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Re: Morse, one more question (from way back)- pot values

Hi Ferd;

Glad to hear that you've got it working!

Hmmm, interesting effect.... I would not try it at 5k, since some sources could have 'heartburn' with that low a load impedence (some may not like 10k, which is why I like a 100kOhm pot).

Okay, the pot should be wired as a voltage divider - in other words, the wiper goes to the plus lead of the load, and one end of the pot goes to ground while the other end of the pot goes to the plus lead of the source as we discussed some time ago. Now, the thing to keep in mind is that the potential that is seen at the load is a function of the ratio of the resistance between the "wiper to ground" and the "source to ground" legs of the pot. Thus, if the 'top' (source + to wiper) is 3k and the 'bottom' (wiper to ground) is 7k, then you would have 70% of the voltage that you would measure between the source and the ground, since 7k/(7k+3k) = .7.

Something you could do would be to look at the resistances in each part of each pot (with it unhooked from source and load of course) at the volume settings you noted and see what's going on that way - just diagram it out like the original schematics, and pencil in the resistances you measure each way.

One thing I would also make sure of is that the 2 pots are both audio taper ones - usually linear taper pots are marked something like "B10K" and audio taper ones will be marked with something like "A100K" or maybe 100kAx2 (the 'x2' means stereo).

I wish I could see it first hand; sorry this isn't more illuminating...

Good luck,
Morse


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  • Re: Morse, one more question (from way back)- pot values - Morse 15:49:51 12/07/02 (0)


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