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I'm trying to convert a stereo signal to a mono signal. It would come from a 3.5 mm stereo audio jack. The jack could be from just a dollar store headphone, where I'd cut out the ear pieces, only leaving the 3.5mm jack and the wire that will be stripped, exposing the left, right, and ground wires.

If I wanted to combine the left and right signals into a mono signal to be fed into an amplifier, I'd use the circuit represented by the following diagram:

Stereo to mono circuit diagram.

The above diagram is the most commonly suggested approach to combine the two signals into single signal.

Notice that before both the left and right signals join into a mono signal, a 1k Ω resistor is used in each of the left and right inputs.

What are the 1k Ω resistors for?

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    \$\begingroup\$ To protect whatever is driving those signals. Note teh differential impedance is 2K while the common mode impedance (source impedance to the output) is 500R. \$\endgroup\$
    – user16324
    Feb 21, 2021 at 13:24
  • \$\begingroup\$ In addition to protecting the source drivers, the resistors help prevent distortion. \$\endgroup\$
    – Hot Licks
    Feb 22, 2021 at 0:57
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    \$\begingroup\$ The resistors also reduce the mutual bleed-through from RIGHT to LEFT (and v.v.) if either of the sources is also connected to a different output, e.g. L & R converted to mono and applied to a pre-amp, but L & R individually also applied to a different pre-amp. The resistors keep the L & R somewhat isolated despite the converter. \$\endgroup\$
    – P2000
    Feb 22, 2021 at 5:13

4 Answers 4

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It's not adding so much as averaging. You have a 1:1 voltage divider there, so the open-circuit voltage at "MONO SUM +" will be halfway between the voltages of the two channels.

But why resistors, and why those values? Well, your audio device is capable of a certain amount of current. It expects to be driving headphones or speakers or whatever. If you connect the two channels directly together, then whenever their voltages are different (which is most of the time) they will be trying to pull each other up or down, with absolutely as much current as they're capable of. This isn't good for longevity, or battery life, or audio quality. With 2k of resistance between one channel and the other, the amount of current that they can use to fight each other is limited to around 1mA or less, which is probably fine.

But there is a down side, which is that this circuit is only really suitable for driving a high-impdance load (like the inputs of most amplifiers). The output voltage is the average of the two inputs when the current flowing out through MONO SUM + is pretty close to zero. But if you connect it to a load with an impedance of less than a few thousand ohms (say some 8-ohm speakers) then the voltage at the output will be much lower, most of the power will go into heating those two 1k resistors, and very little of it will go into the load. Again, the resistors make things safe so that you won't actually kill your audio device, but the result will be a low audio level.

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  • \$\begingroup\$ Interesting considerations about the ubiquitous 2-resistor network... a good illusration of the superposition principle. This is a parallel summer since two grounded input voltage sources are connected "in parallel" (but through resistors) to a grounded load. It is interesting to see what will happen if we implement a "series summer" in the case when the load is floating (eg, headphones or a speaker)... ie, we connect the load between the two outputs. The problem is that the two voltage sources would be opposing... and this arrangement would be a subtractor... \$\endgroup\$
    – Circuit fantasist
    Feb 21, 2021 at 22:23
  • \$\begingroup\$ So, the two output voltages should be with inverse phase ... and this is the idea of the bridge amplifier... \$\endgroup\$
    – Circuit fantasist
    Feb 21, 2021 at 22:25
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For many low level audio output devices you can mostly get away with just connecting left and right channels without using series resistors. In effect there are still series resistors (inherently) in each channel output but these of course are invisible to the casual user.

You can't do the same trick with power amplifiers because they don't inherently have series resistors that can limit current.

So, it's the same for batteries, if you connect a 12 volt battery to a 9 volt battery you might get smoke from the 9 volt battery but, when using series resistors, the voltage at the mid-way point (the mono output) is 10.5 volts i.e. the two resistors form an averager circuit and you get 10.5 volts at the mid-point junction.

It's no different for left and right audio channels; the two resistors produce an output that is the average of left and right channels hence, it is mono. The "averaging" is easily proven using Millman's theorem: -

enter image description here

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For two input voltages (left and right audio) the math becomes: -

$$V_{OUT} = \dfrac{\dfrac{V_{LEFT}}{R} + \dfrac{V_{RIGHT}}{R}}{\dfrac{1}{R}+\dfrac{1}{R}} = \dfrac{V_{LEFT}+V_{RIGHT}}{2}$$

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    \$\begingroup\$ I think it's dangerous to suggest that you can sometimes get away without the resistors. Yes, it (sometimes) works, but it assumes something about the things that you're abusing, which may not necessarily be true. Cost-cutting, for just one possible reason. Then we have a noob blaming an expert for breaking their stuff...if they even think that far. \$\endgroup\$
    – AaronD
    Feb 21, 2021 at 17:53
  • \$\begingroup\$ I would insist on ALWAYS including the resistors explicitly, and to NEVER use an off-the-shelf adapter to combine signals because their cost-cutting doesn't have them. (dead short) Always make one instead. If you're not capable of making one, then you need to fix that before you start making non-standard connections. \$\endgroup\$
    – AaronD
    Feb 21, 2021 at 17:54
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    \$\begingroup\$ @AaronD I think I gave sufficient reasons to use resistors especially with the battery analogy. But, if the op's in any doubt these comments will serve as a reminder. \$\endgroup\$
    – Andy aka
    Feb 21, 2021 at 20:18
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    \$\begingroup\$ Even at line level some absurdly fragile devices (particularly sound cards) can get damaged if backfed. \$\endgroup\$
    – chrylis -cautiouslyoptimistic-
    Feb 22, 2021 at 5:05
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The two resistors form a passive summing circuit with weighted inputs (voltage summer). You can use the superposition principle to find the relation between the two inputs and the output.

Here are links to related resources with my contribution:

Non-inverting Op-Amp Audio Mixer

Making a Simple Audio Mixer

Op-amp can add more than two voltages, while discrete transistors can't?

Parallel Voltage Summer

Walking along the Resistive Film

Passive voltage summer animated I created this interactive Flash movie in 2002 by means of Macromedia Flash (now Adobe). Unfortunately, Adobe stopped their Flash Player... but still there is a way to watch Flash movies - Ruffle Flash emulator. You can add their extension to your browser (see also another explanation). Or, you can download the exe file that has an embedded Flash Player.

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You can't tie the channels together since they are not always at the same potential, and you want to add the channels symmetrically so you need the exact same resistance on both.

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  • \$\begingroup\$ So the resistors are "drowning out" the potential differences between the two channels? \$\endgroup\$
    – Sal Rahman
    Feb 21, 2021 at 7:32
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    \$\begingroup\$ more correctly - they are 'mixing' the two signals \$\endgroup\$
    – Kartman
    Feb 21, 2021 at 7:37
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    \$\begingroup\$ Somewhat. When you mix two complex signals together, you loose the phase information between them. Phase differences get converted to amplitude differences. Think of two sine waves 180 degrees out of phase. When you listen to this in stereo, it sounds like a tone possibly with a preference to in one direction. If you mix these two tones together into a mono signal, you'll have the sound of silence. In a stereo audio signal, the phase information helps to create the illusion of directionality. With a mono signal, there is of course no direction information contained in the signal. \$\endgroup\$
    – ScienceGeyser
    Feb 21, 2021 at 7:41

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