operational amplifier - Impedance to voltage range

I'm trying to get a voltage in a range from a potentiometer.

The potentiometer theorically varies between 22k and 40k (R1 in the shematic) but I'm unsure as I currently don't have it in my hands.

I want to "convert" the potentiometer value to a voltage in [0v-5v].

^{simulate this circuit – Schematic created using CircuitLab}

Here, I first convert the impedance to a voltage trough a voltage divider (R2/R1+R2), and then I input this in a differential LIAO.

It should substract Vc from Va and multiply it with R4/43 and I should have : Vout = (R4/R3)(Va-Vc) which would give me a value in the right range.

But the problem is that It actually won't work because the second voltage divider won't just behave as if Va was a regulated voltage source, and with Thevenin theorem, you can find this equivalent circuit (here just for Va - Vb part) :

^{simulate this circuit}

And then I have Vc = 1.91011 Volts while what I'd want there is Vc = 5*(22k/44k)(68k/78k) = 2.17949

How can i do ?

Answer

Your circuit has 7 resistors and produces a nonlinear response of voltage with respect to resistance.

Here is a circuit with only 5 resistors that can operate from a single 5.0V supply (using a rail-to-rail input* and rail-to-rail output op-amp) and which produces a linear voltage proportional to resistance.

^{simulate this circuit – Schematic created using CircuitLab}

The variable resistance sees a constant current of 39.78uA.

To synthesize this circuit for other ranges, pick R1, R2 to something reasonable considering the op-amp input bias current, accuracy required, and current capability of the variable resistor, and overall consumption.

Pick R4 to maintain a constant current.

Pick R5 and R6 to make the output voltage correct at minimum and maximum resistance.

As simple as that! (some algebra omitted, but it's straightforward)

* actually the input range is only 0.87 to 1.59V so many R-R output op-amps will work.