23.2Solid-State Electronics

Electronic Devices: Diodes — American System Physics Revision Notes & Topical Questions

10 min read 7 parts Lesson 4 of 5 1 quiz

📚 American System Physics Hub / Topical Revision

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Lesson summary

Describe how a depletion layer forms at a p-n junction.Resistance = voltage ÷ currentR = V/I (Ω); a diode's resistance is not constant (non-ohmic).Series resistor for a diode/LEDR = (Vsupply − Vdiode) / I.In a series circuit the supply p.d.

Power = voltage × current
LED photon energy ≈ band gap
Forward bias
Reverse bias
Half-wave rectification
Full-wave rectification

Part 1 of 7Key ideas

Key ideas

Resistance

Resistance voltage ÷ current

R = V/I (Ω); a diode's resistance is not constant (non-ohmic).

Series resistor for a diode/LED

R = (V_supply − V_diode) / I.

In a series circuit the supply p.d. is shared: V_supply = V_diode + V_resistor.

Power

Power voltage × current

P = VI (W).

LED photon energy ≈ band gap

E = hf = hc/λ ≈ qV_threshold.

Peak and r.m.s. a.c. voltages: V_rms = V_peak/√2.

Silicon diode forward threshold about 0.7 V; conventional current flows from anode (p) to cathode (n), in the direction of the arrow in the circuit symbol.

Part 2 of 7Learning objectives

Describe how a depletion layer forms at a p-n junction.

Explain forward and reverse bias and sketch the diode I–V characteristic.

Describe half-wave and full-wave rectification and smoothing.

Calculate the series resistor needed for an LED.

Explain how LEDs and photodiodes work.

Part 3 of 7Essential knowledge

Essential knowledge

23.2 Electronic Devices: Diodes concept map: Resistance = voltage ÷ current, Series resistor for a diode/L, Power = voltage × current, LED photon energy ≈ band gap, Forward bias — Visual guide to the key ideas explained in this lesson.

When p-type and n-type regions are joined, free electrons from the n-side diffuse across the junction and fill holes on the p-side.

This leaves positive donor ions on the n-side and negative acceptor ions on the p-side,
forming the depletion layer with an electric field that stops further diffusion.

The depletion layer has no mobile carriers, so it behaves like an insulator. The fixed ions produce a small built-in p.d. (about 0.7 V for silicon) across the junction.

Forward bias

Forward bias the external p.d. pushes electrons and holes towards the junction, opposing the built-in field. Once the applied p.d. exceeds about 0.7 V, the depletion layer is overcome and a large current flows. A series resistor is needed to limit the current and prevent overheating.

Reverse bias

Reverse bias the external p.d. pulls electrons and holes away from the junction, widening the depletion layer, so only a tiny leakage current flows. If the reverse p.d. becomes very large, breakdown occurs and the diode may be damaged.

The I–V graph of a diode is non-linear: zero current in reverse (up to breakdown), almost zero forward current below the threshold, then a steep rise. A diode does not obey Ohm's law.

Half-wave rectification

Half-wave rectification a single diode in series with the load conducts only during the half-cycles when it is forward biased, giving a pulsating d.c. that is zero for half of each cycle.

Full-wave rectification

Full-wave rectification a bridge rectifier of four diodes directs current the same way through the load in both halves of the cycle; two diodes conduct in each half-cycle. Adding a capacitor in parallel with the load smooths the output.

LEDs are made from compounds such as gallium arsenide phosphide.

When forward biased, electrons crossing the junction recombine with holes and drop by an energy about equal to the band gap, emitting a photon.
Different band gaps give different colours; blue LEDs need larger gaps than red.
- (and higher threshold p.d.)

LEDs are highly efficient (little energy wasted as heat), long-lasting and small, so they have replaced filament lamps in many uses.

Photodiodes and solar cells work in reverse

Photodiodes and solar cells work in reverse absorbed photons create electron-hole pairs in the depletion layer; the junction field separates them, producing a current proportional to the light intensity.

Part 4 of 7Process & sequence

Connect a diode in series with a protective resistor (e.g. 100 Ω), an ammeter and a variable d.c. supply; place a voltmeter across the diode.

Set the diode forward biased and increase the p.d. in small steps from 0 V, recording current and voltage each time.

Note that the current stays near zero until about 0.6–0.7 V, then rises very steeply.

Reverse the diode and repeat, recording the (almost zero) current for reverse p.d.s.

Plot current (y-axis) against voltage (x-axis) to obtain the characteristic I–V curve.

Calculate resistance at chosen points using R = V/I, showing low forward and very high reverse resistance.

Formation of the depletion layerwrite these steps in this order; each step is a marking point.

p-type and n-type regions are joined in one crystal.

Free electrons diffuse from n-side to p-side; holes diffuse the other way.

Electrons and holes recombine near the junction.

Fixed + ions remain on the n-side, fixed − ions on the p-side.

The resulting field stops further diffusion: a carrier-free depletion layer forms.

Part 5 of 7Worked examples

Worked examples

350 Ω

A red LED needs 2.0 V and a current of 20 mA. It is to be run from a 9.0 V battery. Calculate the series resistor required and the power dissipated in the resistor.

p.d. across resistor = 9.0 − 2.0 = 7.0 V. R = V/I = 7.0/0.020 = 350 Ω. P = VI = 7.0 × 0.020 = 0.14 W.
What the symbols mean: P = power (W) · I = current (A) · V = potential difference (V)

A green LED emits light of wavelength 530 nm. Estimate its minimum forward p.d.

E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸)/530 × 10⁻⁹ = 3.75 × 10⁻¹⁹ J. V = E/q = 3.75 × 10⁻¹⁹/1.60 × 10⁻¹⁹ = 2.3 V.
What the symbols mean: E = photon energy (J) · h = 6.63 × 10⁻³⁴ J s · c = 3.00 × 10⁸ m s⁻¹ · λ = wavelength (m)

A silicon diode is forward biased in series with a 470 Ω resistor across a 5.0 V supply. Find the current.

p.d. across diode ≈ 0.7 V, so p.d. across resistor = 5.0 − 0.7 = 4.3 V. I = 4.3/470 = 9.1 × 10⁻³ A (9.1 mA).

Part 6 of 7Summary

A diode lets current flow in one direction only.

Forward bias narrows the depletion layer; reverse bias widens it.

Silicon diodes conduct above about 0.7 V forward.

Diodes are used for rectification: converting a.c. to d.c.

LEDs emit photons with energy about equal to the band gap.

Part 7 of 7Compare: Compare

Forward vs reverse bias
Feature	Forward bias	Reverse bias
Connection	p to +, n to −	p to −, n to +
Depletion layer	Narrows (disappears above threshold)	Widens
Resistance	Low (above about 0.7 V)	Very high
Current	Large, rises steeply	Almost zero (tiny leakage)

Diode types and uses
Device	How used	Application
Silicon diode	Forward/reverse	Rectifiers, protection against wrong polarity
LED	Forward biased with series resistor	Indicators, displays, lighting, remote controls
Photodiode	Reverse biased	Light meters, optical fibre receivers, cameras
Solar cell	Unbiased, generates p.d.	Converts light energy to electrical energy

Key terms

Command terms

Identify / Describe (FRQ)

A short, specific statement; describing requires relevant characteristics, not a single word.

Explain / Justify (FRQ)

Give the cause-and-effect reasoning or evidence; a restated fact does not score.

Calculate

Show setup, substitution and the answer with units.

Predict

State the expected outcome; add a reason only if asked to justify.

MCQ technique

Read every option; eliminate distractors that are true statements but do not answer the question; watch for NOT/EXCEPT.

Exam technique

Keywords the mark scheme looks for in this lesson

p-n junction · Depletion layer · Diode · Forward bias · Reverse bias · Threshold (turn-on) voltage · Rectification · Half-wave rectification · Full-wave rectification · Light-emitting diode (LED) · Photodiode · Smoothing capacitor

Learn each definition in the Key words tab exactly; a missing keyword in a definition usually loses the mark.

How to get full marks (examiner guidance)

Answer exactly what is asked; extra correct material does not earn credit, and contradictory statements can cancel credit (AP FRQ).

Use precise vocabulary, never “it/stuff/things”.

Show units and working in calculations.

In MCQs, predict the answer before reading the options, then eliminate.

Words to avoid

a diode has no resistance in forward biasa diode has low resistance in forward bias once the p.d. exceeds about 0.7 V.

reverse bias removes the depletion layerreverse bias widens the depletion layer so almost no current flows.

rectification changes d.c. into a.c.rectification changes a.c. into d.c.

a diode obeys Ohm's lawa diode is non-ohmic; current is not proportional to voltage.

an LED can be connected straight to a batteryan LED needs a series resistor to limit the current.

Last updated: 13 October 2026