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Tapering resistor noob here

Started by insomniac2295, November 15, 2012, 03:10:30 AM

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insomniac2295

So, I've been reading up on tapering resistors in: http://www.geofex.com/article_folders/potsecrets/potscret.htm
I'm interested in turning a Linear wah potentiometer into something resembling a A25K pot. Can this be accomplished via tapering resistors? It seems to me like it is possible, but I could be wrong. The techy-ness of that article is hard for me to understand at certain points. If it will work, what value wah pot should I be looking at getting and what value resistor?

midwayfair

#1
Quote from: insomniac2295 on November 15, 2012, 03:10:30 AM
So, I've been reading up on tapering resistors in: http://www.geofex.com/article_folders/potsecrets/potscret.htm
I'm interested in turning a Linear wah potentiometer into something resembling a A25K pot. Can this be accomplished via tapering resistors? It seems to me like it is possible, but I could be wrong. The techy-ness of that article is hard for me to understand at certain points. If it will work, what value wah pot should I be looking at getting and what value resistor?

Yes -- linear to "almost" log/audio taper conversion is possible.

You need to determine the appropriate slope of the line -- that's your b. It's not a specific variable and RG gives 6 different examples in his graphs. You can solve for b by plugging various values into Rt and R as trial and error until you find a slope that you like, or you can use one of the slopes he's provided and solve for Rt with R = 25 (the desired total pot resistance), then use the closest value you have for the solved Rt.

Edit: Woops, forgot a step. I suck at math. I believe you might have to use a pot that's double R to actually get the final resistance correct, since Rt is across one half of the pot.

Edit edit: Or maybe not. The wiper + lug 3 will still be 25K max.

insomniac2295

So lets get all the parts straight:

Rt is the taper resistor being used
R is the desired pot value (in my case, 25k)
b is the fraction of the pot being used to get the desired taper (in my case, 1/5, or 0.2)
Where does the pot value being used come into play?

insomniac2295

Wait a second, I just stumbled across this: http://www.diystompboxes.com/analogalchemy/emh/emh.html
Will this calculator help a brother out? My head is spinning a bit

midwayfair

#4
Quote from: insomniac2295 on November 15, 2012, 08:24:18 PM
So lets get all the parts straight:

Rt is the taper resistor being used
R is the desired pot value (in my case, 25k)
b is the fraction of the pot being used to get the desired taper (in my case, 1/5, or 0.2)
Where does the pot value being used come into play?

R is the total pot value (my "edit edit"), the actual value of the pot. If you're changing a 25K pot, R = 25K.

When it's a simple variable resistor (lugs 2+3 are soldered together) with a parallel resistor across the entire pot, the taper IS affected, but I'm not sure about the actual shape of the taper/slope for it (though I can tell you it's reverse log ONLY and never log, regardless of the placement of Rt), and in that case it follows the rule that parallel resistors divide, meaning you'd be using a 50K pot and a 50K (well, 51K) resistor. But it's much better to use it as a potentiometer as shown in RG's article than a simple variable resistor.

Basically, I'm saying that my paragraph before the edit was wrong and R is the pot being used but ONLY when it's being used as RG shows. Sorry for the confusion.

[Edited to fix "reverse log" and "log" in P 2]

insomniac2295

Thanks! I think I get it now.
However, I'm starting to see a problem for my original intent for this application. I was going to change the value of a pot to act as a replacement for the Reverb level on an Electro-Harmonix Holy Grail. However, I am just now realizing that going from a linear pot to a logarithmic pot can only be used in a voltage dividing application and I think my application requires variable resistance, so this method won't work.

insomniac2295

Okay, let's see if I have this correct: say I take a 250k linear pot and want to turn it into roughly a 25k logarithmic pot: my tapering resistor could be 28k (11.2% of 250k) and that would give me a pseudo-log taper with a resistance (following the formula for parallel resistors) of 25.18k