What are power-of-ten errors?

Convert k to ohms and μF to farads before using T = 1.1RC.

A 555 monostable has R = 47\,k and C = 100\,μF. Calculate the pulse duration. [3 marks]

WJEC Eduqas style-style practice: A 555 monostable uses a timing resistor of 100\,k and a timing capacitor of 10\,μF. Calculate the duration of the output pulse.

A Component 2 Calculate question on the monostable. Use T = 1.1RC with consistent units: R = 100\,k = 100\,000\, and C = 10\,μF = 10 × 10^-6\,F (1 mark for the equation and units). Substitute: T = 1.1 × 100\,000 × 10 × 10^-6 = 1.1 × 1.0 = 1.1\,s (3 marks for the calculation, the working and the answer in seconds). Markers reward the equation, consistent units and the time. A common error is to mishandle the powers of ten for the kilohm and the microfarad.

WalesElectronicsSyllabus dot point

How does a 555 timer produce a single pulse of a set length?

The 555 timer in monostable mode: how it produces a single output pulse of fixed duration when triggered, and the equation for the pulse duration in terms of the timing resistor and capacitor.

A focused answer to WJEC Eduqas GCSE Electronics on the 555 monostable timer, covering how it produces a single output pulse of fixed duration when triggered, and the pulse duration equation in terms of the timing resistor and capacitor.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-15

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
What a monostable does
How the timing works
The pulse duration equation
Try this

What this topic is asking

WJEC Eduqas wants you to know the 555 timer in monostable mode: how it produces a single output pulse of a fixed length when triggered, and how to calculate that pulse duration from the timing resistor and capacitor using $T = 1.1RC$ .

What a monostable does

"Monostable" means "one stable state": left alone, the output stays low. A trigger kicks it into the temporary high state for a set time, after which it falls back to low and waits for the next trigger. This is exactly what you want for a one-shot timed action, such as a stairwell light that comes on for a fixed period when you press a switch, or a brief alarm pulse.

How the timing works

The 555 uses the RC charging you already know: a bigger resistor or capacitor charges more slowly, so the capacitor takes longer to reach the two-thirds threshold, and the output pulse is longer. The trigger starts the charge; reaching the threshold ends it. This links the abstract timer directly to the charging curve of an RC network.

The pulse duration equation

This is the key calculation for the monostable. The factor $1.1$ comes from the time it takes the capacitor to charge to two-thirds of the supply through the resistor. To find the pulse length, convert $R$ to ohms and $C$ to farads, then multiply. To design for a wanted pulse length, rearrange to find $R$ or $C$ . The most common error is mishandling the powers of ten, so always convert $\text{k}\Omega$ and $\mu\text{F}$ carefully.

Designing a monostable pulse

You need a 555 monostable to produce a pulse of $2.0\,\text{s}$ . You have a $20\,\mu\text{F}$ capacitor. Find the timing resistor needed.

step 1: Write the equation and rearrange

$T = 1.1RC$ , so $R = \dfrac{T}{1.1C}$ .

step 2: Convert the capacitance

$C = 20\,\mu\text{F} = 20 \times 10^{-6}\,\text{F}$ .

step 3: Substitute

$R = \dfrac{2.0}{1.1 \times 20 \times 10^{-6}} = \dfrac{2.0}{22 \times 10^{-6}} = 90\,909\,\Omega \approx 91\,\text{k}\Omega$ .

step 4: State the answer

Use a timing resistor of about $91\,\text{k}\Omega$ (an E24 preferred value) with the $20\,\mu\text{F}$ capacitor for a $2.0\,\text{s}$ pulse. A larger resistor would give a longer pulse, a smaller one a shorter pulse.

Try this

Q1. A 555 monostable has $R = 47\,\text{k}\Omega$ and $C = 100\,\mu\text{F}$ . Calculate the pulse duration. [3 marks]

Cue. $T = 1.1 \times 47\,000 \times 100 \times 10^{-6} = 1.1 \times 4.7 = 5.17\,\text{s}$ (about $5.2\,\text{s}$ ).

Q2. State how many output pulses a monostable produces for one trigger. [1 mark]

Cue. One pulse, of fixed length, then it returns to its stable low state.

Exam-style practice questions

Practice questions written in the style of WJEC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Eduqas style4 marksA 555 monostable uses a timing resistor of

100\,\text{k}\Omega

and a timing capacitor of

10\,\mu\text{F}

. Calculate the duration of the output pulse.

Show worked answer →

A Component 2 Calculate question on the monostable. Use $T = 1.1RC$ with consistent units: $R = 100\,\text{k}\Omega = 100\,000\,\Omega$ and $C = 10\,\mu\text{F} = 10 \times 10^{-6}\,\text{F}$ (1 mark for the equation and units). Substitute: $T = 1.1 \times 100\,000 \times 10 \times 10^{-6} = 1.1 \times 1.0 = 1.1\,\text{s}$ (3 marks for the calculation, the working and the answer in seconds). Markers reward the equation, consistent units and the time. A common error is to mishandle the powers of ten for the kilohm and the microfarad.

Eduqas style3 marksExplain what is meant by a monostable circuit, and describe what happens at the output when it is triggered.

Show worked answer →

A Component 2 Explain question. A monostable circuit has one stable state (its output normally rests low) (1 mark). When it is triggered (by a short input pulse), the output goes high for a fixed length of time, set by the timing resistor and capacitor, and then returns to its stable low state on its own (2 marks for the single timed pulse and the automatic return). Markers reward the one stable state, and the single fixed-length output pulse that then returns low. A common error is to describe a continuous train of pulses, which is the astable.

Related dot points

Sources & how we know this

WJEC Eduqas GCSE Electronics specification (from 2017) — WJEC Eduqas (2017)