Can anyone clarify if I am not understanding this comparator circuit correctly. The author says that if the temperature drops across the ntc thermister, resistance will go up and cause output to go low. Since it is the input signal connected to the inverting pin, would it not increase the voltage drop across it therefore dropping available voltage between r1 and r2. Non inverting would be greater than inverting and therefore output would be high. If I am wrong, please clarify. Thanks again



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  • NTC on lower side of bridge to Vin-
  • Cool causes R to increase and Voltage rise across thermistor causes output low and NON Off with relay off.
  • Excess heat drops Rth value and also Vin- drops voltage
    • when Vin- < Vin+= V+/2 the ref. point, it causes Op Amp out high and NPN ON and Relay ON.
  • THIS MUST BE FOR A/C mode or COOL setting.
  • HEAT setting reverses the output logic.

  • Notice the feedback resistor ratio to Vin+ causes hysteresis of 11k/3M3 which translates into the allowed backlash and temperature error.

  • It also affects the Relay cycle rate so that it is not too fast or too slow. Using a more precise Vcc accuracy and a trimmer, you can reduce this ratio further from 3' C to 0.5'C in terms of the voltage sensitivity per degree.
  • Often this is done with a precision 2.5Vref IC with 0.1% to 0.5% stability in modern inexpensive digital Thermostats.
  • However the accuracy might still be 3'C depending on other tolerances so one adjusts to how it feels, but a 3 'C hysteresis would be excessive unless sensed closer to the A/C source.


  • \$\begingroup\$ Tony, you are making the same mistake. Bullet 2 is wrong. \$\endgroup\$ – Ale..chenski Sep 30 '16 at 2:54
  • \$\begingroup\$ murphy's law watching tv {Blacklist}... "Voltage drop rise" changed to "Voltage Rise" \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 30 '16 at 3:03

Your analysis is incorrect at the very beginning. If thermistor R goes high, the IN- input (R1/R2 point) goes up, not down. Then the comparator output will go low, and relay will disengage.

  • \$\begingroup\$ Thanks for the response. I realized after that I had made my beginner mistake forgetting that I could sum the resistors from r2 down and plug it into the voltage divider formula. The voltage drop across thermistor would be higher, but overall voltage needed at r2 junction would rise. I think :) Thanks for clearing that up. \$\endgroup\$ – Archaeus Sep 30 '16 at 11:46
  • \$\begingroup\$ If the answer clears your question up, why no acceptance nor upvote? \$\endgroup\$ – Ale..chenski Sep 30 '16 at 15:38
  • \$\begingroup\$ Because I have no reputation to vote. That's why. I did say thank you, but I suppose you're a man of profit.. \$\endgroup\$ – Archaeus Oct 2 '16 at 12:40

The resistance of the thermistor will increase.
This will increase the voltage across the thermistor.
This will increase the voltage across the combination of R2, R7, the thermistor, and the variable resistor, which is the same as increasing the voltage at the - input. of the comparator.

If the voltage increases enough, it will now be greater than the voltage at the + input, which will cause the output of the comparator to go low, which will cause the transistor to turn off, which will cause the relay to turn off.


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