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Question

The full scale output of a 10-bit DAC is 5V. The resolution is :

This question was previously asked in
UGC NET 2015 Paper 1 Question Paper (27-Dec-2015)
The correct answer is

5 mV

What resolution means. It is the analogue change produced by a one-count change of the digital input — the height of a single step in the DAC's staircase output:

\(\text{Resolution}=\dfrac{V_{FS}}{2^{n}-1}\)

Step 1 — count the steps. A 10-bit converter has

\(2^{10}=1024\ \text{codes},\ \text{from }0000000000\text{ to }1111111111\)

Those 1024 codes are separated by 1023 steps, which is why the denominator carries the −1.

Step 2 — divide.

\(\text{Resolution}=\dfrac{5}{1023}=4.888\times10^{-3}\ \text{V}\)

\(\approx 4.89\ \text{mV}\approx \mathbf{5\ mV}\)

which is option 1.

The two conventions, and why they do not change the answer. Some texts divide by \(2^{n}\) rather than \(2^{n}-1\):

\(\dfrac{5}{1024}=4.883\ \text{mV}\)

The two differ by one part in a thousand, so both round to 5 mV. The distinction only matters when deciding whether "full scale" means the highest code's output or the full range of the reference — for a 10-bit part the gap is a tenth of a percent.

A fast mental estimate. Since \(2^{10}\approx10^{3}\), a 10-bit converter divides its span into roughly a thousand parts, so the step is about one thousandth of full scale:

\(\dfrac{5\ \text{V}}{1000}=5\ \text{mV}\)

This "10 bits ≈ 0.1 %" rule places the answer without a calculator.

Where the distractors come from. 2.5 mV would need 11 bits, and 10 mV would come from 9 bits — each bit halves or doubles the step. 20 mV corresponds to 8 bits.

Resolution is not accuracy. Resolution is what the converter can distinguish; accuracy is how close the output is to the ideal value, and is limited by offset, gain error and non-linearity. A 10-bit DAC with 5 mV resolution may still be several millivolts out because of those errors, so a data sheet always quotes both.

Hence, the resolution is 5 mV.

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