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

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Ferd

(A)

A while back you helped me out with some pot wiring trouble I was having. Thanks, I finally understand.
Anyway, I tried two values of pots: 100k and 10k
Now the 100k pot gets to about 10% of the travel range before it gets very loud, while the 10k pot seems to get about 25% travel for the same volume.
Is it correct that the lower the value of the pot the more the attenuation? Seems backwards, but I'm a complete amateur.
If that is the case, then I could go with a 5k pot to get even more travel, right?
Thanks for the help, again.
Ferd

Follow Ups:

My two bits on pot taper and the like - tcpip 01:28:38 12/09/02 (0)
Just my two bits in addition to whatever's been said before me,
all of which makes sense to me.
Firstly, it might be a better idea to use a resistor with a linear
pot to make a log pot, rather than getting a so-called "audio
taper" pot, which usually has a strange taper, not quite a log
one. Moreover, if the pot is dual-ganged, audio taper pots can
show more channel tracking mismatch than linear ones.
Secondly, Morse has remarked about the "A" and "B" labels
on pots to indicate taper type. I think label to taper matching
is a little more confusing than this; see the link to Rod
Elliott's Webpage on the subject below.
Tarun
PS: If you are asking questions about pots, it might be a
very good idea for you to go through all the conceptual
articles on Rod Elliott's Website. They're very useful.
- Rod Elliott's well-written article on pots
Re: Morse, one more question (from way back)- pot values - Morse 15:49:51 12/07/02 (0)
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
Re: Morse, one more question (from way back)- pot values - Tiny Daddy 12:36:25 12/07/02 (4)
Do you have a schematic so we can see where the pot is in the circuit?
- Re: Morse, one more question (from way back)- pot values - Ferd 14:07:47 12/07/02 (3)
  Actually, it's the only component so far.
  I'm using a pot to reduce the signal from my cd player (philips 634) to my monoblocks (wave 8's). Kind of an el-cheapo passive preamp. I'll add a switch later.
  Thanks
  - Use a series resistor. - Tiny Daddy 15:34:18 12/07/02 (2)
    Run the signal through a 10K resistor, then to the pot. The resistor will drop some of the signal and the pot controls what's left. You may want to experiment with various resistor values.
    - That'll work.... - Morse 16:00:21 12/07/02 (1)
      Hi Tiny Daddy;
      Yep, that's a nice simple solution that should get the job done.
      Ferd, as far as resistors are concerned, if you don't have any on hand, just buy a few of the Radio Shack 1/4 watt jobs. The carbon films sound fine and are dirt cheap (if you buy a combo pack anyway). The metal films that they carry are okay too, albeit a little pricier (their metals are 1% precision, so they're nicer that way though). DO NOT use metal oxide resistors or wirewound resistors for this....they will not sound good. Don't think that you have to blow the budget on Rikens, Tantalums, or other 'audiophile' resistors either for this particular project....no point in blowing $3.5 or more on each resistor for an office system like you described. $6 for 100 (or $12 for 500!!) assorted 1/4 watt resistors is all you'll need.
      Again, good luck
      Morse
      - Thanks to you both, I'll try the resistor... (nt) - Ferd 17:37:50 12/07/02 (0)
        
        In Reply to: That'll work.... posted by Morse on December 07, 2002 at 16:00:21:
        
        NT

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