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Pots in parallel with resistors

9/5/2003 7:31 PM
Pots in parallel with resistors
Hello Guys,  
Can I put a resistor in parallel with my gain pot to mod its value? Will it produce the same result as changing the whole pot, or would the non linear response of the pot be affected in a detrimental way? I don't mind changing the value altogether, but the pots are CTS circuit board mounted, and I can't find the parts locally.  
I did a search on the net regarding this issue, but couldn't find anything. On the other hand, I have schematics in my collection that feature pots and resistors in parallel. Any Advice?  
Thank you,  
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9/5/2003 11:33 PM
Mark Lavelle

Everything you want to know is in "The Secret Life of Pots":
9/6/2003 8:31 AM
MJ Harnish
As Mark's post says, check out R.G.'s article b/c it answers all your questions. In a nutshell, yes you can use a resistor to mod the value of a pot but it's going to change the taper.
9/6/2003 6:23 PM

Thanks for the info. R.G. Keen is absolutely brillant and I've incorporated some of his designs in my projects. HIs article, however, doesn't seem to cover my question. I don't want to add a tapering resistor between the input and wiper, but between the the two outer legs, or input and ground. Would that change the taper as well? Or would it effectively change the value of the pot. I want to turn my 1M pot into a 500K by adding a 1M resistor in parallel. Would the added resistance be applied the same throughout the course of the pot?  
Thank you,  
9/6/2003 8:02 PM
Mark Lavelle

"I want to turn my 1M pot into a 500K by adding a 1M resistor in parallel. Would the added resistance be applied the same throughout the course of the pot?"
Not quite. With 1M in parallel, when your pot is at:  
100k (10%) you get 90k (18%)  
250k (25%) you get 200k (40%)  
500k (50%) you get 333k (67%)  
750k (75%) you get 430k (86%)  
900k (90%) you get 475k (95%)  
So the taper is different, with the biggest difference being at the lower values of the pot. I'd guess that in most places the diffeerence won't hurt...
9/7/2003 10:19 AM
Carl/Zwengel Amps

Mark, you're about half right here...  
By placing a resistor across the entire pot you get a parallel resistor arrangement where half the voltage goes to ground and the other half goes through the pot. So basically, you get double voltage division...the first as per Ohm's Law through the two parallel resistors, then what's left gets divided again through the pots taper. However, the load IS being changed to the equvalent resistance of the resistor parallel to the pot.  
Bottom line is that you get a load change as well as massive signal attenuation with this setup.  
Did I make that clear? It's kind of late here and I'm not quite as lucid as I should be right now.  
9/7/2003 6:18 PM

Thank you for the reply Carl.  
I want a load change and an attenuation. That is what I am looking for. I guess my question is: will the load change and attenuation be the same by adding the resistor as it would be by changing the pot altogether, and, will the taper be altered? Again, the resistor is between input and ground, and is not a tapering resistor between input and output.  
The question was for intellectual satisfaction more than anything else. I CAN change the pot, but will this solution be the same?  

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