Monitoring The Temperature
The 2 signals from a variable resistor (top) and a thermistor
(bottom) are buffered and then compared.
The amplified signal difference drives the LED and Vout.
The Vout of this circuit follows the temperature:
it increases with rising temperature
and decreases with dropping temperature.
Adjust Vvar so that the diode barely lights.
This in effect adjust Vvar to match Vtherm.
Here's why.
If Vvar matches Vtherm,
then Vvar = Vtherm = V+ = V-
and there is no current through the 3rd comparator feedback loop.
If the thermistor's value is 10 K at room temperature, i.e. same as
the resistor in series with it, then Vout is 6 V.
In this case the diode draws 1.2 mA and barely lights.
Now if the ambient temperature rises, the thermistor's resistance
will decrease, and Vtherm decreases,
let's say from 6 V to 5.95 V.
V+ is still at 6 V since nothing has changed in that part of the circuit, and the op amp will strive to also keep V- at 6 V.
Therefore a current must flow through R1 to boost 5.95 V to 6 V,
causing a 0.05 V gain across R1.
The same current must flow through R2 and adds another 5 V (100x)
bringing the comparator output to 11 V.
This means 2.2 A through the diode, enough to make it turn bright.
A similar argument can be made to show that as the temperature drops the diode goes dark.
It is easy to modify the circuit so that Vout decreases when the temperature rises and increases when the temperature drops.
The following circuit does just that.
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