Lowering pot values

[vendettav]

Hey guys, I know how it works. Though I was looking for this answer and couldn't find one.
Anyway's here's a short tutorial with a question first when doing this, does the tape change to linear (B) or log (A) ?

to lower the pot value all you need to do is wire a resistor around the outer lugs of the pot
the formula for this is Rsum=R1*R2/(R1+R2)
example: if you want a 10k pot from a 20k pot, you gotta take a 20k resistor and wire it. 20k*20k/(20k+20k)=10
to get a value less than the half of your pot you'll need about 1/3rd of the pot's original value to get more thanthe half you'll need less than 1/4th

anyways can somebody please answer the question on the taper?

[petemoore]

  taper what would be another.
   What's it responds variously with potaper.
  The DMM can chart the taper, put a knob/1-10 dial on the pot, start at 0 read one side, move the knob, repeat.
  Measure the wafer/potvalue, the other side can be taken as the subtracted portion, the two Variable Resistances in the VRDivider [3 prong pot] always add to the wafer resistance.
  A chart with 1-10 up the sides, and 0.0ohm through ? 10k...whatever the R value the potwafer measures [lugs 1 and 3].
  Do this at different knob positions until a taper-curve line is of sufficient resolution to connect the dots.
  Explained as a matter of explaining, also may help to consider the variable resistive divider [pot] as 2 resistances to which parallel resistances can be calculated in. Ie throw a value or two +/- the pot value or even...
  Something valued about 1/2 pot value, thrown across wiper and 1 outside lug, or resistor string across the VRD 'string' [essentially the pot is two resistances in series at all times, putting a resistor across a portion of the pot [wiper and a lug] puts a different curve on whatever curve or linear behaviour the wafer had originally.
  When a smallish fixed R is put across a Large VR the small resistance is dominant over 'more of the variable resistor [1/2 of a pot] range] the pot only begins noticably changing anything as it is dialed to low enough resistance [ie near the very end of the wafer] bunching up all the change in R to that end of the shaft rotation.
  So...depending on whatever and how it responds, a bigger pot to 'shave off of' might be in order if available.
  Also series stop resistor might 'bump the range' while this will lose some unwanted 'end' control. Stop resistor>pot, series is what that looks like, obviously the stop resistor is relatively small or 'tendancy gobbles up the pot-range-effect on 'whatever.

[LucifersTrip]

Hey guys, I know how it works. Though I was looking for this answer and couldn't find one.
Anyway's here's a short tutorial with a question first when doing this, does the tape change to linear (B) or log (A) ?

to lower the pot value all you need to do is wire a resistor around the outer lugs of the pot
the formula for this is Rsum=R1*R2/(R1+R2)
example: if you want a 10k pot from a 20k pot, you gotta take a 20k resistor and wire it. 20k*20k/(20k+20k)=10
to get a value less than the half of your pot you'll need about 1/3rd of the pot's original value to get more thanthe half you'll need less than 1/4th

anyways can somebody please answer the question on the taper?

The more you lower the K of the original pot, the more the taper will be weighted towards the beginning portion of the sweep.

It is very easy to see if you plot a graph. Ie, if you measure a 10K linear pot that you made with a 20K pot + 20K resistor, you will see that it is > 5k if you turn it half way.  The more you change the pot, the more the sweep is weighted towards the beginning. Ie, if you made a 10K pot from a 100K pot + 11K resistor, the sweep will be much more heavily weighted towards the beginning of the sweep.

[mwynwood]

If Rsum=R1*R2/(R1+R2), does that mean that:

If I put a 1k resistor over pins 1 and 3 of a 25k pot, I will effectively get a 1k pot?

(25*1)/(25+1) = 25/26 = 0.961k

[LucifersTrip]

yes, but with a massive change in the sweep

remember, 1-3 is just a resistor
http://www.1728.org/resistrs.htm

[mwynwood]

Awesome, thanks 

By flipping that formula, I can work out what resistor to use to get the value I need...

Resistor = (Pot*Result) / (Pot-Result)