4-bit R/2R DAC with 5 V reference For a 4-bit R/2R ladder digital-to-analog converter using a 5 V reference, determine the analog output corresponding to the binary input code 0101 (bit order as wired in typical R/2R diagrams where the rightmost bit is the MSB).

Introduction / Context:
This problem assesses your understanding of how a binary-weighted R/2R ladder digital-to-analog converter (DAC) translates a 4-bit input code into a proportional analog output based on a given reference voltage. Recognizing bit weights and the step size is essential for quick mental calculations in instrumentation and embedded systems.

Given Data / Assumptions:

• 4-bit R/2R DAC.
• Reference voltage Vref = 5 V.
• Input code = 0101 (with the rightmost bit taken as the MSB per the specified wiring note).
• Unipolar operation; output is proportional to digital code.

Concept / Approach:
The ideal unipolar DAC output (non-inverting form) is Vout = (DecimalCode / 2^N) * Vref, where N is the number of bits. The LSB step size is Vref / 2^N.

Step-by-Step Solution:

Verification / Alternative check:
Weights for a 4-bit code are 8, 4, 2, 1. With the stated ordering, 0101 corresponds to 8 + 2 = 10; 10 steps * 0.3125 V per step = 3.125 V.

Why Other Options Are Wrong:

• 0.3125 V: Only one LSB; does not match code weight 10.
• 0.78125 V: Equals 2.5 LSB steps; not a valid 4-bit output level for this code.
• –3.125 V: A negative output would require an inverting configuration or bipolar mode, which is not stated.

Common Pitfalls:

• Confusing MSB/LSB ordering when reading the code.
• Using 2^N − 1 in the denominator; for step size, use 2^N.

Final Answer:
3.125 V