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.!

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 AAC Fanatic!

May 10, 2009
374
88
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 AAC Fanatic!

Nov 28, 2008
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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 AAC Fanatic!

May 10, 2009
374
88
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 AAC Fanatic!

Nov 28, 2008
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Got it. Thanks.

7. Tesla23 AAC Fanatic!

May 10, 2009
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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,

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8. AlainB Active Member

Apr 12, 2009
39
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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
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I'm a wee bit confused - diagram .?

10. eblc1388 AAC Fanatic!

Nov 28, 2008
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You are not the only one.  File size:
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11. AlainB Active Member

Apr 12, 2009
39
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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 AAC Fanatic!

May 10, 2009
374
88
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 AAC Fanatic!

Aug 10, 2008
887
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(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 AAC Fanatic!

May 10, 2009
374
88
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:

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

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 AAC Fanatic!

May 10, 2009
374
88
Presented without guarantees, courtesy of Maxima:
Rp = pot resistance
Rmin = min res (= 1520 here)
Rmax = max res (=1660 here)

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

19. Tesla23 AAC Fanatic!

May 10, 2009
374
88
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:

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

20. eblc1388 AAC Fanatic!

Nov 28, 2008
1,543
104
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. File size:
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