Definition
A diode is a two-terminal semiconductor device that allows current to flow predominantly from anode to cathode when forward-biased and blocks current when reverse-biased, up to its reverse voltage limit.
Learning objectives
- Identify the anode and cathode and explain forward and reverse bias.
- Calculate the series resistor or current using the forward voltage drop.
- Recognise rectifier, freewheeling, Zener, LED and protection roles.
- Test a diode in and out of circuit with a meter and interpret the result.
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Progressive explanation
Start with the plain-language idea, then move into the engineering detail. Each level stands on its own, so you can stop at the depth you need.
Beginner The core idea
A diode is an electrical one-way valve. The anode is the entry side and the cathode is the exit side, marked by a band on the package. When the anode is more positive than the cathode by enough voltage, current flows and the diode drops a small voltage across itself. When the polarity is reversed, the diode blocks current, apart from a tiny leakage current.
Diodes protect circuits from reverse polarity, convert AC into pulsating DC, suppress the voltage spike from a coil and produce light in an LED. The forward voltage drop is a real cost: the diode gets warm in proportion to current times that drop, and the circuit must be designed for it.
Diodes protect circuits from reverse polarity, convert AC into pulsating DC, suppress the voltage spike from a coil and produce light in an LED. The forward voltage drop is a real cost: the diode gets warm in proportion to current times that drop, and the circuit must be designed for it.
Intermediate How it behaves in a circuit
The forward voltage of a conducting diode depends on its material and type: silicon junctions typically sit in the region of a few tenths of a volt to about a volt at useful currents, while Schottky diodes have a lower forward drop and specialised types differ again. These are general tendencies, not fixed numbers: the actual drop depends on current, temperature and the specific part, so always use the datasheet for a design.
In reverse bias, the diode blocks until its reverse voltage rating is exceeded. Exceeding that rating can cause avalanche breakdown, which may be destructive unless the part is designed for it. A rectifier turns AC into pulsating DC by conducting on one half-cycle and blocking the other; a bridge rectifier arranges four diodes so that both half-cycles contribute. A freewheeling diode gives an inductive load a safe path when its current is interrupted, and a Zener or TVS diode is designed to limit voltage in a controlled way.
In reverse bias, the diode blocks until its reverse voltage rating is exceeded. Exceeding that rating can cause avalanche breakdown, which may be destructive unless the part is designed for it. A rectifier turns AC into pulsating DC by conducting on one half-cycle and blocking the other; a bridge rectifier arranges four diodes so that both half-cycles contribute. A freewheeling diode gives an inductive load a safe path when its current is interrupted, and a Zener or TVS diode is designed to limit voltage in a controlled way.
Advanced Limits, assumptions, engineering detail
The junction follows an exponential current–voltage relationship, so small voltage changes produce large current changes; the thermal voltage and ideality factor set the slope. Temperature affects the forward drop, and reverse recovery charge limits how fast a diode can switch, producing loss and ringing in high-frequency converters. Capacitance, leakage and avalanche behaviour matter in signal and protection design.
Diode selection balances forward loss, reverse leakage, recovery, surge current and thermal impedance. In parallel operation, diodes share current poorly unless they are matched or ballasted, because the forward voltage has a negative temperature coefficient at high current. In series, voltage sharing needs balancing networks. Reliability also depends on surge events: inrush current, inductive kickback and lightning-induced transients can exceed steady-state ratings by orders of magnitude.
Diode selection balances forward loss, reverse leakage, recovery, surge current and thermal impedance. In parallel operation, diodes share current poorly unless they are matched or ballasted, because the forward voltage has a negative temperature coefficient at high current. In series, voltage sharing needs balancing networks. Reliability also depends on surge events: inrush current, inductive kickback and lightning-induced transients can exceed steady-state ratings by orders of magnitude.
Formulas and variables
Series resistor for a known forward drop
R = (Vs − Vf) / I
R Series resistance · Ω
Vs Supply voltage · V
Vf Diode forward voltage at the chosen current · V
I Desired current · A
A 12 V supply, a 2.0 V LED drop and 20 mA give R = (12 − 2.0) / 0.020 = 500 Ω.
Diode power dissipation
P = Vf × I
P Power dissipated in the diode · W
Vf Forward voltage · V
I Forward current · A
A diode with Vf = 0.7 V at 2 A dissipates 1.4 W.
Junction current–voltage relationship
I = Is × (e^(V / (n × Vt)) − 1)
I Diode current · A
Is Reverse saturation current · A
V Voltage across the junction · V
n Ideality factor · dimensionless
Vt Thermal voltage, about kT/q · V
The exponential term explains why a small increase in forward voltage can multiply the current.
Units of measurement
| Quantity | Symbol | Unit |
|---|---|---|
| Forward voltage | Vf | volt (V) |
| Reverse voltage | Vr | volt (V) |
| Current | I | ampere (A) |
| Power | P | watt (W) |
Worked examples
Worked example
LED series resistor
Problem. An LED with a forward drop of 2.0 V is to run from 12 V at 20 mA. Find the series resistor and its dissipation. What happens with a nearby standard value of 470 Ω?
Solution. R = (12 − 2.0) / 0.020 = 500 Ω. With 470 Ω: I = (12 − 2.0) / 470 = 0.0213 A = 21.3 mA. Resistor power = (12 − 2.0)² / 470 ≈ 0.213 W.
Use about 500 Ω, or 470 Ω giving 21.3 mA; choose a resistor rated above 0.25 W with margin.
Worked example
Reverse-polarity protection drop
Problem. A 12 V supply feeds a 100 Ω load through a silicon diode with Vf = 0.7 V. Find the load current and the diode dissipation.
Solution. The diode and load are in series: I = (12 − 0.7) / 100 = 0.113 A. Diode power = 0.7 × 0.113 = 0.079 W.
Load current ≈ 113 mA and diode dissipation ≈ 79 mW; the diode costs 0.7 V of the supply.
Practical applications
- Rectifying AC into pulsating DC in power supplies and alternators
- Protecting electronics against reverse polarity and transient spikes
- Freewheeling inductive loads such as relays, solenoids and motors
- Lighting with LEDs, clamping with Zener/TVS diodes and demodulating signals
Automotive applications
Alternators use rectifier diodes to turn three-phase AC into DC, and freewheeling diodes protect drivers for injectors, relays and solenoids. Diodes also provide reverse-polarity protection, isolate circuits, suppress load-dump transients and form the basis of LED lighting. Under-hood diodes must survive vibration, thermal cycling and repetitive surge events; a diode that tests fine at low current can still fail under real load.
Industrial applications
Industrial rectifiers, freewheeling circuits, snubbers, welding supplies and surge-protection networks rely on diodes. In motor drives, freewheeling and braking diodes carry large currents and must be thermally managed. Diode failure often presents as a short circuit, which then stresses upstream fuses and contactors, so diagnostics should check both the diode and the protection coordination.
Measurement procedure
- Isolate the circuit and verify dead before testing; a diode test on a live circuit is meaningless and may damage the meter.
- Use the meter’s diode-test mode: a good silicon junction shows a forward drop in one direction and no conduction in the other, but the exact reading depends on the meter current and the part.
- Desolder one end when a parallel path could produce a misleading reading; in-circuit results are only a first check.
- For power diodes, compare the forward voltage drop under load with the expected value and check for overheating or a shorted junction.
- Test LED polarity with the diode mode and a current-limited source; never connect an LED directly across a supply without a resistor.
Common mistakes
- Assuming every diode has exactly the same forward drop; it depends on type, current and temperature.
- Connecting an LED without a series resistor or current-limited driver, which destroys it almost immediately.
- Ignoring reverse voltage rating and allowing a transient to exceed it.
- Replacing a rectifier diode with a part of lower current, lower reverse voltage or slower recovery than required.
Safety considerations
- Diodes in mains rectifiers and high-voltage supplies must be treated as live until the circuit is isolated and the capacitors are discharged.
- A shorted diode can overheat and fail open or violently; replace it and investigate the cause rather than only the symptom.
- Use eye protection when working around high-current rectifiers and battery systems; a reverse-connected battery can produce an arc.
- In electric vehicles, diodes and power modules in the traction inverter are part of the high-voltage system; only qualified personnel should service them.
Frequently asked questions
Does a diode drop the same voltage at all currents?
No. The forward voltage rises with current and falls with temperature, and different diode types have very different drops. Use the datasheet value at the actual operating current and temperature, not a remembered rule of thumb.
What happens if I connect a diode backwards?
If the reverse voltage stays within its rating, it blocks and the circuit simply does not conduct. If the reverse voltage exceeds the rating, it may break down, conduct heavily and be damaged. Some diodes are designed to break down safely, such as Zener and TVS types, but an ordinary rectifier is not.
Can I test a diode without removing it?
Often you can get a useful indication, but parallel paths can make a good diode look shorted or a bad one look normal. For a reliable result, disconnect at least one terminal and repeat the test.
Why does my diode get hot?
It dissipates Vf × I, so a diode carrying several amps can produce several watts. Check the current, the actual forward drop and the thermal path. A diode that is hot beyond its rating may be undersized, poorly cooled or passing more current than expected.