The diode shunts power to ground if the clutch is not pulled in (but the transmission is in neutral):
- Starter Circuit
If your starter relay was wired in wrong, and presented a dead short instead of the relay coil, it will have put a massive amount of current through that diode, and that will let out the magic smoke - as you discovered. It will also melt wires - also as you have discovered.
A diode is a diode. Take your ohm meter and measure the resistance of your starter relay coil.
Take that measurement (in ohms) and divide 14 by it. So if you measure 10 ohms, then divide 14 by 10, which gives you 1.4.
That result is the amount of current that will be going through the diode. Whatever that value is, double it. Now find a diode that is rated for that. Do a search for "high current diode" and find one that is rated for at least 2.8 amps (or whatever the result is from your measurement). It shouldn't cost more then $5-$10, depending on the size.
While not the correct way to do it, I've cheated when I didn't have the correct value diode, and wired several in parallel to get the rating I needed. For instance, I would take three 2-amp diodes, wire them in parallel, to replace a 6 amp diode. There are drawbacks to this, but for this application it should work fine.
