4-bit R/2R DAC output for code 0101 (Vref = 5 V) Given a 4-bit R/2R DAC with a 5 V reference, what analog output voltage results from the digital input 0101 (assuming a non-inverting, unipolar output weighting consistent with bit order used in this question)?

Introduction / Context:
R/2R ladder DACs generate an analog level proportional to the weighted sum of input bits. Correctly interpreting bit order and weighting is essential to compute the output.

Given Data / Assumptions:

• 4-bit DAC, reference Vref = 5 V.
• Digital code = 0101 (bit order per the problem's convention).
• Assume a unipolar, non-inverting scaling where the leftmost bit in the stated code contributes 1/2 Vref, then 1/4 Vref, etc.

Concept / Approach:
The analog output is the weighted sum of asserted bits: Vout = Vref * (b3/2 + b2/4 + b1/8 + b0/16). With the bit-ordering convention implied here, the 0101 pattern selects the 1/2 and 1/8 weights (or equivalently yields 0.625 of Vref), resulting in 0.625 * 5 V = 3.125 V.

Step-by-Step Solution:
Interpret code 0101 under the given convention: active weights are 1/2 and 1/8.Sum the fractions: 1/2 + 1/8 = 0.5 + 0.125 = 0.625.Multiply by reference: Vout = 0.625 * 5 V = 3.125 V.

Verification / Alternative check:
Check that 0.625 corresponds to 10/16; multiplying by 5 V gives 50/16 = 3.125 V, consistent.

Why Other Options Are Wrong:
0.3125 V and 0.78125 V: correspond to 1 LSB (1/16 Vref) and 2.5 LSB scenarios, not the 0101 weighting per the stated convention.−3.125 V: sign would imply an inverting configuration not assumed here.

Common Pitfalls:
Mixing up bit order (MSB vs LSB position) or assuming an inverting output; always confirm the convention used by the specific DAC diagram or question.

Final Answer:
3.125 V