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I am new to using potentiometers and I just wanted to see if this is possible,

I know you can use resisters to alter the value of the potentiometer and starting values and so forth.

I have a 5 V supply and from the wiper I would like a range of 0.5 to 4.5 V, or even 0.5 - 5 V.

It's the lower end I have issues with, when it is at 100% my device does not detect the "sensor".

I would like the minimum voltage to be no lower that 0.5 V.

Is this possible without complicated circuitry?

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    \$\begingroup\$ Is this pot being used as a divider or variable resistor? \$\endgroup\$
    – DKNguyen
    May 24 at 14:29
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    \$\begingroup\$ Certainly. Just put mechanical stops blocking each end of the range. \$\endgroup\$
    – Mark
    May 25 at 0:41
  • \$\begingroup\$ I hope you are aware that you will only get a voltage out of that, but no current (unlike the 5V power supply). You'll need to feed that voltage into a high impedance input. This circuit will not give you a voltage regulated power supply. \$\endgroup\$ May 27 at 8:12

5 Answers 5

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Certainly!

Two resistors are needed to limit both upper and lower range. One resistor at the bottom if you just want to limit the lower level.

You won't end up with exactly 0.500 and 4.500 V without adjusting the values to odd ones (out of E12 series), but close enough for your application?

schematic

simulate this circuit – Schematic created using CircuitLab

As pointed out by vu2an, the mathematically ideal values for R1 and R2 is 1.25 k\$\Omega\$ for a 10 k\$\Omega\$ pot. Downside is that 1.25 k\$\Omega\$ is not even in E192 series.

As pointed out by charmer, if you need precise 0.5 to 4.5 V range, possibly with 10 % tolerance on your 10 k\$\Omega\$ pot (the elephant in the room), you need an arrangement like this:

schematic

simulate this circuit

VR1 and VR3 are pots in rheostat configuration and R1, R2, R6 and R7 are to desensitize VR1 and VR2 adjustments to give you more fine control. Probably way overkill. Feel free to adjust values to give you more room for adjustment (remove R1, R2, R6 and R7 completely) or finer control.

Happy poting!

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    \$\begingroup\$ @LorenzoMarcantonio Near the bottom. Makes no sense not to have it next to the resistor symbol \$\endgroup\$
    – winny
    May 24 at 14:34
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    \$\begingroup\$ @Dangermike Cheap 1 kΩ trim pots instead of the 470 Ω resistors would allow a bit of tweaking of the min and max voltage output. Or 2 kΩ in light of vu2nan's comment. \$\endgroup\$ May 24 at 17:14
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    \$\begingroup\$ Hi Winny, R1 = R2 = R3 / 8 =1.25 kΩ. \$\endgroup\$
    – vu2nan
    May 24 at 17:15
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    \$\begingroup\$ Six Components and Two Manual Adjustments, production department would love you :P \$\endgroup\$
    – crasic
    May 25 at 17:09
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    \$\begingroup\$ @carsic They’ll hate it for sure, but if this is for mass production, it would call for a different solution. OP seem more interested in a DIY-bench friendly solution with components found around. \$\endgroup\$
    – winny
    May 25 at 19:40
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Is this possible without complicated circuitry?

It's possible.

enter image description here

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Yes, you can just add resistors to the ends, for example with a 10k\$\Omega\$ pot the nominal current through the element for 0.5~4.5V is 4V/10k = 400\$\mu\$A, so the resistors at either end will be 0.5V/0.4 = 1.25k\$\Omega\$ (you might use 1.3K 5%). That's likely all you need.


Following comments are for more general situations:

That may well be an acceptable approach for relatively small resistances relative to the element. For larger resistances consider shunting the element with a third resistor to reduce the effect of element tolerance (often as bad as +/-20%). For example, if you shunt a 10k\$\Omega\$ +/-20% element with 499\$\Omega\$ +/-1% then the overall tolerance of the pair is better than +/-5%. Of course that consumes more current from the supply.

Another approach that works well if the voltage difference from the rails is high enough is to buffer a voltage divider with rail-to-rail op-amps as so:

schematic

simulate this circuit – Schematic created using CircuitLab

This has the advantage of low output impedance with low power consumption and almost no sensitivity to pot tolerance.

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Put two resistor in series with the pot, one on each side. The one on the ground side 'lifts' the low value, the one on the high side lower the top value.

Just do a voltage divider with three resistors (the low, the whole pot winding and the upper).

It's quite frequent that some analog signal is transmitted in a 0,5 to 4,5V range, you're doing it correctly.

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Building on the answers already given, here is a solution that reduces the variation of the upper and lower limit voltages due to tolerance or drift in the potentiometer's resistance, but without adding adjustable or active components.

schematic

simulate this circuit – Schematic created using CircuitLab

This will only be suitable if the load on the output has a high resistance compared to the 100k pot and if you don't mind that the potential divider draws nearly 5 mA - it may not be what you want in a battery-powered application, for example. Of course you can choose your own component values to adjust these trade-offs.

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