If Vref is the output reference value and Vfs is the ideal full-scale output voltage ...

If $V_{ref}$ is the output reference value and $V_{fs}$ is the ideal full-scale output voltage when all digital inputs are 1. Which of the following equations is correct?

A. $V_{ref} = V_{fs}\left(1-\frac{1}{2^n}\right)$

B. $V_{fs} = V_{ref}\left(\frac{1}{2^n}-1\right)$

C. $V_{fs} = V_{ref}\left(1-\frac{1}{2^n}\right)$

D. $V_{ref} = V_{fs}\left(\frac{1}{2^n}-1\right)$

The correct answer is C:

$\boxed{V_{fs}=V_{ref}\left(1-\frac{1}{2^n}\right)}$

For an n-bit DAC, the largest digital input is:

$111\ldots111 = 2^n-1$

The DAC output is:

$V_o=V_{ref}\frac{\text{Digital input}}{2^n}$

At the maximum input:

$V_{fs}=V_{ref}\frac{2^n-1}{2^n}$

Separate the fraction:

$V_{fs}=V_{ref}\left(\frac{2^n}{2^n}-\frac{1}{2^n}\right)$

Therefore:

$\boxed{V_{fs}=V_{ref}\left(1-\frac{1}{2^n}\right)}$

So Option C

Notice that full-scale output is slightly less than $V_{ref}$ .

For example, for an 8-bit DAC with $V_{ref}=5V$ :

$V_{fs}=5\left(1-\frac{1}{256}\right)$

$V_{fs}=5(0.99609375)$

$\boxed{V_{fs}=4.9805V}$

So even though $V_{ref}=5V$ , the maximum ideal DAC output is 4.9805 V, not exactly 5 V.

Easy memory trick:
For an $n$ -bit DAC:

$\boxed{V_{fs}=V_{ref}\frac{2^n-1}{2^n}}$

$\boxed{V_{fs}=V_{ref}\left(1-\frac{1}{2^n}\right)}$

Both are the same formula.

Further Reading: GATE EC Previous Year Papers with Solutions