I'd like to use a ceramic heating resistor to keep a 4-5 cu ft insulated box above freezing on cold nights. A 15w heater belt is too much so I was thinking about using either a 5w or 10w heating resistor. Seemed like something I could eventually design a circuit to control the heat with a DS18B20 sensor and an Arduino.

Does the following look correct or am I missing something?

The resistor I looked at, a Riedon UAL-5, is rated for 1500 VAC. I assume I can run it at 120 VAC which is the RMS value for house AC. I=P/V so 5w/120v = 0.0417 amps. R = V/I so 120v/.04a = 3000 ohms and that would be the size of the resistor I need to generate 5w?

Does the current need to be rectified?

PDF of a part I was considering which will mount nicely on a heat sink:


Thanks, Dan

EDIT1: I've done some empirical tests with a 15w heater belt that I borrowed off of a friend and I know the insulation is good enough that 5-10w will be enough. It's the calculation for the resistor and question on rectifying that I was sure of. Not to mention that I can always move up in resistor wattage.

I appreciate the answers about area heating and insulation but I purposely didn't ask about that. I guess I should have said ignore the insulation and area heating as I can deal with that. What I really wanted was to understand the calculation and then I would buy the size of resistor I thought was appropriate. Now that I think about it I'll probably go 10 or 2x5 on separate switches.

I was going to use a grounded electrical box ventilation hold to contain the resistor and heatsink along with shrink tubing on the connections. I am aware of the cautions around using 120v.

I appreciate all the answers. Thanks Barleyman. Barley?

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  • \$\begingroup\$ Without knowing the thermal resistance of the insulating box, you are guessing. If ΔT('C)/W matches your intuition, you may be lucky. Otherwise you should measure the temp rise ΔT, for a fixed W then regulate or compensate depending on ΔT. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 22 '18 at 3:03
  • \$\begingroup\$ Yes, that would be the right value to get 5W. No, it doesn't need to be rectified. \$\endgroup\$ – user253751 Feb 22 '18 at 3:16
  • \$\begingroup\$ As a general rule 1.5 watts of heat raises the local ambient temperature about 10 to 15 degrees C. That would cover about 4 cubic inch's, depending on insulation. \$\endgroup\$ – Sparky256 Feb 22 '18 at 3:46
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    \$\begingroup\$ what about a heater for lizards ... get one at a pet shop ... or a coffee cup heater \$\endgroup\$ – jsotola Feb 22 '18 at 7:15
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    \$\begingroup\$ If you know that 15W is enough, why not just put together something to regulate the temperature? Thermistor and a comparator and something to turn on the heater when it gets too cold in the box. That way, you can also handle nights when it gets colder than expected. \$\endgroup\$ – JRE Feb 22 '18 at 12:12

Your calculations are correct and no, for generating heat there's no need to rectify anything. Short form of power is \$ P = \frac{u^2}R \$ so in other words \$\frac{120^2}{3000} = 4.8W\$ which is what you wanted.

That being said, connecting live wires to an exposed resistor is dangerous, do you have skill/background to make a robust connection that's not going to let the wires come loose etc? 5V power supply wires shorted creates sparks and possibly smoke depending on the power supply, loose live wires can create a dead hobbyist.

A much safer alternative would be to get 5V or 12V power supply and use that to generate the heat. You don't need to get a single chunky resistor either, you can daisy chain cheap 1W or 2W resistors with reasonable derating (say, 50%) to achieve the same heating. Solder them to a breadboard for extra robustness.

Do not use breadboard for live circuits unless you really know what you're doing and have everything inside insulated case with cable strain reliefs et al.

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