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I have a pair of expensive in-ear monitors (earbuds). Each side utilizes six balanced-armature drivers and a passive crossover.

The right side is quieter than the left side. The headphone wires are detachable, so measuring across the mini coaxial port where the headphones connect to the cable, the right side's internal resistance is 15.60 ohms while the left side is 15.16 ohms. This imbalance in impedance might explain the imbalance in volume between each side. In Windows and Android software I can choose to increase the volume of the quieter right side. I have to increase the software's volume slider by 6dB in order for both sides to feel balanced.

Normal headphones have matching impedance on both sides. Something inside the right side is causing an increase in impedance and all the solder joints still seem solid, so I think the issue may be something internal to one of the six drivers. The 6dB difference is simply what the audio-control software in Windows and Android say I need. It is not actually the difference measured at the audio ports of the headphones.

You can see photos here:

Photos of Internals

Would I be able to take a headphone cable and on the left side solder a 0.45 ohm resistor inline to balance out the impedance of both sides, and thus fix the volume balance issue?

Would adding a resistor affect the quality of the analog signal reaching the drivers? Would it change the phase or timing of the signal, causing weird imbalances in the music coming out of the headphones?

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    \$\begingroup\$ it wouldn't change any aspect of the signal except the amplitude, but it seems odd that there's such a large loudness difference from a 3% difference in resistance... \$\endgroup\$ – BeB00 Oct 1 '18 at 19:48
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    \$\begingroup\$ make an adapter that swaps the left and right headphones ..... see if the problem stays or if it moves \$\endgroup\$ – jsotola Oct 1 '18 at 19:53
  • \$\begingroup\$ Well, to be clear, the 6dB difference is not measured at the audio ports of the headphones. In Windows and Android software I can choose to increase the volume of the quieter right side. I have to increase the volume slider by 6dB in order for both sides to feel balanced. \$\endgroup\$ – fuzzybabybunny Oct 1 '18 at 19:54
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    \$\begingroup\$ If the drivers are jacked up enough to make 6dB of difference in volume, then I expect that it is already causing all the things you are worried about - but since all you can hear is the difference in volume, I don't expect it will matter if you put in a resistor. I don't think it'll help much, though. \$\endgroup\$ – JRE Oct 1 '18 at 19:56
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    \$\begingroup\$ That's what I'm saying. Something is seriously wrong, but all you hear is the difference in volume. All the other things you are afraid will happen if you install a resistor are already happening because of whatever is wrong - and none of it matters. The only part of it you can detect is the difference in volume. Human ears are wonderful things, but they are not precision measurement devices. \$\endgroup\$ – JRE Oct 1 '18 at 20:14
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The DC resistance you measure with a multimeter has little to do with the impedance and/or frequency response across the full audio range. If you want to diagnose this properly, you need to hook them up to a network analyzer.

Just putting a resistor in series with the lead will increase the apparent output impedance of your amplifier, which will likely have a serious effect on frequency response. (I.e., the "damping factor" will go way down.) I wouldn't recommend it.

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  • \$\begingroup\$ Since these are tiny balanced armature drivers and the amplifiers used on these are only ~1.2V max per channel with power output around 45mW, would an inline resistor affect frequency response much? I understand that damping factor for instance is often applied to subwoofers. \$\endgroup\$ – fuzzybabybunny Oct 1 '18 at 21:12
  • \$\begingroup\$ Maybe, maybe not. But the more important point is that a 0.45 ohm resistor isn't going to give you the 6 dB decrease in power you're expecting. For that, you'd need a 15 ohm resistor. \$\endgroup\$ – Dave Tweed Oct 1 '18 at 21:37

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