Binary-weighted DAC (4-bit) resistor sizing: If the least significant bit (LSB) of a four-bit binary-weighted resistor D/A converter uses a 200 Ω resistor, determine whether the resistor for the next more significant bit should be 100 Ω.

Introduction / Context:
Binary-weighted D/A converters generate an analog output by summing currents (or voltages) that are proportional to binary bit weights. Each more significant bit must contribute exactly twice the effect of the next less significant bit. The easiest way to achieve this with a resistor-weighted DAC is to halve the resistance for each step toward the most significant bit (MSB), thereby doubling the current for a given reference voltage. This item checks your understanding of that geometric scaling rule.

Given Data / Assumptions:

• Architecture: binary-weighted resistor DAC (not an R-2R ladder).
• Four-bit resolution: bits are LSB, next bit, next bit, MSB.
• LSB resistor value is 200 Ω, driven by the same Vref as other bits.
• Ideal switches and op-amp summing node assumed for clarity.

Concept / Approach:
For a resistor-weighted DAC that sums currents into a virtual ground, bit current is I_bit = Vref / R_bit when the switch is ON. To double the contribution when moving one position toward the MSB, we must double the current. With a constant Vref, doubling current requires halving the resistance. Hence, if the LSB uses 200 Ω, the next bit should use 100 Ω, then 50 Ω, and so on, reaching the smallest value at the MSB.

Step-by-Step Solution:

Verification / Alternative check:
If the four resistors are 200 Ω (LSB), 100 Ω, 50 Ω, and 25 Ω (MSB), their ON currents form 1:2:4:8, exactly matching binary weights. Summing these currents via an op-amp produces a linear transfer characteristic (ignoring switch/amp nonidealities).

Why Other Options Are Wrong:

Common Pitfalls:
Confusing the binary-weighted scheme with the R-2R ladder (which does not use powers-of-two resistors) and assuming Vref changes per bit. In a proper design, Vref is fixed and resistor values enforce weighting.

Final Answer:
Correct