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Is it possible to use this single supply, non-inverting, summing op-amp configuration to sum multiple arbitrary DC voltages at unity gain?

I've used the inverting summing op-amp config on bipolar power supplies to sum AC and DC input voltages (paired with a second op-amp to 'un-invert' them) but this is my first time trying to do the same thing with a single supply, non-inverting config - and I'm struggling with how the R values are interacting.

Needs to be this config as it will be running off a single 12V supply but I don't need anything past whole number precision for the DC summing and how close the op-amp can get to GND or VCC is also not that important at this stage.

In this first example - I get the expected output. +5V at input R3 and +2.5V at input R4 gives me +7.5V at VOut.

Schematic 1

But in the second example - if I disconnect either one of these inputs - I start getting a gain of 2x at the output i.e. just +5V at input R3 gives me +10V at Out.

Schematic 2

How can I configure this so adding or removing outputs will always give me a gain of 1x?

Thanks

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  • \$\begingroup\$ A disconnected (floating) input is not 0v, it's whatever voltage is at the other side of the input resistor (the opamp's non-inverting input). If you grounded the "unused" input instead you'll see greater success. \$\endgroup\$
    – brhans
    Commented Mar 28, 2023 at 17:18
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    \$\begingroup\$ But note that your configuration only really works for 2 inputs with equal weighting, where the sum of the input voltages is the same value as the average of the input voltages multiplied by the number of inputs (2 inputs, so gain of 2). \$\endgroup\$
    – brhans
    Commented Mar 28, 2023 at 17:21
  • \$\begingroup\$ R1 (feedback resistor) needs to be zero ohms, for your second example. \$\endgroup\$
    – Vernon
    Commented Mar 29, 2023 at 7:49

5 Answers 5

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The inverting summer sums current. The non-inverting summer attempts to sum voltage but the source resistances of the other inputs form a voltage divider.

If you choose the resistors so that a loss of a factor of 3 occurs then the non-inverting gain of 3 will compensate.

When a non-inverting input is removed, it must be connected to zero volts I order to maintain the voltage divider.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Marked this as the answer because it most clearly showed how my original circuit was dividing by the number of inputs then gaining by that same number at the output - which gives me the same result as a unity gain summer - just not in the way I'd originally assumed. Thanks. \$\endgroup\$
    – formerlolz
    Commented Mar 29, 2023 at 9:42
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Your analysis of the first circuit is incorrect: it is not working as you think.

R2 and R4 form a two-resistor voltage divider between the 5 V and 2.5 V inputs. The voltage at the non-inverting input is 3.75 V. With two equal-value feedback resistors, the non-inverting gain is 2. Thus, the output is 2 x 3.75 - 7.5 V. If you had almost any other input voltages, the output would not have been what you expected. You lucked out one of the one combinations that masked the true action of the circuit, where 2 x the average equaled 1 x the sum.

The most simple way to have all inputs have equal gain is to have all inputs drive the inverting input through individual resistors. This is the classic inverting summer circuit. In this circuit, the non-inverting input is grounded.

With approximations, you can sum inputs into the non-inverting input without the large interaction you have now, but each input will require its own 2:1 divider.

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    \$\begingroup\$ That circuit works for any pair of input voltages (within the opamp's input range), not just the ones the OP picked. For example 1v & 2v gives an average of 1.5v (from the input resistor voltage divider action) multiplied by gain of 2 produces 3v (the sum of 1v & 2v). Similarly 3v & 4v averages to 3.5v, which becomes 7v after gain of 2. \$\endgroup\$
    – brhans
    Commented Mar 28, 2023 at 17:24
  • \$\begingroup\$ @brhans So using my first example - for any two inputs it should still output their sum - even though it's doing it via the average of the 2 inputs multiplied by a gain of 2? Am I understanding that correctly? Thanks. \$\endgroup\$
    – formerlolz
    Commented Mar 28, 2023 at 17:40
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    \$\begingroup\$ @robobozo as long as you connect your unused inputs to 0V and not leave it floating. \$\endgroup\$
    – Designalog
    Commented Mar 29, 2023 at 7:25
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Is it possible to use this single supply, non-inverting, summing op-amp configuration to sum multiple arbitrary DC voltages at unity gain?

No, that method just produces a weighted average. And, it's still a weighted average if dual supplies are used. You only need to think a little about it to see what I mean; if both input resistors are the same value and both inputs are the same voltage (say 5 volts), then the voltage at the non-inverting input is also 5 volts (and not 10 volts).

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Take the following scenario:

schematic

simulate this circuit – Schematic created using CircuitLab

A little bit of Ohm's and Kirchhoff's lawyering. By Ohm's law, the currents \$I_1\$, \$I_2\$ and \$I_3\$ are:

$$ \begin{aligned} I_1 &= \frac{V_1-V_Q}{R_1} \\ \\ I_2 &= \frac{V_2-V_Q}{R_2} \\ \\ I_3 &= \frac{V_2-V_Q}{R_3} \end{aligned} $$

By KCL, we have:

$$ \begin{aligned} I_Q &= I_1 + I_2 + I_3 \\ \\ \end{aligned} $$

Putting all that together, we get:

$$ \begin{aligned} I_Q &= \frac{V_1-V_Q}{R_1} + \frac{V_2-V_Q}{R_2} + \frac{V_3-V_Q}{R_3}\hspace{20pt}\text{[1]} \\ \\ \end{aligned} $$

Now let's examine the special case where all the resistances are equal, and \$I_Q\$ is zero. This latter condition sounds odd, but it's not breaking any rules. The algebra is still correct even if we don't connect OUT to anything! So, given:

$$ \begin{aligned} I_Q &= 0 \\ \\ R_1 = R_2 = R_3 &= R \end{aligned} $$

Equation 1 becomes:

$$ \begin{aligned} 0 &= \frac{V_1-V_Q}{R} + \frac{V_2-V_Q}{R} + \frac{V_3-V_Q}{R} \\ \\ 0 &= V_1 + V_2 + V_3 - 3V_Q \\ \\ V_Q &= \frac{V_1 + V_2 + V_3}{3} \end{aligned} $$

That means that at the junction of any number of equal resistances is the average of the potentials at the other ends of the resistances. Neat. Actually, I could have pointed you to the principle of superposition, which says exactly that, in a round-about way.

The only problem is, this only works if you draw no current from that junction, or if you do, that it be negligible compared to \$I_1\$, \$I_2\$ and \$I_3\$.

So, to obtain \$V_{OUT} = V_1 + V_2 + V_3\$, all we need to do is multiply \$V_Q\$ by 3, without drawing appreciable current:

schematic

simulate this circuit

Of course, weird things happen if the sum is close to or less than zero, and you should take care to avoid that.

You may ask why we rarely do this, and perhaps it's because with a proper inverting, summing amplifier, each input resistance is connected to a virtual ground. No input can influence any other, and the input resistance is well defined.

Here though, each input is connected to all others, via resistances, and you lose that isolation. If the sources of those input potentials have appreciable output resistance (even remotely comparable to \$R\$), then this arrangement will cause one input to affect the others.

So, if you do do this, you must ensure that each source's output impedance is very low, compared to \$R\$.

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What is the summer here?

In both inverting and non-inverting configurations, the op-amp does not sum anything; it is the same resistor circuit that does it.

Building op-amp voltage summers

The philosophy of this simple circuit can be revealed by following its evolution from passive to active.

STEP 1: Why not series summer?

The simplest way to sum voltages is to connect, according to KVL, the voltage sources in series; so the loop is the simplest voltage summer. The problem, however, is that some of them are "floating" (not connected to the ground). With two voltage sources this problem is solved by an amplifier with a differential input, but with more sources it cannot. To illustrate it, in the schematic below, I have set the op-amp gain equal to 1. Thus the op-amp acts as a fixed gain amplifier only converting the differential input voltage to an output voltage referenced to ground.

schematic

simulate this circuit – Schematic created using CircuitLab

STEP 2: Current summer

In contrast, the simplest way to sum currents is to connect, according to KCL, the current sources in parallel. All they and the load can be grounded; so the node is the simplest current summer.

schematic

simulate this circuit

STEP 3: Parallel voltage summer

So, we can convert the voltages into currents, sum them and then convert back the current into voltage. Simply speaking, what we need to do is to connect the voltage sources through resistors to a common point and use it as an output. In the schematic below, I have used more convenient values ​​of voltages and resistances.

schematic

simulate this circuit

STEP 4: Op-amp non-inverting summer

Then we can buffer and amplify the voltage sum by any amplifier with high input resistance. It can be with a simple single-ended input like the OP's non-inverting amplifier.

schematic

simulate this circuit

STEP 5: "Inverting summer man"

The problems of this resistor summer are caused by the non-zero voltage of the common point. But this voltage is vital to us simply because it is the output voltage! This is a kind of paradox because on the one hand we want this voltage to exist and on the other hand it prevents us.

"Measurement by compensation". Such technical controversies are resolved through powerful inventive ideas as "compensation" - to measure a quantity, we annihilate it with another equal in value "anti-quantity" and measure the latter.

Implementation. To apply it in our (OP's case), we add another (third) input and connect the compensating "input" voltage source V3 through another resistor R3.

schematic

simulate this circuit

Rearranging the schematic. Next, to show the connection with the next op-amp circuit, we transfer Vin3 and R3 to the right and rename Vin3 to Vout.

schematic

simulate this circuit

Our task in this fun "game" (aka "negative feedback") is to set Vout so that to zero the common point voltage (it is no longer output; it is, as they say, a "virtual ground"). To do this, we connect a null indicator NI (sensitive voltmeter), open Vout parameters and start changing Vout until we zero the indicator. Then we can use Vout. As you can see, I have managed to zero the voltage sum (shown by the null indicator) by adjusting Vout to -3 V.

STEP 6: Op-amp inverting summer

This job does not require special intelligence, and therefore we can assign it to an op-amp. Thus we get the famous circuit of the op-amp inverting summer.

schematic

simulate this circuit

Conclusion

The OP's non-inverting summer and the op-amp inverting summer above contain the same resistor summing circuit that sums the voltages; the first has two inputs (resistors) and the second has three inputs (resistors) one of which is used as a circuit output. The op-amp does not sum the voltages; it only buffers and amplifies the summer output or produces the compensating voltage.

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