I have an AC single phase motor (120V/60Hz) connected to mains (plus a power resistor in series, and capacitor and voltage controlled resistor in parallel) The motor has a NTC thermistor attached to it. I'd like to know how to design a circuit that uses the thermistor to cut off the AC current when the motor gets too hot.

In my research I've read different things about using relays or BJT's but haven't been able to come to a definite solution. My best guess is that I use the thermistor to divide the voltage from a DC source and use that to control a relay. I read here (Relay control of AC motors, handling turn-off kickback) about some problems with just having the relay. So I'd like to know what you'd suggest.

I'd like this solution to fit on a proto board if possible, if that helps with scope.

If it helps, I found that the thermistor is about 8 kOhms when the motor is at room temperature and about 4 kOhms when the motor gets too hot to handle. I don't need you to design the circuit for me but an arrow in the right direction would be grand.

Thank you!

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    \$\begingroup\$ A more reliable and repeatable method would use the themistor as part of a voltage divider driving a comparator. Use a variable resistor to drive the second input of the comparator, that way you can set the temperature trip point manually on the fly. That'll give you a nice controlled and repeatable cut-off when the motor overheats. The comparator should be able to drive a BJT of MOSFET without much difficulty, which would in turn drive a relay etc etc etc. \$\endgroup\$ – Sam Feb 9 '17 at 8:08

Here's one way to do this with a few transistors and resistors:

R1 is the thermistor, which goes lower in resistance as its temperature rises. When that happens, Q1 is turned on, which turns on Q2, which turns on the relay, which turns off the motor. The normally closed contact of the relay (pins 1 and 4 in this example) are in series with the motor on the hot side of the AC line.

R5 provides a little positive feedback to get snap action instead of fading the relay on and off. This is better for the relay and for overall operation.

R1 and R2 form a voltage divider. The relay will be energized (motor turned off) when the output of this divider gets to about 600 mV. In this example, that means the thermistor trip resistance is about 73 kΩ. You will need to adjust R2 to set the trip point for whatever thermistor you actually get.

Much lower thermistor values, with accordingly lower R2 values, will work, but be aware of self-heating. At the trip point, the thermistor will have about 4.4 V across it. The dissipation of the thermistor is therefore

  W = 19.4 V2 / R

where R is the resistance of the thermistor in Ohms, and W its dissipation in Watts. You want to keep W below a few 10s of mW. Working backwards using 25 mW as the maximum dissipation, that means the thermistor must be 776 Ω or more at the trip point. Fortunately thermistors of a few kΩ are easy to find.

As I said above, R5 sets the hysteresis size. You probably need to experiment to find what provides snap action without causing a overly large hysteresis band. I'd start with the current thru R5 about 1% that of R2.

As others have said, use a cheap wall wart that runs from the same power the motor is driven with to get the 5 V.

  • \$\begingroup\$ Hi Olin! Thank you for the response. I have been working with that idea for the last few weeks. I don't understand when you say that R5 should start with about 1% of R2. In your diagram R5 is 10000% of R2. Did you mean the other way? \$\endgroup\$ – ABradford Mar 14 '17 at 22:10
  • \$\begingroup\$ @ABrad: Sorry, I meant the current thru the resistor, not the resistance. That means R5 should be about 100x R2. I updated the answer. \$\endgroup\$ – Olin Lathrop Mar 15 '17 at 11:22

Here's how I would do it (similar to Sam's comment above):

-use a wall wart DC power supply and a 7805 voltage regulator to get 5V

-find a normally-closed (NC) relay that can handle your motor voltage and current

-find a 5V comparator chip that can output enough current to drive the relay coil

-build a voltage divider that goes 5v - thermistor - 6kohm resistor - GND (the point between the resistor and thermistor is your "sense" voltage. The 6kohm isn't important; you just want something kind of close to the thermistor resistance.)

-hook up a potentiometer (maybe in the 1k to 50k range; not important) in a voltage divider setup to produce a variable "reference" voltage

-hook up the comparator chip so it powers the relay only when "sense" voltage is < "reference" voltage

Also, you probably want an additional feature: the motor should stay off until it cools down a lot, rather than come back on as soon as it cools down a tiny bit. This is usually called "hysteresis" or a "bistable alarm". One way to accomplish it is to add a feedback resistor from the comparator output to one of the voltage nodes. I.e. if you choose to make the comparator output 0V when the temperature is too high, you could connect a resistor from its output to the "reference" voltage. That will force the "reference" setpoint lower, so the temperature has to go lower before the relay will come back on.


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