From what I have read here, the 60A rectifier would have worked just fine with a proper heat sink.
The diode junctions probably have an absolute max temperature of 150C. Let's assume an ambient temperature of 40C. This means the junction can rise 110C above ambient.
Let's also assume the regulator needs to dissipate 80W of heat. This means that for every watt of heat the temperature can rise by no more than 110/80 degrees. This is 1.375 degrees C per watt.
Now we need to look at how we can make all the thermal resistances add up to less than 1.375 degC/W.
First is the thermal resistance from the junction to the case of the rectifier.
Here is a 50A bridge, and
here is the datasheet.
The typical forward voltage drop at 40A is a shade over 1V. Note that this is at 20degC, as the temperature rises this voltage will fall (which will reduce dissipation -- a good thing)
Note the thermal resistance junction to case is 2 deg C/W. This part would not be suitable.
Let's try a "beefier" bridge.
Here is one, and
here is the datasheet.
The first thing to notice that at 25C the device can dissipate 140W. This doesn't mean that it can do this without a heatsink, it simply means that if you keep the case at 25C the device can safely dissipate 110W. This is actually a fairly fine margin.
The thermal resistance from the junction to the case is 1.15 degC/W, and the typical resistance to the heatsink is a further 0.1 degC/W. This eats up 1.25 degC/W of our 1.375degC/W budget.
You would have to find a heatsink rated at 0.125degC/W. This would be *huge*
However, we can do better. Imagine we could use the diode bridge we looked at first, but at half the current.
Now it dissipates 40W and or allowance becomes 110/40 or 2.75 degC/W. Allow another 0.1 degC/W for the connection to the heatsink and we now have 0.65degC/W remaining.
How does this help? Well, if we place two of these rectifiers in parallel, we can get by with smaller heatsinks. However, for this to work we need them both to be mounted to the same heatsink! At what does it's thermal resistance need to be? Easy, 0.65/2 degC/W. So you want a heatsink rated at about 0.3degC/W
This is still a massive heatsink, but it is probably a third of the size of a 0.125degC/W heatsink.
Can we do better? Sure. If there is a marge enough piece of metal that we can connect the rectifiers to, one that remains at about ambient temperature, we can use this to simply conduct the heat away. We can also use a fan to pass air over the heatsink (does the original case have a fan in it?) or we can even water cool the heatsink.
We can also assume that at higher wind speeds the ambient temperature will be lower, and that the same wind will more effectively cool the heatsink. We can also assume that it won't be operating at full power for very long. All of these things allow us to reduce the size of the heatsink.
