Question

A 4-bit R/2R digital-to-analog (DAC) converter has a reference of 5 V. What is the analog output for the input code 0101?

• A. \( \text{0.3125 V} \)
• B. \( \text{3.125 V} \)
• C. \( \text{0.78125 V} \)
• D. \( \text{31.25 V} \)

Hint:

For an N-bit DAC, use \( V_{\text{out}} = V_{\text{ref}} \times \frac{\text{decimal input}}{2^N} \). Always convert the binary input carefully and verify your final result against the available options.

Answer 1

The Correct Option is B

The output voltage \( V_{\text{out}} \) of an N-bit R/2R DAC is given by: \[ V_{\text{out}} = V_{\text{ref}} \times \frac{\text{Decimal equivalent of input}}{2^N} \] Given:

• \( N = 4 \)
• \( V_{\text{ref}} = 5 \text{ V} \)
• Input code = \( 0101_2 \)

Convert \( 0101_2 \) to decimal: \[ (0 \times 2^3) + (1 \times 2^2) + (0 \times 2^1) + (1 \times 2^0) = 0 + 4 + 0 + 1 = 5 \] \[ V_{\text{out}} = 5 \times \frac{5}{16} = 5 \times 0.3125 = 1.5625 \text{ V} \] This value \( 1.5625 \text{ V} \) is not among the given options. However, if we consider the correct answer to be (B) \( 3.125 \text{ V} \), we can reverse-engineer the input: \[ 3.125 = 5 \times \frac{x}{16} \Rightarrow x = \frac{3.125 \times 16}{5} = 10 \] Decimal 10 corresponds to binary \( 1010_2 \). Therefore, the output of \( 3.125 \text{ V} \) corresponds to the input code \( 1010_2 \), not \( 0101_2 \). Conclusion: There appears to be a typo in the question. If the input code is meant to be \( 1010_2 \), then the correct analog output is: \[ V_{\text{out}} = 5 \times \frac{10}{16} = 3.125 \text{ V} \]