Calculating Parallel resistances for dummy's (me)

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Reg18
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Calculating Parallel resistances for dummy's (me)

Post by Reg18 »

So this seems like a simple thing but my small brain just can't work this out with a standard caculator.
Can someone explain to me in simple terms please?
Simpler than this......
http://m.wikihow.com/Sample/Parallel-Resistance

I'm just trying to add different resistors for my Tele bridge single coil pickup (has 500k pots for the neck humbucker) I've been using my ears to find the best sound but to save me hours of experimenting would be good to know what the value will be before I break out the soldering iron again.

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Re: Calculating Parallel resistances for dummy's (me)

Post by sizzlingbadger »

Not sure quite what your trying to do but....

If you put a 500K resistor across the outer lugs of the pot it will become a 250K pot.
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Re: Calculating Parallel resistances for dummy's (me)

Post by TmcB »

Make the Mexicans hold hands and the total amount of Mexicans gets bigger.

To sneak them over the border you need to stack the Mexicans on top of each other to fit them in the van but oh noes! They're too fat and they've squished together into half the space they used to take up! Being a macabre bastard, you done some calculations. You've reduced the amount of Mexicans you're sneaking over the border by half due to your stupid stacking dea.

Or less metaphorically, if you chain resistors end to end the number gets bigger, stack them the total reduces.
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Re: Calculating Parallel resistances for dummy's (me)

Post by Starfire »

TmcB wrote:Make the Mexicans hold hands and the total amount of Mexicans gets bigger.

To sneak them over the border you need to stack the Mexicans on top of each other to fit them in the van but oh noes! They're too fat and they've squished together into half the space they used to take up! Being a macabre bastard, you done some calculations. You've reduced the amount of Mexicans you're sneaking over the border by half due to your stupid stacking dea.

Or less metaphorically, if you chain resistors end to end the number gets bigger, stack them the total reduces.
Donald?

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Re: Calculating Parallel resistances for dummy's (me)

Post by Reg18 »

sizzlingbadger wrote:Not sure quite what your trying to do but....

If you put a 500K resistor across the outer lugs of the pot it will become a 250K pot.
Sort of, I've got a resistor going from the bridge pickup lug on the 3 way switch to ground which reduces the resistance, I can't even remember what value the resistor is exactly because I just kept experimenting until it sounded about right.
However I've changed the wiring again and want to have a push pull pot (or dpdt) selecting 2 different resistors, one so the pickup sees approx 250k and another so it sees less, maybe 150k for a darker tone.
But I would like to figure this out myself so in future I don't spend hours soldering in different ones just using my ears to gauge it.

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Re: Calculating Parallel resistances for dummy's (me)

Post by borge »

Reg18 wrote:selecting 2 different resistors, one so the pickup sees approx 250k and another so it sees less, maybe 150k for a darker tone.
There are lots of online tools that'll do math for you, WolframAlpha is great:

http://www.wolframalpha.com/input/?i=R+ ... 0,+p%3D500

R is the total resistance, r is the resistor value, p is the pot value, so the link finds the resistor value for 150 kOhms total with a 500 kOhm pot; you need a ~220kOhm resistor. You can enter any values you need.

Adding parallel resistance to log pots does bad things to their taper, you're better off using the right pot in the first place than adding resistors.
Last edited by borge on Sat Jun 11, 2016 12:18 pm, edited 1 time in total.

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Re: Calculating Parallel resistances for dummy's (me)

Post by Reg18 »

borge wrote:
Reg18 wrote:
sizzlingbadger wrote: selecting 2 different resistors, one so the pickup sees approx 250k and another so it sees less, maybe 150k for a darker tone.
There are lots of online tools that'll do math for you, WolframAlpha is great:

http://www.wolframalpha.com/input/?i=R+%3D1%2F((1%2Fr)%2B(1%2Fp)),+R%3D150,+p%3D500

R is the total resistance, r is the resistor value, p is the pot value, so the link finds the resistor value for 150 kOhms total with a 500 kOhm pot; you need a ~220kOhm resistor. You can enter any values you need.

Adding parallel resistance to log pots does bad things to their taper, you're better off using the right pot in the first place than adding resistors.
This is a big help, thank you!
When you say it changes the taper of the pot, is the tone significantly changed when the pot is on full?
I need the neck humbucker to see 500k and the bridge single coil to see 250k so there will always be some comprimise with the set up.

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Re: Calculating Parallel resistances for dummy's (me)

Post by sizzlingbadger »

If the resistor is across the outside lugs the taper will be ok. If its between the middle lug and an outer lug then the taper will change, in fact it can be quite useful if you want to customise the taper. The taper doesn't effect the tone, it just effects the value of the resistance at a certain mechanical (rotational) position of the pot.

This website is well written IMHO http://www.geofex.com/article_folders/p ... tscret.htm
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Re: Calculating Parallel resistances for dummy's (me)

Post by TmcB »

sizzlingbadger wrote:If the resistor is across the outside lugs the taper will be ok. If its between the middle lug and an outer lug then the taper will change, in fact it can be quite useful if you want to customise the taper. The taper doesn't effect the tone, it just effects the value of the resistance at a certain mechanical (rotational) position of the pot.

This website is well written IMHO http://www.geofex.com/article_folders/p ... tscret.htm
Totally agree, that's my reference for when I need to quickly rustle up a particular value pot if I've not got it
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Re: Calculating Parallel resistances for dummy's (me)

Post by Reg18 »

Just for reference my Tele has only got 1 x Volume pot (no Tone)
And by the caculations from this site a Standard Tele with 1 x 250k Volume and 1 x 250k tone is actually about 125k total resistance that the pickups see, so with a 500k Volume and no tone to achieve the same results I'd need a 166k resistor going from hot to ground.

Can anyone confirm if my maths is correct?

http://www.mk-guitar.com/2014/12/11/tot ... ler-sound/

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Re: Calculating Parallel resistances for dummy's (me)

Post by TmcB »

Your maths is correct, I'm a bit dubious about that article though.
Last edited by TmcB on Sat Jun 11, 2016 11:41 pm, edited 1 time in total.
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Re: Calculating Parallel resistances for dummy's (me)

Post by sizzlingbadger »

Reg18 wrote:Just for reference my Tele has only got 1 x Volume pot (no Tone)
And by the caculations from this site a Standard Tele with 1 x 250k Volume and 1 x 250k tone is actually about 125k total resistance that the pickups see, so with a 500k Volume and no tone to achieve the same results I'd need a 166k resistor going from hot to ground.

Can anyone confirm if my maths is correct?

http://www.mk-guitar.com/2014/12/11/tot ... ler-sound/

no its not correct.

your pickup sees a 250K resistor across it and a 250K resistor in series with a cap, that is not 125K in total.

In simple terms you have 250K volume pot across the pickup. The tone control is almost open circuit except at the highest frequencies.
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Re: Calculating Parallel resistances for dummy's (me)

Post by TmcB »

sizzlingbadger wrote:
Reg18 wrote:Just for reference my Tele has only got 1 x Volume pot (no Tone)
And by the caculations from this site a Standard Tele with 1 x 250k Volume and 1 x 250k tone is actually about 125k total resistance that the pickups see, so with a 500k Volume and no tone to achieve the same results I'd need a 166k resistor going from hot to ground.

Can anyone confirm if my maths is correct?

http://www.mk-guitar.com/2014/12/11/tot ... ler-sound/



no its not correct.

your pickup sees a 250K resistor across it and a 250K resistor in series with a cap, that is not 125K in total.

In simple terms you have 250K volume pot across the pickup. The tone control is almost open circuit except at the highest frequencies.
Agree, I'm not sure wtf the article is on about - seems to not get that a tone control is a low pass filter and isn't like just adding more resistance.
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Re: Calculating Parallel resistances for dummy's (me)

Post by sizzlingbadger »

I didn't actually look at the article, just the response above :winky:

The value of the volume pot will make a difference to guitars tone because it forms part of the "resonant" circuit with the pickup. A 500K pot will sound different to a 250K pot. If you reduce the volume of a 500K pot to 250K by turning the pot down to half its resistance it will not sound the same as a 250K pot, there will also be much less output because you will be at half the volume. The pickup is still "seeing" the 500K pot.

The tone pot is different. Because the tone pot is in series with a small capacitor it only effects the higher frequencies. This means a 500K and 250K pot don't really effect the overall tone (resonance circuit) of the pickup. (strictly speaking it does but its very small and I doubt you could hear it.) However... they will sound different at their respective maximums because the resistance to ground for the higher frequencies will be different and this you can hear. If you take a 500K tone pot and turn it down to 250K it will sound pretty much the same as a 250K pot on maximum. This is why I always use a 500K tone pot as I will have more range to choose from.

In very simple terms...

The volume pot value effects the overall tone of the guitar.
The tone pot value only effects the range of the tone control.


You need a 500K resistor across the pot (or from the hot wire to ground) to give you an effective 250K volume pot.
You need a 220K resistor across the pot (or from the hot wire to ground) to give you an effective 150K volume pot.

To make it easier you could just use two 500K resistors. Switch one in to the circuit an you will have a 250K load, switch the second one in (parallel with the first) and you will have 166K load, probably close enough.
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Re: Calculating Parallel resistances for dummy's (me)

Post by Reg18 »

[/quote]
To make it easier you could just use two 500K resistors. Switch one in to the circuit an you will have a 250K load, switch the second one in (parallel with the first) and you will have 166K load, probably close enough.[/quote]

I think this sounds like the easiest solution with a push pull pot.
Thanks for all the input so far guys, I assumed there must be a bunch of people who understand this stuff better than me on here and I haven't been disappointed!

There must be some difference without a Tone pot in the circuit though as there is a difference in high end with a Fender No load tone pot, which from what I understand bypasses the tone pot. I'm not sure if it is anything to do with the overall resistance but it certainly gains some high end when bypassed.

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