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I'd like to build a simple audio mixer which takes signals from two or more inputs, combines them and outputs to the headphhones or speakers. I followed the other question: Making a Simple Audio Mixer [fig 1], but using 4.7kΩ resistors seems too much to drive headphones. On the other hand, I considered using 20Ω resistors instead, but I'm concerned that the inputs feeding high current to each other may damage them. Is my concern justifiable? What would be an optimal resistance in such case?

schematic

simulate this circuit – Schematic created using CircuitLab

Another option I considered was to stick with the high input impedance and use a summing operational amplifier circuit to provide current for the headphones/speakers [fig 2]. Which op-amp would be suitable here? I'm not aiming for expensive amps with excellent sound quality, it's rather a hobby project. I've considered LM386 but since its gain is 20-200, wouldn't it mean that 1V pk-pk would be amplified to 20V pk-pk, so I'd have a lot of signal clipping?

schematic

simulate this circuit

warownia1
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  • You could reduce the gain in the summing stage or add a volume pot after it, then feed your signal to the LM386 if you're worried about clipping. – Bort Dec 19 '19 at 13:21
  • Search for an opamp that can drive the required voltage into a 32 ohm load and runs from a single 5 volt supply is my recommendation in the absence of info on part (a) of your question. – Andy aka Dec 19 '19 at 13:21
  • Several audio PAs are commercially available (either class AB or class D) which would be better than a generic op amp for driving a 32 ohm load. Most have BTL (bridge tied load) outputs which can add 6dB to your peak output while eliminating the 220uF output capacitor. – Cristobol Polychronopolis Dec 19 '19 at 13:43
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    You want *dirt simple*? Tie the grounds from **IN_1** & **IN_2** together. The 32-ohm speaker goes direct from **IN_1** to **IN_2**. In other words, the "optimal" resistors in your circuit are 0 ohms. Each of your signals have their own volume control?...turn the volume down a bit. A disadvantage - both sources have to be active. Beware of Class D drivers! – glen_geek Dec 19 '19 at 14:22
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    I am not an audiophile but here are some general considerations about the simple passive versions.To make the first OP circuit almost "dirt simple":), remove R3; it is not necessary. Do the same with C1 and C3 (if the amplifier outputs allow it). Thus the two (low resistive) R form a "parallel voltage summer" (divider). The speaker load is grounded, the summer inputs can be weighted (scaled). The @glen_geek circuit is a series voltage summer (according to KVL)... more precisely speaking, this is a subtractor. The speaker load is floating; the summer inputs have the same gain of 1 (not scaled). – Circuit fantasist Dec 19 '19 at 18:48
  • Thanks for all the comments. @glen_geek your solution seems the easiest, but I'd like to be able to add more than two inputs. What's wrong with D-class drivers? I though they are common in consumer electronics. – warownia1 Dec 19 '19 at 19:15
  • "I'd like to be able to add more than two inputs." - how many inputs would you like to have? – Bruce Abbott Dec 19 '19 at 19:17
  • @Circuitfantasist IN_1, IN_2 will be connected to audio devices such as mobile phone, music player, pc sound card, console directly. I just want to make sure whether it's safe to connect them together through i.e. 20Ω-100Ω resistor or pot. – warownia1 Dec 19 '19 at 19:21
  • @BruceAbbott I'll start with two but it will be no more than 4 – warownia1 Dec 19 '19 at 19:22
  • @warownia1, you cannot connect more than two sources (audio outputs) in series (in a bridge configuration) since they are grounded; you can connect them only in parallel through resistors (according to your first circuit). Are the inputs of these audio devices (phones, players, etc.) analog type? If yes, there is no problem. Only, if they are low resistive, the resistors should have low resistance (more than the speaker resistance). Also note the resistive summer of two equal resistors has weighted inputs - VOUT = VIN1/2 + VIN2/2 (according to the superposition principle). – Circuit fantasist Dec 19 '19 at 21:29
  • @warownia1 A class-D output *might not have a ground connection*. Some are bridge output...these *already* pull the trick I mentioned. If you're unsure, summing two outputs to your headphone is risky. And I wouldn't try it with a headphone any lower than 32 ohms (too much peak current might flow). – glen_geek Dec 19 '19 at 23:54

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