I am given a DC line with a variable voltage along a set scale: let's say 0-10V. I would like to send an output DC line with a voltage reversed along that scale.
For example:

  • 10 Volts in becomes 0 Volts out,
  • 9 Volts in becomes 1 Volt out,
  • 5 Volts in becomes 5 Volts out,
  • 0 Volts in becomes 10 Volts out.

Note that the polarity is not reversed.

My question is: is there device I can buy or a relatively simple wiring scheme that can get me there? My intention is to take a signal from a PID tension-controller that controls a brake, and instead power a motor when needed. Oof, good question on the current. Let's say it is one amp.

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    \$\begingroup\$ 1) If you want an answer, ask a question. 2) Need more context; What exactly are you trying to achieve, and what for? \$\endgroup\$ – Dampmaskin Sep 21 '17 at 14:37
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    \$\begingroup\$ Also mention how much current is needed. 0-10V is a standard control voltage for LED dimmers so are we talking about that? \$\endgroup\$ – peufeu Sep 21 '17 at 14:42
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    \$\begingroup\$ I guess OP wants us to design a circuit for that. That's not what we do here. You can propose a circuit and we can comment on that and suggest improvements. But since you have no experience with electronics all I can tell you is: forget this. If this needs to be done get someone with experience in electronics. There are simply hundreds of things you would need to know even if we gave you a circuit design. \$\endgroup\$ – Bimpelrekkie Sep 21 '17 at 14:42

If you need the voltage just for indication or measurement then a simple DC differential amplifier with unity gain will do the job, because the output voltage is 10V minus input voltage: \$V_{out} = 10V - V_{in}\$. :


simulate this circuit – Schematic created using CircuitLab

Note that this circuit is not easy to build as it looks even for an experienced person because of the need of a negative supply (-12V). Also, generating the reference voltage ("10V" to the left-side leg on R2) can be problematic (e.g. voltage divider or from an outer generator).

  • \$\begingroup\$ Hi Rohat! Thanks for the answer! I'm having difficulty running the simulation for that circuit, but it if it as described, I think we can do it. \$\endgroup\$ – RexPrestige Sep 21 '17 at 15:12

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