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How are analogue signals converted to digital so a microcontroller can process them?

Analogue and digital signals, the binary number system, analogue-to-digital and digital-to-analogue conversion, resolution and quantisation.

A CCEA A-Level Technology and Design answer on analogue and digital signals, binary numbers, analogue-to-digital (ADC) and digital-to-analogue (DAC) conversion, and how the number of bits sets the resolution and the size of each quantisation step.

Generated by Claude Opus 4.812 min answer

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

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What this dot point is asking

CCEA expects you to distinguish analogue from digital signals, use the binary number system, explain analogue-to-digital (ADC) and digital-to-analogue (DAC) conversion, and calculate resolution and the number of levels from the number of bits. Resolution and "number of levels" calculations are common.

The answer

Analogue and digital signals

Binary and number of levels

Conversion and resolution

Worked example: choosing ADC bits for a measurement

Examples in context

Example 1. Digital thermometer. A thermistor's analogue voltage is digitised by an ADC so the microcontroller can display a number; more ADC bits mean a finer temperature reading.

Example 2. Audio CD. Sound (analogue) is sampled and quantised to 16-bit values; 16 bits give 65,536 levels, fine enough that quantisation error is inaudible, the resolution idea applied to music.

Try this

Q1. How many discrete levels does a 6-bit ADC provide? [1 mark]

  • Cue. 26=642^6 = 64 levels.

Q2. A 12-bit ADC covers 0 to 5 V. Find its resolution. [2 marks]

  • Cue. 212=40962^{12} = 4096; resolution =5/40961.2 mV= 5/4096 \approx 1.2\ \text{mV}.

Q3. Why must a microcontroller use an ADC to read an analogue sensor? [2 marks]

  • Cue. The microcontroller is digital (works in binary) but the sensor's output is a continuous analogue voltage; the ADC converts it to a binary number the microcontroller can process.

Exam-style practice questions

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

CCEA 20196 marksExplain the difference between analogue and digital signals, and explain why an analogue-to-digital converter is needed when a microcontroller reads a temperature sensor. State how many discrete levels an 8-bit ADC provides.
Show worked answer →

An analogue signal varies continuously and can take any value within a range (the voltage from a thermistor divider, which changes smoothly with temperature). A digital signal has only discrete levels (in a binary system, two: logic 0 and 1), and information is represented as binary numbers.

A microcontroller works in binary and can only process digital values, but a temperature sensor gives a continuously varying analogue voltage. An analogue-to-digital converter (ADC) is therefore needed to sample the analogue voltage and convert it into a binary number the microcontroller can read and process.

An 8-bit ADC provides 28=2562^8 = 256 discrete levels (codes 0 to 255).

Markers reward the continuous (analogue) vs discrete (digital) contrast, the reason an ADC is needed (the microcontroller is digital but the sensor is analogue), and 28=2562^8 = 256 levels.

CCEA 20214 marksA 10-bit ADC measures a voltage range of 0 to 5 V. Calculate the resolution (the voltage represented by one step), and state what happens to the resolution if more bits are used.
Show worked answer →

The number of steps for an nn-bit ADC is 2n2^n, and the resolution (size of one step) is the range divided by the number of steps. (Using 2n2^n steps.)

For a 10-bit ADC over 0 to 5 V:

steps=210=1024,\text{steps} = 2^{10} = 1024,

resolution=5 V10240.0049 V4.9 mV.\text{resolution} = \frac{5\ \text{V}}{1024} \approx 0.0049\ \text{V} \approx 4.9\ \text{mV}.

So each step represents about 4.9 mV.

If more bits are used, there are more steps (2n2^n grows), so each step is smaller and the resolution is finer (the digital value follows the analogue signal more closely, with less quantisation error).

Markers want 210=10242^{10} = 1024, the resolution of about 4.9 mV (5 V / 1024), and the statement that more bits give finer resolution.

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