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This is circuit is proposed for a thermomether based on the TSP102 sensor (linear PTC). How can I determine the equation for the output voltage? My main question is about the op amp. I know that two voltages dividers are the inputs of the op amp. What would the equation be for the op amp output?

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    \$\begingroup\$ Have you done the current calculations yet? \$\endgroup\$ – Ignacio Vazquez-Abrams Jul 11 '16 at 2:45
  • \$\begingroup\$ Yes, I have them. \$\endgroup\$ – Blue_Electronx Jul 11 '16 at 3:32
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    \$\begingroup\$ Then calculate Vo/Vin and...done? \$\endgroup\$ – Daniel Jul 11 '16 at 4:33
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One way to simplify the problem is to use the Thevenin equivalent circuits for those voltage dividers you have there. That is, you could draw the circuit in this manner:

schematic

simulate this circuit – Schematic created using CircuitLab

Here $$V_{th1}=\frac{R_1}{R_{sensor}+R_1}V_{cc}$$ \$R_1\$ is just your 680 ohm resistor in series with your 500 ohm potentiometer. \$V_{cc}\$ is the supply voltage.

Your \$R_{th1}\$ is simply: $$R_{th1}=R_1||R_{sensor} $$

The same procedure is done for \$V_{th2}\$ and \$R_{th2}\$.

You should get \$ V_{th2}=\frac{V_{cc}}{2}\$ and \$R_{th2}=\frac{R}{2}\$. Where \$R=1.2k\Omega\$.

Now, you can use \$V^-=V^+\$ and start your opamp analysis. In the end, you should obtain the equation for the opamp as a differential amplifier. I'll let you do the math, but your final solution for \$V_{out}\$ should be (if I didn't mess up somewhere):

$$ V_{out}=V_{th2}-\frac{R_f}{R_{th1}}(V_{th1}-V_{th2})$$

Where \$R_f\$ is your \$5k\Omega\$ feedback potentiometer.

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