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I have a problem understanding the following Wheatstone bridge circuit. It originates from a temperature controller, the voltage V_out is supplied to a PID regulator and hence can be seen as an error signal. V_out is zero when the thermistor resistance matches the value of the reference resistor.

Wheatstone bridge with a thermistor.

I'd like to understand the purpose of the OP-AMP in the Wheatstone bridge and how I can model it on paper. Additionally, what is the (dis-)advantage to the more common way of e.g. this circuit?

Common wheatstone bridge.

Thank you!

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On your hand drawn circuit all the opamp does is generate a negative supply the same magnitude as the positive supply. Vout will be a continuous analog signal that is zero when the Rtherm = Rref.

The second circuit uses the opamp as a comparator to give a discrete signal indicating whether the temperature is above or below balance. It then operates the relay to control the heater.

One comment about the first circuit - it is not good design practice to put a capacitor directly on the output of an on-amp. It may cause incorrect operation, oscillation, or even damage to the on-amp.

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  • \$\begingroup\$ 2nd schematic: Ah right, instead, I meant having an IN-AMP at the same position, which should do the same as the 1st circuit. 1st schematic: Good to know, that the capacitor (you probably mean the 22uF) is bad practice. What would be its purpose, and can I just omit it? \$\endgroup\$ – A. Impertro Aug 19 '17 at 14:22
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I understand the 2nd circuit but not the 1st one. Explain how the 1st circuit can even work because I do not see how it can even work. It simply makes no sense to me.

The 2nd circuit a reference voltage using R2 and R3 and (the opamp) compares that to a varying (over temperature) voltage made by R1 and the thermistor.

When the resistance of the thermistor rises due to low temperature, the voltage at the + input of the opamp rises. When it becomes larger than the reference voltage at the - input of the opamp, the output of the opamp becomes "high" (switched to positive supply rail) which will switch on Q1 and therefore the heating element.

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  • \$\begingroup\$ Thanks for your comment. I have now simulated the first circuit in LTSpice and it behaves as expected. Since I posted the question, I have found that the OP-AMP does a "common mode suppression", see e.g. link Fig.4. Still, there are very few resources about this circuit but it seems to have very good characteristics. I am currently trying to calculate the output voltage as a function of the resistances by treating the OP-AMP as ideal and neglecting the capacitances. Because of the inverting amp, the node between the two R_w is VGND. \$\endgroup\$ – A. Impertro Aug 18 '17 at 21:35

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