Binary-Weighted 4-Bit DAC: Given the LSB input resistor is 32 kΩ, find the MSB input resistor value.

Introduction / Context:
Binary-weighted resistor DACs implement digital-to-analog conversion using input resistors scaled by powers of 2. The most significant bit (MSB) contributes the largest current (smallest resistor), while the least significant bit (LSB) contributes the smallest current (largest resistor). This question asks you to back-calculate the MSB resistor from a given LSB resistor.

Given Data / Assumptions:

• DAC type: 4-bit binary-weighted, op-amp summing configuration.
• Resistor scaling: MSB → R, next → 2R, next → 4R, LSB → 8R.
• Given: LSB resistor = 32 kΩ.

Concept / Approach:

In a binary-weighted DAC, each less significant bit uses a resistor that doubles compared to the next higher bit. Therefore, with 4 bits, the LSB resistor equals 8R if the MSB equals R. Solve 8R = 32 kΩ to find R, which equals the MSB resistor value.

Step-by-Step Solution:

Verification / Alternative check:

Check the set: R = 4 kΩ, 2R = 8 kΩ, 4R = 16 kΩ, 8R = 32 kΩ, which is a correct power-of-two progression for a 4-bit binary-weighted DAC.

Why Other Options Are Wrong:

• 32 kΩ: That is the LSB resistor, not the MSB.
• 16 kΩ and 8 kΩ: These correspond to intermediate bits (4R and 2R respectively) in the scaling sequence.
• 64 kΩ: Would be 16R, not used in a 4-bit binary-weighted scheme.

Common Pitfalls:

• Confusing R-2R ladder DACs with binary-weighted DACs.
• Reversing the scaling (assigning 8R to the MSB).

Final Answer:

4 kΩ