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I'm currently trying to get a 200kΩ resistance from scrapped materials or new ones but there's a limited range. (There isn't a good supply of components in my city and we can't spare too much time at the moment)

Is there any inverse formula to get the combinations of resistors where their final resistance value is N ?

We don't want a lot of resistors in series but if we lack a solution we will go after that.

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    \$\begingroup\$ What do you mean by "inverse formula"? Just take the normal ones and rearrange, basic math. And for series resistors its trivial addition \$\endgroup\$ – PlasmaHH Sep 25 '17 at 14:15
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    \$\begingroup\$ I guess he means a formula where he puts in 200 k and gets out all the combinations which make that. Problem is, there are an infinite number of solutions: 2 in series: 1k + 199k, 2k+198k ... 100k+100k etc. Not showing 3 in series or even parallel resistors or combinations of that. If you know how to calculate the value of two resistors in series and two in parallel, it becomes obvious how the relations work. \$\endgroup\$ – Bimpelrekkie Sep 25 '17 at 14:18
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    \$\begingroup\$ Kelvin-Ohms is not a meaningful measure of anything related to this question. \$\endgroup\$ – Olin Lathrop Sep 25 '17 at 14:31
  • \$\begingroup\$ I made this ask because components supply are very limited here. And we want to use the least possible ammount of components so the ideal way would be (for us) using parallel resistors \$\endgroup\$ – Joe Sep 25 '17 at 14:35
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    \$\begingroup\$ See: electronics.stackexchange.com/questions/325303/… \$\endgroup\$ – jonk Sep 25 '17 at 16:45
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Easiest way to catalog and do this quickly is to stop using resistance values and use conductance instead.

Mark the resistors you have with their conductance value and then add those to get whatever conductance value you want.

e.g. 200k = 5uS (S = seimens not seconds)

=> 1Meg = 1uS , and 250k = 4uS

=> 1M||250k = 200k

People seem to forget that the traditional parallel resistor equation does exactly the above.
enter image description here

That is; convert each value to a conductance (\$G = 1/R\$), add them together, then convert them back (\$R=1/\Sigma G\$).

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There are two ways to combine resistance.

In series, the resistances simply add. 10 kΩ, 33 kΩ, and 15 kΩ in series yields a overall 58 kΩ resistance.

In parallel, you get the reciprocal of the sum of the reciprocals. To compute 10 kΩ, 33 kΩ, and 15 kΩ in series, you compute:

    1/(10 kΩ) + 1/(33 kΩ) + 1/(15 kΩ) = 0.000197/Ω

The reciprocal of that is 5.08 kΩ, which is the overall resistance of the three individual resistances in parallel.

You can reverse any of these formulas to find the resistance to add in series or in parallel to get a particular desired value.

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No. You can start with the value you want to end with, 200k, in the case you cite, but there are countless combinations of resistors that would get you there - there's no unique answer. You might be best off by starting with a value near what you want, and trimming it up with other resistors, in series, or down, with resistors in parallel.

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I can see that you are only interested in finding a combination of resistors and not necessarily working out the combination yourself.

In this case I would recommend some online tools like the Resistor Finder on the www.electronicscalculator.co.uk webpage.

You can plug in the desired value and the tool will output a combination of the two serial or parallel resistors that will produce the desired value.

If you only have let's say 6 resistor values in your set you could go under the advance settings and only plug those values in which case the tool will only use those values (of course you could save the list in a text file so you don't have to type them in all the time).

If let's say that your desired value cannot be achieved with only two resistors with the desired accuracy I would suggest to identify the most common value that you own and subtract that from the desired value and use the resulting value with the tool.


enter image description here

enter link description here

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