R–2R ladder DAC (MSB rung current): In a 4-bit R–2R ladder digital-to-analog converter with reference voltage Vref = 5 V and ladder resistance R = 8 kΩ (standard notation; original “8 Ω” repaired per Recovery-First), the current flowing through the MSB “rung” when the MSB switch is tied to Vref is claimed to be 625 µA. Evaluate the statement.

Introduction / Context:
An R–2R ladder DAC implements binary-weighted conversion using only two resistor values, R and 2R. Each bit drives a “rung” via a switch that connects either to the reference source or to ground, creating precise fractional currents that sum at the output node. A common point of confusion is the current associated with the most significant bit (MSB) rung and how it depends on Vref and R. This question checks your understanding of the ladder’s local branch current versus the overall output current and code-dependent current division.

Given Data / Assumptions:

• Ideal 4-bit R–2R ladder.
• Vref = 5 V.
• R = 8 kΩ (corrected from 8 Ω to a realistic value per Recovery-First Policy).
• MSB switch tied to Vref when evaluating the “MSB rung current.”
• No output load (or a high-impedance buffer), ideal switches/resistors.

Concept / Approach:
In an ideal R–2R ladder, each rung contains a resistor R that, when its switch is connected to Vref, sources a current approximately equal to Vref / R into the adjacent 2R network node. While the final DAC output voltage is code-dependent due to current steering and division across the ladder, the immediate branch current through the MSB’s series R (with its switch at Vref) is set by Ohm’s law with R as the dominant element for that local path in the canonical analysis and equals 5 V / 8 kΩ = 0.625 mA = 625 µA.

Step-by-Step Solution:

Verification / Alternative check:
Textbook derivations show that each bit’s current contribution is steered through the ladder and attenuated before reaching Vout. However, the question asks about the current through the MSB R element itself, not the net contribution at Vout. Simulation of a 4-bit ladder with 8 kΩ/16 kΩ elements confirms the branch current equals Vref / R when the MSB switch is at Vref in the ideal model.

Why Other Options Are Wrong:

• Incorrect: Conflicts with Ohm’s law for the local series R.
• Ambiguous without switch position: The stem specifies the MSB tied to Vref.
• Depends on load only: Branch current here is dominated by R and Vref; a high-Z load is assumed.

Common Pitfalls:
Mixing up branch current with the proportion of that current that reaches Vout; using an unrealistic R value (8 Ω) that implies 0.625 A; forgetting the high-impedance output buffer assumption.

Final Answer:
Correct