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I built an audio mixer, and tested it with two different pairs of headphones. When I test it with one pair of headphones (I'll call Headphones #1), there is no noise (except for a few "pop" sounds when it first connects). When I test it with the second pair of headphones, there is very audible noise/crackling.

The only difference in specs that I'm aware of between the two pairs of headphones is the impedance. Headphones #1 is about 30 ohms, and Headphones #2 is lower impedance at around 8 ohms.

Given the below circuit, in which V1 and V2 are input audio signals, and the output is connected to a headphone channel, can you see a reason why noise would occur with Headphones #2? Why would a lower impedance headphone be subjected to audible noise with this circuit, or could it be due to a different characteristic?

The 1k resistor at the output is a discharge resistor I put in to prevent popping.

enter image description here

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  • \$\begingroup\$ What op amp are you using? \$\endgroup\$
    – vofa
    Jan 28, 2018 at 4:40
  • \$\begingroup\$ It's an LME49721, datasheet: ti.com/lit/ds/snas371c/snas371c.pdf \$\endgroup\$
    – donut
    Jan 28, 2018 at 4:43
  • \$\begingroup\$ What is the nominal peak-to-peak voltage of the input signals? \$\endgroup\$
    – vofa
    Jan 28, 2018 at 4:56
  • \$\begingroup\$ The peak to peak voltage is about 100 mV. It's music playing from a phone. \$\endgroup\$
    – donut
    Jan 28, 2018 at 5:16

1 Answer 1

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This is a preamp IC. Its output current capablity is too low for 8 Ohm load and 100mV output voltage. I guess that strong bass peaks cut off everything else momentarily. The headphones are complex loads, 8 Ohms is the nominal value, the actual impedance is heavily frequency dependent and can be at some frequencies much lower.

You should use a speaker amp IC.

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  • \$\begingroup\$ Wow thank you for the insight, I just noticed in the "description" section of the datasheet that the output current capability is ±9.7mA, which I didn't see in the spec section. So it has been trying to draw (nominally!) 100 mV/8 ohms = 12.5 mA... \$\endgroup\$
    – donut
    Jan 28, 2018 at 5:53
  • \$\begingroup\$ @donut Careful: If it's 100mVpp, then the peak voltage is half that and so is the peak current. At 100mVpp (±50mVpp) it would draw ±6.25mA. Still, I think this is the right answer and your signal level is probably higher than 100mV. \$\endgroup\$
    – vofa
    Jan 28, 2018 at 6:04
  • \$\begingroup\$ Would this IC be a suitable replacement: ti.com/lit/ds/symlink/tpa6111a2.pdf It's a 150-mW Stereo Audio Power Amplifier \$\endgroup\$
    – donut
    Jan 28, 2018 at 6:06
  • \$\begingroup\$ @donut The pinouts are not compatible. Look at pins 3 and 5. That IC should work, but you will need a different circuit to put it in. Section 10.2 in the datasheet gives a lot of guidance on how to use it. \$\endgroup\$
    – vofa
    Jan 28, 2018 at 6:13
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    \$\begingroup\$ I know the signal inversion can't be heard- I just need the signal to be in phase with the original signal for my application. The output dc blocking capacitance is that size because I needed a very low cutoff frequency. Thank you! \$\endgroup\$
    – donut
    Jan 28, 2018 at 7:15

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