I looked around at other questions, and either I didn't understand what all I was reading, or it didn't answer my question so if there is an article that I've missed feel free to redirect me.

I'm fairly new to electronics, and came in through Arduino so I'm pretty fuzzy when it comes to the actual electronics behind how things work.

I'm trying to create a heating device for heating a chest freezer that I'm trying to convert to a incubator. I have the "coding" side done with Arduino, now I'm trying to build the electronics. I have some relays that I'll be using to turn things on.


I have a 12v 6a DC power supply that is powering my entire project. I'm trying to use 4 x 10W 12ΩJ Cement resistors. How should they be configured to produce my needed heat.

So I thought to myself, "I have 12v * 6a = 72w power supply, if I'm using 4 x 10w resistors 72 watts would be too much for them." I hooked them up this way though, the resistors in series, and they barely heat up, so I must not know how these things work even though I thought I did.

Is there a good way to calculate how much power a resistor should get for me to reach a specific temperature. Is there a calculator, or anything that can be used, or a formula, or is it resistor specific?

If I connect Just 1 or 2 in series they get ALLOT hotter, but then I'm worried about burning out the resistors.

Any help would be great, even if its a bunch of "Go look and read over here." comments.

Thanks all.


The space being heated is about 1.5 Cubic meters, I don't know the insulation type, its a freezer, and the other unit I'm looking to make this for is a wine cooler.

The reason for using 12v DC is incase of prolonged power outage that it can be easily run off of a battery back up, or a solar panel / battery system.

The space can be heated with a 40w light bulb fairly well. There are other commercial devices like this, that use a similar setup.

The temperature I'm hoping to achieve varies depending on the egg, but no more then 100F.

As far as the 72 WATTS I was just saying that is what my DC supply is putting out, It would be much more convenient to run it on less obviously.

  • \$\begingroup\$ Before we try to straighten this out, can you describe the incubator (size, insulation, etc.), what temperatures you hope to achieve. Then tell us how you figured out that 72 W would be sufficient to heat it up. You could do an experiment by putting a 60 W (incandescent) bulb in there and see if it heats up enough. Also, why are you running it all from a battery rather than mains. Please edit and improve your question rather than post the answers in comments. \$\endgroup\$
    – Transistor
    Dec 29, 2015 at 22:01

2 Answers 2


First concentrate on each of your 12 ohm resistors. Their max rated dissipation is 10 watts. The quickest way to find volts and current from power and resistance is max_voltage = sqrt(max_power * resistance) (or you could simply calculate current and hence power for each of several applied voltages). In this case, sqrt(10*12) is roughly 11v.

As you have a 12v supply, you need 2 in series, where the 24ohm total will give you 0.5A, and so 3 watts per resistor. With a 10w rated resistor, if you feel it's getting too hot at 3w, that may mean it's not heatsinked sufficiently, or your expectations need recalibrating. A 10w resistor would need to be big to be air-cooled, or to get very hot. If it looks like it can be bolted down, do bolt it to a heatsink. If it's ceramic with no flat surfaces, then it's supposed to run very hot in air.

Two parallel strings of two series resistors will give you a total of 12 watts. Is that enough for your application?

[hooligan alert] In my world (turn it up until it blows up, then back off a bit), 12v on a 12ohm resistor will give you 12 watts, which a 10w resistor will probably survive for a while, as long as you keep it cool. Four such resistors in parallel would give you 48 watts, enough? [/hooligan alert]

  • \$\begingroup\$ Thank you for this, I really appreciate the help. They are ceramic, and I'm using a 12v pc fan to run air past them. I too would often try your "hooligan" method, but I didn't want to ruin anything as I'm on a tight budget. The resistors themselves are rated to 155C, way to hot for me to "feel" if they were too hot, and in theory my arduino will be turning them on and off to keep about a cubic meter of air at the right temperature, and its insulated so, I don't think they should get too hot "I guess" before they get turned off again. Thank you! \$\endgroup\$ Dec 29, 2015 at 22:36

I think temperature will be very tricky to calculate, because you can have a very high temp. over a small area, and it's probably not the result you want.

Can you convert your heat requirement to BTU?

Then it becomes a very straight forward calculation, 1 BTU = ~1055 wattseconds.

If you hook your resistors in series, you'll dissipate 3 watts. Current = Volts / Resistance = 12/48 = .25 amp watts = Volts * current = 12 * .25 = 3 watts

Here is a site for that.


If you hook the resistors in parallel you'll ~ 48 watts, but you should check that, because I just ran the numbers quickly in my head. Remember the "sum of reciprocals calculation for resistors in parallel.

Here is a site that has some calculators for thermal units, etc.


  • \$\begingroup\$ It just occurred to me that if you can get your power dissipation values close, close everything up, and see if the temp in the box reaches your target range in < 4 hours. If it does, then it's just a matter of 'calibrating' with your software to maintain the correct temp. I assume you have a feedback system - some way to measure the temp in the box, thermo-couple, etc. If you don't get close to your target, you need a bigger power supply. Also, the 48 watts I came up with agrees with the 2nd answer, so that's prob. close. \$\endgroup\$ Dec 29, 2015 at 22:13
  • \$\begingroup\$ Thanks for this, and for the links. Yes I have an arduino with a temp sensor on it, and admitedly, with the 4 in series it was working "slowly" as a 5" by 8" by 2" box ( fairly small ) changed temperature from 76F to 93F in the course of about 20 minutes. Thanks for your work and calculations up there and the links. I also feel a bit safer after seeing what you've said that I'm not going to burn something out. Thank you! \$\endgroup\$ Dec 29, 2015 at 22:39
  • \$\begingroup\$ What is your goal temperature? (100F?) I actually think it will be a little safer if your resistor network produces more heat, and it's switched off for a longer time. (lower duty cycle) If you go with the 12w configuration, I think you'll find the temp easier to regulate. If you're taking the cover off, or if the ambient temp drops, you've got a little more head room of heating capacity to bring the box back to the spec temperature. \$\endgroup\$ Dec 29, 2015 at 23:00
  • \$\begingroup\$ Yeah around 100F is the goal. I did test today with two series of two resistors and was able to heat that same little box to about 135F in under 5 minutes. Based on application needs, I'd say you are right, it would be better to burn hotter faster for a short period of time. Originally I thought to have two modes, a maintain mode which produced less heat and could average heat loss over a time frame and then compensate proactively, and a burst heating mode, for exactly as you mentioned, say if the door was opened. Thanks again for the help Cheers! \$\endgroup\$ Dec 30, 2015 at 5:50

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