Question

What is the output voltage \( V_{{out}} \) for the circuit shown below? 

 

Hint:

In a summing amplifier configuration, the output voltage is the weighted sum of the input voltages, where the weights are determined by the ratios of the feedback resistor to the input resistors.

Answer 1

In this case, we have a summing amplifier configuration, where the output voltage \( V_{{out}} \) is given by the formula: \[ V_{{out}} = - \left( \frac{R_f}{R_1} V_1 + \frac{R_f}{R_2} V_2 + \frac{R_f}{R_3} V_3 \right) \] where:
\( R_f = 10 \, k\Omega \) is the feedback resistor
\( R_1 = 5 \, k\Omega \), \( R_2 = 4 \, k\Omega \), and \( R_3 = 2 \, k\Omega \) are the resistors associated with \( V_1 \), \( V_2 \), and \( V_3 \), respectively
\( V_1 = 2 \, {V} \), \( V_2 = 2 \, {V} \), and \( V_3 = 1 \, {V} \)
Step-by-Step Calculation:
1. Calculate the individual contributions to the output voltage: \[ \frac{R_f}{R_1} = \frac{10 \, k\Omega}{5 \, k\Omega} = 2 \] \[ \frac{R_f}{R_2} = \frac{10 \, k\Omega}{4 \, k\Omega} = 2.5 \] \[ \frac{R_f}{R_3} = \frac{10 \, k\Omega}{2 \, k\Omega} = 5 \] 2. Multiply each by the corresponding voltage: \[ 2 \times V_1 = 2 \times 2 \, {V} = 4 \, {V} \] \[ 2.5 \times V_2 = 2.5 \times 2 \, {V} = 5 \, {V} \] \[ 5 \times V_3 = 5 \times 1 \, {V} = 5 \, {V} \] 3. Sum the contributions: \[ V_{{out}} = - \left( 4 \, {V} + 5 \, {V} + 5 \, {V} \right) = -14 \, {V} \] Thus, the output voltage \( V_{{out}} \) is \( -14 \, {V} \).