mistracking pots...
Nov 3, 2001 at 9:23 PM Thread Starter Post #1 of 4

Braver

Will upgrade headphoneswhen there's a MX600.
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well, I was thinking of making a pot-module, like a volume pot in a box that you put before the amp. saves space when portable cause I wont need the pot (and my tin isn't big enough for one), but enables you to use the amp without adustable outputs.

thing is, the cheap 10kOhms pot I got mistracks between 0.40 kOhms and 0.80 kOhms. how bad is this?

if it's not too bad, I'm going to use it with the lower resistance on the right channel as my right ear is not as good as my left one (volleyball seems to be the thing that does it, smashing sometimes makes my ear ring. I'll play volleyball till I'm loosing limbs tho).

so, whaddya think? is it like "whoa thats bad!" or more like "should be okay"?
 
Nov 4, 2001 at 5:11 AM Post #2 of 4
I'll assume you're measuring it backwards; low side to wiper. Otherwise you'd never have it turned up that loud... on a given generic 10k log pot, it might not even get down to 800 ohm.

Yea that's a bad spot; since you'd really be using the other side of that pot then 800 would be ~8000 on the high side, so yea this is just barely turned up... just a "notch" or two.

The thing is, almost all pots mistrack here... some better than others. But .10 or 10.00, they all do. The situation will dictate how bad this mistracking drives ya nuts.
 
Nov 4, 2001 at 5:58 AM Post #3 of 4
Wait, just re-read that. You mean mismatched between channels at a specific point? Same thing applies, the lower it is, the more sensitive it's gonna seem. It's rare for a multigang pot to track perfectly. I don't think it's possible. A stepped attenuator or a digital potentiometer is the only way...

If you wanna know at a specific point you'd take a voltage divider and figure it:
dBV = 20*log(Vin*(Rlo/(Rhi+Rlo))

Then you can see how much it'll matter at a specific point in the rotation.

This is just the input attenuation! You have to multiply in the gain because ultimately this is what makes it to your ears; the amp's output.
 
Nov 4, 2001 at 10:30 AM Post #4 of 4
yeah, what I meant was that the resistance of one side is higher by 0.40 kOhms to 0.80 max.

just gonna solder it together, who knows, maybe it works.
 

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