What are power-of-ten errors?

Convert k and μF before substituting into the frequency equation.

A 555 astable has R_1 = 1.0\,k, R_2 = 10\,k and C = 1.0\,μF. Calculate the frequency. [3 marks]

WJEC Eduqas style-style practice: A 555 astable uses R_1 = 1.0\,k, R_2 = 6.8\,k and C = 10\,μF. Calculate the output frequency.

A Component 2 Calculate question on the astable frequency. Use f = 1.44(R_1 + 2R_2)C with consistent units (1 mark for the equation). The bracket is R_1 + 2R_2 = 1.0 + 2 × 6.8 = 1.0 + 13.6 = 14.6\,k = 14\,600\, (1 mark). With C = 10 × 10^-6\,F: f = 1.4414\,600 × 10 × 10^-6 = 1.440.146 = 9.9\,Hz (2 marks for the substitution and the answer in hertz). Markers reward the equation, the R_1 + 2R_2 term and the frequency. A common error is to forget to double R_2.

WalesElectronicsSyllabus dot point

How does a 555 astable generate a continuous stream of pulses, and how do we find its frequency and mark-space ratio?

The 555 timer in astable mode: how it produces a continuous square-wave output, the equations for frequency and period, and the meaning and calculation of the mark-space ratio.

A focused answer to WJEC Eduqas GCSE Electronics on the 555 astable timer, covering how it produces a continuous square-wave output, the equations for frequency and period, and the mark-space ratio.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-15

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

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Quick answer

An astable circuit has no stable state: it switches its output continuously between high and low on its own, producing a square wave. In a 555 astable, a timing capacitor repeatedly charges (through $R_1$ and $R_2$ ) and discharges (through $R_2$ ), giving the steady oscillation. The frequency is $f = \dfrac{1.44}{(R_1 + 2R_2)C}$ , and the period is $T = \dfrac{1}{f}$ . The mark is the time the output is high and the space is the time it is low; the mark-space ratio is the ratio of these two times. Because the capacitor charges through $R_1 + R_2$ but discharges through only $R_2$ , the mark is normally longer than the space. The astable is used wherever a continuous clock or flashing/beeping signal is needed.

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What this topic is asking
What an astable does
How the oscillation works
Frequency and period
Mark-space ratio
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What this topic is asking

WJEC Eduqas wants you to know the 555 timer in astable mode: how it produces a continuous square-wave output (a steady stream of pulses), how to calculate its frequency and period, and what the mark-space ratio means and how to find it.

What an astable does

"Astable" means "not stable in either state": the output never settles, but flips back and forth on its own. This continuous square wave is the opposite of the monostable's single pulse. An astable is used as a clock to drive counters, and to make things flash or beep at a regular rate. The rate is set by the timing components.

How the oscillation works

The capacitor's voltage cycles between one-third and two-thirds of the supply: while it charges (through $R_1 + R_2$ ) the output is high (the mark); while it discharges (through $R_2$ only) the output is low (the space). Because charging goes through more resistance than discharging, the high time is longer than the low time. Each full charge-discharge cycle is one period of the square wave.

Frequency and period

The frequency tells you how many pulses occur per second. Larger resistors or a larger capacitor give a lower frequency (a longer period), because the capacitor takes longer to charge and discharge. The most common slip is forgetting to double $R_2$ in the bracket. The period $T = \dfrac{1}{f}$ is the time for one complete cycle, the mark plus the space.

Mark-space ratio

The mark-space ratio describes the shape of the square wave: a ratio of 1 means equal high and low times (a symmetrical square wave), while a larger ratio means the output is high for longer than it is low. For the 555, the high time depends on $R_1 + R_2$ and the low time on $R_2$ , so the mark is always somewhat longer than the space. To get close to a 1:1 (50%) waveform, choose $R_1$ much smaller than $R_2$ so the two paths are similar.

Finding frequency, period and mark-space behaviour

A 555 astable uses $R_1 = 2.2\,\text{k}\Omega$ , $R_2 = 10\,\text{k}\Omega$ and $C = 4.7\,\mu\text{F}$ . Find the frequency and period, and say whether the mark or the space is longer.

step 1: Find the bracket term

$R_1 + 2R_2 = 2.2 + 2 \times 10 = 2.2 + 20 = 22.2\,\text{k}\Omega = 22\,200\,\Omega$ .

step 2: Find the frequency

Convert $C = 4.7 \times 10^{-6}\,\text{F}$ . Then $f = \dfrac{1.44}{22\,200 \times 4.7 \times 10^{-6}} = \dfrac{1.44}{0.1043} = 13.8\,\text{Hz}$ .

step 3: Find the period

$T = \dfrac{1}{f} = \dfrac{1}{13.8} = 0.072\,\text{s}$ (about $72\,\text{ms}$ ).

step 4: Mark or space

The capacitor charges through $R_1 + R_2 = 12.2\,\text{k}\Omega$ (high time) but discharges through only $R_2 = 10\,\text{k}\Omega$ (low time), so the mark (high) is longer than the space (low). The frequency is about $14\,\text{Hz}$ with a period of about $72\,\text{ms}$ .

Try this

Q1. A 555 astable has $R_1 = 1.0\,\text{k}\Omega$ , $R_2 = 10\,\text{k}\Omega$ and $C = 1.0\,\mu\text{F}$ . Calculate the frequency. [3 marks]

Cue. $R_1 + 2R_2 = 21\,\text{k}\Omega = 21\,000\,\Omega$ ; $f = \dfrac{1.44}{21\,000 \times 1.0 \times 10^{-6}} = \dfrac{1.44}{0.021} = 69\,\text{Hz}$ .

Q2. State what the period of a square wave represents. [1 mark]

Cue. The time for one complete cycle (the mark plus the space), $T = \dfrac{1}{f}$ .

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 astable uses

R_1 = 1.0\,\text{k}\Omega

R_2 = 6.8\,\text{k}\Omega

and

C = 10\,\mu\text{F}

. Calculate the output frequency.

Show worked answer →

A Component 2 Calculate question on the astable frequency. Use $f = \dfrac{1.44}{(R_1 + 2R_2)C}$ with consistent units (1 mark for the equation). The bracket is $R_1 + 2R_2 = 1.0 + 2 \times 6.8 = 1.0 + 13.6 = 14.6\,\text{k}\Omega = 14\,600\,\Omega$ (1 mark). With $C = 10 \times 10^{-6}\,\text{F}$ : $f = \dfrac{1.44}{14\,600 \times 10 \times 10^{-6}} = \dfrac{1.44}{0.146} = 9.9\,\text{Hz}$ (2 marks for the substitution and the answer in hertz). Markers reward the equation, the $R_1 + 2R_2$ term and the frequency. A common error is to forget to double $R_2$ .

Eduqas style3 marksExplain what is meant by the mark-space ratio of an astable output, and state how to make the mark and the space more nearly equal.

Show worked answer →

A Component 2 Explain question on mark-space ratio. The mark is the time the output is high and the space is the time it is low; the mark-space ratio is the ratio of these two times (1 mark). For a 555 astable the mark (high time, set by charging through $R_1 + R_2$ ) is longer than the space (low time, set by discharging through $R_2$ ), so the ratio is greater than 1 (1 mark). To make the mark and space more nearly equal, make $R_1$ small compared with $R_2$ , so the charging and discharging paths are more similar (1 mark). Markers reward mark = high time, space = low time, and reducing $R_1$ relative to $R_2$ to even them out. A common error is to swap mark and space.

Related dot points

Sources & how we know this

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