Resistor in parallel w/ pot

[gutsofgold]

I have a 250k pot, and I want it to work as a 100k.
Assuming (R1 x R2)/(R1 + R2) is the new resistance, I need ~190k ohm resistor to get this?

And it would go across lugs 3 and 2, no?
Thanks!

[Solidhex]

I believe you place it on the outside lugs...

--Brad

[zombiwoof]

To parallel a resistor, it goes across the two outside legs, in other words 1 and 3.  If you put it from wiper to an outside leg, you are making it a tapering resistor, changing the taper of the pot.

Al

[5thumbs]

I have a 250k pot, and I want it to work as a 100k.
Assuming (R1 x R2)/(R1 + R2) is the new resistance, I need ~190k ohm resistor to get this?

And it would go across lugs 3 and 2, no?
Thanks!

You can accomplish what you're after with tapering resistors as well.  Check out http://www.diystompboxes.com/analogalchemy/emh/emh.html.  There are calculators there for creating custom value linear and log pots with larger-value pots and resistors.

You didn't say if you were trying to create a linear or log taper pot, but if you want to make a 100K linear pot from a 250K linear pot, you'd need to put an 82K resistor between lugs 1+2, as well as lugs 2+3.  This will give you a 99K linear pot.  (To get a 100K linear pot, you'd need to use 83K resistors, which I could not locate in a quick search.)

For a 100K log (audio) pot with a 90% taper, you'd need a 250K linear pot.  Solder a 320K resistor between lugs 1+2 (or 2+3, your pick), then solder a 10K resistor between the other lugs.  If you want a different taper percentage, visit the aforementioned page, pick your taper and it'll tell you what you need to use to get the resistance and taper you're looking for.

Good luck!

[gutsofgold]

So many variables   

I need to turn this pot into a 100k, http://www.allparts.com/store/ep4485-000,Product.asp
I'm assuming it is a linear taper pot so I might as well keep it linear. Thanks!

[zombiwoof]

So many variables   

I need to turn this pot into a 100k, http://www.allparts.com/store/ep4485-000,Product.asp
I'm assuming it is a linear taper pot so I might as well keep it linear. Thanks!

I don't know if you need both sections of that pot to be 100k, but Allparts also has a Dano concentric pot that has one 100k section:

http://www.allparts.com/store/electronics-miscellaneous-pots-ep-4868-000,Product.asp

Al

[gutsofgold]

Yup, both need to be 100k (one is the tone pot for a Big muff, the other is a blend control). That 1Meg/100k would make things easier, just need a resistor on the 1Meg than. Thanks !

[gutsofgold]

Ok so using this calculator here...

http://www.diystompboxes.com/analogalchemy/emh/emh.html


I am told I will need two 83k Ohm resistors. Where do I place them tho?

[zombiwoof]

Ok so using this calculator here...

http://www.diystompboxes.com/analogalchemy/emh/emh.html


I am told I will need two 83k Ohm resistors. Where do I place them tho?

Just like the diagram, the arrow is the wiper (middle lug), the dots at each end are the other two lugs, so the resistors each go from the wiper to the outside lug.  In other words, one goes from lug 1 to lug 2, the other goes from lug 2 to lug 3.

Al

[John Lyons]

Let me see if I can get this right.
(All of the diagrams above are shown as voltage dividers as in a volume pot.)



(Fig A) With a linear pot and both fixed resistors the same value, the pot will be linear correct?
Won't fig D be linear also, starting with a linear pot?

With the same set up as Fig A the fixed resistors can be altered to make linear, log or anti log correct?
The fixed resistors would alter the curve of the taper and also allow you to make an S taper pot.

Once I get all this straight I need to make a diagram of all the mutations...sheesh!

John

[flo]

Not to steal the thread but what does the single taper (as a percentage) value mean in: Tapered Potentiometer, http://www.diystompboxes.com/analogalchemy/emh/emh.html
In http://www.geofex.com/article_folders/potsecrets/potscret.htm is stated that the taper op a pot has a "curve" of the "fraction of total rotation" against "resistance division ratio". This I understand. So can the whole curve of the linear pot with resistors be calculated in: Tapered Potentiometer, http://www.diystompboxes.com/analogalchemy/emh/emh.html

[John Lyons]

The percentage of the taper is the slope of the pot taper/curve I beleive.
I assume 90% is an extreme log taper and 0% is linear...?

I'm not sure how the percentage relates to Log vs reverse log though.

In anyone can confirm or deny my grasp of the taper diagams above I'd appriciate it.

Thanks

John

[John Lyons]

"please allow me to bump thee"
anyone guess that lyrical quote?

[dschwartz]

too much information here for somethin that simple.. just put a 190k resistor between outer lugs ..the taper will remain the same...

[5thumbs]

"please allow me to bump thee"
anyone guess that lyrical quote?

That's from 'The Humpty Dance' by Digital Underground.  Do I win some of puretube's free chips for answering first? 

As for the taper percentage question, I think 50% taper is linear, < 50% is reverse-log and > 50% is log.  Somebody please correct me if I'm incorrect, but that's how I grokked it.

[flo]

So is this taper percentage value, the percentage of the total taper width when the pot is set halfway or something?
So for a log pot this is 10% and for a linear pot this is 50% and for a reverse pot 90%?

[stinki]

Beware weary internet traveler.

There are some rather imaginative theories about parallel resistance here.

No, putting equal values resistors across pins 1+2 and 2+3 does not magically create a smaller value pot with a linear curve. Why wouldn't anybody actually try it to find out (or just think about how parralel resistance works when one resistance is constantly changing) before making a broken "calculator" for this?

For example, this magical calculator tells me that a 50K linear pot can be turned into a 25K linear pot by putting 25K resistors across pins 1+2 and 2+3. LOL. No. That creates a pot with 25K resistance at *center* position, and about 12.5K at the outer pins. Perfect!

[iainpunk]

welcome to the forum
that's correct, there is some sketchy maths going on here.
cheers

[anotherjim]

Well, at extremes the pot shorts out one resistor. I make that 17k.
What d'yer think of this?
https://www.diystompboxes.com/brisance/old/?R1=25000&R2=25000&VR=50000

[Phend]

OK, so many pot - resistor in parallel topics here.
Aka, changing 10K pot into an Approx 1K pot for instance.

So I connected a 10k pot (1.20R to 9.52K per DMM) and a 978R to my BB.
(978R in // with 9.52K = 887R)

I put the 978R resistor across 1 and 3.
Measuring across 3 and 2,  I get:
1.20R going Up constantly to 2.7K.
Then Drops constantly to 890R.
So this isn't going to be a 10K pot turned into an Approx 1K pot using this method.

Lets try two near equal 978R resistors across 1-2 and 2-3.
Measuring across 3 and 2 :
Goes Up near constantly from 1.20R to 890R.
This looks better.

Now I will try two ~1.2K resistors to get closer to 1K......

Moral of the story? BB it and measure.