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Assume I have a linear potentiometer that, say, has a range of 0 to 10Kohm. Then assume that:

0deg = 0 ohms

10deg = 370 ohms

270deg = 10Kohm

My question is, how do I show 370 ohms while the potentiometer is ≤ 10 degrees, but as soon as it gets > 10 degrees, ONLY the potentiometer value is shown (e.g. 11deg would show about 407 ohms)? That is, IGNORE the potentiometer when it is ≤ 10 degrees, but then continue as normal past 10 degrees.

I could do it with an active design, but was wondering if there was a way to do it with just passive components that for some reason I'm not seeing..

Note that the reason this has to be done with external components (rather than just change the potentiometer range to be 370-10K) is that the additional floor needs to be switchable.

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    \$\begingroup\$ XY problem #1 \$\endgroup\$ Aug 8 '17 at 19:35
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    \$\begingroup\$ Edit your question to explain what the real problem is rather than what your solution is. There may be a much simpler solution. There's a schematic button on the editor toolbar. A schematic saves many words and much guessing on our part. \$\endgroup\$
    – Transistor
    Aug 8 '17 at 19:40
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    \$\begingroup\$ I think you'd better spend more time talking about what all this is getting used for. \$\endgroup\$
    – jonk
    Aug 8 '17 at 19:40
  • \$\begingroup\$ pots usually unreliable for absolute value in low % just before 0, so this is a poor assumption \$\endgroup\$ Aug 9 '17 at 1:49
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    \$\begingroup\$ Are you just outputting a voltage with the pot? If yes, then that's easier to design for (output some fixed voltage for the first 10 degrees of rotation). Or do you really need Ohms (as in resistance that stays at 370 Ohms for the 10 degrees)? \$\endgroup\$ Aug 10 '17 at 6:05
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I essentially need to take a wild guess at how you're using your pot. The simplest passive(ish) circuit that I can think of, assuming that you are using the pot to sweep through 5V:

schematic

simulate this circuit – Schematic created using CircuitLab

You'll find that at the bottom of its range, the potentiometer will appear to produce a voltage flattening out to approximately that produced by 10deg deflection. There are much better ways to do this (i.e. op-amp circuit).

Another note: I assume that the next stage is high-impedance; otherwise, this is going to have problems.

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  • \$\begingroup\$ Are you sure this works at 5 V when the diode voltage drop is taken into account? As written, and according to CircuitLab's DC sweep simulator, it has no effect, and tweaking the voltage divider higher produces a quite non-flat lower end. (Of course, if the supply voltage (which was not specified in the question) is much higher it becomes insignificant.) \$\endgroup\$
    – Kevin Reid
    Aug 9 '17 at 21:08
  • \$\begingroup\$ I in fact simulated this with a different simulator, PartSim (since CircuitLab asks for a license to simulate). In PartSim, the Shottky forward drop is very low at this current, approx. 127mV. Assuming that that holds true, then yes, the diode should take effect. \$\endgroup\$
    – Reinderien
    Aug 9 '17 at 21:11

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