Impedance balancing of headphones + pot
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 5th August 2012, 01:24 AM #1 jsirois   diyAudio Member   Join Date: Aug 2012 Impedance balancing of headphones + pot I'm building a volume controller with a potentiometer, and I'm having thoughts with impedances. I understand that: Headphone jack (low imp) >> pot. (high imp) is fine. but wouldn't having the pot at say 15kOhm for example, load down the headphones? pot (15kohm) >> phones (55ohm) == bad?? Now the output impedance is much higher than the input. So all these questions are racing to me. Does only the impedance the headphone jack out sees matter? (pot + phones) I.e. as long as the load the headphones jack sees is acceptable (above 8 ohms or so), sound isn't altered Or Couldn't I technically throw another < 50ohm resistor between the potentiometer and the phones to balance the impedances? Last thing... someone mentioned to me "buffers". could someone just go over that super quick run-down
 5th August 2012, 03:11 AM #2 benb   diyAudio Member   Join Date: Apr 2010 It would be that the 55ohm headphones "load down" the 15kohm potentiometer. The volume would be very low over most of the pot's range, and would suddenly jump to full volume over a very small range at the end. A 50 ohm resistor in series would do little except lower the maximum volume by 3dB. You can do the calculations yourself by learning a little bit of circuit analysis with resistor networks. The pot is two resistors in series that add up to 15k ohms, and the one that goes to ground also has a 55 ohm resistor across it. As you turn the pot, the point 3dB below max volume will be when the top resistor is 55 ohms (and the bottom resistor 14,945 ohms, presuming the pot is an exact 15,000 ohms). As you might guess from the resistor ratio, this point is where the control is turned 99+ percent of the full on position. A buffer is just an anplifier. Most amplifiers amplify both voltage and current, but buffer usually means it amplifies current only, and keeps the voltage the same. A buffer has a high input impedance (maybe 10k to 1 meg) and a low output impedance (maybe 100 or 10 or 1 ohm), and will "match" a higher impedance volume pot with the headphones. The buffer should, of course, have enough voltage and power output available to drive the headphones to the maximum desired volume.
 5th August 2012, 06:20 AM #3 jsirois   diyAudio Member   Join Date: Aug 2012 That was very informative, but I'm not sure exactly addressed what I meant. 15kOhm was a bad example I'm just speaking in terms of impedance matching, does the fact that the pots "output" impedance is higher than the phones causes an "unmatched" signal, or is sound not affected by this, or is it negligible. I'm probably going to end up using a 250ohm 1watt logarithmic pot
 5th August 2012, 07:01 AM #4 Mooly   diyAudio Moderator     Join Date: Sep 2007 It will affect sound quality in some ways because the headphone is a (slightly) reactive load. This means that its impedance varies with frequency just like a speaker does. So the inclusion of a resistive divider will modify the frequency response of the phones. Best advice is to try with fixed resistors first and see if you can hear any difference. A 250k log pot (stereo... two gang) might be tricky to obtain, but I haven't looked

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