Describe the LED as a light-emitting one-way component and calculate the series resistor needed to set a safe LED current

What this dot point is asking

SEAB wants you to describe a light-emitting diode (LED) as a diode that gives out light when it conducts, and to calculate the series resistor that sets a safe current through it. The central insight is that an LED is a diode, so it conducts only one way and has a forward voltage drop, and that because its current rises steeply once it turns on, a series resistor is always needed to limit the current.

The answer

What an LED is

A light-emitting diode is a diode made from a material that emits light when current flows through it in the forward direction. Like any diode it conducts one way only: with the anode positive and the cathode negative. The cathode is marked by the shorter lead and a flat on the side of the case. When forward biased and conducting, the energy released by the current appears as light rather than only heat, which makes LEDs efficient indicators and lamps.

The LED forward voltage

An LED does not light until the forward voltage reaches its turn-on value, which depends on the colour but is typically about $2\ \text{V}$ (higher for blue and white). Once conducting, it holds roughly this forward voltage across itself, just as a silicon diode holds about $0.7\ \text{V}$. This forward drop must be subtracted from the supply when sizing the series resistor.

Why a series resistor is essential

Past its turn-on voltage, an LED's current rises very steeply for a small rise in voltage. Connected straight across a supply, the current would be far too large and the LED would burn out at once. A series resistor solves this: it drops the leftover voltage and, by Ohm's law, fixes the current at a safe value (commonly $10$ to $20\ \text{mA}$ for an indicator LED). The resistor and the LED carry the same current because they are in series.

Sizing the series resistor

The method is always the same:

1. Subtract the LED forward voltage from the supply voltage to find the voltage across the resistor: $V_R = V_S - V_{LED}$.
2. Divide by the chosen LED current: $R = \dfrac{V_R}{I}$.
3. Choose the nearest larger preferred resistor value, which keeps the current at or just below the target.

Examples in context

Example 1. A power-on indicator. Almost every appliance has a small LED that glows when it is switched on. The designer picks a current of about $10\ \text{mA}$ for a clear glow, then uses $R = (V_S - V_{LED})/I$ to choose the series resistor for the supply voltage. The same three-line calculation appears on every indicator in the device.

Example 2. A seven-segment display. A digital clock displays numbers using segments made of LEDs. Each segment needs its own current-limiting resistor sized the same way, so that all the segments glow equally brightly and none is overdriven. Multiplying one LED calculation across many segments is routine electronics design.

Try this

• Cue. An LED of forward voltage $2.0\ \text{V}$ runs at $20\ \text{mA}$ from a $6.0\ \text{V}$ supply. Find the series resistor. Voltage across resistor $= 6.0 - 2.0 = 4.0\ \text{V}$, so $R = 4.0 / 0.020 = 200\ \Omega$.

• Cue. State how to identify the cathode of an LED. The cathode is the shorter lead, on the side with the flat on the case; it is the negative terminal.

• Cue. Explain what happens to an LED connected across a supply with no series resistor. Nothing limits the current, so an excessive current flows and the LED is destroyed almost at once.