Topic: Analog Electronics
Output voltage $V_{0}$ of the circuit shown in figure below. (The input voltages are $V_{1}=2.5~V$ and $V_{2}=1~V$)
- A. 4.0 V
- B. -4.0 V
- C. -4.5 V
- D. 4.5 V
Answer: C. -4.5 V
Explanation: Using superposition or nodal analysis at the op-amp terminals:
Voltage at non-inverting terminal $V_+ = V_2 \times \frac{1.5\text{k}}{3.3\text{k} + 1.5\text{k}} = 1 \times \frac{1.5}{4.8} = \frac{1.5}{4.8}\text{ V}$.
Output voltage $V_0 = -V_1 \left(\frac{2.2\text{k}}{1\text{k}}\right) + V_+ \left(1 + \frac{2.2\text{k}}{1\text{k}}\right)$.
$V_0 = -2.5(2.2) + \left(\frac{1.5}{4.8}\right)(1 + 2.2) = -5.5 + \frac{1.5}{4.8}(3.2) = -5.5 + 1 = -4.5\text{ V}$.