Is the cathode for schottky diode hotter than the anode, as is common for most diodes, or is it reverse polarity? A schottky is a metal-semiconductor junction, with the metal at the anode. So it should be better for heatsinking, even if the cathode is actually larger, right (i.e. reverse polarity in package nomenclature)?
I'm specifically interested in the 1N5819 and 1N5822 leaded DO-41 and DO201 package.
I have never seen a diode with a thermal pad on the anode.
Always on the cathode if one exists.
This is a Vishay SS2PH10 Schottky Barrier Rectifier used in my Buck LED driver in a DO-220AA package.
The cathode has a much larger pad for thermal dissipation.
And the temperature measurement point is the cathode.
A schottky is a metal-semiconductor junction, with the metal at the anode. So it should be better for heatsinking
It may be true that the metal is going to conduct heat better but metal characteristics are not going to be affected much by heat.
The metal anode could be smaller than the cathode contact metal.
Source: Electronic Notes
The cathode also has a metal layer.
Source: Electronic Notes
Source: Electronics Maker
Which Method is Best?
Even though method 3 has the lowest RΦja method 1 may be the better method.
The RΦja listed is for leads only.
Method 1 would allow an additional thermal path from the case to the board.
Method 2 could be interesting if the vector pins were copper with large vertical fins.
Kind of like this:
There might be differences, but if your concern is just heatsinking, you have to rely on the thermal mode given by the manufacturer. According to the thermal model in the datasheet, they don't specify a different thermal resistance from junction to the anode or from the junction to the cathode.
This is the model you have to use to make your thermal calculations. RtL and rtJ are the thermal resistances for junction and lead.
I wouldn't assume that the metal used to form the junction's anode is actually the metal used to form the diode's anode lead. This seems very unlikely to me. There's probably a thin bond wire connecting the two, so the thermal conduction may be no different than a normal PN diode. Check the datasheet or ask the manufacturer.
LEDs have an ultra-thin gold wire whisker bond to the anode so the heat sink is the cathode but since the wire bond is large in these diodes, they don’t have to look small which is why they conduct heat well to both terminals. Some power diodes now use aluminium instead of copper for wire bonds or direct terminals.
