Two mechanisms can cause it:
- Avalanche breakdown: minority carriers accelerated by the field gain enough energy to knock valence electrons free; each new pair is accelerated in turn.
- Zener breakdown: in a very thin, heavily doped junction the field is strong enough to tear electrons out of their bonds directly.
Diodes built to work in breakdown (avalanche or Zener diodes) have a useful property: once broken down, the voltage across them stays almost constant whatever the current. Placed in reverse across a load, with a resistor in series, one keeps the load voltage steady even when the supply or the load changes.
A 6.0 V Zener diode is fed from a 10 V supply through a 200 Ω resistor, with a 1.0 kΩ load across the diode. Find the currents in the resistor, the load and the diode.
- 1Diffusion across a p-n junction leaves fixed ions, a field from n to p, a barrier, and a carrier-free depletion layer.
- 2Unbiased: diffusion current (p → n) = drift current (n → p).
- 3Forward bias lowers the barrier: current grows rapidly. Reverse bias raises it: only µA drift current.
- 4Dynamic resistance ΔV/Δi = 1/slope of the i–V curve.
- 5Photodiode: light makes pairs (reverse current ∝ intensity). LED: recombination emits photons.
- 6Zener/avalanche breakdown holds the voltage nearly constant: used for regulation.
Check yourself
Videos, simulations & further reading (7)
Rectifiers
Turning AC into one-way current.
Wall sockets supply alternating voltage, but phone chargers and radios need direct current. A diode, which only lets current through one way, is the first step in converting one into the other.
Half-wave rectifier
A rectifier turns an alternating voltage into a one-way (direct) voltage. The simplest has an AC source, one diode and a load resistor R in series. Let the source voltage be V = V₀ sin ωt.
- Positive half-cycle: the diode is forward-biased, so its resistance Rj is small. The current is i = V₀ sin ωt/(R + Rj) ≈ (V₀/R) sin ωt, and nearly the whole source voltage appears across R.
- Negative half-cycle: the diode is reverse-biased and its resistance is huge, so the current is practically zero.
The load current always flows the same way, but only for half of each cycle: half-wave rectification.
Full-wave rectifier
To use both halves, take the AC from a transformer secondary with a centre tap C. Each end, A and B, goes through its own diode (D₁ and D₂) to a common point E, and the load sits between E and C.
- When A is positive relative to C (and B negative), D₁ conducts and D₂ blocks: current flows A → D₁ → E → load → C.
- Half a cycle later B is positive: D₂ conducts and D₁ blocks: current flows B → D₂ → E → load → C.
Either way the current goes through the load from E to C. The output is a train of positive humps, twice as many as before.
Average output
Averaged over a cycle, a half-wave output gives V₀/π, and a full-wave output 2V₀/π. Real silicon diodes need about 0.7 V before they conduct well, which trims a little off every hump. In practice a large capacitor across the load smooths the humps into a nearly steady voltage.
A 12 V (peak) AC supply feeds a 1.0 kΩ load through an ideal diode in half-wave mode. Find the peak load current and the average load voltage. What would the average be with a full-wave circuit giving the same peak?
- 1A diode passes current in the positive half-cycles only: half-wave rectification.
- 2Two diodes with a centre-tapped transformer conduct alternately: full-wave rectification.
- 3Output is pulsating DC; average V₀/π (half) or 2V₀/π (full) for ideal diodes.
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Videos, simulations & further reading (2)
Junction Transistors
A small current controlling a large one.
Sandwich a very thin slice of one type of semiconductor between two of the other type and a tiny current can control a big one. That is a transistor: the amplifier, the oscillator and, billions of times over, the computer chip.
Structure
A junction transistor has three layers. In an n-p-n transistor a thin p-type layer sits between two n-type layers; in a p-n-p it is the other way round.
- Base (middle): very thin (about 1 μm) and lightly doped.
- Emitter: heavily doped; it supplies the charge carriers.
- Collector: moderately doped, and larger, to collect them. The collector–base junction has a larger area than the emitter–base one.
Each layer has a terminal, so the transistor is a three-terminal device. In normal use the emitter–base junction is forward-biased and the collector–base junction is reverse-biased. In the symbol, an arrow on the emitter shows the direction of current through the emitter–base junction: outward for n-p-n (current from base to emitter), inward for p-n-p.
How it works (n-p-n)
The forward bias makes the emitter pour electrons into the base. The base is so thin, and has so few holes, that almost none of the electrons recombine there. Instead they reach the collector–base junction, where the reverse bias’s field sweeps them into the collector. They return to the emitter through the external batteries.
So the collector current IC is nearly the whole emitter current IE. The small difference, from electrons that leave through the base terminal, is the base current IB, typically 1–5 % of IE. Holes play a small part too, but the lightly doped base keeps it minor. By Kirchhoff’s junction rule:
Why a small base current controls the collector current: a few electrons from the emitter recombine with holes in the base, and some holes diffuse into the emitter. If those lost holes were not replaced, the base would charge negatively and choke off the flow from the emitter. The base current replaces them. Increase it a little and the collector current rises a lot.
Three ways to connect it
One terminal is shared between the input and output circuits and kept at zero potential: common-emitter (the most used), common-base or common-collector.
The common-emitter amplifier
A battery biases the base–emitter junction forward; another holds the collector positive. The signal vi is added in series in the base circuit, and the output is taken across a large load resistor RL in the collector circuit.
- The forward-biased base–emitter junction has a small input resistance RBE, so the signal changes the base current by ΔIB = vi/RBE.
- The collector current changes by ΔIC = βΔIB.
- The voltage across the load changes by vo = RLΔIC.
A change of 5.0 mA in emitter current causes a change of 4.9 mA in collector current. What change in base current gives the same 4.9 mA collector change? Find α and β.
In a common-emitter amplifier, adding a 15 mV signal to the base–emitter voltage changes the base current by 10 μA and the collector current by 1.5 mA. The load is 4.0 kΩ. Find β, the input resistance, the transconductance, the voltage gain and the power gain.
The oscillator
An oscillator produces an alternating voltage of chosen frequency using only DC supplies. It combines a common-emitter amplifier with an LC circuit that resonates at
Part of the output is fed back to the input through the LC network. The amplifier boosts it, part is fed back again, and so on: the circuit sustains itself with no external signal. Only the frequency ν₀ that the LC network favours builds up, and changing L or C tunes it.
Characteristics
For an n-p-n transistor in common-emitter mode, the base circuit is the input and the collector circuit the output.
- Input characteristic (IB against VBE): like a forward-biased diode. Almost no current until VBE passes the barrier, then a rapid rise, though IB stays in microamperes because most carriers go on to the collector.
- Output characteristics (IC against VCE, one curve per IB): IC rises steeply at small VCE, then levels off at a value set mainly by IB. Larger base currents give higher curves.
- 1n-p-n or p-n-p: thin, lightly doped base; heavily doped emitter; larger collector.
- 2Normal use: emitter–base forward-biased, collector–base reverse-biased.
- 3I_E = I_B + I_C; α = I_C/I_E ≈ 0.95–0.99; β = I_C/I_B = α/(1 − α) ≈ 20–100+.
- 4Common-emitter amplifier: current gain β, voltage gain βR_L/R_BE, power gain β²R_L/R_BE; g_m = ΔI_C/ΔV_BE.
- 5Oscillator = amplifier + LC feedback at ν₀ = 1/2π√LC.
Check yourself
Logic Gates & Binary Adders
Computing with 0s and 1s.
Switches are either open or closed; a lamp is on or off; a statement is true or false. Build circuits that compute with just two values, 0 and 1, and you can make a machine that adds, and then one that does anything a computer does.
A world of 0s and 1s
A logical variable can take only two values, written 0 and 1. A logical function of such variables is fully described by listing its value for every combination of inputs: its truth table.
AND, OR, NOT with switches
Let A and B be two switches (1 = closed) and C a lamp (1 = on), all powered by one battery.
- Switches in series: the lamp lights only if both are closed. C = A AND B, written A·B or AB.
- Switches in parallel: the lamp lights if either (or both) is closed. C = A OR B, written A + B.
- A switch across the lamp: closing it short-circuits the lamp. The lamp is on exactly when the switch is open: B = NOT A, written .
| A | B | A AND B | A OR B | NOT A |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 1 |
| 0 | 1 | 0 | 1 | 1 |
| 1 | 0 | 0 | 1 | 0 |
| 1 | 1 | 1 | 1 | 0 |
Here “+” means OR, not ordinary addition.
Construct the truth table of X = (A + B)·, i.e. (A OR B) AND (NOT A). What simpler function is it?
Show that Y = is 1 for every input.
Logic gates
| A | B | X |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
AND and OR from diodes
AND: two diodes run from the output X to inputs A and B, and a resistor connects X to +5 V. If either input is at 0 V, its diode is forward-biased and (being ideal) connects X to 0 V: X = 0. Only when both inputs are at 5 V do both diodes stop conducting; then no current flows through the resistor and X sits at 5 V. That is AND.
OR: the diodes point from A and B to X, and the resistor goes from X to 0 V. If both inputs are 0 V nothing flows and X = 0. If either is at 5 V, its diode conducts and lifts X to 5 V. That is OR.
NOT from a transistor
Diodes alone cannot invert. Use an n-p-n transistor with its emitter at 0 V and its collector joined to +5 V through a resistor RC; the output X is taken at the collector, and the input A drives the base.
- A = 0: no base–emitter bias, so no current anywhere, no drop across RC, and X = 5 V, i.e. 1.
- A = 1: the base–emitter junction is forward-biased, a large collector current flows, RC drops nearly the full 5 V, and X ≈ 0.
NAND, NOR and XOR
| A | B | NAND = NOT(AB) | NOR = NOT(A + B) | XOR |
|---|---|---|---|---|
| 0 | 0 | 1 | 1 | 0 |
| 0 | 1 | 1 | 0 | 1 |
| 1 | 0 | 1 | 0 | 1 |
| 1 | 1 | 0 | 0 | 0 |
- NAND is an AND followed by a NOT; NOR is an OR followed by a NOT. Their symbols add a small circle (a “bubble”) at the output.
- XOR (exclusive OR) is 1 when exactly one input is 1. It can be built from AND, OR and NOT gates using A XOR B = .
Universal gates
Every logic gate can be built from NAND gates alone (or NOR gates alone), so each is called a universal building block:
- NOT: join both inputs of a NAND: NOT(A·A) = .
- AND: a NAND followed by a NAND-as-NOT, which inverts back to AB.
- OR: invert each input with a NAND-as-NOT, then NAND them: .
Adding with gates
Computers store numbers in binary, using only the digits 0 and 1: 1, 10, 11, 100, 101… stand for one, two, three, four, five. The addition rules are 0 + 0 = 0, 0 + 1 = 1, 1 + 0 = 1 and 1 + 1 = 10 (“one zero”, i.e. two). When the answer has two digits, the left one is the carry C and the right one the sum S.
| A | B | Carry C | Sum S |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0 | 1 | 0 | 1 |
| 1 | 0 | 0 | 1 |
| 1 | 1 | 1 | 0 |
Compare with the gate tables: C = A AND B and S = A XOR B. One AND gate and one XOR gate make a half adder.
Adding multi-digit numbers needs columns that add three bits: two digits plus the carry from the right. A full adder does it with two half adders: add the first two bits (sum S₁, carry C₁), then add the third bit to S₁ (sum S, carry C₂). C₁ and C₂ can never both be 1, so the final carry is C₁ OR C₂.
Add the binary numbers 1011 and 0110 column by column, and check in decimal.
- 1Logical variables take values 0 and 1; a truth table fully defines a logical function.
- 2AND (series switches), OR (parallel switches) and NOT (shorting switch) are the basic operations.
- 3Diodes make AND and OR gates; a transistor makes NOT.
- 4NAND and NOR are universal: any gate can be built from either.
- 5Half adder: C = AB, S = A XOR B; a full adder combines two half adders and an OR.