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So, I'm kinda new to the field and the time has come to design my first PCB heater. The main concept is to drive it using a Raspberry Pi and a temperature sensor as feedback.

Here is the whole procedure I followed in steps.

  1. Since my AC/DC adapter outputs 12V/3A I need a heater around 4 ohms in order to be efficient.
  2. By using this trace width calculator and applying the following values I managed to get a result of 3.73 ohms and a voltage drop of 11.2V.

    current: 3Amps
    thickness: 35um
    temp rise: 65 (I want to be able to reach 90C)
    ambient temp: 25
    trace length: 7m
    
  3. Given that I want a heater of 13mm width I calculated the number of traces that will be created. Which is: \$trace\_length/heater\_width = 7m/13mm = 7000/13 = 538\ traces\$

  4. Then by multiplying \$number\_of\_traces\ *\ trace\_separation = 538\ *\ 0.12mm(120um) = 64.56mm \ heater's\ length.\$

  5. Finally, I have a heater with the following characteristics:

    length: 64.5mm
    width: 13mm
    trace width: 1.14mm
    trace separation: 0.12mm
    

I would like to mention here that in calculator's page there are results for internal and external layers. For my calculations, I used the values of the internal layers and I'm not sure if I did well. So what do you think? Is this procedure and calculations reasonable? Would you suggest a different approach?

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The calculator temperature rise figures are not going to be accurate- you'd be better off to use another method to estimate the rise of an isothermal plate the size of your PCB from the total wattage.

Copper has a positive temperature coefficient of about +0.39%/K so you'll have about 25% less power at 90°C, however you also should never exceed the maximum current output of your power supply, so you probably should aim for a bit higher than 4 ohms. The copper thickness is also not all that well controlled, so unless you can adjust the voltage a bit you might want to be a bit cautious. You could add some taps to trim the cold resistance.

I get more like 3\$\Omega\$ @ 25°C from first principles using \$\rho\$ = 1.72E-8 \$\Omega-\text m\$ for the resistivity of copper for 7m of 1.14mm wide 34.8um thick trace. That's probably worth looking into.

I didn't look at the other calculations- there will be some effect from the shape of the corners (sharp vs. radiused).

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  • \$\begingroup\$ So, you suggest me to increase the total length of the trace in order to achieve a higher total resistance? e.g from 7m to go at 8m? Sorry for being that noobish but I'm just trying to understand the whole idea. Thanks \$\endgroup\$ – J. Doe Sep 22 '19 at 9:05

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