Thermal power dissipation for a diode with heatsink

Question

I have a circuit that has a diode that needs to support 3A through it. The supply is at 7.2V and is therefore ~22W.

I have currently selected a schottky diode show here. Key specs are it can support up to 5A, has thermal resistance junction to case of 12K/W, and a max junction temp of 150C. The heatsink I plan on using has a thermal resistance of about 3C/W.

I am trying to do some thermal calculations to make sure the device can support my usecase. I am using an online calculator such as this one and input the specs from above. Two things I have questions on:

1. The units are in K/W. Looking online, it seems people treat C/W and K/W the same, and I am not sure why as they are different units; however if you convert the K to C values, you will get all negative values. Once again more confusion.

2. What do I use for power here? Using 22W (system power) gives ridiculous results and states that the heat sink will not work, so I am assuming that I use the max forward voltage at max current (~.36V) times the current I plan on running it at (3A) and take .36 * 3 = 1.8W. This then was input to the calculator showing a junction temp lower than what the diode is spec'd to (i.e. I can use it). Am I correct using this power?

Answer 1

Temperature differences in Degrees C and K are the same. Using K for differences helps keep it straight that we are talking about differences such as rise above ambient, and not the actual temperature.

You use the power dissipated in the diode to calculate the temperature rise- Vf multiplied by the current, just as you conjectured.

Answer 2

Spehro is (of course) totally correct in what he said, but I'd like to add:

• As I read the datasheet the forward voltage at 3A is 0.34V, the 0.36V is for 5A. But being conservative won't hurt you.

• How are you going to use this diode with a heatsink? The 12K/W you quote is the resistance to the soldering point of the cathode TAB, so the normal way to heatsink is to add a large PCB copper area. I don't think a copper area of 3C/W is realistic: this calculator uses 1000 (cm^2)*C/W, so you'd need ~ 300 cm^2, but I think their calculation only works for smaller areas. You could check this question How do I determine the area of copper needed on a PCB to provide adequate heatsinking for a power SMD MOSFET?