Difficulty: Easy
Correct Answer: 4-bit D/A converters
Explanation:
Introduction / Context:
Binary-weighted DACs implement each digital bit with a resistor weighted in powers of two. This simple idea runs into real-world limitations as resolution increases. Understanding why helps learners choose the right DAC topology for a given application.
Given Data / Assumptions:
Concept / Approach:
As bit count N grows, the largest resistor must be 2^(N-1)*R. Tight matching across a wide range becomes difficult and expensive. Parasitics and switch resistance distort higher-order bits. Thermal drift and tolerance stack-up degrade linearity and monotonicity. Consequently, beyond a few bits, the R/2R ladder topology (only two values, R and 2R) is preferred for scalability.
Step-by-Step Solution:
List required resistor values for N = 4: R, 2R, 4R, 8R (manageable).For N = 8: values up to 128R are needed; matching across 7 octaves is difficult.Switch on-resistance adds a fixed series element that disproportionately affects small currents in high-value branches.Industry and teaching labs typically limit binary-weighted demos to about 4 bits for predictable accuracy.
Verification / Alternative check:
Compare integral and differential nonlinearity expectations: for loose 1% resistors, errors quickly exceed 1 LSB above 4 bits; even 0.1% can struggle at higher N without trimming.
Why Other Options Are Wrong:
R/2R ladder DACs (option A) are a different topology and scale well.
8-bit binary-weighted (option C) is theoretically possible but impractical without precision trimming.
Op-amp comparators (option D) are not DACs.
Current-steering DACs (option E) are used for high resolution, not a limitation case.
Common Pitfalls:
Assuming resistor tolerance alone determines accuracy; switch on-resistance and reference stability also matter.
Final Answer:
4-bit D/A converters
Discussion & Comments