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Author Topic: Potentiometer Hack / Trick  (Read 8040 times)

Offline jadaro2600

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Potentiometer Hack / Trick
« on: February 19, 2010, 04:37:09 PM »
Potentiometers can be hacked, I suppose, this wouldn't really qualify as do it your self but:

Any three lead potentiometer can be hacked to provide a lesser resistance ( 1/4th that of it's over all ).

The basic idea is that you hook up both ends of the over-all resistance and take the path from the variable lead as one side and use one of the other leads.  This puts both paths in parallel.

The result is a potentiometer with a variable resistance exactly 1/4th that of the over all resistance.  The maximum resistance of the potentiometer is found in the middle position ( figure B ).

A 10k pot becomes a 2.5k pot, a 1m pot becomes a 250K pot, etc.

This can further be altered by biasing the variable lead and/or the over-all leads, as well as adding a resistor up front or at rear.

This is an old trick, really.   The diagram is just a simple concept / schematic.

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Potentiometer Hack / Trick
« on: February 19, 2010, 04:37:09 PM »

Offline MrMag

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Re: Potentiometer Hack / Trick
« Reply #1 on: February 19, 2010, 08:58:53 PM »
Not to sure about this. I don't remember seeing this in use before. One problem you will have is that you only have control 1/2 way. Once you reach the mid point, the voltage would start to increase. I haven't tried it but I think a 1K pot would become a 500 ohm one, not a 250.

I have seen the wiper tied to one of the ends of the rheostat but not in this configuration.

Offline jadaro2600

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Re: Potentiometer Hack / Trick
« Reply #2 on: February 20, 2010, 04:43:13 AM »
Not to sure about this. I don't remember seeing this in use before. One problem you will have is that you only have control 1/2 way. Once you reach the mid point, the voltage would start to increase. I haven't tried it but I think a 1K pot would become a 500 ohm one, not a 250.
   Technically, you have as much control on one half as the other, the resistance is symmetrical at midpoint, and asymmetrical in any other position, you would just need to know where the midpoint is to know where them maximum resistance is; I can see who this may be deceptive of the device's normal operating parameters, but it is technically a resistance hack.

Well, if you measure, it is like having two ( on a 1k rheostat ) 500 ohm resistors.  Since they are in parallel, the current would flow from the ends to the variable terminal.  this would give you resistance control from midpoint ( variable ) to end terminals on both sides.  This essentially puts the device in parallel mode.  Having it resist as 500 ohms at it's center, the parallel resistance is 1/2 that of the center resistance.  ..so 1/4th resistance, maximum at the center, and less resistance on either side.

Why would the voltage increase? ..a 1k rheostat would become a 0-250 ohm rheostat, 0 ohms would be just that, no resistance.

This would only work with a three-lead potentiometer - to enact a base resistance, place a resistor with the selected value up front of the potentiometer.

I think this works well for deriving multiple min/max values from a single potentiometer.

I have seen the wiper tied to one of the ends of the rheostat but not in this configuration.

Placing a resistor tied to the wiper would be doing the same thing, but the resistor would be a variable and the wiper would be a variable.

Say that the wiper was R1 and the bias resistor to the wiper was R2: on a 1k potentiometer...

The math for parallel resistors ( two of them ) is (R1*R2)/(R1+R2) ...biasing the wiper of a 1k ( R2 ) rheostat with a 500 ohm resistor ( R1 ) would give you:

Wiper at midpoint: 500*500 / (500+500) = 250 ohms
at 3/4 of 1k: 500*750 / (500+750) = 300 ohms
at 1k resistance: 333.33 ohms...

0 - 333.33 ohms over all.

This essentially yields the same results, it just uses both far end terminals for parallel resistance.  Biasing the wiper would result in larger variable ( 0 - Rx ) resistance but also result in smaller over-all resistance too. :)

 

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