Operational Amplifier (4)

Electricity and Magnetism Level 4

Consider the operational amplifier circuit above, where the OpAmp is connected to n n input voltage sources V i V_i each with a serial resistance R i R_i . Find the non-inverting input voltage V + V_+ in terms of V i V_i and R i R_i .

∑ i = 1 n V i R i ∑ i = 1 n 1 R i \frac{\sum_{i = 1}^{n}\frac{V_i}{R_i}}{\sum_{i = 1}^{n}\frac{1}{R_i}} 1 n ∑ i = 1 n V i \frac{1}{n}\sum_{i = 1}^{n}V_i ∑ i = 1 n V i R i ∑ i = 1 n R i \frac{\sum_{i = 1}^{n}V_i R_i}{\sum_{i = 1}^{n}R_i} ∑ i = 1 n V i \sum_{i = 1}^{n}V_i

Charley Shi by Charley Shi , 19, New Zealand

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2 solutions

Chew-Seong Cheong
Nov 11, 2020

Since the current flowing into the non-inverting input is zero, we have:

V 1 − V + R 1 + V 2 − V + R 2 + V 3 − V + R 3 + ⋯ + V n − V + R n = 0 \begin{aligned} \frac {V_1-V_+}{R_1} + \frac {V_2-V_+}{R_2} + \frac {V_3-V_+}{R_3} + \cdots + \frac {V_n-V_+}{R_n} & = 0 \end{aligned}

∑ i = 1 n V i − V + R i = 0 ∑ i = 1 n V i R i − ∑ i = 1 n V + R i = 0 V + ∑ i = 1 n 1 R i = ∑ i = 1 n V i R i ⟹ V + = ∑ i = 1 n V i R i ∑ i = 1 n 1 R i \begin{aligned} \sum_{i=1}^n \frac {V_i-V_+}{R_i} & = 0 \\ \sum_{i=1}^n \frac {V_i}{R_i} - \sum_{i=1}^n \frac {V_+}{R_i} & = 0 \\ V_+ \sum_{i=1}^n \frac 1{R_i} & = \sum_{i=1}^n \frac {V_i}{R_i} \\ \implies V_+ & = \boxed {\frac { \sum_{i=1}^n \frac {V_i}{R_i}}{ \sum_{i=1}^n \frac 1{R_i}}} \end{aligned}

2 Helpful 1 Interesting 2 Brilliant 0 Confused
Steven Chase
Nov 11, 2020

The current into the non-inverting input is zero.

∑ i = 1 n V i − V + R i = 0 ∑ i = 1 n V i R i = ∑ i = 1 n V + R i = V + ∑ i = 1 n 1 R i \sum_{i=1}^n \frac{V_i - V_+}{R_i} = 0 \\ \sum_{i=1}^n \frac{V_i }{R_i} = \sum_{i=1}^n \frac{V_+ }{R_i} = V_+ \sum_{i=1}^n \frac{1 }{R_i}

Finally:

V + = ∑ i = 1 n V i R i ∑ i = 1 n 1 R i \large{V_+ = \frac{\sum_{i=1}^n \frac{V_i }{R_i}}{\sum_{i=1}^n \frac{1 }{R_i}}}

1 Helpful 1 Interesting 2 Brilliant 0 Confused

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