3
\$\begingroup\$

I am trying to make sense of the difference between 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?

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

\$\endgroup\$
4
  • 1
    \$\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
    Commented Dec 21, 2019 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\$ Commented Dec 21, 2019 at 20:18
  • 2
    \$\begingroup\$ Nope! and AC/DC has nothing to do with it. See Jasen's answer below. \$\endgroup\$
    – Oldfart
    Commented Dec 21, 2019 at 20:36
  • \$\begingroup\$ In future, when designing opamp circuits, you may want to consider using Rf (instead of R5). Though not universal, it's a fairly common convention. \$\endgroup\$ Commented Mar 6, 2023 at 17:05

5 Answers 5

3
\$\begingroup\$

I am summing 3 signals from sources and only 1 is generating sinusoidal signal with 100 mV amplitude I get 33 mV at the output of non-inverting configuration and in inverting case I have 100mV at the output. Why?

In the top (inverting) amplifier, the summing junction is a virtual ground. 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 (non-inverting) amplifier, the three inputs flow into a three-way voltage divider so given that the resistors are equal they get averaged; 0,0,100 mV gets you 33.3 mV. The amplifier just buffers that gives a 33.3 mV output.

The 470 k does basically nothing, so I'm going to ignore it.

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

\$\endgroup\$
3
  • 2
    \$\begingroup\$ The 470 k resistor provides a path for the input current of the non-inverting input... \$\endgroup\$ Commented Dec 21, 2019 at 21:39
  • 1
    \$\begingroup\$ The 470k is certainly not doing nothing. All inputs are AC coupled. Without it the DC potential of the input is not defined. Without it you can easily end up with the circuit running with several volts of DC offset and lose a lot of headroom. \$\endgroup\$
    – danmcb
    Commented Mar 7, 2023 at 9:36
  • 1
    \$\begingroup\$ When analysing the circuit I assumed an ideal op-amp. real parts have input bias current so the circuit needs it. but for analysis it only less than 0.1% effect on the output - basically nothing. \$\endgroup\$ Commented Mar 7, 2023 at 11:02
2
\$\begingroup\$

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.

\$\endgroup\$
2
\$\begingroup\$

@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...


Finally, let's formulate a "golden rule" for converting a passive summing circuit into an active inverting circuit:

  1. Add another input (resistor).

  2. Supply it with a compensating voltage that neutralizes the output voltage of the passive summing circuit.

  3. Use the compensating voltage as the output voltage.

See also a Wikibooks story about the passive summing circuit and download an interactive Flash movie (exe file with embedded Flash player) about the op-amp inverting summer.

\$\endgroup\$
1
\$\begingroup\$

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

\$\endgroup\$
1
  • \$\begingroup\$ The "passive mixing circuit" can be best analyzed by the help of the superposition principle. \$\endgroup\$ Commented Feb 18, 2021 at 20:24
1
\$\begingroup\$

TL;DR: The primary advantage of using a an inverting amplifier for audio mixers is that it gives us virtual ground, which solves the OP's mix point amplitude reduction problem.

Adding to the existing answers, I'd like to offer a slightly different perspective; virtual ground and backfeeding. To explain the difference between inverting and non-inverting opamp configurations in terms of a practical application, we tan talk about two mixer types; passive and active. For context, I'm drawing upon knowledge from the article, Audio Signal Mixing by Rod Elliott (ESP).

Edit: Rod actually uses the word "crosstalk" in his articles, but most people seem to agree that crosstalk instead refers to signal coupling between lines, therefore I will instead use the word "backfeeding" as suggested by others. This is purely academic, as in most cases, backfeeding into a source signal is not a concern since there's almost always some impedance/resistance blocking any noise.

Note: This answer mainly explains why you would use an inverting instead of a non-inverting amplifier for audio mixing.

Passive mixer

Passive mixers are cheap and require no power source, but at the mix point they introduce backfeeding and reduce the signal amplitude (as OP describes).

schematic

simulate this circuit – Schematic created using CircuitLab

Passive mixer - input backfeeding

As in Rod Elliott's design, the opamp isn't actually part of the passive mixer, but after the passive mixer stage, you would usually want to recover the signal amplitude. For the sake of example, we're using non-inverting; if we were to use an inverting amplifier, it would turn the mixer into an active one (that would be a good thing, but we'll get to that).

When you add a non-inverting recovery amp after a passive mixer, the passive mixer remains this way because a non-inverting amplifier is high impedance (incidentally, also making it not very good at passing current, so not ideal for audio mixers). The non-inverting input of the opamp does nothing to the voltage at the mix point, so the mix point has backfeeding and voltage problems remain.

You could say that despite the backfeeding, low signal amplitude, and current limit, one advantage of non-inverting that the input and output signals are buffered and in-phase, but that's not much of a consolation for our audio mixer application. Also, opposite-phase audio signals sound the same.

Active mixer

What makes an active mixer so, is the inverting amplifier and its virtual ground.

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?

To answer the OP's comment, the inverting amplifier (used in an active mixer) solves the mix point voltage issue with virtual ground.

Note: Virtual ground confuses a lot of people new to opamps; huh, why is the inverting input 0 V? (Dave's opamp video is great)

At the mix point (connected to the opamp inverting input), the voltage is driven to 0 V (via the opamp output and feedback resistor), which eliminates the backfeeding since the mix at 0 V cannot interfere with other inputs. Virtual ground also solves the mix point amplitude problem, again, because the voltage is always driven to 0 V; the output current is now the sum of the two input currents (despite being 0 V at the opamp inverting input).

schematic

simulate this circuit

Active mixer - no input backfeeding

Side note on current: An inverting amplifier also hypothetically allows more current to be passed (as current flows through the feedback resistor and not from the opamp output). An LM324, for example, can source -30 mA and sink 20 mA at \$1\ V_{ID}\$, so for non-inverting this is the typical maximum. An inverting amplifier is limited instead by the feedback resistor, which can be whatever we want. An inverting amplifier could be used as a low current amplifier (though it's certainly not a common type of current amplifier). That said, the feedback resistor would have to be at such a low value (e.g. 20 ohms at 1 V) to pass even 50 mV. It would be nice if we could demonstrate this in CircuitLab/SPICE, but when simulated, exceeding the LM324's output current is of no consequence.

The general scheme is shown in Figure 4, and the opamp is an inverting stage, with all mixed signals connected to the inverting input. While it's not commonly described as such, an opamp connected this way is a current amplifier (inverting). Whatever current flows into (or out of) the input is balanced by the current flowing through the feedback resistor (R4), such that the difference between the two inputs is close to zero. In essence, the opamp causes the instantaneous current I4 to be exactly equal and opposite to the sum of instantaneous currents I1, I2 and I3.

Side note on phase: though an inverted sound wave will still sound the same, if for some reason you want the input and output signal of your mixer to be in-phase, then you add a 2nd inverting amplifier. Often in an active mixer you'll find inverting pre-amps on the inputs, as well as the summing/recovery inverting amplifier after the mix point. The two together produce in-phase input and output signals. Some amp designs introduce ~50k resistance at the input to match the impedance of audio sources and dynamic microphones, which creates the need for a pre-amp. Pre-amps are also useful for increasing weak signal sources.

Note: LM324 should not be used for an audio amp, use a low noise/distortion audio opamp instead.

\$\endgroup\$
8
  • \$\begingroup\$ Good explanation! I just didn't understand what you mean by saying that "non-inverting also allows more current to be passed (this current goes through the feedback resistor, and is inverted)"... \$\endgroup\$ Commented Mar 6, 2023 at 21:30
  • 1
    \$\begingroup\$ Whoops! That's because I meant "inverting also allows more current to be passed" \$\endgroup\$ Commented Mar 7, 2023 at 9:23
  • 1
    \$\begingroup\$ "Crosstalk" in an audio mixer means a signal finding its way into other paths by coupling. (For instance we might drive channels 1 to 8 at 10kHz with high amplitude into the LEFT bus and then see how much of that we can measure on the RIGHT output. On an analogue mixer with input modules the L and R virtual earth points are physically long and actually very sensitive to crosstalk. Designers go to a lot of trouble to minimise it (e.g. by using adjacent cores of a ribbon cable as "shields". What you are referring to in this example is more "backfeeding" of the signal which is not a concern. \$\endgroup\$
    – danmcb
    Commented Mar 7, 2023 at 9:49
  • 1
    \$\begingroup\$ indeed, he is using the term "crosstalk" in a way that is not usual. I worked (many years ago) for Soundcraft (a mixer manufacturer) as a test tech. There is a pretty widely accepted definition of "crosstalk" in a mixer, and it's not really what he describes. However the mechanism he talks about could lead to crosstalk - if the affected input channel is driving a number of summing busses (which it generally is). But he doesn't explain that at all, and IMO he really should. Basically if the output of ch A has a bit of the signal on ch B, that (by itself) is of no real consequence. \$\endgroup\$
    – danmcb
    Commented Mar 7, 2023 at 13:30
  • 1
    \$\begingroup\$ en.wikipedia.org/wiki/Crosstalk - read the section on audio systems. \$\endgroup\$
    – danmcb
    Commented Mar 7, 2023 at 13:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.