Topic: Analog Circuits
Given the output for the following non-inverting summing amplifier, the relation between $R_{f}$ and R in the circuit is
(The diagram shows a non-inverting summer with 4 input voltages $V_{1}$ to $V_{4}$ connected via identical resistors $R$ to the non-inverting terminal. The output is $V_{o}=(V_{1}+V_{2}+V_{3}+V_{4})/2$.)
- A. $R_{f}=R$
- B. $R_{f}=4R$
- C. $R_{f}=2R$
- D. $R_{f}=R/2$
Answer: A. $R_{f}=R$
Explanation: Voltage at the non-inverting node A is $V_{A} = \frac{R/3}{R + R/3} \dots$ wait, by KCL at the non-inverting node: $\frac{V_{1}-V_{A}}{R} + \frac{V_{2}-V_{A}}{R} + \frac{V_{3}-V_{A}}{R} + \frac{V_{4}-V_{A}}{R} = 0 \Rightarrow 4V_{A} = V_{1}+V_{2}+V_{3}+V_{4} \Rightarrow V_{A} = \frac{V_{1}+V_{2}+V_{3}+V_{4}}{4}$.
For a non-inverting amplifier, $V_{o} = \left(1 + \frac{R_{f}}{R}\right) V_{A}$.
$V_{o} = \left(1 + \frac{R_{f}}{R}\right) \frac{V_{1}+V_{2}+V_{3}+V_{4}}{4}$.
Given $V_{o} = \frac{V_{1}+V_{2}+V_{3}+V_{4}}{2}$, we have $\frac{1 + R_{f}/R}{4} = \frac{1}{2} \Rightarrow 1 + \frac{R_{f}}{R} = 2 \Rightarrow R_{f} = R$.