# PCB heater power considerations - sanity check

I'm designing a PCB heater capable of heating to 55°C in a 20°C environment.

Given the trace length/width that I decided to use, I used the Saturn PCB Design tool to calculate what amperage I would need to get the desired result. The tool estimated that the trace will have a resistance of 1.5 Ohm, a voltage drop of 8.5 V and 5.7 A will be drawn to reach the wanted temperature, effectively using 48.5 W of power.

All of this means that, considering heating element only, I would (at least) need an 8.5 V PSU capable of delivering 5.7 A. Of course 8.5 V PSUs don't exist, and even if they do I would get something more capable of that to have a safety margin. So let's say I use a 12 V PSU: to achieve the 48.5 W needed it would only have to draw 4 A of current, but the resitance in the circuit would always be 1.5 Ohm, effectively drawing 8 A of current (too much).

To solve this, and to have some control on the temperature, I can add an N-Channel MOSFET and regulate the Gate voltage (with PWM + RC filter) to add the right resistance to the circuit and get the desired power draw, ranging from 0 to 48.5 W.

The MOSFET I selected is BUK9M156-100E, which is suitable for microcontroller projects and can withstand ~100 V and more than 8 A (if thermals are managed correctly).

By looking at the graph below I understand that the Vgs range suitable for my application is between 2 V and 2.8 V, which of course needs to be verified with the actual hardware in hand.

The voltage after the heating element will be 12 - 8.5 = 3.5 V, which by going into the MOSFET will be dissipated into 4^2 * 0.15 Ohm = 2.4 W. This considering a max load situation of 4 A @ 12 V.

2.4 W seems pretty doable to me with good PCB design, also the MOSFET is rated for 36 W.

Here is the schematic for the heater portion, I still need to select many components and do the rest so bear with me.

The rest of the circuit consists of an ESP32, IRM-60-12 AC/DC brick, AP2112K LDO for the ESP32 and an INA226 to monitor current from a shunt resistor.

Being my first high power project I'd like a sanity check, do these calculations make sense?

• Can you draw a schematic example with the tool (edit the question) Commented Nov 12, 2022 at 22:17
• @VoltageSpike you mean a schematic of the circuit? Commented Nov 12, 2022 at 22:24
• Looks good, thanks Commented Nov 12, 2022 at 22:45
• is the INA226 used in the heating circuit? Commented Nov 12, 2022 at 22:47
• @VoltageSpike I haven't made the sch yet, but yes I will be using it with a shunt to monitor the current going through the heater. I have yet to decide where to put the shunt, before or after the heater+MOSFET pair. I need to study the matter before taking a decision. Commented Nov 12, 2022 at 22:51

If you use a MOSFET in linear mode it will create a hotspot on your PCB. If the goal of a pcb heater is to make a flat surface at a homogenous temperature, that could be a problem.

Note the 36W rating of this MOSFET is given with the board magically cooled to 25°C. If you put a copper pour on the drain, you will have to calculate its thermal impedance. 2.5W is quite a lot for such a small part, unless the PCB is cooled by contact through a squishy pad.

The simulation provided by Saturn Toolkit most likely rests on a number of prior assumptions which are not clear. For example a PCB mounted vertically will have much better convection cooling than if it is mounted horizontally, and if you use it to heat something, that would also have to be taken into account. So the power calculated by Saturn should be taken with a grain of salt.

Since the 4mm trace width is quite large, I'd recommend working with 24V and using thinner traces to adjust the resistance value to something easier to drive with simple PWM. Also if you use the cheap Chinese manufacturers, you have to read the docs, because quite often the copper is only 18µm, which will double your resistance.

If you don't want to use PWM due to noise concerns, then you'd have to drive the MOSFET with an opamp, and use the copper trace as the current sense resistor. There is no need for a separate sense resistor in this case. Driving the gate directly with a microcontroller DAC would require a fast control algorithm, otherwise the MOSFET will go in thermal runaway quickly.

• I'm indeed using an heatsink on back of the MOSFET with a squishy pad on exposed copper. Anyway your idea of upping the voltage to lower the amps and thus less power loss no matter what MOSFET/gate driving technique use is awesome. So now with 24 V, to obtain the 48.5 W theoretical needed power (I agree not to trust Saturn pcb too much and plan for more), I would only need 2 A. With a 3 mm trace width its even less. Now with 150 mOhm RdsOn I get 0.6 W power loss on the MOSFET. Combined with the heatsink and all this should now be a lot better, right? Commented Nov 13, 2022 at 19:47

First thing, you don't want any capacitance on the FET, or a voltage divider. The fet needs to be fully off or fully on. If it isn't power will be dissipated in the FET and that is undesirable and could break the FET.

Lets think about the fully on situation of the FET. 150mΩ*8A^2 is 9.6W. If for some reason this circuit get's stuck on even in the fully on situation, it will experience a 40C temp rise (~9.6W * 4.16K/W (junction resistance)= ~40K or 40C) which isn't bad considering it can take 175C but you might not want to handle that much power being dissipated in your design in one spot.

Really you want to rougly calc the power being dissipated in the FET and junction temp, because the temperature of the part should not be exceeded.

The other problem is when switching, transition where the resistance is near the middle (1.5Ohms for the FET) so that would be 3Ohms total and 1.5Ω*3A^2=13.5W for the FET, which it could handle being in this state also, it would probably be in the 55C range at 13.5W.

So the FET could handle anything you could throw at it. It's probably better to have the FET switch on and off as fast as possible. Also no current shunt is necessary, you already have a good idea of how much power is going into the circuit (9.6W) and lets say a duty cycle of 50% is used... then you'll get 4.55W, and a 10% duty cycle would be 0.96W.

A better way for feedback would be to get a chip temp monitor or an NTC thermistor and measure that. You can then run a PID and do temperature control.

• I would prefer not to use the MOSFET as a switch because I will never need it to be fully on and I wouldn't want to introduce any switching noise in the circuit. Also I saw many examples of using a MOSFET as a load driver in an heater circuit, so I would feel confident about that. All I'm worried are the calculations in the original post, you didn't complain so I would assume they are sound. I indeed will add an NTC thermistor to have over temp protection, forgot to mention it. I agree that the shunt is not necessary, but for educational purposes I would like to use it to learn about it. Commented Nov 12, 2022 at 23:41
• The fully on was only for heat analysis its important to look at the edge cases. The NTC would also be used for temp control if you used a digital control loop. Iearning PID's would be very useful as they are used everywhere. Commented Nov 12, 2022 at 23:52
• I will indeed be implementing a PID after having tried a basic, manual non-PID solution. Just to make sure, is it correct to say that with a 12 V PSU, if the heater has voltage drop of 8.5 V the FET will receive 3.5 V as Vds? Commented Nov 13, 2022 at 0:25