Practical resolution limit of a binary-weighted resistor DAC: up to how many bits is this approach generally feasible before resistor ratio spread becomes impractical?

Difficulty: Easy

Correct Answer: 8 bits

Explanation:


Introduction / Context:
Binary-weighted resistor DACs require resistor values in powers of two. As resolution increases, the ratio between the smallest and largest resistors grows quickly, stressing tolerance, temperature tracking, and switch on-resistance requirements. This item asks for the commonly cited practical upper limit in bits.



Given Data / Assumptions:

  • Binary-weighted topology (R, 2R, 4R ... 2^(n-1)R).
  • General-purpose, discrete or monolithic resistors with realistic tolerances.
  • Focus on practical, economical implementation rather than extreme custom trimming.


Concept / Approach:
The resistor spread grows exponentially with resolution. For 8 bits, the largest element is 128R relative to the least significant bit branch. Beyond around 8 bits, achieving low integral and differential nonlinearity with untrimmed parts becomes hard and expensive. Hence designers typically shift to R/2R ladders for higher resolution.



Step-by-Step Solution:

Relate resolution to resistor spread: n bits → largest branch is 2^(n-1)R.Check feasibility: matching, temperature coefficients, and switch errors accumulate with spread.Industry practice: about 8 bits is a common, practical ceiling without heavy trimming.Therefore the best choice is 8 bits.


Verification / Alternative check:
Reference application notes consistently recommend R/2R ladders at or above 8 bits due to manufacturability and linearity advantages.



Why Other Options Are Wrong:
10 bits: possible only with tight trims, special processes; not generally practical.

2 bits / 4 bits: these are trivial and not a “limit.”



Common Pitfalls:
Confusing absolute tolerance with ratio matching; the main issue is exponential spread across branches, not just nominal value accuracy.



Final Answer:
8 bits

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