Alternative Wheatstone bridge for thermistor - Interpretation needed

Question

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.

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?

Thank you!

Answer 1

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.

Answer 2

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.