If i parallel a 82k resistor with a 25k linear pot....

[daz]

It'll change the pot to about 20k. The question is, will the taper still be linear?

[The Dude]

Short answer? No. Here's some info kindly provided by RG that will give you more detail.

The Secret Life of Pots

Edit: That doesn't mean don't do it. Sometimes results are undesirable, sometimes not. It depends on the circuit.

[loudthud]

How you can sometimes do parallel resistors in your head:

You want something in parallel with 25K to make 20K. Note that 25 is 100/4 and 20 is 100/5. So the answer is 100K.

If you can't find the common numerator, (1/20K) - (1/25K) = (1/100K)

Use the 1/X key: 20K {1/X} - 25K {1/X} = {1/X}

[R.G.]

Actually, the answer is "depends on how you connect it". That point was not clear from the OP's question.

If you connect an 82K resistor across the two ends, hot/CW and cold/CCW, then you have not changed the pot's appearance to the outside world at all, all you've done is load down whatever drives the pot with an additional 82K to drive. If you connect the 82K from wiper to hot, or wiper to cold, you greatly change the effective taper of the pot as seen from the outside world at the wiper, and also change the loading on whatever drives the pot as the pot is turned, so you again have to take into account the impedance of the source driving the pot and the load on the wiper. The stuff in "The Secret Live of Pots" talks about the end-to-wiper connections a lot, probably none about putting a resistor across both end terminals of the pot.

It took me a long time to learn to automatically worry about the source impedance driving a circuit and the load impedance on the output. These two things are almost never on a schematic, but they change the circuit's action, sometimes a lot.

And finally, good observations, LT. I never scribble out R1*R2/(R2+R2) to find parallels any more. I add the conductances, which is 1/R, then take the inverse of the sum. It's as easy on most scentific calculators (about $9-12 at Frys, for example) as:

- enter R1
- press 1/x
- press +
- enter R2
- press 1/x
- press =
- press 1/x

This relies on the calculator to to the 1/x on the X entry, not the stuff in the hidden "Y" that will be used when the "=" key is pressed.

You can enter many R's and then 1/x and get the parallel of many resistors.

I also like using this backwards, as what I usually want to know is what resistor do I need to parallel some fixed resistor to get some smaller value.

If a resistor in a circuit is, say, 75K, and I want to find a resistor to parallel it down to 35K, I do

- enter 35 (the "k" can just be understood, or you can enter "35000" if you like)
- press 1/x
- press "-"
- press 75 (or 75000 if you're using 0's, not remembering them)
- press 1/x
- press "="
- press 1/x, and the result is in k to get 65.625K

Resistors and inductors add in series. Capacitance is really a conductance, so they add in parallel.

Conductances (1/R, 1/L) also add in parallel, and inverse capacitance (1/C) add in series.

[Justin Thomas]

Time to bust out my old TI regularly jnstead of punching in 1/((1/R1)+(1/R2)) = into my dumb-phone...
Thanks!

[The Dude]

Not sure how dumb your phone is, but, depending on your OS, most phones have a scientific calculator option in the calculator menu. You might have a look.

Edit: Or a button somewhere to switch back and forth.

[Justin Thomas]

It's a 4-year-old LG slider with a full button keyboard... and it was obsolete when I got it.

Justin

[d95err]

Is there really a significant difference between 25k and 20k in your circuit? I'd say thay in 99 of 100 circuits, the difference would not matter at all.

So, are you sure you can't just use the 25k pot as it is?

[daz]

Is there really a significant difference between 25k and 20k in your circuit? I'd say thay in 99 of 100 circuits, the difference would not matter at all.

So, are you sure you can't just use the 25k pot as it is?

Yes, but i don't want to discuss that i just want to know what would happen to the taper. If it's not going to remain linear, what will it do....become like a audio? More exaggerated audio? It will be used as a variable resistor, that is center and one side used, and the 82k would be strapped across the outside lugs.

By the way, i have done this in guitars and the taper didn't seem to change, but then it was a much closer match to 1/2 the value, IE:220k across a 250k. Is that a deciding factor....how close the match is between the pot and resistor? I guess i should just try it and measure it now that i think about it !

[d95err]

If the difference is that small, the change in taper will probably not be noticeable.

[J M Fahey]

It'll change the pot to about 20k.

No it will not, sorry.

The question is, will the taper still be linear?

Yes.

[daz]

No it will not, sorry.

I used "electronics assistant" to calculate it. I added 2 resistors in parallel, a 25k and a 82k. Here's the result i copy/pasted from the app itself. 19.1588785046729 Kilohms

[R.G.]

Yes, but i don't want to discuss that i just want to know what would happen to the taper. If it's not going to remain linear, what will it do....become like a audio? More exaggerated audio? It will be used as a variable resistor, that is center and one side used, and the 82k would be strapped across the outside lugs.

If you're strapping an 82K - well, any resistor, of any value at all - across the two outside lugs, the effective taper of the pot, that is, the fraction of the end-to-end voltage on the outside lugs that appears from one lug to the wiper, does not change in any way. However, adding a paralleled resistor across the outside lugs may very well load down the voltage applied to the pot, and so the pot's output will still be linear taper, but a smaller voltage because the additional resistor lugged down the thing that was supplying the pot.

If you strap a resistor between wiper and either end lug, then it changes the effective taper. How much it changes the taper depends on how big or smaller the added resistor is compared to the pot. Obviously a resistor that's BIG compared to the pot resistance will have almost no effect. Imagine putting a 10M resistor across any two lugs of the pot. No change for pots 1M and lower. A 10 ohm resistor - big changes. So the ratio of added resistor to pot DOES matter, a lot, if it's between the wiper and either other leg. The math for figuring that out is what went into "The Secret Life of Pots".

By the way, i have done this in guitars and the taper didn't seem to change

That's right, it didn't. It just loaded down the pickups, etc.

Also notice that whatever is connected to the wiper changes how much the paralleled resistor changes the pot's action.

It is **impossible** to really figure out what is happening without knowing what drives the pot, and what loads it.

But the taper will not change if the added resistor is from outside lug to outside lug. The extra loading on the driving circuit may change it's tone balance, which is what I think you're after, but it will not change the pot's taper.

[daz]

It is **impossible** to really figure out what is happening without knowing what drives the pot, and what loads it.

But the taper will not change if the added resistor is from outside lug to outside lug. The extra loading on the driving circuit may change it's tone balance, which is what I think you're after, but it will not change the pot's taper.

Well, it's like this. I am using it as a presence and i found that i like the sound various points down to 20k resistance and don't like what happens lower than that, plus i want the end of the pot's throw to be that point i like so i don't have to guess at it and make sure it's pointed right where i like it. I can just turn it all the way down. But i DO also like to add more highs so i DO use it brighter too. I just want the sweet spot to be the stop point on the knob and a 82k across a 25kpot will yield the 220k where i like it at the end of it's travel. I might string a few together depending on the pot's actual value (haven't gotten it yet) to get exactly 20k.

[J M Fahey]

I used "electronics assistant" to calculate it. I added 2 resistors in parallel, a 25k and a 82k. Here's the result i copy/pasted from the app itself. 19.1588785046729 Kilohms

So ?????????

The calculation you made shows two resistors in parallel, in this case a fixed one ... and another fixed one, since you are taking end to end track resistance.

Might do if you want to compare a a 20k pot set on 10 with a 25k pot, set on 10 , but with an 82k resistor in parallel .
At any other setting from 1 to 9 they will not be the same, hence my comment.

But a potentiometer is not a fixed resistor , itīs a variable one , and obviously you need a pot there or would simply use a fixed 20k resistor there and call it a day.

Do we agree so far?

So, IF adjustment is implied, letīs see what we have at different settings, and see whether they are the same or not.

Many Math problems include something called "the trivial solution", (trivial as in "silly") ,what we might call the "duh???" solution in everyday language, the one youīll say about: "hey dude , is it a joke? I already knew that!!! it does not help me" ... yet you can not say itīs *wrong* , it IS a solution even if nothing new.
As in: "what tube does replace a 6L6 and works exactly the same?"
Trivial solution: "another 6L6" .

Here both solutions match perfectly ... at 2 settings and those settings only, so they might be called "trivial solutions" to your doubt.:
a) both match on 10 , you just did the Math , both have 20k total. (wonīt complain about the tiny difference, thatīs not the point here)
b) both match on 0 ... both show 0 ohms.

so far so good, but itīs an adjustable control, what will happen if we set it to numbers different from "trivial solutions" at 0 and 10?

Ok, we might do it at every setting from 0 to 10 but to simplify letīs split the difference and calculate at 5.
You can repeat it at all other values ifnyou wish.

* A 20 k pot set to 5 will show 10k , period.

* A 25k pot, set to 5, and with an 82 k resistor soldered end to end, will show:
10 k (half track)- in parallel with - 10k (the other half track) in series with 82k=9k

Not the same. (hence my comment).