Difficulty: Easy
Correct Answer: 8 bits
Explanation:
Introduction / Context:There are two classic DAC core topologies: binary-weighted resistor DACs and R–2R ladders. While binary-weighted designs are conceptually simple, they demand precise resistor ratios that become increasingly difficult to realize as resolution increases.
Given Data / Assumptions:
Concept / Approach:Each additional bit doubles the resistor ratio range. By 8 bits, the largest/smallest resistor differ by 128:1. Maintaining accurate ratios across that span with discrete parts or even on-chip passive elements becomes challenging and expensive. Errors lead to DNL/INL problems and potential non-monotonic behavior. Thus, practical binary-weighted DACs rarely exceed about 8 bits; for higher resolution, the R–2R ladder is preferred because it uses only two resistor values, easing matching requirements.
Step-by-Step Solution:
List required values up to N bits: R, 2R, 4R, ... , 2^(N−1)R.Note ratio span at N=8: 2^7 = 128:1 between MSB and LSB resistors.Consider tolerance/temperature drift and switch on-resistance impact larger for small resistors.Conclude practical limit ≈ 8 bits before DNL/INL degrade.Verification / Alternative check:Historical practice: discrete binary-weighted DACs (audio/motion control) top out near 8 bits; higher-resolution DAC ICs predominantly use R–2R or segmented architectures.
Why Other Options Are Wrong:
Common Pitfalls:Assuming “more bits” just means adding more resistors; ignoring that resistor and switch errors scale unfavorably with bit count in this topology.
Final Answer:8 bits
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