Topic: Analog Circuits
In a given op-amp circuit, an input current $I_{in}$ flows into a node connecting the non-inverting terminal and a resistor $R_s$ to ground. The output $V_0$ is fed back to the inverting terminal via a voltage divider consisting of $R_A$ and $R_B$. Find the output voltage $V_0$.
- A. $(R_A/R_B) \times I_{in} \times R_s$
- B. $I_{in} \times R_s \times (1 + R_A/R_B)$
- C. $(R_B/R_A) \times I_{in} \times R_s$
- D. $I_{in} \times R_s \times (1 + R_B/R_A)$
Answer: B. $I_{in} \times R_s \times (1 + R_A/R_B)$
Explanation: The voltage at the non-inverting terminal is $V_p = I_{in} \times R_s$. By the virtual short concept, the inverting terminal voltage $V_n$ equals $V_p$. The feedback network is a simple voltage divider, meaning $V_n = V_0 \frac{R_B}{R_A + R_B}$. Equating the two, we get $I_{in} \times R_s = V_0 \frac{R_B}{R_A + R_B}$. Solving for $V_0$ yields $V_0 = I_{in} \times R_s \times (1 + \frac{R_A}{R_B})$.