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I have a device that is controlled by connecting a resistor between two of its connectors. The resistor can take any value between 0 and X, where X is an odd value. (E.G. X=1523ohm)

Does anybody have an idea how I can realise a circuit whose resistance can be varied manually between 0 and X? There must not be the possibility that due to user error the resistance is larger than X!

Many thanks.

Edit 1

To clarify: I do not have a variable resistor but am looking for a circuit that has a variable resistance with an odd upper limit. Variable resistors have standard upper limits, e.g. 100 ohm, 1kOhm, etc. but not 1523 ohm...

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  • \$\begingroup\$ Schematic, please. Also what does the part data sheet say? \$\endgroup\$ – Brian Carlton Jul 9 '12 at 23:15
  • \$\begingroup\$ @BrianCarlton I think there was a misunderstanding. I do not have a variable resistor but am looking for a circuit that has a variable resistance with an odd upper limit. Such a variable resistor cannot be bought. \$\endgroup\$ – ARF Jul 9 '12 at 23:32
  • \$\begingroup\$ It would help to have the device details (model number, datasheet, etc) if you have them. \$\endgroup\$ – Oli Glaser Jul 9 '12 at 23:41
  • \$\begingroup\$ What is step of resistance changing? 1ohm, 1kOhm...? \$\endgroup\$ – George Gaál Jul 10 '12 at 0:04
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Like gbarry says. The next higher value you find in the Digikey offering is 2 kΩ, so you would need a 6386 Ω resistor in parallel. That is between one end and the wiper, not across the full resistance. 6386 Ω is also an odd value, so you may want to compose that using different standard values as well.

Now the 1523 Ω, how precise does that have to be? You represent it in 4 significant digits, that's 0.1 %. Potmeters are often 20 %, more precise ones can be pretty expensive. So our 2 kΩ can be anywhere between 1600 Ω and 2400 Ω. In the first case your parallel resistor should be 31650 Ω, in the second 4170 Ω. That's a wide range. You could use a 33 kΩ trimming potmeter. Use a multiturn is you want to get the 1523 Ω really precise.

enter image description here

Note that placing another resistor over the potmeter's wiper distorts its linear characteristic.

enter image description here

The greenish curve shows the resistance as a function of the rotation in % for a 2.2 kΩ resistor parallel to a 5 kΩ linear potmeter. The purple curve shows that resistance for the 2 kΩ potmeter plus the 6386 Ω, so that's not so bad after all. The rule for the most linear characteristic is to have the ratio resistor/potmeter as large as possible.

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  • \$\begingroup\$ If one uses a ganged pot, and the resistance of the two wipers in parallel is less than the target resistance, one may improve linearity by adding a series resistor to one of the wipers. For example, if one wanted a resistance of 510 ohms and one had a twin 1K pot, one could put a 40 ohm resistor in series with one of the pots and then put the other pot in parallel with that (resistor-plus-pot) combination. Of course, one would have to have a ganged pot for that to work. \$\endgroup\$ – supercat Jul 10 '12 at 17:06
  • \$\begingroup\$ @supercat - Sorry, man, I read it three times, and I don't understand what you mean. But that could also be my rusty brain :-) \$\endgroup\$ – stevenvh Jul 10 '12 at 17:21
  • \$\begingroup\$ It is possible to buy assemblies with two or more potentiometers fixed to a current shaft. If one could find e.g. an assembly with two 2.5K pots, one could wire the wiper of each pot to the end of rotation that should represent maximum resistance, and wire a 1.4K resistor between them. Then wire the other end of each pot together, and use the resistance between that end and the wiper of one of the pots. \$\endgroup\$ – supercat Jul 10 '12 at 17:44
  • \$\begingroup\$ Alternatively, it may be possible to find pots which can slide onto a shaft; in that case, it may be possible to combine two or more different pots onto a single shaft. In that case, just pick some pots which can be wired in series, parallel, or some combination to yield a desired voltage. \$\endgroup\$ – supercat Jul 10 '12 at 17:45
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Choose a variable resistor--probably the next value up from 1523 ohms would be best. Then, calculate a resistor in parallel with the maximum setting that gives you 1523 ohms. Now you are at 1523 at max, and can adjust it down to zero.

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