potentiometer in parallel with a normal resistor

[Orc General]

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

 

[OBones]

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

 

[Spehro Pefhany]

On Mon, 9 May 2005 11:50:32 -0700, the renowned "Orc General"
<orc_general@hotmailNOSPEM.com> wrote:

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

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
Embedded software/hardware/analog Info for designers: http://www.speff.com

 

[OBones]

Spehro Pefhany wrote:

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.

Or one could use a logarithmic potentiometer

 

[Keith Williams]

In article <d5o0lf$9qt$1@reader1.imaginet.fr>,
obones_gfd_@_gfd_altern.org says...

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

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

 

[Frank Bemelman]

"OBones" <obones_gfd_@_gfd_altern.org> schreef in bericht
news:d5o1lo$a79$1@reader1.imaginet.fr...

Spehro Pefhany wrote:

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.

Or one could use a logarithmic potentiometer

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)

 

[Spehro Pefhany]

On Mon, 9 May 2005 23:21:49 +0200, the renowned "Frank Bemelman"
<f.bemelmanq@xs4all.invalid.nl> wrote:

"OBones" <obones_gfd_@_gfd_altern.org> schreef in bericht
news:d5o1lo$a79$1@reader1.imaginet.fr...
Spehro Pefhany wrote:

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.

Or one could use a logarithmic potentiometer

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

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

 

[Pooh Bear]

Orc General wrote:

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.

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