Energy bands: conductors, insulators and semiconductors
Copper wire carries current easily, rubber stops it, and the silicon chip in a phone does something in between and under control. Link to earlier classes : conductors and insulators from Chapter 1 and energy levels from Chapter 12.
In a solid the closely packed atoms split the allowed energy levels into bands. The band filled with the outer (valence) electrons is the valence band; the band above it, in which electrons can move freely, is the conduction band. The gap between them is the energy gap E_g.
Rule 1: Classification by bands — In a conductor the two bands overlap (or the conduction band is partly filled), so electrons move freely. In an insulator the gap is large (E_g greater than about 3 eV) and no electron can cross at normal temperatures. In a semiconductor the gap is small (silicon 1.1 eV, germanium 0.67 eV) and some electrons cross as the temperature rises.
- Easy : Diamond (E_g = 5.4 eV) is an insulator; silicon (1.1 eV) is a semiconductor.
- Medium : Light can lift an electron across the gap if hf ≥ E_g; for silicon λ ≤ 1240 ÷ 1.1 = 1127 nm, and for germanium λ ≤ 1240 ÷ 0.67 = 1851 nm.
- Exam : Diamond needs λ ≤ 1240 ÷ 5.4 = 230 nm, in the ultraviolet, so it is transparent to visible light.
Rule 2: Resistivity and temperature — A semiconductor has a resistivity between that of conductors and insulators (about 10⁻⁵ to 10⁶ Ω m). When it is heated, more electrons cross the gap and the resistivity falls, the opposite of a metal. Each electron that crosses leaves behind a vacancy called a hole, which behaves as a positive carrier.
- Easy : Metals have a resistivity below about 10⁻⁸ Ω m, insulators above about 10⁸ Ω m.
- Medium : In a semiconductor both electrons and holes carry current.
- Exam : At 0 K a pure semiconductor behaves like an insulator.
In the figure : Energy-band diagrams of a conductor, a semiconductor and an insulator, with the energy gap marked.
Conductor: no gap. Semiconductor: small gap. Insulator: large gap; heating lowers the resistivity of a semiconductor.
Saying that a semiconductor does not conduct at all. It conducts a little, and conducts more when warm.
Memory clue : "Gap: none, small, large = conductor, semiconductor, insulator."
Spot the difference
| Solid | Energy gap | Example |
|---|---|---|
| Conductor | zero | copper |
| Semiconductor | about 1 eV | silicon, germanium |
| Insulator | more than 3 eV | diamond |
Practice questions
- 1Define the energy gap.
- 2Name the band in which electrons can move freely.
- 3What is a hole?
- 4Find the longest wavelength that can excite an electron across the gap of silicon (1.1 eV).
- 5Find it for germanium (0.67 eV).
- 6Find it for diamond (5.4 eV).
- 7Why does the resistivity of a semiconductor fall when it is heated?
- 8State the approximate range of resistivity of semiconductors.
Show answers (try them yourself first)
- 1The gap between the top of the valence band and the bottom of the conduction band.
- 2The conduction band.
- 3The vacancy left in the valence band by an electron; it acts as a positive charge carrier.
- 4λ ≈ 1127 nm.
- 5λ ≈ 1851 nm.
- 6λ ≈ 230 nm.
- 7More electrons gain enough energy to cross the gap, so there are more carriers.
- 8About 10⁻⁵ to 10⁶ Ω m.
Intrinsic and extrinsic semiconductors
Every transistor, LED and solar cell is made from silicon or germanium into which a tiny amount of another element has been mixed on purpose. Link to earlier classes : valency and the periodic table from Class 10.
A pure semiconductor is intrinsic; it has equal numbers of electrons and holes, n_e = n_h = n_i. Adding a small amount of impurity, called doping, makes it extrinsic and greatly increases its conductivity. Silicon has four valence electrons.
Rule 1: n-type and p-type — Doping with a pentavalent atom (phosphorus, arsenic, antimony) gives an n-type semiconductor: each atom gives a free electron (a donor), electrons are the majority carriers and holes the minority. Doping with a trivalent atom (boron, aluminium, indium) gives a p-type semiconductor: each atom makes a hole (an acceptor), holes are the majority carriers.
- Easy : Phosphorus in silicon gives an n-type crystal.
- Medium : Boron in silicon gives a p-type crystal.
- Exam : Both types are electrically neutral overall; the extra carriers are balanced by the fixed ions.
Rule 2: Mass-action law — In thermal equilibrium, n_e n_h = n_i². Doping with donors raises n_e and lowers n_h so that the product stays the same.
- Easy : For silicon, n_i = 1.5 × 10¹⁶ m⁻³ at room temperature.
- Medium : An n-type sample with n_e = 5 × 10²² m⁻³ has n_h = (1.5 × 10¹⁶)² ÷ 5 × 10²² = 4.5 × 10⁹ m⁻³.
- Exam : So electrons are the majority carriers (5 × 10²²) and holes the minority carriers (4.5 × 10⁹).
n-type: pentavalent donors, majority electrons. p-type: trivalent acceptors, majority holes. n_e n_h = n_i².
Thinking that an n-type crystal is negatively charged. It is neutral.
Memory clue : "n for Negative carriers (electrons), p for Positive carriers (holes)."
Spot the difference
| Feature | n-type | p-type |
|---|---|---|
| Dopant | pentavalent (P, As) | trivalent (B, Al) |
| Majority carrier | electron | hole |
| Impurity atom is called | donor | acceptor |
Practice questions
- 1What are the majority carriers in a p-type semiconductor?
- 2Name a pentavalent impurity used for doping silicon.
- 3State the relation between n_e, n_h and n_i.
- 4An n-type silicon sample has n_e = 5 × 10²² m⁻³ and n_i = 1.5 × 10¹⁶ m⁻³. Find n_h.
- 5Name the majority and minority carriers in that sample.
- 6Name two dopants for a p-type semiconductor.
- 7Is an n-type crystal negatively charged? Explain.
- 8What is doping and why is it done?
Show answers (try them yourself first)
- 1Holes.
- 2Phosphorus (or arsenic, antimony).
- 3n_e n_h = n_i².
- 4n_h = n_i² ÷ n_e = 4.5 × 10⁹ m⁻³.
- 5Majority : electrons; minority : holes.
- 6Boron and aluminium (or indium).
- 7No; the free electrons are balanced by the positive donor ions, so the crystal is neutral.
- 8Adding a small amount of impurity to a pure semiconductor to increase its conductivity in a controlled way.
The p–n junction and the semiconductor diode
The tiny valve that allows current in one direction only is inside every phone charger, radio and LED. Link to earlier classes : electric potential from Chapter 2 and the previous section.
A p–n junction forms when a p-type and an n-type region are joined in one crystal. Electrons diffuse from n to p and holes from p to n. This leaves behind a thin region without free carriers, the depletion region, with a built-in barrier potential (about 0.7 V for silicon and 0.3 V for germanium). A p–n junction used as a device is a diode.
Rule 1: Biasing — In forward bias (p side to the positive terminal), the applied voltage reduces the barrier and narrows the depletion region, so a large current flows once the voltage exceeds the knee voltage. In reverse bias the barrier increases, the region widens, and only a tiny leakage current (microamperes) flows.
- Easy : A silicon diode in forward bias drops about 0.7 V.
- Medium : With a 5 V battery and a 430 Ω resistor in series, I = (5 − 0.7) ÷ 430 = 10 mA.
- Exam : A germanium diode with 5 V and 470 Ω gives I = (5 − 0.3) ÷ 470 = 10 mA.
Rule 2: The I–V characteristic — The graph is not a straight line, so the diode does not obey Ohm's law. It shows a sharp rise in the forward direction and almost nothing in reverse until the breakdown voltage is reached.
- Easy : A diode conducts in forward bias and blocks in reverse bias: it is a one-way valve.
- Medium : In reverse bias the current is nearly constant (the reverse saturation current).
- Exam : The resistance in forward bias is small and in reverse bias very large.
In the figure : The I–V curve of a p–n diode: a steep rise beyond the knee voltage in forward bias and a tiny current in reverse bias.
Forward bias: barrier smaller, current large. Reverse bias: barrier larger, current tiny.
Mixing up the terminals. Forward bias means p to positive, n to negative.
Memory clue : "Positive to P: Passes. Positive to N: Not."
Spot the difference
| Bias | Depletion region | Barrier | Current |
|---|---|---|---|
| Forward | narrower | lower | large |
| Reverse | wider | higher | very small |
Practice questions
- 1What is the depletion region?
- 2What is the barrier potential of a silicon junction?
- 3Does a diode obey Ohm's law?
- 4Find the current in a silicon diode in series with 430 Ω across a 5 V battery in forward bias.
- 5Find the current for a germanium diode with 470 Ω and 5 V.
- 6What current flows in reverse bias?
- 7What happens to the width of the depletion region in forward bias?
- 8Why does a diode conduct only in one direction?
Show answers (try them yourself first)
- 1A thin region at the junction with no free charge carriers.
- 2About 0.7 V.
- 3No; its I–V graph is not a straight line.
- 4I = (5 − 0.7) ÷ 430 = 10 mA.
- 5I = (5 − 0.3) ÷ 470 = 10 mA.
- 6A tiny leakage current of microamperes.
- 7It becomes narrower.
- 8In forward bias the barrier is lowered so carriers cross easily; in reverse bias it is raised and almost no carriers cross.
The diode as a rectifier
The charger of a mobile phone turns the alternating mains supply into the direct current the battery needs. Link to earlier classes : alternating current from Chapter 7 and the diode from the previous section.
A rectifier converts alternating current into direct current by using the one-way conduction of a diode. A half-wave rectifier uses one diode and passes only one half of every cycle. A full-wave rectifier uses two diodes (with a centre-tapped transformer) or four (a bridge) and passes both halves, with the same direction of current in the load.
Rule 1: Output of the rectifiers — Half-wave: V_dc = V_m ÷ π and output frequency = input frequency. Full-wave: V_dc = 2 V_m ÷ π and output frequency = twice the input frequency. The output is pulsating, so a filter (a capacitor across the load) is used to smooth it.
- Easy : A 12 V rms input has V_m = 12 × 1.414 = 17.0 V.
- Medium : Half-wave : V_dc = 17.0 ÷ π = 5.4 V. Full-wave : V_dc = 2 × 17.0 ÷ π = 10.8 V.
- Exam : For a 50 Hz supply, the half-wave output has a ripple of 50 Hz and the full-wave output 100 Hz.
Rule 2: Filtering — A capacitor charges to the peak and discharges slowly through the load while the diode is off, so the output becomes nearly steady. A larger capacitor or a faster ripple gives a smoother output, which is why full-wave rectifiers are easier to filter.
- Easy : The capacitor is connected in parallel with the load.
- Medium : A bridge rectifier needs four diodes, a centre-tap rectifier two.
- Exam : In practice the output peak is lower than V_m by the diode drop (0.7 V for silicon in half-wave, 1.4 V in a bridge).
In the figure : The input AC waveform, the half-wave output and the full-wave output.
Half-wave: f_out = f_in. Full-wave: f_out = 2f_in. Filter = capacitor in parallel.
Believing that a rectifier gives perfectly steady DC. It gives a pulsating DC that needs filtering.
Memory clue : "Half-wave: half the waves, same frequency. Full-wave: double the frequency."
Spot the difference
| Feature | Half-wave | Full-wave |
|---|---|---|
| Diodes | 1 | 2 or 4 |
| Output frequency | f | 2f |
| Average output | V_m ÷ π | 2V_m ÷ π |
Practice questions
- 1What is a rectifier?
- 2What is the output frequency of a half-wave rectifier for a 50 Hz input?
- 3What is the output frequency of a full-wave rectifier for a 50 Hz input?
- 4Find the peak voltage of a 12 V rms supply.
- 5Find the average DC output of a half-wave rectifier for that supply.
- 6Find the average DC output of a full-wave rectifier for that supply.
- 7What is the purpose of a filter capacitor?
- 8How many diodes are used in a bridge rectifier?
Show answers (try them yourself first)
- 1A circuit that converts AC into (pulsating) DC.
- 250 Hz.
- 3100 Hz.
- 4V_m = 12 × 1.414 ≈ 17.0 V.
- 5V_dc = V_m ÷ π ≈ 5.4 V.
- 6V_dc = 2V_m ÷ π ≈ 10.8 V.
- 7To smooth out the pulsating output into a nearly steady DC.
- 8Four.