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I need to record audio with a device that only has a mic-in which is heavily amplified. I built a potential divider with R1 = 10 kOhm and R2 = 4.7 kOhm to get a factor of about 30%. I chose the resistors relatively large to prevent blowing the sound card which is signal comes from.

Is it better to use larger resistors not to damage the output device?

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  • \$\begingroup\$ To get good frequency response, it will may be necessary to use some combination of resistors and capacitors. If you have a scope, I would suggest that you output a signal which contains a mixture of two equal-amplitude sinewaves at very different frequencies (e.g. 100Hz and 5,000Hz) and view the signal captured by the microphone. I'd suggest starting with a divider that has 0.01uF and 10K in series, and 0.1uF and 1K between the output and ground. You should be unlikely to damage either the input or the output with those values. Set input and output gain to the middle and see what you get. \$\endgroup\$
    – supercat
    Jun 13 '12 at 15:38
  • \$\begingroup\$ Ideally, if the peak level of your sinewave combo is close to the peak level of your output, your input should on close inspection look like two non-distorted sinewaves summed together. If you zoom out, you should see a "fat" 100Hz sine wave, where the "bottom" of the upward peak is roughly at ground, and likewise for the "top" of the downward peak. The vertical "width" of the line should be about the same at the peaks as in the middle. If the "fat stroke" is too thin, decrease the series cap; if it's too fat, increase the parallel cap. \$\endgroup\$
    – supercat
    Jun 13 '12 at 15:41
  • \$\begingroup\$ If you've ever played music while plugging or unplugging the headphones, you've applied a dead short to the output, because of the way the contacts are arranged. It should survive that fine, so don't worry about blowing up the headphone amp. As long as your total resistance is greater than a headphone, it will be fine. \$\endgroup\$
    – endolith
    Jun 13 '12 at 17:54
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I would use resistor values that total roughly what the impedance of the headphones are, which is what the amp is supposedly designed to work with. Generally loading less (higher resistance) does no harm, but some amps may not have the specified frequency response without some loading.

There are various types of "headphones". The large ones that completely cover the ears are really small speakers and have low impedance, just like other speakers. The in-ear types can by dynamic, which can have a few 100 Ω down to speaker impedance, or piezo, which can have much higher impedance. If in doubt, I'd load a "headphone" amp with around 600 Ω. That should work well enough.

However, I would use substantially less than a gain of 30%. Microphone signals are a few millivolts at best, with headphone signals a volt or more. I'd try about 1000:1 voltage divider to start with, then adjust as needed. A good ratio is when both the headphone and microphone amps are at about 1/3 volume.

So if you need concrete numbers to start with, try 1 kΩ followed by 1 Ω to ground. Adjust the top resistor (the 1 kΩ) up or down by factors of 2 or so to get roughly the right attenuation ratio.

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  • \$\begingroup\$ Shouldn't I multiply both resistors by ~20 to make sure I won't blow up my output devices if I plug it wrong side round? \$\endgroup\$
    – bot47
    Jun 13 '12 at 16:00
  • \$\begingroup\$ @MaxRied: Then you should make them 100 W resistor in case you accidentally plug them into a wall outlet too. Even a dead short shouldn't hurt a properly designed headphone amp. You could add 8 Ohms in series with the mic port if you're really that worried about this. If I was doing this, I'd put good labels on the ends. \$\endgroup\$ Jun 13 '12 at 16:10
  • \$\begingroup\$ "but some amps may not have the specified frequency response without some loading" You probably want the better frequency response and lower distortion that you get with large resistors, though. \$\endgroup\$
    – endolith
    Jun 13 '12 at 17:51
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I think it would be better to use an amplifier; instead of amplifying the signal you decrease it to 30%.

I'm not sure how well the voltage divider would work. So I can't say too much on it. It looks like you're connecting the sound card output into the sound card input.

If your concern is blowing the sound card from either too high of voltage or too much current, then you would want to keep the resistors higher. The voltage doesn't impact most devices as much as current.

You really need to know what kind of voltage is coming out of the sound card, and use the voltage divider to figure how much voltage is going in.

I think I would put a 1 MΩ potentiometer on the divider to make the output 1%, and slowly change it until I like it. You could put a 10 kΩ resistor on the divider, making it a 1% voltage output.

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  • \$\begingroup\$ I would combine what the Olin's values where, but I still like the turn pot approach. \$\endgroup\$ Jun 13 '12 at 15:47

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