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05-07-2012, 11:01 PM #1
 Class 1 Laser Join Date: Nov 2011 Location: Vermont, USA Posts: 86 Rep Power: 9
skir0987
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Potentiometer valuation

I have a potentiometer that goes from about 0-50 k ohm, and I'd like to reduce it's range to ~0-100 ohm. Is there any way to do this with added resistors? I've tried combinations of series and parallel but I can't seem to figure it out. Maybe one of the electronics gurus here could help?

Thanks,
Shep

EDIT: Just for clarification, this is to give a 12v fan a range from ~5v to 12v. I measured empirically that 100ohm brings the speed down to about where 5v is.

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Last edited by skir0987; 05-07-2012 at 11:03 PM.

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05-08-2012, 12:38 AM #2
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Gryphon
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Re: Potentiometer valuation

Hmm, thats a bit tricky. If you put resistors in parallel you will only add to the total resistance at full swing. The only way to reduce resistance is to put extra resistors in parrallel with existing ones, this increases the number of paths current can take, thus lowering total resistance.
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Last edited by Gryphon; 05-08-2012 at 12:40 AM.

05-08-2012, 12:41 AM #3
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Re: Potentiometer valuation

The only way I see would be to to use a 100 ohm resistor parallel to the middle pin and the pin you are using. This would give a range of about 0-100 ohms, but it would be terribly non linear and practically only a switch.

All ways I can think of would either not allow something near 0 or near 100. Also there are better ways of controlling a fan then a potentiometer. If you run it with this low (5V) the efficiency would more then halve... Also if your fan drains a lot current the pot can get very hot.
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05-08-2012, 12:46 AM #4
 Class 4 Laser Join Date: Sep 2008 Location: Quebec, Canada Posts: 15,692 Rep Power: 39059
lasersbee
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Re: Potentiometer valuation

Buy another pot of the value you want....
They are not expensive...

Jerry
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05-08-2012, 12:47 AM #5
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Gryphon
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Re: Potentiometer valuation

Alright i think i got it. Just use a 100 ohm resistor in parallel with the pot. When the pot is at 50K ohms, the 2 resistors in parallel come out to about ~102 ohms total. Then when you ramp down the resistance of the pot, current will be biased throught the pot all the way till it hits nearly zero resistance in which case the 2 resistors total to under 1 ohm.

EDIT: Jerry's got one killer idea!
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I see................wait, what now?

(3x) 405nm
(6x) 445nm
(4x) 532nm
(1x) 543nm
(1x) 632.8nm
(1x) 635nm
(3x) 660nm
SSY-1 Laser

Last edited by Gryphon; 05-08-2012 at 12:48 AM.

05-08-2012, 01:32 AM #6
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Cyparagon
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Re: Potentiometer valuation

Quote:
 Originally Posted by Gryphon Just use a 100 ohm resistor in parallel with the pot. When the pot is at 50K ohms, the 2 resistors in parallel come out to about ~102 ohms total. Then when you ramp down the resistance of the pot, current will be biased throught the pot all the way till it hits nearly zero resistance in which case the 2 resistors total to under 1 ohm.
Not quite. 50kΩ || 100Ω is 98.8Ω. 0Ω || 100Ω is 0Ω.

The problem as stated above is the linearity. With 100Ω in parallel, The total resistance stays above 90 until the last 2% of the pot's rotation. This is pretty much useless. Just use a 100Ω pot
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Last edited by Cyparagon; 05-08-2012 at 01:36 AM.

05-08-2012, 03:05 AM #7
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skir0987
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Re: Potentiometer valuation

After some experimenting I believe you're right about nonlinearity. I couldn't come up with any combination to create a nice smooth transition. I may end up just using a switch to toggle between 5 and 12v... or get a 100Ω pot like mentioned

Shep
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