I've got a very simple circuit. A MOSFET (TK15D60U) is powering a 3 to 4 Ohm heat bed by 19.6 volts. From my calculations this results in 6.5Amps. The power adapter seams to have its specs at max 6.32A but seems to function fine. The MOSFET should be able to handle 15Amps continous drain. The drain should be able to handle 600v, so that 19.6Vdc should be fine.
The gate is driven by a 25Vdc via a 10k resistor to the gate and 100k resistor from the gate to ground (to allow to leak away to discharge the gate, I thought it was a good thing). So effectively there is (25vdc/110k)*100k=22.7Vdc voltage at the gate. The Gate-source voltage is 30v so that is below max.
What I can read from the datasheet is that the Drain-source ON resistance @Vgs 10Vdc @ 10Amps is about 0.15Ohm. Lets take 0.2. I never really thought about the ON resistance. I thought, the gate would be fully open, so resistance should be minimum and will handle it without a problem. But when the MOSFET is active, it becomes too hot to touch within 10 seconds or so.
Now this part is something I am not so familiar with. Hence the question. Is it true that the formula P=(I2)*R (Watt=Ampere squared * resistance) yields (6.5A*6.5A)*0.2Ohm=6.5Watt that is dissipated via the MOSFET?
6.5Watt is a lot if you ask me on a TO-220. Is it not? How large should the heatsink be to handle that? Should I actively cool it using a fan? All I have now is this very tiny heatsink. This topic is probably a laugh for the experts around here.