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I am trying to make sense of the difference in non-inverting and inverting op-amp based audio "mixers". The circuits are basically summing op-amp amplifiers. Here is the part which I do not understand. (Figure from simulations below)

I simulated both and the main difference is that in the non-inverting case, the other sources act like shorts, but in inverting configurations they do not. Why is that so? Can someone explain?

So basically if I am summing 3 signals from sources and only one is generating a sinusoidal signal with 100 mV amplitude, I get 33 mV at the output of the non-inverting configuration and in the inverting case, I have 100 mV at the output. Why?

simulation

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    \$\begingroup\$ "The circuits are basically summing op amp amplifiers" No they are not. The bottom one is a 1x amplifier and the adding/mixing is done by the resistors. \$\endgroup\$ – Oldfart Dec 21 '19 at 20:10
  • \$\begingroup\$ According to this source it is: electronics-tutorials.ws/opamp/opamp_4.html At least it would be if we would be dealing with DC, but I can understand that in case of AC signal things are different, right? \$\endgroup\$ – Jüri Bogatkin Dec 21 '19 at 20:18
  • \$\begingroup\$ Nope! and AC/DC has nothing to do with it. See Jasen's answer below. \$\endgroup\$ – Oldfart Dec 21 '19 at 20:36
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I am summing 3 signals from sources and only 1 is generating sinusoidal signal with 100mV amplitude I get 33mV at the output of non-inverting configuration and in inverting case I have 100mV at the output. Why?

In the top amplifier the summing junction is a virtual ground so the feedback current exactly matches the current into the virtual ground and the output is the inverted sum of the input voltages (modified by the resistor ratio, - here unmodified because all the resistors are the same)

In the bottom amplifier the 470K does basically nothing, so I'm going to ignore it,

The three inputs flow into a three-way voltage divider so given that the resistors are equal they get averaged so 0,0,100mV gets you 33.3mV

the amplifier just buffers that gives a 33.3mV output.

If you want 100mV from the non-inverting amplifier configure it to have a gain of 3. eg: 2K from output to inverting input and 1K from there to ground.

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    \$\begingroup\$ The 470 k resistor provides a path for the input current of the non-inverting input... \$\endgroup\$ – Circuit fantasist Dec 21 '19 at 21:39
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The key thing to remember with opamps is that with negative feedback, in a stable configuration, the + and - inputs will have the same voltage.

In the inverting, this means the - is also a ground, and it behaves like a transimpedance (current) amplifier, where the input currents are given by the Vins and the resistance. In the non-inverting case, you have a voltage averager followed by a voltage follower amplifier with gain 1.

Oldfart's comment may be pedantically correct but from a practical and useful perspective you have two summing amplifier arrangements, one with a gain of -1 and one with a gain of 1/3.

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I will also answer my own question (I want to leave the original question as it is). After the answers of Jensen which helped me the most and also thanks to Oldfart and David, here is my summery. Basically it is possible to simplify the circuits to understand what is going on. In non-inverting case the circuit simplifies to just resistor divider. The V3 is a voltage source and if the other sources V1 and V2 are not outputting a signal they are ground connections. Therefore in "non-inverting" case the schematic simplifies like this:

schematic

simulate this circuit – Schematic created using CircuitLab

If we use the voltage divider formula we get that Vop_amp = V3 * (R2,3 / (R4 + R2,3)) = 100mV * (0.5k / 1.5k) = 100mV * 0.333 = 33.3mV Which is exactly what I see in the simulations.

Meaning that it is a passive mixing circuit like Oldfart said.

In case of inverting circuit the part before the op amp can be simplified like this: enter image description here

Like Jasen said:

In the top amplifier the summing junction is a virtual ground so the feedback current exactly matches the current into the virtual ground and the output is the inverted sum of the input voltages (modified by the resistor ratio, - here unmodified because all the resistors are the same)

I also found a webpage about audio mixing circuits where exactly the same topic is explained (and the circuits are very similar). Some explanations on the this website get maybe little bit complicated, but overall a very good material for the same topic.

https://sound-au.com/articles/audio-mixing.htm

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@Jüri Bogatkin, I like your answer and will only enrich it with more "philosophy". I will try to explain not only HOW the circuit is made but WHY it is made in this way.

If you look closely at the two circuits, you will find they are the same 4-input resistor summer + op-amp. The only difference is in the input voltage of the fourth input - in the non-inverting circuit 0 V (ground) is applied to R1 (the ground is the "source") while in the inverting circuit VOUT is applied to R5 (the op-amp output is the source). In other words, the op-amp output voltage has replaced the zero ground voltage at the lower end of R1. What is the point of this?

The output voltage of the humble resistor summing network is undesired since it decreases the input currents and makes them interdependent. So, we have to remove (zero) it... but we need it since this is the output voltage (the sum).

The clever trick is to remove this voltage by equivalent but opposite voltage and use the latter as output voltage. The op-amp does this work. It "observes" the voltage of the summing point and "pulls" it up or down in the opposite direction (like a "tug of war") until zero this voltage. Thus this point is always "virtual ground" and the op-amp output voltage represents its voltage (with an opposite sign).

The op-amp output voltage is always equal to the voltage drop across R5 and compensates it. Figuratively speaking, it acts as a "negative resistor" with resistance -R5. The result is zero resistance... as though the summing point is connected by a piece of wire to ground... but still there is output voltage at the op-amp output...

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