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Binary-weighted DAC — In a classic binary-weighted-resistor DAC, the least significant bit (LSB) input connects to which resistor value?

Difficulty: Easy

Correct Answer: connects to the largest resistor

Explanation:


Introduction / Context:
Binary-weighted DACs implement bit weights using resistors in powers of two (R, 2R, 4R, 8R, ...). The mapping of each bit to a resistor value determines the contribution of that bit to the output current or voltage.

Given Data / Assumptions:

  • Most significant bit (MSB) must contribute the largest weight.
  • Smaller resistance in a current-summing DAC yields more current (greater weight), all else equal.
  • Thus resistor magnitude is inversely related to bit weight.


Concept / Approach:
To realize binary weighting, the MSB is tied to the smallest resistor (highest weight). Conversely, the least significant bit is connected to the largest resistor (lowest weight). This ensures that each less significant bit contributes half the current (or voltage) of the previous bit.

Step-by-Step Solution:

Assign MSB → smallest resistor (weight = 1/2 of full-scale step basis inverted by R).Assign LSB → largest resistor (1/2^N of MSB weight).Confirm summing network yields binary-weighted output.


Verification / Alternative check:

Standard DAC texts show LSB on the largest resistor for binary-weighted designs.


Why Other Options Are Wrong:

connects to the smallest resistor: That would make the LSB too large in weight.supplies the least voltage: Varies by topology; the defining feature is resistor value/weighting.connects to a 1 kΩ resistor: Arbitrary value; depends on design scale.


Common Pitfalls:

Confusing binary-weighted DACs with R/2R ladders, which use only two resistor values.


Final Answer:

connects to the largest resistor
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