I assume the heatsink is there for when it's connected the correct way around and current is flowing through it? if it is I think I can get away without one as normally it's only 500ma bursting to 2.5A for a few hundred ms, also it is connected to a large ground plane that will help things.
I assume the heatsink is there for when it's connected the correct way around and current is flowing through it? if it is I think I can get away without one as normally it's only 500ma bursting to 2.5A for a few hundred ms, also it is connected to a large ground plane that will help things.
You can deduce all you need from the datasheet. If you assumed 2.5A worst case (I know this is more than you expect), Fig.3 says the fwd voltage is about 0.9V. Assume 1.0V. This is a dissipation of 2.5W. The mechanical Specifications table says max. thermal resistance is 15 deg/W with modest (8mm^2) pad size. This means the temperature rise should not exceed (2.5W*15deg/W) 37.5 degrees C. The maximum junction temperature is 150 C. Therefore, you can very conservatively tolerate an ambient temperature of about 110 C, which is above the boiling point of water.
If you're just trying to protect the device against being hooked up backwards, just use a reverse biased diode across the power and fuse the input (which you should do anyways). Then the diode does nothing in normal use but will short out the power source and blow the fuse if hooked up wrong.
And the reverse-biased rectifier across the fused line is the only way to do it without incurring voltage drops with the exception of more exotic crowbar circuits or circuits using MOSFETs in-line with the supply.
The series diode should be a Schottky type if you want a simple circuit that minimizes voltage drop.
If you can stand all sorts of voltage drop, put a bridge rectifier in series with the supply and ground leads and you can connect the power up either way and the device will work fine with only two diode drops in series. Again, Schottkys will help minimize that.