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How is a digital code turned back into an analogue voltage, and how is the smooth signal recovered?

Digital-to-analogue conversion: the summing-amplifier (weighted-resistor) DAC, the R-2R ladder DAC, the output equation, and reconstruction with a low-pass filter.

A focused answer to WJEC A-Level Electronics digital-to-analogue conversion, covering the summing-amplifier weighted-resistor DAC, the R-2R ladder DAC, the output voltage equation, and how a low-pass reconstruction filter smooths the stepped output.

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

Once a signal has been processed digitally, it usually has to become analogue again to drive a speaker, motor or display. WJEC expects you to describe the summing-amplifier (weighted-resistor) DAC and the R-2R ladder DAC, use the output equation, compare the two, and explain reconstruction with a low-pass filter. The binary-weighting argument and the R-2R advantage are standard exam content.

The answer

The weighted-resistor (summing) DAC

The output equation

The output is proportional to the digital value, so stepping the code from 0 up to its maximum steps the output up in equal voltage increments. The most significant bit contributes the largest step.

The R-2R ladder DAC

Reconstruction with a low-pass filter

The DAC output is a staircase of voltage steps, which contains high-frequency components that were not in the original signal. A low-pass reconstruction filter removes these, smoothing the steps into the continuous analogue waveform.

Examples in context

Example 1. Playing back digital audio
A music player's DAC turns each 16-bit sample into a voltage, stepping through values thousands of times a second, and a low-pass reconstruction filter smooths the staircase into the analogue waveform that drives the headphones. Without the filter you would hear high-frequency artefacts from the steps.
Example 2. Generating a waveform from a microcontroller
A microcontroller can output a changing digital code to an R-2R DAC to synthesise a sine or sawtooth wave. The R-2R ladder is favoured here because it needs only two resistor values, so it can even be built cheaply from ordinary resistors on a breadboard.
Example 3. Setting a control voltage
A DAC lets a digital system set a precise analogue control voltage, for example to bias an amplifier or set a motor speed reference. The output equation makes the relationship exact: each increment of the digital code raises the control voltage by one predictable step.

Try this

Q1. State the main advantage of the R-2R ladder DAC over the weighted-resistor DAC. [1 mark]

  • Cue. It uses only two resistor values (R and 2R), which are far easier to match precisely, especially in an integrated circuit.

Q2. Explain why a low-pass filter is placed on the output of a DAC. [2 marks]

  • Cue. The DAC output is a staircase of steps containing high-frequency components; the low-pass filter removes these, smoothing the steps into a continuous analogue waveform.

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.

WJEC Eduqas 20215 marksA 4-bit weighted-resistor DAC uses a summing amplifier. The most significant bit feeds a 10kΩ10\,\text{k}\Omega resistor, and each less significant bit feeds a resistor double the previous one. With a feedback resistor and a logic high of 5.0V5.0\,\text{V}, explain why the resistors are binary-weighted and state the relative contribution of each bit.
Show worked answer →

In a weighted-resistor DAC each bit feeds the virtual earth of a summing amplifier through its own resistor, so each bit's current is its voltage divided by its resistor.

For the output to follow the binary value, each bit must contribute in proportion to its place value: the most significant bit (worth 8) must contribute twice the next bit (worth 4), and so on.

Because current is inversely proportional to resistance, doubling the resistor halves the current. So the MSB has the smallest resistor (10kΩ10\,\text{k}\Omega), the next bit 20kΩ20\,\text{k}\Omega, the next 40kΩ40\,\text{k}\Omega and the LSB 80kΩ80\,\text{k}\Omega, giving currents in the ratio 8:4:2:1, exactly the binary place values.

Markers reward the virtual-earth summing of bit currents, the inverse current-resistance relationship, and the 8:4:2:1 weighting matching the binary places.

WJEC Eduqas 20194 marksState one disadvantage of the weighted-resistor DAC and explain how the R-2R ladder DAC overcomes it. Also explain why the DAC output is passed through a low-pass filter.
Show worked answer →

The weighted-resistor DAC needs resistors over a very wide range of values (for many bits the largest is thousands of times the smallest), and these must be precisely matched, which is hard and expensive to manufacture accurately.

The R-2R ladder DAC uses only two resistor values, R and 2R, repeated, which are far easier to match precisely on a chip, so it scales to many bits with good accuracy.

The DAC output is a series of voltage steps (a staircase), which contains high-frequency components not in the original signal. A low-pass reconstruction filter smooths these steps into the continuous analogue waveform, removing the stepping.

Markers reward the wide-range matched-resistor problem, the two-value R-2R solution, and the smoothing role of the low-pass reconstruction filter.

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