The spec sheet had no thermal specs.

I found another diode with similar package (SMA).

The calculations will be based on it's specs below.

**Specs for SMA Package**

**Here are some precalculated values**

This seems really confusing ..

If we use,

- Just the minimum pad size
- One ounce copper
- 50% duty cycle

we get $100^{\circ} C$ per watt

If worst case is 100% duty cycle we now get, $200^{\circ}$ per watt

Thus at 0.5 watts we get a final temp of $125^{\circ}C$ assuming $25^{\circ}C$ ambient

Let's do the calculations manually,

For the 'leads'

In [3]:

```
:opt no-lint
import Text.Printf
degCperWatt = 645*125/11.8
putStrLn $ printf "DegC per Watt for Leads Min Pad Size %8.2f deg C per watt" degCperWatt
```

DegC per Watt for Leads Min Pad Size 6832.63 deg C per watt

Well, the leads are no help .. !

Now let's try junction to ambient ..

Spec says $270^\circ C$ per watt,

So at 0.5 watts we get a final temp of $160^\circ C$.

I tried an online calculator (not completely sure I'm using it properly) and got,

At least we're in the ballpark ..

There are two questions remaining ..

- What is the leakage vs temperature curve
- How big of a copper pad should we use

Three choices from Jay for 'analysis' but I found an interesting axial diode that's cheap.

Uses DO201AD Axial Package .. And it's cheap!

If we mount the diode with one centimeter leads on each side and about 3mm in the air we can achive about $80^{\circ}$ C per watt. We could probably live with 5mm lead length on both sides.

At 0.5 Watts,

$40^{\circ} (diode) + 25^{\circ} C (ambient) = 65^{\circ} C$

If we assume absolute worst case is $100^{\circ}$ C then the

reverse leakage power is about 50mw .. No problem

I consider this a success.