Pull-up and pull-down resistors that give a switch a defined logic level, and the current-limiting (series) resistor that protects an LED, including calculating its value.

What this topic is asking

WJEC Eduqas wants you to know two practical uses of resistors at the edges of a logic system: a pull-up or pull-down resistor that gives a switch a definite logic level, and a current-limiting resistor in series with an LED to stop it being destroyed. You must be able to explain why each is needed and to calculate the LED series resistor.

Why a floating input is a problem

A floating input is unreliable: it can read high or low at random and respond to interference, so the circuit behaves unpredictably. The fix is to add a resistor that gently "pulls" the input to a known voltage whenever the switch is not pressed, while still allowing the switch to override it when pressed.

Pull-down and pull-up resistors

In both cases the resistor and the switch form a kind of potential divider for the logic input. With a pull-down, the open switch leaves the input held low through the resistor, and pressing the switch connects it firmly to the supply; the resistor then limits the current that flows from the supply to $0\,\text{V}$. With a pull-up, the open switch leaves the input held high, and pressing connects it to $0\,\text{V}$. A typical pull resistor is around $10\,\text{k}\Omega$: large enough to waste little current, small enough to beat the noise.

Current-limiting a light-emitting diode

Like any diode, an LED does not obey Ohm's law: once it is conducting, its voltage barely changes while the current can rise without limit, so it would quickly burn out. The series resistor takes up the difference between the supply voltage and the LED's forward voltage, and because it does obey Ohm's law, it fixes the current. Forgetting the resistor is a classic way to destroy an LED.

Calculating the LED series resistor

The resistor and LED are in series, so the same current flows through both, and the voltage across the resistor is the supply voltage minus the LED's forward voltage (the series voltage rule). Dividing that voltage by the working current gives the resistance. You then round up to the nearest preferred value, which slightly reduces the current, keeping the LED safe.

Try this

Q1. State the logic level at a switch input that uses a pull-up resistor when the switch is open. [1 mark]

• Cue. Logic 1 (high): the pull-up holds the input at the supply when the switch is open.

Q2. An LED with forward voltage $2.0\,\text{V}$ runs at $20\,\text{mA}$ from a $12\,\text{V}$ supply. Calculate the series resistor. [2 marks]

• Cue. $V_R = 12 - 2.0 = 10\,\text{V}$; $R = \dfrac{10}{0.020} = 500\,\Omega$ (use $470\,\Omega$ or $510\,\Omega$).