Binary-weighted-input DAC with inverting op-amp feedback A binary-weighted-input DAC uses an op-amp with feedback resistor Rf = 12 kΩ. If 50 µA flows through Rf into the summing node, what is the output voltage of the inverting stage?

Introduction / Context:Binary-weighted DACs often use an inverting op-amp summer. The digital code generates a current proportional to the code, which is then converted to a voltage by the feedback resistor. The sign and magnitude of the output are governed by op-amp inverting configuration rules.

Given Data / Assumptions:

• Feedback resistor Rf = 12 kΩ.
• Current into the summing node through Rf is I = 50 µA (flows from output through Rf into the node).
• Ideal op-amp assumptions: virtual ground at summing junction, linear operation.

Concept / Approach:For an inverting current-to-voltage conversion, Vout = – I * Rf. The negative sign arises because current entering the summing node through Rf must be supplied by the op-amp output with opposite polarity.

Step-by-Step Solution:Use formula → Vout = – I * Rf.Substitute → I = 50 µA = 50 × 10^-6 A; Rf = 12 kΩ = 12 × 10^3 Ω.Compute → Vout = – (50 × 10^-6) * (12 × 10^3) = – 0.6 V.Sign check → Inverting stage outputs negative voltage for positive input current into the summing node.

Verification / Alternative check:Dimensional check: amperes * ohms = volts. 50e-6 * 12e3 = 0.6 V; sign negative as expected for inverting mode.

Why Other Options Are Wrong:

• 0.6 V and 0.1 V: Ignore the inversion and/or magnitude.
• –0.1 V: Wrong magnitude; would correspond to I = 8.33 µA.

Common Pitfalls:Confusing current direction or forgetting the negative sign of the inverting configuration.

Final Answer:–0.6 V