Friday, Oct 07, 2016 at 14:30
.
Justin, I'm afraid that you are somewhat confused, maybe someone is confusing you?
The component in question is certainly not what is commonly described as a "VRD" (Voltage Reduction Device). Such devices are much more complex than the simple 2-lead component circled in the photo. It is clearly a simple diode or a zener diode and from its location relative to the 12Vdc input pins I would expect it to be across the input in reverse bias in order to protect the circuit board from the destruction of having the 12V presented to the fridge with reverse polarity. This is a common protection method which shunts the reverse dc and causes the external protection fuse to blow thus preventing reverse polarity to the board. Of course, if there is not a suitably rated protection fuse (10 to 20A) in the supply then a heavy current will flow and destroy the diode and consequently the circuit board.
It is probable that this is what has happened to your fridge. The fact that it still works on 240Vac calls into question that any damage was caused by dampness and the application of 240V.
Your power calculations and conclusions are all wrong. The fridge will draw about 50W on AC or DC so if you wish to run it from a 12Vdc-to-240Vac inverter then one rated at 200W or more should handle it. It also would need to be a "Pure Sinewave" inverter to do the job. A "Modified Sine Wave" inverter MAY work but then again it may not. Certainly a "Square Wave" one will not work. However, be aware that due to efficiency losses of about 20-25% in the inverter you will be using more 12V power than with the original configuration. Incidentally, I have no idea of what you are referring to as "variable voltage
inverters on ebay for under $10". Anything fitting that description and price has no application here.
All in all, if the fridge only cost you $120 then I would not spend further on it. Keep it as a 240Vac fridge for the Christmas drinks and invest in a new fridge for the vehicle.
FollowupID:
874751