I have a hall effect single axis joystick that outputs 0-5V from backwards to forwards (2.5V at center). I need it to output 5V when back, 0V at center, and 5V when forward. I believe this can be done with op amps and/or resistors but haven't found a great solution yet. Can someone help provide a solution?

 

A key part of the circuit would probably be an absolute value circuit. That and a little offset should get you there.

https://www.ti.com/lit/an/sboa068/sboa068.pdf

https://www.analog.com/en/analog-di...lue-from-absolute-value-circuit-diff-amp.html

 

but haven't found a great solution yet.

Doesn't look logical in a linear output.
0 volt default = 5 volt desired
5 volt default = 5 volt desired

 

I have a hall effect single axis joystick that outputs 0-5V from backwards to forwards (2.5V at center). I need it to output 5V when back, 0V at center, and 5V when forward.

What power supplies do you have available?

With a 5V supply I have a circuit that goes to about 4.5V output max.
Would that be okay?:

 

What power supplies do you have available?

Right now just 24V and 5V.

 

Offset and scale this output with another opamp and you are there:
Any rail-to-rail opamp will work. The input is a ramp from 0 to 5V.

 

Below is the sim of a precision rectifier with offset circuit, using rail-rail op amps supplied with 5V.
Since all the resistors are the same value, you could use a matched resistor array.

Its max output with 0v input is a little over 4.5V due to the forward drop of diode D1.
Is that acceptable?
Edit: If not, then you could use an LM317 or LM78xx to generate a 6-12V supply for the circuit from the 24V supply to get the full 5V output (with a corresponding change is some of the resistor values).

 

Offset and scale this output with another opamp and you are there

Added -5 volt reference and summing amp circuit for full 5 volt output at 0 volts Vin.

 

Below is my previous circuit with an LM317 regulator powered by the 24Vdc supply, configured to generate V+ = 10Vdc, which allows a full 0 to 5V output for both 0V and 5V inputs (if needed):
Resistor R7 was added, and several resistor values changed to accommodate the new voltage.

 

My version using a MCP6004 or two MCP6002 with 5 volt Vcc and rail to rail output.

 

My circuit with the added scaling and offset.

Needs only 5V supply and two opamps. The feedback and divider for the second opamp need to have a variable resistor to calibrate it for 0V and 5V.

 

My version using a MCP6004 or two MCP6002 with 5 volt Vcc and rail to rail output.

Below is the LTspice sim of your circuit.
I didn't have the MCP6004 model, so I used the LMV358 dual CMOS RR out, (LMV324 quad).

My initial sim with the values for R1 and R3 on your schematic (R2 and R9 below) gave a small asymmetry at the output.
I assume you just made a mistake in the calculation for those, as you need 1.25V from the divider.
The values below give 1.25V, giving the desired output.

 

Theoretical 1.25 is correct. After actually breadboarding the circuit needed to reduce to 1.20 volts for zero output at 2.5 volt in.

 

Theoretical 1.25 is correct. After actually breadboarding the circuit needed to reduce to 1.20 volts for zero output at 2.5 volt in.

That's a fairly significant 4% difference.
Do you know why that was?
Is it from resistor tolerances or op amp offset?

 

sghioto, your circuit can be simplified some by eliminating one op amp and its two resistors (below).
But that really only saves the two resistors, since you already have the four op amps:
Edit: Since the resistors are all 10kΩ or multiples of that, you could use 10kΩ matched resistor array(s) to minimize the number of parts.

 

Do you know why that was?
Is it from resistor tolerances or op amp offset?

I believe it was from output offset. I'll check again.

 

I believe it was from output offset. I'll check again.

Double check all the resistor values.

 

Double check all the resistor values.

All the resistors were 5% rated but all read within 1%. Was actually using two MCP602 chips and swapping the two around improve the offset.
Then I wired your mod of my circuit and with 0 volts at the + input on U3 had a 100mv offset on the output. That was the same amount I read before on my circuit with the same chip. Replaced U3 with another chip and offset dropped to appx 5mv.

 

My circuit with the added scaling and offset.

You need to lighten the blue trace on your sim graph as it's essentially invisible on my computer (or maybe that's just my computer ).

My only quibble with your design is that is depends upon the matching of two diode forward-drops over current and temperature, as they are not inside a feedback loop.

 

@crutschow

Would you mind confirming the operation of your circuit in post #7.

When V(in) is at 0 volts U1 output is at 5 volts trying to place the minus input at 2.5 volts.
As V(in) rises the output of U1 lowers to try to keep the minus input at 2.5 volts.
When V(in) exceeds 2.5 volts the upper output transistor goes open and the lower transistor closes completely, but the diode blocks any further feedback at that point and the input to U2 starts to follow V(in) exclusively.

U2:
When the voltage at the plus input is 5 volts the output goes to 5 volts to place the minus input at 5 volts.
Then when the input is at 2.5 volts the output must be 0 volts to place the minus input at 2.5 volts.

Is that essentially correct? If not, can you please explain.

Thanks