A 555 monostable has R = 100\ k and C = 10\ μF. Find the output pulse duration. [2 marks]

WJEC Eduqas Eduqas 2022-style practice: A 555 monostable must produce a single output pulse of 0.5\ s when triggered, using a 4.7\ μF capacitor. Calculate the timing resistor required.

Monostable pulse duration (up to 5 marks): the 555 monostable output pulse lasts T = 1.1\,R C. Rearranging for the resistor, R = T1.1\,C = 0.51.1 × 4.7 × 10^-6 = 0.55.17 × 10^-6 = 9.7 × 10^4\ ≈ 97\ k. Markers reward the monostable relation T = 1.1 R C, the rearrangement, and the resistor value of about 97\ k (a 100\ k preferred value gives about 0.52\ s).

EnglandElectronicsSyllabus dot point

How does the 555 timer generate a continuous square wave or a single timed pulse?

Timing circuits: the 555 timer in astable mode (frequency, period and duty cycle) and monostable mode (pulse duration), and oscillators for clock generation.

An Eduqas A-Level Electronics answer on timing circuits: the 555 timer in astable mode with its frequency, period and duty cycle, the monostable mode producing a single timed pulse, and the role of oscillators in generating a clock signal for digital systems.

Generated by Claude Opus 4.814 min answerUpdated 2026-06-02

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

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What this dot point is asking

Eduqas wants you to use the 555 timer in astable mode (frequency, period, duty cycle) and monostable mode (pulse duration), and explain oscillators for clock generation. Timing circuits set the heartbeat of a digital system.

The answer

The 555 astable

Duty cycle

The 555 monostable

Oscillators and clock generation

Designing a 555 astable for a target frequency

Design a 555 astable to flash an LED at about $2\ \text{Hz}$ , using $R_2 = 47\ \text{k}\Omega$ and $C = 4.7\ \mu\text{F}$ , then find $R_1$ .

step 1: Rearrange the frequency formula

$f = \frac{1.44}{(R_1 + 2R_2)C}$ , so $R_1 + 2R_2 = \frac{1.44}{fC}$ .

step 2: Substitute

$R_1 + 2R_2 = \frac{1.44}{2 \times 4.7 \times 10^{-6}} = \frac{1.44}{9.4 \times 10^{-6}} = 1.53 \times 10^{5}\ \Omega = 153\ \text{k}\Omega$ .

step 3: Solve for R1

$R_1 = 153\ \text{k}\Omega - 2(47\ \text{k}\Omega) = 153 - 94 = 59\ \text{k}\Omega$ (a $56\ \text{k}\Omega$ or $62\ \text{k}\Omega$ preferred value gives close to $2\ \text{Hz}$ ).

Examples in context

The 555 timer is one of the most-used chips in electronics: as an astable it flashes LEDs, generates tones and provides a simple clock; as a monostable it debounces switches, sets timed delays (a stairwell light that stays on for a fixed time) and stretches pulses. Its astable output clocks the counters and flip-flops of the earlier topics, and a crystal oscillator does the same job precisely for a microcontroller, tying the timing topic to the rest of the digital module.

Try this

Q1. State the equation for the output frequency of a 555 astable. [1 mark]

Cue. $f = \frac{1.44}{(R_1 + 2R_2)C}$ .

Q2. A 555 monostable has $R = 100\ \text{k}\Omega$ and $C = 10\ \mu\text{F}$ . Find the output pulse duration. [2 marks]

Cue. $T = 1.1RC = 1.1 \times 100000 \times 10 \times 10^{-6} = 1.1\ \text{s}$ .

Q3. State why a crystal oscillator is used instead of a 555 for a microcontroller clock. [1 mark]

Cue. Its frequency is far more accurate and stable (set by a quartz crystal's resonance).

Exam-style practice questions

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

Eduqas 20206 marksA 555 astable uses

R_1 = 1.0\ \text{k}\Omega

R_2 = 10\ \text{k}\Omega

and

C = 100\ \text{nF}

. Calculate the frequency of the output and state whether the duty cycle is above or below 50 per cent.

Show worked answer →

Frequency (up to 4 marks): the standard 555 astable frequency is $f = \dfrac{1.44}{(R_1 + 2R_2)C}$ . Here $R_1 + 2R_2 = 1000 + 2(10000) = 21000\ \Omega$ , so $f = \dfrac{1.44}{21000 \times 100 \times 10^{-9}} = \dfrac{1.44}{2.1 \times 10^{-3}} = 686\ \text{Hz}$ .

Duty cycle (up to 2 marks): in the standard astable the capacitor charges through $R_1 + R_2$ and discharges through $R_2$ , so the high time exceeds the low time and the duty cycle is above 50 per cent.

Markers reward $f \approx 686\ \text{Hz}$ from $\frac{1.44}{(R_1 + 2R_2)C}$ and recognising the duty cycle is greater than 50 per cent.

Eduqas 20225 marksA 555 monostable must produce a single output pulse of

0.5\ \text{s}

when triggered, using a

4.7\ \mu\text{F}

capacitor. Calculate the timing resistor required.

Show worked answer →

Monostable pulse duration (up to 5 marks): the 555 monostable output pulse lasts $T = 1.1\,R C$ . Rearranging for the resistor, $R = \dfrac{T}{1.1\,C} = \dfrac{0.5}{1.1 \times 4.7 \times 10^{-6}} = \dfrac{0.5}{5.17 \times 10^{-6}} = 9.7 \times 10^{4}\ \Omega \approx 97\ \text{k}\Omega$ .

Markers reward the monostable relation $T = 1.1 R C$ , the rearrangement, and the resistor value of about $97\ \text{k}\Omega$ (a $100\ \text{k}\Omega$ preferred value gives about $0.52\ \text{s}$ ).

Related dot points

Sources & how we know this

Eduqas GCE AS/A Level Electronics specification (A410QS) — WJEC Eduqas (2017)