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I'd like to connect one pair of headphones to the headphone jack on two audio sources simultaneously. I realize I can't just wire two audio sources in parallel since they'll then be driving each other. Commercial solutions appear to all involve amplification and volume control, which my two sources already have.

How can I combine audio signals from the headphone outputs of two different devices while still presenting just the impedance of the headphones to each device?

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There are a few methods:

A simple passive resistive mixer is basic, but a bad solution for a couple of reasons:

Passive Mixer

One is that in order to keep a low impedance output you need to use low value resistors and this loads each output excessively, plus creates a voltage divider between the outputs. Each output in the above example would see a 150 ohm load (e.g. the leftmost output will see R1 || (R2 + R3))

So we can buffer the signal:

Mixer Buffered Non Inv

This solves the loading issue (now each output sees 3.3k which isn't as bad), but not the voltage dividing issue. Say we have 3 inputs of 1V pk-pk. With all three plugged in, the contribution of each output will be a maximum of 333mV. This is okay (as we can add a gain of 3 to the opamp to compensate), as long as we don't unplug one of the signals.
If we unplug one of the signals, we change the loading on the other two and the voltage divider changes. The signal voltage from each will now be 500mV. If we unplug another then the full 1V pk-pk will be output.
So the output level of each channel is greatly affected by change of the other inputs - not just unplugging, imagine using volume controls.

A solution to this problem is the active inverting opamp mixer:

Mixer Inv

This is a current amplifier, and uses a virtual ground at the summing point to prevent any interaction between the channels. The feedback resistor R1, matches the sum of the currents flowing through R3, R5, and R6 (in order to keep the inverting input at 0V)
This means that the output voltage is simply (I(R3) + I(R5) + I(R6)) * R1.
If we remove an input, the voltage contribution from the other inputs stays the same.
So this is the best simple mixing circuit out of the three shown.

Try simulating the above circuits in SPICE to get a feel for what's going on.

The ESP pages linked to by Shimofuri are an excellent source of such information.

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  • \$\begingroup\$ please remove the diagram from ESP as it is under copyright. \$\endgroup\$ – shimofuri Jul 27 '12 at 13:15
  • \$\begingroup\$ Thanks for the detailed explanation. I thought headphone outputs were made to drive ~16-ohm loads. Why all the effort to increase the impedance seen by the headphone outputs? \$\endgroup\$ – ArgentoSapiens Jul 27 '12 at 17:51
  • \$\begingroup\$ Because (apart from the passive mixer) they are not driving the headphones directly anymore, it's better to keep the maximum voltage level by not loading them too much. There is no point wasting power by doing otherwise. Then the opamp (maybe attached to an output amp like the LM386) can drive the headphones. The passive mixer may suit your requirements if you want to keep it simple, but you can't avoid the devices loading each other. \$\endgroup\$ – Oli Glaser Jul 27 '12 at 18:11
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A non-inverting summing opamp circuit is the simplest mixer you can construct. The value of the input resistors is the impedance presented to the devices. Please check out this link from one of the best audio resource in the Web (https://sound-au.com/Elliott Sound Products) for a comprehensive discussion on audio mixing.

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You're looking for what is commonly known as a "combiner" (the opposite of a splitter) or in the audio world it is referred to as a "mixer". You can get them for fairly cheap, but sound quality (maximum amplitude, frequency range) might suffer. If you get a powered one it will sound better.

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  • 2
    \$\begingroup\$ The product you linked "connects a high impedance, line level, unbalanced output signal to a low impedance microphone level balanced input," whereas my application is connecting two low-impedance, headphone-level unbalanced output signals to a low-impedance, headphone-level unbalanced input. \$\endgroup\$ – ArgentoSapiens Jul 26 '12 at 21:16
  • \$\begingroup\$ @ArgentoSapiens - Added another link for you. This one is all headpone jacks. \$\endgroup\$ – Joel B Jul 27 '12 at 15:15

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