A 555 astable runs at 200\ Hz. Find its period. [1 mark]

WJEC Eduqas Eduqas 2019-style practice: A 555 astable uses R_1 = 2.2\ k, R_2 = 6.8\ k and C = 10\ μF. Calculate the frequency of the output.

Use the astable frequency equation f = 1.44(R_1 + 2R_2)C. Find R_1 + 2R_2 = 2200 + 2(6800) = 2200 + 13600 = 15800\. Substitute: f = 1.4415800 × 10 × 10^-6 = 1.440.158 = 9.1\ Hz. Markers reward the R_1 + 2R_2 term (with the factor of 2 on R_2), the capacitance in farads, and the frequency of about 9.1\ Hz. The usual error is omitting the factor of 2.

EnglandElectronicsSyllabus dot point

How does a 555 timer in astable mode produce a continuous square wave, and what sets its frequency?

The 555 astable: producing a continuous square-wave output, the frequency and period equations, the duty cycle, and using an astable as a clock or flasher.

An Eduqas GCSE Electronics answer on the 555 timer in astable mode: how it free-runs to give a continuous square wave, the frequency and period equations, why the standard duty cycle exceeds 50 per cent, and using an astable as a clock, flasher or tone generator.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-02

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

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

Eduqas wants you to use the 555 timer in astable mode: how it free-runs to give a continuous square wave, the frequency and period equations, the duty cycle (and why the standard one is above 50 per cent), and using an astable as a clock, flasher or tone generator. The astable is the standard simple oscillator of the course.

The answer

The astable as an oscillator

Frequency and period

Duty cycle

Uses of the astable

Designing a 555 astable to flash an LED

Design a 555 astable to flash an LED at about $1\ \text{Hz}$ using $R_1 = 10\ \text{k}\Omega$ and $C = 47\ \mu\text{F}$ , then find $R_2$ .

step 1: Rearrange the frequency equation

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

step 2: Substitute the values

$R_1 + 2R_2 = \dfrac{1.44}{1 \times 47 \times 10^{-6}} = \dfrac{1.44}{4.7 \times 10^{-5}} = 3.06 \times 10^{4}\ \Omega \approx 30.6\ \text{k}\Omega$ .

step 3: Solve for R2

$2R_2 = 30.6 - 10 = 20.6\ \text{k}\Omega$ , so $R_2 = 10.3\ \text{k}\Omega$ (a $10\ \text{k}\Omega$ preferred value gives close to $1\ \text{Hz}$ ).

Examples in context

The 555 astable is everywhere in the course: it flashes warning LEDs, sounds buzzers and, most importantly, provides the clock that drives the counters and seven-segment displays of the sequential module. Its frequency depends on the same RC product as the time delays of the previous module, tying the two together. For precise timing a crystal oscillator is preferred (used in microcontrollers), but the simple, cheap 555 astable is the standard choice for flashers, tones and slow clocks.

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 astable runs at $200\ \text{Hz}$ . Find its period. [1 mark]

Cue. $T = \frac{1}{f} = \frac{1}{200} = 5.0\ \text{ms}$ .

Q3. State why the standard 555 astable has a duty cycle above 50 per cent. [2 marks]

Cue. The capacitor charges through $R_1 + R_2$ but discharges through only $R_2$ , so the high time is longer than the low time.

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 20195 marksA 555 astable uses

R_1 = 2.2\ \text{k}\Omega

R_2 = 6.8\ \text{k}\Omega

and

C = 10\ \mu\text{F}

. Calculate the frequency of the output.

Show worked answer →

Use the astable frequency equation $f = \dfrac{1.44}{(R_1 + 2R_2)C}$ .

Find $R_1 + 2R_2 = 2200 + 2(6800) = 2200 + 13600 = 15800\ \Omega$ .

Substitute: $f = \dfrac{1.44}{15800 \times 10 \times 10^{-6}} = \dfrac{1.44}{0.158} = 9.1\ \text{Hz}$ .

Markers reward the $R_1 + 2R_2$ term (with the factor of 2 on $R_2$ ), the capacitance in farads, and the frequency of about $9.1\ \text{Hz}$ . The usual error is omitting the factor of 2.

Eduqas 20224 marksExplain why the output of a standard 555 astable has a duty cycle greater than 50 per cent, and state one way to make it closer to 50 per cent.

Show worked answer →

Why over 50 per cent (up to 3 marks): the timing capacitor charges through $R_1 + R_2$ (output high) but discharges through only $R_2$ (output low). Because the charging resistance is larger, the charge (high) time is longer than the discharge (low) time, so the output is high for more than half of each cycle and the duty cycle exceeds 50 per cent.

Making it nearer 50 per cent (1 mark): make $R_1$ very small compared with $R_2$ , or add a diode across $R_2$ so charge and discharge use separate, equal paths.

Markers reward the unequal charge and discharge resistances giving a longer high time, and a valid method to equalise them.

Related dot points

Sources & how we know this

WJEC Eduqas GCSE (9-1) Electronics specification (C490) — WJEC Eduqas (2017)