You write: “If your goal is unity gain, then you’re just wasting an op-amp.”
That is incorrect. You might need a buffer.

I’m glad you found it useful.

Thank you, Hans, for your comment. Your form shows the sum of currents in the non-inverting input, multiplied by the equivalent resistance of all the resistors in the same input. I see that it renders a different view of how the circuit works, especially for the N inputs. I will think of a way to include this in the article.

Nice article, I like how you use superposition. The essential dependencies can be made clearer if you rewrite the formulas somewhat. In particular, your equation (1) can be written as

Vout = (1 + Rf2/Rf1) . (V1 . 1/R1 + V2 . 1/R2) . R1||R2

Likewise, for 3 inputs, equation (8) can with a little work be rewritten as

Vout = (1 + Rf2/Rf1) . (V1 . 1/R1 + V2 . 1/R2 + V3 . 1/R3) . R1||R2||R3

and for n inputs (I hope the alignment shows as intended)

n n
Vout = (1 + Rf2/Rf1) . Sigma(Vj . 1/Rj) / Sigma(1/Rj)
j=1 j=1

I feel this form better reflects the application of this circuit as weighted sum, where the weight of each input is simply the conductance if its input resistor divided by the total conductance of all input resistors.

Rgds, Hans

Intractable? Maybe. As with anything involving math, for some people will be easy for some other will be difficult. Take a look at Matt’s comment above. He proposes a spreadsheet to play with values. I can also tell you that a Mathcad file will also make it easy. I wrote the N input article to have fun with it and to show people that it is possible to find a general formula. I did not find this anywhere else.

For 2, 3, or 4 inputs this is really easy. I used this circuit many times in my career to mix signals, each one at a different amplitude level. You are right that if all resistors are equal you get the signals’ average. I described it in this article, MasteringElectronicsDesign.com: Using the Summing Amplifier as an Average Amplifier. However, this is not the only application. A more important application is creating a weight function where some inputs have more weight than others and this is what this article is about.

However, the special case where all input resistors Ri are equal is A LOT simpler to calculate, and appears to give the result, for V1 only, Vp=V1/N. Hence, Vp=sum(Vi/N)=1/N * Sum(Vi). Effectively, the input to non-inverting summing amplifier reduces to *the average* of input signals.

Thank you. Please click the like button.