Yes, it's possible to lower the voltage by adding a diode. The PS current will decrease and you won't get to the 60A foldback.
The fix is not pretty. The objective is to keep the junction temperature under 150C.
data sheetLooking at the 100 C performance curve, we get a 2.1V diode drop at 60A. The max power the diode dissipates is 2.1V * 60A = 126W.
The thermal resistance of the diode is 0.8C/W. We need to calculate the heatsink thermal resistance so that the junction temp is at 150C when dissipating 126W.
The diode package as a thermal resistance of 0.8C/W. The temperature rise within the package is 0.8C/W * 126W = 100C.
If ambient temperature is 25C, that leaves just 25C for the heatsink. The thermal resistance of the heatsink is:
Theta_HS = 25C/126W = 0.2 C/W
That is extremely small. This won't work, the diode will burn up. You'll have to parallel multiple diodes together.
Let's take 4 diodes in parallel. Each diode conducts 60A/4 = 15A. The diode drop is 1.3V at 15A, at 100C. Power is 1.3V * 15A = 20W.
Temp rise in diode package package is; 0.8C/W * 20W = 16C
Theta_HS = (150C - 25C - 16C) / 20W = 5.5C/W
That's doable.
Google 5 C/W heatsink
You need to wire up 4 of these in parallel. You'll get about 1.3V drop.
