Background: Designing voltage summer without op-amp and Operational Amplifier Adder

I know that op-amp can provide virtual ground for the circuit, and can have benefits in eliminating crosstalk. But the latter part confuses me.

Suppose that voltage sources are modeled as independent voltage source plus resistor. In such a case, resistor can be added to existing resistors to provide total resistance value, and voltage laws to calculate voltage works, so crosstalk does not affect the sum value, as long as resistance value of each source is quite different.

Is the whole concern regarding crosstalk related to different sources having different output impedance? Or am I understanding something completely wrong?

To summarize, I do understand that adding voltage makes the output voltage higher than input voltages, and current flows from high voltage to low voltage. But Kirchoff's voltage laws are still there unless we are talking about distributed circuits, and the calculation using same output impedance for different voltage sources results in an accurate-enough sum. So why active summer over passive summer?


2 Answers 2


The concern about backfeeding/crosstalk only matters if:

  • The source(s) in question have enough output impedance to detect the unwanted signal
  • The signal splits off from the source(s) and goes somewhere else in addition to the summer

If either of those is false, then you don't care about backfeeding/crosstalk.

If you don't care as defined above, and you want an average voltage, then the passive adder is usually the better solution. If you do care, or if you actually want to add and not average, then you need active.

The reason you need active to actually add signals is because the result could easily be higher than all of the inputs individually. Since you can't get that signal swing from the inputs, it has to come from somewhere else, like a power supply. That's pretty much active by definition. You could feed a passive averager into an amplifier, but that's two separate modules that retain all of their own properties, both good and bad.

You might know this, but it's not clear from your question: The output from an active adder is negative, or inverted. A positive sum is presented as a negative voltage and vice-versa. This is how KCL and KVL still work. If you really want a non-inverting summer, then you need to invert all the inputs (sometimes easier) or invert the output as a separate inverting amplifier/summer with one input.

It often helps my understanding of a particular concept to think of something I already know as a special case of what I'm trying to understand. In this case, a voltage divider can be thought of as a passive averager between the signal and ground. Adjusting the divider ratio and translating that back into a passive summer should show you how to make a weighted average.

  • \$\begingroup\$ Let's say that source signal goes only to the summer and not anywhere else. But I still want to obtain sum. You said that if I want to add, then I need active. But one of the conditions you presented is false, so it seems passive is fine. Am I misunderstanding something? \$\endgroup\$
    – Lay Kruzvk
    Commented Sep 10, 2015 at 13:29
  • \$\begingroup\$ @LayKruzvk: If one of those conditions is false, then you don't care about crosstalk. That's different from wanting to add or average. [copy/paste]: If you don't care as defined above, and you want an average voltage, then the passive adder is usually the better solution. If you do care, or if you actually want to add and not average, then you need active. \$\endgroup\$
    – AaronD
    Commented Sep 10, 2015 at 15:11
  • \$\begingroup\$ OK. So I want to add, not just average, and thus need an active summer regardless of conditions. But still, why in this case? Why do we require ative? \$\endgroup\$
    – Lay Kruzvk
    Commented Sep 10, 2015 at 15:32
  • \$\begingroup\$ @LayKruzvk: When you average mathematically, first you add and then you multiply by 1/count. The 1/count scale factor is the only difference between adding and averaging, and it guarantees that the output is never bigger than any input. If you want the output to exceed one or more of the inputs (straight addition), then you need somewhere else for that output to come from, which is almost the definition of active. \$\endgroup\$
    – AaronD
    Commented Sep 10, 2015 at 16:07

One thing to consider active summer over the passive summer is the output impedance. Your active summer will have a low output impedance which will not be affected by loading effects. In the other hand your passive summer can be affected significantly by loading effects. An example of this is if you are feeding the sum of the voltages to an ADC, the active summer will be able to drive the sample and hold circuit of the ADC much better. While the passive summer due to the higher output impedance you might have issues with the sample and hold circuit not being driven properly and then consideration of the aperture time of the sample hold circuitry.


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