Sensors in voltage divider configuration

[Aldobrandi]

I am building a simple thermostat circuit using a thermistor and to get started I've looked at plenty of exemples. I understand how they work (most of them using an op-amp in Shmitt Trigger configuration) but I'm curious as to why the sensor (the thermistor in my case) is always used in a voltage divider configuration: why can't you simply use the thermistor on its own on the op-amp inverting input ? Here's one of the many circuits I've looked at and which uses this voltage divider configuration : http://pcbheaven.com/circuitpages/Simple_Reliable_Thermostat/

Thanks a lot !

[alm]

Thermistors have resistance. Op-amps expect a voltage at their inputs. How do you convert resistance to voltage?

[Aldobrandi]

Thanks alm for your quick reply.

Please forgive me if I am being dense, but I think I still need clarification on this topic and to be sure I am understood I have thrown together a partial circuit (I know this is not a working circuit, I've only included the parts relevant to my question).

Since the thermistor's resistance is proportional (inversely in the case of a NTC) to temperature, would the voltage dropped across R1 decrease as the temperature increases (because the resistance will decrease as the temperature increases) ?

Please explain to me like I'm 5 why my circuit wouldn't work  Thanks for your patience and your help !

[Aldobrandi]

To follow up I decided to draw an equivalent circuit with a new resistor, R0, forming a voltage divider with the thermistor, but where R0's value is infinitely small. Would such a circuit work ?

[MikeK]

Your first circuit won't work, because the op-amp input is essentially tied to 12V.  And then dangerous things happen when the sensor resistance goes to zero...You're shorting the supply voltage to ground.

A voltage dividier is a basic ciruit whereby two resistors produce a fraction of the input voltage.  And that's what your second circuit is doing.  As R1 goes up the voltage at the op-amp input goes down.  And as R1 goes down the voltage at the op-amp input goes up.

Vout = Vin x (R0/(R0+R1))

[Aldobrandi]

Thank you so much MikeK for your detailed explanation. As we say in French, I am the kind of person who understands quickly as long as you explain slowly! This is much appreciated.

[Jeff1946]

You should put a resistor in series with the thermistor.  It will work best if the resistor has the same resistance as the thermistor at the control temperature.  Proof of this is left to the reader.