Weighted-resistor DAC design: In a 4-bit weighted-resistor DAC, if the MSB input is applied to a 20 kΩ resistor, determine the resistor value used for the LSB branch (assume ideal binary weighting).

Introduction / Context:
Weighted-resistor DACs implement binary weighting by assigning different resistor values to each bit input. The MSB has the largest weight (largest contribution), which corresponds to the smallest resistor. The LSB has the smallest weight and therefore the largest resistor.

Given Data / Assumptions:

• 4-bit DAC with bits b3 (MSB), b2, b1, b0 (LSB).
• MSB branch uses 20 kΩ.
• Ideal binary weighting: each lower-significance bit contributes half the previous bit’s current.

Concept / Approach:
In a simple weighted-resistor summing DAC, branch current is inversely proportional to branch resistance for a common reference voltage: I ∝ 1/R. To achieve a binary current ratio, each successive bit’s resistor doubles relative to the previous, so that I_LSB = I_MSB / 8 in a 4-bit system (since 2^(4−1) = 8).

Step-by-Step Solution:
Given R_MSB = 20 kΩ for b3.Binary weighting requires R_b2 = 40 kΩ, R_b1 = 80 kΩ, R_b0 = 160 kΩ.Therefore, the LSB resistor is 160 kΩ.

Verification / Alternative check:
Check current ratios: I_MSB : I_b2 : I_b1 : I_LSB = 1 : 1/2 : 1/4 : 1/8, which meets binary weighting when resistors are 20 kΩ, 40 kΩ, 80 kΩ, 160 kΩ.

Why Other Options Are Wrong:
2.5 kΩ and 5 kΩ would overweight the LSB. 80 kΩ is insufficient; it corresponds to b1. 320 kΩ would underweight the LSB beyond binary ratio.

Common Pitfalls:
Mixing up R-2R ladder (uniform R values) with weighted-resistor DACs (varying R values).

Final Answer:
160 kΩ