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potentiometer in parallel with a normal resistor

Does Ohm's law work the same way for a potentiometer if one is to put another resistor in parallel with the pot? Suppose I have a pot with a maximum resistance ohm of 500OHM. But I want to reduce this maximum to a lower value say around 85ohm. According to ohm's law Resistance(total) = 1/( (1/Resistor1) + (1/Potentiometer)). Therefore I could substitute Resistor1 with a 100Ohm resistor. And now I can vary the Potentiometer resistance and I would get a range from 1Ohm to 83.3Ohm as shown below. I am assuming that you never really get a 0ohm potentiometer because this formula would blow up due to a divide by zero. Is this behaviorial assumption correct? Sorry its been a while since I cracked open that physics text book.

Resistance(total) = 1/( (1/Resistor1) + (1/Potentiometer)) Resistor1=100ohm Potentiometer=500ohm max

1/((1/100)+(1/500)) 83.333333333

1/((1/100)+(1/400))

80.000000000

1/((1/100)+(1/300))

75.000001875

1/((1/100)+(1/200))

66.666666666 1/((1/100)+(1/100)) 50.000000000

1/((1/100)+(1/50))

33.333333333

1/((1/100)+(1/40))

28.571428571

1/((1/100)+(1/30))

23.076923254

1/((1/100)+(1/20))

16.666666666

1/((1/100)+(1/10))

9.090909090

1/((1/100)+(1/5))

4.761904761

1/((1/100)+(1/2))

1.960784313

1/((1/100)+(1/1)) .990099009

Reply to
Orc General
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You can also use the formula R = R1*R2/(R1+R2), which works fine if either resistor is zero.

And if you want to know what resistor to put in parallel, you can use

Rp = 1/(1/R - 1/Rpot)

Eg. pot 500 ohms, you want 85 ohms, so 102.4 ohms.

Check: R = 102.4 * 500/(602.4) = 84.99 ohms

Now, the problem with what you are doing is this-- look at the linearity of the resulting compound pot:

Pot % Rpot R Ideal pot value

0 0 0.00 0.00 5 25 20.09 4.25 10 50 33.60 8.50 15 75 43.29 12.75 20 100 50.59 17.00 25 125 56.29 21.25 30 150 60.86 25.50 35 175 64.60 29.75 40 200 67.72 34.00 45 225 70.37 38.25 50 250 72.64 42.50 55 275 74.62 46.75 60 300 76.34 51.00 65 325 77.87 55.25 70 350 79.22 59.50 75 375 80.44 63.75 80 400 81.53 68.00 85 425 82.52 72.25 90 450 83.42 76.50 95 475 84.24 80.75 100 500 84.99 85.00

As you can see the last 5% of pot rotation changes the parallel value as much as half the rotation of an ideal pot. So the pot setting might be finicky, unstable, noisy etc. if it happens to be down in that range.

Best regards, Spehro Pefhany

--
"it\'s the network..."                          "The Journey is the reward"
speff@interlog.com             Info for manufacturers: http://www.trexon.com
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Reply to
Spehro Pefhany

Also remember that the combined resistance will be a non-linear function of pot "setting". With an 85ohm resistor in parallel with a

5K pot, 100% "setting" the combination would be 83.5ohms. At 90% "setting" it would be 83.4ohms. At 10% pot "setting" the resultant is an 85 ohm resistor in parallel with 500ohms, or about 72ohms, which is about 86% of the total resistance. 
--
  Keith
Reply to
Keith Williams

Well, he might not be able to get an 85 ohm pot off the shelf as they usually are only offered in the 1 2 5 sequence of values, but 100 ohms shunted with 567 ohms would be a heckofa lot better than his 500 ohms shunted with whatever it was.

Best regards, Spehro Pefhany

--
"it\'s the network..."                          "The Journey is the reward"
speff@interlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com
Reply to
Spehro Pefhany

Well, the formula wouldn't blow up, as it can be written as:

R1*Rp/(R1+Rp)

But you are right, you never really get 0, that would be supra conductive material ;-)

As to your calculation, yes you are right, but keep in mind the power dissipated through the resistor and potentiometer: P = R*I*I

Reply to
OBones

Or one could use a logarithmic potentiometer

Reply to
OBones

Let's not forget that old trick of using a potentiometer with the correct value ;)

--
Thanks, Frank.
(remove \'q\' and \'invalid\' when replying by email)
Reply to
Frank Bemelman

In that case please start with a 100 ohm pot ! It'll have a far better 'law'.

Pot values are typically +/- 20% btw. It's very poor practice to entirely rely on the track value for critical settings.

Graham

Reply to
Pooh Bear

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