So, I've been reading up on tapering resistors in: http://www.geofex.com/article_folders/potsecrets/potscret.htm (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?
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 (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 R
t 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 R
t.
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 R
t is across one half of the pot.
Edit edit: Or maybe not. The wiper + lug 3 will still be 25K max.
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?
Wait a second, I just stumbled across this: http://www.diystompboxes.com/analogalchemy/emh/emh.html (http://www.diystompboxes.com/analogalchemy/emh/emh.html)
Will this calculator help a brother out? My head is spinning a bit
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 R
t), 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]
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.
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