Skip to main content
EnglandElectronicsSyllabus dot point

How do flip-flops count clock pulses in binary and divide a frequency?

Counters: chaining flip-flops to count clock pulses in binary, frequency division by each stage, and the modulus of a counter.

An Eduqas GCSE Electronics answer on counters: how chained flip-flops count clock pulses in binary, how each stage divides the frequency by two, the modulus (number of states) of a counter, and using counters to divide frequency and count events.

Generated by Claude Opus 4.813 min answer

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

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

Jump to a section
  1. What this dot point is asking
  2. The answer
  3. Examples in context
  4. Try this

What this dot point is asking

Eduqas wants you to explain how flip-flops are chained to count clock pulses in binary, how each stage divides the frequency by two, and the modulus (number of states) of a counter. Counters are the most common sequential circuit and the bridge from a clock to a useful count or a divided frequency.

The answer

Counting in binary

Frequency division

The modulus of a counter

Uses of counters

Examples in context

Counters are everywhere a digital system needs to keep count or divide time. A digital clock divides a crystal oscillator down to one pulse per second and counts seconds, minutes and hours; a frequency divider produces several timing signals from one clock; a people-counter or a production tally increments on each sensor pulse. The binary count drives a decoder and seven-segment display (the next topic), and the clock that steps the counter is the 555 astable or crystal oscillator from the timing module.

Try this

Q1. State by what factor a single flip-flop divides a frequency. [1 mark]

  • Cue. By two (it toggles at half the input rate).

Q2. How many states does a counter of 5 flip-flops have? [1 mark]

  • Cue. 25=322^5 = 32 states (counting 00 to 3131).

Q3. A 2 kHz2\ \text{kHz} clock drives 4 flip-flops in a chain. Find the output frequency of the last stage. [2 marks]

  • Cue. fout=200024=200016=125 Hzf_\text{out} = \frac{2000}{2^4} = \frac{2000}{16} = 125\ \text{Hz}.

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 20194 marksA counter is built from 4 flip-flops. State the maximum number of different states it can have, and explain how this relates to the number of flip-flops.
Show worked answer →

Number of states (up to 2 marks): each flip-flop stores one bit (two values), and nn flip-flops give 2n2^n combinations, so 4 flip-flops give 24=162^4 = 16 states (counting 00 to 1515).

Explanation (up to 2 marks): the flip-flops together hold an nn-bit binary number; with 4 bits the count runs from 00000000 to 11111111, which is 1616 different values before it rolls over to 00 again.

Markers reward 24=162^4 = 16 states and the link to an nn-bit binary number counting from 00 to 2n12^n - 1.

Eduqas 20224 marksA 1 kHz1\ \text{kHz} clock is applied to a chain of 3 flip-flops, each dividing the frequency by two. Calculate the output frequency of the last flip-flop.
Show worked answer →

Each flip-flop divides the frequency by two, so a chain of 3 divides by 23=82^3 = 8.

Output frequency: fout=fin23=10008=125 Hzf_\text{out} = \dfrac{f_\text{in}}{2^3} = \dfrac{1000}{8} = 125\ \text{Hz}.

Markers reward the division by 2n2^n (here 23=82^3 = 8) and the output frequency of 125 Hz125\ \text{Hz}. The common error is dividing by 33 (the number of stages) instead of 232^3.

Related dot points

Sources & how we know this