I have two systems I am trying to get to work together.

One system has two inputs, input 1 takes an output from a potentiometer with a voltage ranging from 0.9 V - 4.5 V using a supply reference voltage of 5 V. The second input takes an output from a potentiometer ranging from 0.45 - 2.25 V. The output of the second potentiometer is always half the output from the first potentiometer. For safety reasons these inputs must remain independent.

These potentiometers are always physically locked together so that when one moves, the other one does too. The voltage from pot 2 is always half of pot 1 unless there is a failure. We will call this potentiometer assembly 1.

I am trying to interface with a different potentiometer, due to physical constraints I cannot use a potentiometer assembly other than this one. We will call this potentiometer assembly 2. This potentiometer assembly works just like the assembly 1 but the value of output 2 is inverted, meaning as the pot is actuated the output #2 goes from 4.5 V to 0.9 V, the oposite of what is needed. Reversing the input wires of the pot does not work. If you add up the voltage outputs of assembly 2 they will always add up to 5 V unless something has failed.

So here is a description of what I need: a circuit that creates two independent outputs from two independent inputs. input number one will receive 0.9 - 4.5 V. Input number two will receive 4.5 - 0.9 V. The voltages of the inputs will always add up to 5 V unless something has failed. failure of the input voltages to add up to 5 V should cause the outputs to go to 0 V. output number one voltage should be equal to input number one. Output number 2 should always output half the voltage of output number one unless the combined voltage from the inputs deviates from 5 V. If a deviation happens then the circuit should be in a fail state and drop all output voltage to 0

  • \$\begingroup\$ Can this just be solved with a voltage buffer? What does "independent" mean in this case? \$\endgroup\$ Commented Mar 30, 2023 at 21:15
  • \$\begingroup\$ Hello Lance and welcome ... can I ask if this is a university question? \$\endgroup\$
    – jonathanjo
    Commented Mar 30, 2023 at 21:17
  • 4
    \$\begingroup\$ If one is at 0.9 when the other is at 4.5 that adds up to 5.4 not 5, so you've already got a problem before you even start. What limits are there on the deviation from 5 V? \$\endgroup\$
    – GodJihyo
    Commented Mar 30, 2023 at 21:21
  • 1
    \$\begingroup\$ You could save a lot of words by drawing a graph of the transfer functions. Plot 0 - 330° on the X-axis (most pots only turn that much) and plot traces of what outputs you want on each wiper. \$\endgroup\$
    – Transistor
    Commented Mar 30, 2023 at 21:27
  • 1
    \$\begingroup\$ A graph will make it easier to see the whole picture and to calculate any resistor dividers, etc. A video doesn't make a great user requirement specification. \$\endgroup\$
    – Transistor
    Commented Mar 30, 2023 at 21:52

2 Answers 2


Does this meet your requirements? The output is 5V for inputs of 0.9 to 4.5 volts, and 4.5 to 0.9 volts.

Inverse Voltages

  • \$\begingroup\$ I believe this will do it, I will have to cut the output down by half but I know how to do that. Thanks! I will test this weekend and get back to you. \$\endgroup\$
    – Lance
    Commented Mar 31, 2023 at 13:48

As I understand your requirements, you will bypass one signal and short this to GND in case of a fault. The other signal needs to be inverted using a reference of 2.7 V as center, divided by 2 and also shorted to GND on faults. This will emulate the properties of your "assembly 1".

The fault condition is a deviation in the mean value of both input signals combined from 2.7 V within some small margin.

I included two buffer amplifiers because the circuit would affect a potentiometer as voltage source. Buffer output amplifiers may also be needed.

If rail to rail input and output opamps are used, the circuit can operate from a single 5 V supply.

This circuit should behave like that (not physically tested):


simulate this circuit – Schematic created using CircuitLab


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.