Resistor maths - some basic help

Discussion in 'General Electronics Chat' started by axeman22, Jun 10, 2009.

  1. axeman22

    Thread Starter Active Member

    Jun 8, 2009
    53
    0
    Hi All,

    I know my resistor theory but this is just doing my head in so thought I'd ask the forum..

    I have a 500Ω Pot and what I actually need is a 'resistor' which can be variable from 1520Ω-1660Ω. I'm thinking that with a series resistor of X value and another resistor in parallel to the pot of Y value I could possibly achieve this by making up a small circuit which represents a single resistor for all intents and purposes.. it's just that the math is doing my head in, simple I know.

    appreciate any comment on what X and Y should be in order to make this work.!

    Thanks in advance.
     
  2. creakndale

    Active Member

    Mar 13, 2009
    68
    7
    X = 1.52K ohms
    Y = 194.44 ohms

    For X: when the 500 ohm pot is at one stop it's 0 ohms. So you need a 1.52K in series with the pot. 1.52K + 0 = 1.52K

    For Y: when the 500 ohm pot is at the other stop it's ~500 ohms so you need a parallel resistor to bring it's resistance to 140 ohms (1.66K - 1.52K = 140 ohms)

    The math to calculate the 500 ohm pot's parallel resistor to result in 140 ohms:
    Y = 1 / (1/140 - 1/500)
    Y = 194.44 ohms

    creakndale
     
  3. Tesla23

    Active Member

    May 10, 2009
    318
    67
    That's one way but it gives a very non-linear response and the resistance actually peaks at about 1693 ohms.

    Another way is to also use two resistors, make one connection to the pot wiper, the other connection has a 2249Ω resistor to one end of the pot and a 4189Ω resistor to the other. The attached plot shows the difference in the resistance characteristic.

    Sorry - this just piqued my interest...
     
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  4. eblc1388

    Senior Member

    Nov 28, 2008
    1,542
    102
    Your plot is great but I can't figure out how the resistors are connected and which two points varies from 1520~1660Ω.

    Can you post a sketch or describe the connections of resistors with reference to:

    500Ω POT pins( P1, P2 and W). Thanks.
     
    Last edited: Jun 10, 2009
  5. Tesla23

    Active Member

    May 10, 2009
    318
    67
    You want a resistor that varies from 1520 to 1660 ohms. The ends of this resistor are A and B. A connects to the wiper on the pot. B has a 2249Ω resistor to one end of the pot and a 4189Ω resistor to the other.
     
  6. eblc1388

    Senior Member

    Nov 28, 2008
    1,542
    102
    Got it. Thanks.
     
  7. Tesla23

    Active Member

    May 10, 2009
    318
    67
    I just realised that I may have unfairly judged your solution, I had the 194Ω across the pot and the wiper the output with nothing else connected to it.

    If you connect the pot as a 500Ω variable resistor across the 194Ω then the resistor does stay within the bounds (series 1 on the attached plot), but not very linear. I hope you get what I mean,
     
  8. AlainB

    Active Member

    Apr 12, 2009
    39
    0
    Hi,

    Could you clarify a point for me please?

    Where will be connected the other end of the 4189 Ohms resistor.

    Alain
     
  9. axeman22

    Thread Starter Active Member

    Jun 8, 2009
    53
    0
    I'm a wee bit confused - diagram .?
     
  10. eblc1388

    Senior Member

    Nov 28, 2008
    1,542
    102
    You are not the only one. :)

    [​IMG]
     
  11. AlainB

    Active Member

    Apr 12, 2009
    39
    0
    About linearity, one must know that all pots are not equal. Some are linear and some others are not. Audio pots are not linear.

    "The best way to indentify them is to use a DMM and measure them with the taper exactly in the middle position - if both sides show the same value (exactly 50% of the pot´s value) it´s a linear pot, otherwise it´s an audio pot."​

    The page from where come this quote:
    http://www.singlecoil.com/docs/pot.pdf

    Telsa23, It would be nice to post your calculations method. Knowing the values needed is good but knowing how to do it is even better.:)

    Alain ​
     
    Last edited: Jun 10, 2009
  12. AlainB

    Active Member

    Apr 12, 2009
    39
    0
    Hi,

    I hate to see my question vanishing without an answer. This is pretty interesting stuff (to me anyway). Are these calculations completely unattemptable by mortals with typical low maths knowledge like me?

    Alain;)
     
    Last edited: Jun 12, 2009
  13. Tesla23

    Active Member

    May 10, 2009
    318
    67
    It's fairly straightforward, if the resistances are x and y then

    x(y+500)/(x+y+500) = 1520
    y(x+500)/(x+y+500) = 1660

    I was lazy and solved them iteratively in Excel, if you did it manually - you'd get a quadratic, the solutions are:

    x=(377000√17+1861000)/(325√17+179)
    y=(1625√17+5875)/3
     
  14. AlainB

    Active Member

    Apr 12, 2009
    39
    0
    Thanks Tesla23,

    My notions of algebra are very far. At least 45 years away. Could you solve the first equation for me please, step by step:

    x(y+500)/(x+y+500) = 1520

    It is unfortunately too complicated for me and I can't find any relevant litterature that would help me. The second one, I will probably be able to solve it myself after seeing the steps involved in the first one.

    Thanks!

    Alain
     
  15. hobbyist

    Distinguished Member

    Aug 10, 2008
    764
    56
    (R1 x R2) / (R1+R2) = RT

    R1=500

    RT=140

    solve for R2.

    (R1 X R2) = (RTR1 + RTR2)

    (R1 x R2) - RTR1 - RTR2 = 0

    R2 (R1 - RT) - RTR1 = 0

    R2 = RTR1 / (R1 - RT)

    R2 = (140 x 500) / (500 - 140) = 194.44
     
  16. Tesla23

    Active Member

    May 10, 2009
    318
    67
    You have to solve them as a pair of simultaneous equations, rewrite them as:

    xy+500x = 1520(x+y+500)
    xy+500y = 1660(x+y+500)

    subtract them to get rid of the nasty xy term, this gives you an equation relating x and y which you can reorganise to
    y = αx + β

    substitute this into the first equation gives you a quadratic in x which you can solve, throw away the -ve solution.

    Of course I didn't do this, I was lazy and I used Maxima (free download from http://maxima.sourceforge.net/),

    simply entering:
    eq1:x*(500+y)/(500+x+y)=1520;
    eq2:y*(500+x)/(500+x+y)=1660;
    solve([eq1,eq2],[x,y]);

    and you get:

    $$[[x=\frac{377000\,\sqrt{17}-1861000}{325\,\sqrt{17}-179},y=-\frac{1625\,\sqrt{17}-5875}{3}],[x=\frac{377000\,\sqrt{17}+1861000}{325\,\sqrt{17}+179},y=\frac{1625\,\sqrt{17}+5875}{3}]]$$

    to get this evaluated enter:
    float(solve([eq1,eq2],[x,y]));

    $$[[x=-264.0712453053809,y=-275.015547209566],[x=2248.446245305381,y=4191.682213876233]]$$

    but I guess that is cheating!

    If you do download Maxima there is a useful tutorial here
     
  17. AlainB

    Active Member

    Apr 12, 2009
    39
    0
    Thank you Tesla23!

    All this I must say is "foreign language" to me.:)

    What I would like to do ultimately is an Excel sheet to compute the values and give answers for differents scenarios.

    something like this:

    Inputs:
    VR.................500
    Low..............1520
    High..............1660

    Result:
    R1................2249
    R2................4189

    Not as easy as I thought it would be!

    Alain
     
    Last edited: Jun 13, 2009
  18. Tesla23

    Active Member

    May 10, 2009
    318
    67
    Presented without guarantees, courtesy of Maxima:
    Rp = pot resistance
    Rmin = min res (= 1520 here)
    Rmax = max res (=1660 here)

    $$R1=-\frac{-{Rp}^{3}+\sqrt{{Rp}^{2}+4\,Rmax\,Rmin}\,\left( {Rp}^{2}-Rmax\,Rp\right) +Rmax\,{Rp}^{2}-2\,Rmax\,Rmin\,Rp}{Rp\,\sqrt{{Rp}^{2}+4\,Rmax\,Rmin}-{Rp}^{2}-2\,Rmax\,Rmin+2\,{Rmax}^{2}},R2=\frac{Rp\,\sqrt{{Rp}^{2}+4\,Rmax\,Rmin}-{Rp}^{2}+2\,Rmax\,Rp}{2\,Rp+2\,Rmin-2\,Rmax}$$

    you can probably simplify it a bit, particularly the R1 expression
     
  19. Tesla23

    Active Member

    May 10, 2009
    318
    67
    Just to show you what I did, all I had to do was to change my Maxima script to:

    eq1:x*(Rp+y)/(Rp+x+y)=Rmin;
    eq2:y*(Rp+x)/(Rp+x+y)=Rmax;
    solve([eq1,eq2],[x,y]);

    and the result was:

    [[x=-\frac{{Rp}^{3}+\sqrt{{Rp}^{2}+4\,Rmax\,Rmin}\left( {Rp}^{2}-Rmax\,Rp\right) -Rmax\,{Rp}^{2}+2\,Rmax\,Rmin\,Rp}{Rp\,\sqrt{{Rp}^{2}+4\,Rmax\,Rmin}+{Rp}^{2}+2\,Rmax\,Rmin-2\,{Rmax}^{2}},y=-\frac{Rp\,\sqrt{{Rp}^{2}+4\,Rmax\,Rmin}+{Rp}^{2}-2\,Rmax\,Rp}{2\,Rp+2\,Rmin-2\,Rmax}] \\ [x=-\frac{-{Rp}^{3}+\sqrt{{Rp}^{2}+4\,Rmax\,Rmin}\,\left( {Rp}^{2}-Rmax\,Rp\right) +Rmax\,{Rp}^{2}-2\,Rmax\,Rmin\,Rp}{Rp\,\sqrt{{Rp}^{2}+4\,Rmax\,Rmin}-{Rp}^{2}-2\,Rmax\,Rmin+2\,{Rmax}^{2}},y=\frac{Rp\,\sqrt{{Rp}^{2}+4\,Rmax\,Rmin}-{Rp}^{2}+2\,Rmax\,Rp}{2\,Rp+2\,Rmin-2\,Rmax}]]

    I think the second pair are the ones you want (I just subbed in our values to the 'y' equation).
     
  20. eblc1388

    Senior Member

    Nov 28, 2008
    1,542
    102
    AlainB,

    Are you thinking this method would be a general solution suitable for every case with a total resistance range changes less than the range of the POT?

    It turns out if the required range change is less than some 20~30% of the total range of the POT, the circuit would not perform satisfactorily and there is a higher than wanted resistance before dropping back to the wanted value.

    The mathematical solution above only produces a solution that fits the two extreme end points of the target range but there could be a maxima in between those two points.

    Therefore the implementation should be checked in Excel before using it in your application.

    I will illustrate this with an example.

    [​IMG]
     
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