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Potentiometer Loading
6/5/2017

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D Feucht
6/12/2017 7:14:35 PM
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Re: Loading a linear pot can be USEFUL
GS & Aubrey,

Thanks for your substantial additions as comments to my article. GS, perhaps you might do a sequel article to this one and put in your algebra. I submit articles to Steve Taranovich as MS Word files in which the math is done with the MS Word math editor. All comes out okay this way.

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Mark M
6/11/2017 11:08:06 AM
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Thanks
It is a very interesting subject

Thanks

Mark

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antedeluvian
6/11/2017 7:41:39 AM
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Re: Loading a linear pot can be USEFUL
@GSKrasle I am travelling at the moment and so I haven't even had a chance to consider you previous post. I don't have you email, but since mine is fairly well publicized, we could start with that

a k a g a n  a t  e m p h a t e c  d o t  c o m

- remove all the spaces and replace at and dot with their symbols

-Aubrey

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GSKrasle
6/10/2017 5:44:27 AM
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Re: Loading a linear pot can be USEFUL
Ugh: I wish I could edit that!

If you can privately e-mail me, I can give you the spreadsheet and plots. I SHOULD revisit it and make it prettier, but it serves.

I should also mention that I have seen the curves for the commercial 'log' (actually 'exponential') pots, and they are poor curve-fits, certainly worse than my solution.

Try THIS though!
https://xkcd.com/356/
I worked a related problem in three dimensions, and the answer is 376.73.

I still haven't worked-out its relation to ^(?=(?!(.)\1)([^\DO:105-93+30])(?-1)(?<!\d(?<=(?![5-90-3])\d))).[^\WHY?]\$

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GSKrasle
6/9/2017 4:39:17 PM
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Re: Loading a linear pot can be USEFUL
Aubrey,

It's somewhere on EETimes or Planet Analog, I'm sure, but hete's a synopsis:

I enjoy algebra, so I tried the solution. I'm not CERTAIN I'm correct, (because nobody cares to check me), but the solution is usable.

I used iterations of the solver in Excel2007 to least-squares curve-fit to simple exponantial and log curves.

The "meat" is the algebraic expression in col G for Vo/Vi:
=((1/B16)+(1/D16)-(((C16+B16)*((C16*F16)/(C16+F16)))/(B16*C16*(B16+((C16*F16)/(C16+F16))))))/((1/D16)+(1/E16)+(((C16+B16)*((B16*C16)/(B16+C16)))/(B16*C16*(F16+((B16*C16)/(B16+C16))))))

A: pot position index B: upper pot R (to Vi), C: lower one (to GND), D: R from Vi to Vo, E: R from Vo to GND, F: R in wiper to Vo, G: Vo/Vi. The rest calculates the target curve and the squared difference; I used to solver to minimize the summed squared difference by futzing the values of the other three resistors.

If someone with better editorial access could intervene, I'll try to add some images:

Here are the optimal fits to the example curves:

Here is the top of the table I ended-up with (however badly it formats here:
 nicr: R1+R3: 1.0E+3Ω SoS: AdjFact: 0.01 1.0E+3 0.077669 41.5043 n RP1: RP2: R1: 1.0E+9Ω R2: 943.5E+0Ω R3: 000.0E+0Ω Vo/Vi Exp adjExp Delt^2 Delt Delt^2, Adj 0 1000 0 1.0E+9 943.5E+0 000.0E+0 #DIV/0! 0 0 #VALUE! 0.01 990 10 1.0E+9 943.5E+0 000.0E+0 0.009896 0.01005 0.005849 1.64E-05 0.004047 0.00068 0.02 980 20 1.0E+9 943.5E+0 000.0E+0 0.019593 0.020201 0.011757 6.14E-05 0.007836 0.002549

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antedeluvian
6/7/2017 2:11:41 AM
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Re: Loading a linear pot can be USEFUL
@GSKrasle

If you add a resistor in series with the wiper, and after that a resistor to the upper terminal, and one to the lower,

Ilearned about this technique at a Maxim seminar as a method to combat thermal effects. I described it in my blog on potentiometers a while ago.

The algebra and curve-fitting for this five-resistor circuit (two values being linked by the constitutive relation that their sum be constant) is more challenging than you'd think

I tried several times to develop a model for this and failed. I even planned on using it as fodder for one of my ideas on the use of Excel in Electronics, but my skills and patience were insufficient.

I have posted my math online before

I would really like to see this. Could you provide a pointer, please?

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GSKrasle
6/6/2017 5:30:31 PM
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Loading a linear pot can be USEFUL
The nonlinear effects of loading the wiper of a pot can actually be USEFUL: If you add a resistor in series with the wiper, and after that a resistor to the upper terminal, and one to the lower, then by judicious choice of values, the output voltage may be made to approximate either a log or an exponential transfer function vs position, and to do it better (and obviously cheaper) than the 'log-' or 'audio-' pots that are available. For ganged pots, the tracking is also improved.

The algebra and curve-fitting for this five-resistor circuit (two values being linked by the constitutive relation that their sum be constant) is more challenging than you'd think, and is surely the nonrecurring cost that persuades manufacturers to keep buying the worse and more-expensive products, a recurring cost and an admission of inadequate Engineering.

I have posted my math online before, both the algebraic solution of the circuit and the values for curve-fits, but I would love to see other derivations, if just to confirm my own.

I also hope you will address the power-rating issues with pots: ones in lab-drawers often come to have have the ends of their ranges ruined by overcurrent. And also the practice of not leaving the third terminal of a variable-resistor open, so as to reduce noise due to a dirty/'iffy'" wiper contact.

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