Using A Tapering Resistor to Get Reverse Log Pot

Frequently asked question regarding resistors, potentiometers, types, requirements, ratings etc.
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jfromel
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Post by jfromel »

Solidhex wrote:Ah

I spoke too soon. To replicate a reverse audio voltage divider you need to use a tapering resistor that is 20% of the larger value running in parallel. Read this: http://www.geofex.com/article_folders/p ... tscret.htm and this: https://www.diystompboxes.com/smfforum/ ... ic=70732.0
Someone specifically notes that to make a 100k reverse log pot you need to place a 100k resistor across lugs 3 and 2 of a 500k linear pot. 100k being 20% of 500k and the parallel value of 500k and 100k being 100k. This is specifically related to the reverse audio voltage divider like the sustain knob in a Buzzaround.
So the reason the approximation of the revlog wasn't as good as the actual 100kc pot is probably because of those wrong pot/tapering resistor values.

--Brad
Resistance in parallel is always less than the smallest value. Placing a 100K resistor in parallel with a 500K will yeild a maximum resistance of about 83K, if you want to end up with 100K of resitance at about 85% taper you would use a 127K resistor in parallel.

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Post by joegore »

Wow guys -- thanks for the tips and links.

Okay, bear with me, 'cause I'm even worse at math than at soldering. Am I understanding this right?

To create a reverse log pot of a specific value, you take a pot of approximately five times the desired resistance, and then use a resistor with approximately 20% the value of the larger pot to "step it down" to the desired value?

If, for a C100K pot, you take a 500K and bridge lugs two and three with a 127K resistor, does that mean that:

C50K approximately equals B250K bridged with a 47K resistor across lugs two and three?

C10K equals B50K bridged with a 10K resistor across lugs two and three?

C5K approximately equals B25K bridged with a 4.7K resistor across lugs two and three?

C1K equals B5K bridged with a 1K resistor across lugs two and three?

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Solidhex
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Post by Solidhex »

jfromel wrote:
Solidhex wrote:Ah

I spoke too soon. To replicate a reverse audio voltage divider you need to use a tapering resistor that is 20% of the larger value running in parallel. Read this: http://www.geofex.com/article_folders/p ... tscret.htm and this: https://www.diystompboxes.com/smfforum/ ... ic=70732.0
Someone specifically notes that to make a 100k reverse log pot you need to place a 100k resistor across lugs 3 and 2 of a 500k linear pot. 100k being 20% of 500k and the parallel value of 500k and 100k being 100k. This is specifically related to the reverse audio voltage divider like the sustain knob in a Buzzaround.
So the reason the approximation of the revlog wasn't as good as the actual 100kc pot is probably because of those wrong pot/tapering resistor values.

--Brad
Resistance in parallel is always less than the smallest value. Placing a 100K resistor in parallel with a 500K will yeild a maximum resistance of about 83K, if you want to end up with 100K of resitance at about 85% taper you would use a 127K resistor in parallel.

Yeah I should have mentioned that. You can't get any value with any taper doing it that way. With the standard pot values available you have to either stray from the pot value to get the taper you want or have a less than logarithmic with the taper... Thanks for mentioning that.
I don't think this is uncommon for the Buzzaround circuit. I've noticed it a little but not enough to be bugged by it in my builds. A reverse audio taper would be cool there.
Don't forget that you can take a garden variety audio taper pot and wire it backwards to get reverse audio. Your control will function backwards but you will be able to dial in the fuzz better.

--Brad

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Post by joegore »

So can anyone tell me whether I was correct when I speculated:
joegore wrote:Wow guys -- thanks for the tips and links.

Okay, bear with me, 'cause I'm even worse at math than at soldering. Am I understanding this right?

To create a reverse log pot of a specific value, you take a pot of approximately five times the desired resistance, and then use a resistor with approximately 20% the value of the larger pot to "step it down" to the desired value?

If, for a C100K pot, you take a 500K and bridge lugs two and three with a 127K resistor, does that mean that:

C50K approximately equals B250K bridged with a 47K resistor across lugs two and three?

C10K equals B50K bridged with a 10K resistor across lugs two and three?

C5K approximately equals B25K bridged with a 4.7K resistor across lugs two and three?

C1K equals B5K bridged with a 1K resistor across lugs two and three?

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Post by Alex Frias »

If the pot maximun resistence is what we are trying to get, then according to the formula:
R1 x R2 / R1 + R2 = Rx
500k x 127K / 500k + 127k = 101.28
I think you can use 120k or even 150k with not much notable difference...

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Solidhex
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Post by Solidhex »

Yo

Like Jfromel and me mentioned when using a tapering resistor you can either get a 100% reverse log taper OR the parallel value you need. You can't get both at the same time so you have to make a compromise. If you do the math with a 500k pot... take 20% of 500k which is 100k then we go: 500 times 100, divided by 500 plus 100. You end up with what's supposed to be a perfect reverse log taper pot BUT the value of the pot will be 83.3K. Not the 100k we want. You have to up the value of the tapering resistor until you get close enough to the 100k you want. The taper will start to flatten out into more of a linear response as you do this so you need to strike a balance.

--Brad

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Post by Alex Frias »

OK

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Post by joegore »

Okay, can someone help fill in the blanks so the math numbskulls among us can print it out? (Actually, judging by what I read here, I may be the sole math numbskull on FSO, so let's just say it's for my benefit.) :oops:

The best way to simulate a C100K pot is to take a B500K pot and put a 127K resistor across lugs 2 and 3.
The best way to simulate a C50K pot is to take a ___ pot and put a ___ resistor across lugs 2 and 3.
The best way to simulate a C25K pot is to take a ___ pot and put a ___ resistor across lugs 2 and 3.
The best way to simulate a C10K pot is to take a ___ pot and put a ___ resistor across lugs 2 and 3.
The best way to simulate a C5K pot is to take a ___ pot and put a ___ resistor across lugs 2 and 3.
The best way to simulate a C1K pot is to take a ___ pot and put a ___ resistor across lugs 2 and 3.

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Post by SPeter »

Hi,
Here is the simple formula:

X = 1.25xA ( x=multiplying , /=dividing )

Where:
X is the value of the Resistor we should put across lugs 2 and 3.
A is the value of the RevLog pot we need.


Example:

We need 25KOhm RevLog pot !
Therefore A = 25KOhm.
X = 1.25xA = 1.25x25KOhm = 31.25KOhm !

Check up:

(5xA)x31.25/5xA + 31.25 =
=(5x25)x31.25/5x25 + 31.25 =
=125x31.25/125 + 31,25 =
=3906.25/156.25 = 25!

That's it, so simple!
Enjoy!
:D

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joegore
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Post by joegore »

Wow, I'm impressed! Math even I can understand. :applause:

Thanks!

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Post by GeekMacDaddy »

SPeter wrote: X is the value of the Resistor we should put across lugs 2 and 3.
A is the value of the RevLog pot we need.


Example:

We need 25KOhm RevLog pot !
Therefore A = 25KOhm.
X = 1.25xA = 1.25x25KOhm = 31.25KOhm !
sorry for the mini hijack. does this take into account the size of the pot used? for instance, would the same formula apply if I wanted a C25k pot made out of a B50k? or does this just apply to pots that are the same value, aside from the taper, which to my unmathematical mind makes more sense?

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SPeter
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Post by SPeter »

GeekMacDaddy wrote:
SPeter wrote: X is the value of the Resistor we should put across lugs 2 and 3.
A is the value of the RevLog pot we need.


Example:

We need 25KOhm RevLog pot !
Therefore A = 25KOhm.
X = 1.25xA = 1.25x25KOhm = 31.25KOhm !
sorry for the mini hijack. does this take into account the size of the pot used? for instance, would the same formula apply if I wanted a C25k pot made out of a B50k? or does this just apply to pots that are the same value, aside from the taper, which to my unmathematical mind makes more sense?
If you mean the value of the pot, yes it does,the formula can't be applied!
Out of B50K pot according to the formulas you can make 10K ReLog Pot - 5x10K=50K! In this case you need 12.5K resistor! For 25K RevLog you should calculate accordingly! in short you need 5x25K=125K Lin Pot and 25Kx1.25=31.25K resistor soldered on lugs 2 and 3!

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Post by GeekMacDaddy »

Thanks for the clarification.

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