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I have just completed my first ever electronic creation, a pair of headphones. It composes of a TRS stereo jack joined to four leads (left, right and two for common ground) braided nicely along the way to a 1k potentiometer along the two common ground leads. The braid then carries on to the right speaker (connected by the right lead and one of the ground leads), from there two leads (left and the other ground lead) go over the head band to the left speaker.

The gadget works almost perfectly, almost as it has one problem that I cannot fix. When the potentiometer is set to minimum resistance, the sound is perfect and I'm happy. However, trying to decrease the volume using the potentiometer results in only one of the speakers go quiet plus the overall sound quality seems off.

I hypothesized a couple of causes but I have no idea what the actual fault may be. The most likely cause in my opinion is putting the potentiometer on the ground leads instead of connecting a double gang (or maybe double pole single throw(?)) potentiometer to control simultaneously through right and left leads. On the other hand, putting potentiometer on common ground should affect both speakers equally nevertheless.

I don't know what to do so I'm asking the experts to give a helping hand.

EDIT: For those who requested a schematic, I hope I displayed the situation clearly in this one.

schematic

simulate this circuit – Schematic created using CircuitLab

The reason for the double ground wire between potentiometer and the TRS plug is just so the braid look uniform plus to hopefully reduce resistance along the wire (increased cross-sectional area).

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  • \$\begingroup\$ There are three signal paths in a stereo headphone audio circuit. \$\endgroup\$ – Ignacio Vazquez-Abrams Jun 29 '14 at 0:22
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    \$\begingroup\$ Can you post the schematics of your circuit? \$\endgroup\$ – Ricardo Jun 29 '14 at 0:54
  • \$\begingroup\$ Is your headphone stereo? Did you use a double pot like these? Did you wire each channel to a separate set of pot terminals? That's because the pot controls the volume on both channels, but each channel must remain separate. \$\endgroup\$ – Ricardo Jun 29 '14 at 0:58
  • \$\begingroup\$ @Ricardo: "... a 1k potentiometer along the two common ground leads." \$\endgroup\$ – Ignacio Vazquez-Abrams Jun 29 '14 at 2:28
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    \$\begingroup\$ I agree with you that everything is stated in the question, but your objective is to get an answer, and you will find that posting a schematic will produce one. It is much easier to think when confronted with an actual schematic than by having to first mentally construct one from a English-language description. It's up to you. \$\endgroup\$ – Marquis of Lorne Jun 29 '14 at 9:47
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A stereo headphone audio circuit has two low-impedance paths: from signal left to ground, and from signal right to ground. However, it also has a third high-impedance path from signal left to signal right.

Normally the parasitic third path does not affect the overall system very much, since signal left and signal right prefer to return on the much lower-impedance path to ground. But when you increase the impedance on ground, say by adding a potentiometer and subsequently turning it up, you cause the third path to become much more prominent. The exact effect from doing so varies, but in general it could be said that it "turns the audio to sh*t".

Using a double-gang potentiometer instead will also increase the impedance on the normal return paths, but the added impedance on the third path will be double that added to the normal paths, reducing its effect overall and maintaining audio fidelity.

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  • \$\begingroup\$ "The exact effect from doing so varies…" But, in general, it ends up taking the difference of the left and right audio signals and sending it to both speakers. This can have interesting effects on music — it'll often strip out vocals, for instance, because they're usually centered. \$\endgroup\$ – duskwuff -inactive- Jul 30 '14 at 2:42
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Having a single common pot won't work for two reasons:

  1. Using your circuit, you cannot completely turn the volume down to -ꝏ dB. When the pot resistance is 0Ω, the input signal appears fully across the load (the headphones), and when the pot resistance is at its maximum value of 1kΩ, you are still going to have some signal across the load.
  2. Audio signals are alternating currents. When the pot is set to a value other than 0Ω, current can flow directly between the Left and Right coils, which means you will get all sorts of very audible phase cancellation artifacts. You definitely want to keep the two circuits independent, like this:

schematic

simulate this circuit – Schematic created using CircuitLab

A 1K pot is usually a good choice for most headphones. The fixed resistor is only needed if you will be connecting attenuator between two amplifier stages. If connecting just headphones, the circuit will work without R1 and R2.

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    \$\begingroup\$ Welcome to EE.SE. Note that the question was from four years ago. The OP hasn't been on the site since 2014. \$\endgroup\$ – Transistor Nov 29 '18 at 18:26

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