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|4/30/2003 10:33 PM|
||figuring out resistance level|
I have read RG Keen's Secret Life of Pots article, found here: http://www.geofex.com/Article_Folders/potsecrets/revlog.gif
I am using a 500k Ohm potentiometer, and was wondering what level of resistor I should be using in parallel to obtain b = 8. Would this just be 500k/8? Somewhere around 60k ohm resistor? By the looks of the graph the 8 level of b would be a steeper and quicker climb to the peak level, where as a lower level would be a smoother climb, right? For instance with b=0 it shows a straightened diagonal from beginning to end. My goal is to make the volume pot peak very quickly, so if I understand correctly, wiring it as a reverse log, and with a 60k Ohm or so resistor, it will do this?
|5/1/2003 12:28 AM|
The original level of potentiometer should come into play here as well, right? If there are differences between a 250k, 500k, and 1meg ohm potentiometer, then shouldnt the equation have another variable?
|5/5/2003 12:47 PM|
If what you want is to convert a 500K linear pot into a 500K reverse log, you must solder a 100K resistor between the center ans CW lugs.
If you can read spanish go to my page:
It's based upon the article "The secret life of pots" by R.G. Keen:
but what's new in there is that there is a spreadsheet file: TapeRes.zip (Lotus 123 format but can be used with Excel). Load it and "see" graphically how different tapering resistors affect the pot's taper.
|5/5/2003 8:17 PM|
Yes, I read RG's article thoroughly. I'm wondering where you got the 100k value? If it is using the same equation as on RG Keen's site, then your value of b was 5, which RG showed as the uppermost standard value for such a pot taper...Although I wanted something that was even more extreme than that, so wouldn't a lesser value such as 60k do the job?
|5/6/2003 9:33 AM|
Hmm... I think I misunderstood your initial request I've reread it and now I see that what you want is a "steeper" or "more abrupt" antilog pattern. 100K as tapering resistor is to "convert" a 500K lin pot into a more or less standard 500K antilog pattern (wiper to CW lug)
If what you want is indeed a steeper rise curve, use my aforementioned spreadsheet (http://www.pisotones.com/Potes/Potes.htm ) to "see" graphically the effect of other values for the tapering resistor. Change cells C1, C2 & C3:
R1 = total pot value = 500
R2 = tapering resistor for alog pattern = 100
R3 = "9e9" i.e. infinity or no resistor at all.
See the shape of the green curve in the graphic. Change R2 to 60 (your proposed value) and watch the curve to become steeper. Play with the values.
Hope it helps.
|5/6/2003 3:23 PM|
I cannot open the spreadsheet. All I have at home is Microsoft Works, and works views Lotus files, as well as Excel, but it won't read your file...
Therefore I don't know what cells C1, C2, and C3 represent?
|5/7/2003 10:54 AM|
Right. I have translated it into English and exported to Excel format:
Hope it's self explanatory.
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