# Serpentine PCB heater size calculation

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?

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.