QB_PH_SEC8_018: Operational Amplifiers & Analog Circuits

An ideal summing amplifier circuit has three input resistors $R_1 = 10\text{ k}\Omega$, $R_2 = 20\text{ k}\Omega$, $R_3 = 50\text{ k}\Omega$ connected to voltages $V_1 = +2.0\text{ V}$, $V_2 = -4.0\text{ V}$, and $V_3 = +5.0\text{ V}$ respectively. If the feedback resistor is $R_f = 100\text{ k}\Omega$, calculate the output voltage $V_{\text{out}}$ in Volts.

Correct Answer

### Step-by-Step Derivation & Verification: 1. **KCL at Inverting Virtual Ground Node ($V_- = 0$)**: $\frac{V_1 - 0}{R_1} + \frac{V_2 - 0}{R_2} + \frac{V_3 - 0}{R_3} + \frac{V_{\text{out}} - 0}{R_f} = 0$ 2. **Output Voltage Expression**: $V_{\text{out}} = -R_f \left[\frac{V_1}{R_1} + \frac{V_2}{R_2} + \frac{V_3}{R_3}\right]$ 3. **Evaluation of Branch Currents**: - $I_1 = \frac{2.0\text{ V}}{10\text{ k}\Omega} = +0.20\text{ mA}$ - $I_2 = \frac{-4.0\text{ V}}{20\text{ k}\Omega} = -0.20\text{ mA}$ - $I_3 = \frac{5.0\text{ V}}{50\text{ k}\Omega} = +0.10\text{ mA}$ $\sum I_{\text{in}} = 0.20 - 0.20 + 0.10 = 0.10\text{ mA}$ 4. **Output Voltage**: $V_{\text{out}} = -R_f \sum I_{\text{in}} = -(100\text{ k}\Omega)(0.10\text{ mA}) = -10.0\text{ V}$ 5. **Official Answer**: -10.