Gary,
I'd like to voice a concern about the description of the system you have. You say:
...My take is that there's too much current being carried by the contacts during closure or opening...
...A very easy way to fix this would be to replace the 0000 sized wire going to and from the solenoid with some wire of a much smaller gauge, sized to carry solely the current that the alternator can produce...
...The theory is that the wire adds enough resistance in the circuit that the current thru the solenoid is somewhat limited to what the alternator can produce...
What it sounds like you are suggesting is diliberately using an undersized wire, to "current limit" the link between the two banks. What I don't see here is any mention of a fuse... This is the concern I have.
A wire, no matter how undersized, will attempt to transmit the full power demand of the circuit regardless of its size. If a wire is replicating a "current limiting" effect, this means that the resistance of the wire is preventing further increases in power transfer. How this happens is that at a specific point in the increasing slope of transfer, the wire becomes saturated and the electrons instead of flowing through the wire like a stream - are forced through like a 1500psi pressure-washer hose. This causes the conductor to become hot and its resistance to increase, which eventually (assuming that the load is not continuously too high) will balance and "limit" the ammount of current through the conductor.
Unfortunately this heating of the conductor can melt an insulating jacket well before the current and temerature reach equilibrium, or cause the conductor itself to open (think
electrical fire here please). This is the whole point of a fuse - it has a smaller gauge conductor which is designed to open when a sustained current transfer throught its package exceeds the rating of the part. By having a small device in a known location, one can control the failure of the long cable run (ensuring that the fuse opens before the cable gains heat which is too much to dissapate safely). Without a fuse, the heat will build up in a cable where the conductor cannot radiate the heat fast enough to keep the temperaure from compromising the integrety of the cable - and an open will result. The other risk, is that the insulating jacket is rated for a specific temperature - if this is exceeded internally, the jacket will melt losing the insulating characteristics (thinning, or dripping off of the cable) and the conductor can short against an exposed ground surface causing arcing, welding, or (if close enough to a wood structure) a major fire.
What you are hitting at Gary, but not recommending - is the use of a current limiting resistor - however, I would recommend that one design the bank-tie with a device like this doing the limiting - not the cable! Just like a fuse, this allows us to select a regulation point and locate that part in a known and suitable location.
The math is very simple for a current limiting resistor - one need to know the expected worst-case voltage and current to be limited to and do some simple math:
(Resistor Minimum Value in Ohms) = (Voltage) / (Current limit set-point in Amps)
(Resistor Minimum Power Rating in Watts) = (Voltage) x ("Current limit set point" x 1.25)
For example if you are expecting a differecne between a 12Volt Chassis Battery and a 12Volt House battery of 4Volts, and you want to limit your current to 40Amps the math would be:
4Volts/40Amps =
0.1Ohms (100 milliohms)
at
4Volts x 40Amps =
160WattsWhat you can see here is that value is in the order of milliohms - an undersized conductor can easily have several milliohms of resistance per foot, so the risk is real.
Now that's the subject of current limiting resistors - but just like with high power LEDs (current limiting resistors are fine for an LED with a drive current of 0-20mA), a current limiting resistor quickly errodes the efficiency of a system. As you can see in the above math, the resistor will actually convert 160Watts of your precious power into heat while it is protecting the wire! This is like running your headlights. To maximize the transfer of energy from one bank to the other - one must design the system to support the full surge load of the banks (this can easily mean a current number somewhere in the odds of >500Amps).
Back to the example of the LED, lets assume a 3.25Volt 15mA blue LED (because I think they're cool). If we want to run that LED from a 28.8Volt supply, we need a current limitting resistor which will consume the difference between the 28.8volt supply and the 3.25Volt rating of the LED.
28.8 - 3.25 = 25.55Volts
To pick the current limiting resistor, we do the same math as we did for the bank tie:
25.55Volts / 15Milliamps = about 1.7KOhms
With a power rating of at least:
25.55 x 15mA = 0.383Watts
With a resistor, you always pick the next part UP in size - so this would be a 1.7KOhm 1/2Watt resistor. Looking at the efficiency, we see that the LED consumes (3.25Volts x 15mA =) 0.049Watts and the resistor consumes 0.383Watts out of the total power consumed (0.383 + 0.049 = 0.432Watts) the LED only consumes 11.343% of the total power used for the circuit!! That's almost 10% efficiency and 90% waste as heat!
This same lesson can be applied to the battery bank tie circuit with a wire acting as a current limitting device - you'll probably lose more in the wire loss that you gain with the transfer itself, if you have an under-designed circuit. This has a significant impact for those who boondock or full-time. Careful selection of components and special attention to design have a vast impact in the power consumption of the final system.
And please everyone - Always place a fuse with a value picked to protect the wire it is attached to (and place the fuse within 8-inches of the circuits +supply). And place a fuse or fuse block any time you have a transition from one wire size to another - to protect the smaller of the two wires! Only you, can prevent electrical fires in your rig.
Cheers!
-Tim
