Forum Discussion
LipschitzWrath
Aug 14, 2018Explorer
What is your concern? Splitting amps? I don't get it.
I have a 4ga cable running to each battery positive AND negative? Is this a problem?
I don't believe I have any circuit protection bypassed. The converter output goes to one side of the breaker, the batts are on the opposite. So the converter has to go through the breaker to charge the batteries.
The converter is on the same side of the breaker as the DC panel. There may not be any protection there, but if there isn't now, there never was. I don't know there is a large fuse downstream somewhere (I haven't seen one), or maybe they let the branch fuses take care of that. The one exception here is that when I used the "return" wire from the DC panel as a second "feed", the converter can now dump it's current into the DC panel without any circuit protection via the old "return" wire. Now, if the old "feed" wire has no circuit protection, then this is no different and all I did functionally was double the number of feed wires to the DC panel without any protection bypass.
Phrased a different way, here's what's connected to the breaker
"Input" - Batteries (only)
"Output" - Converter ouput, DC panel feeds
Is this in error?
EDIT - If you go to the link earlier in this thread: http://www.smartgauge.co.uk/batt_con.html
I used method 3. I don't see what I did wrong.
I have a 4ga cable running to each battery positive AND negative? Is this a problem?
I don't believe I have any circuit protection bypassed. The converter output goes to one side of the breaker, the batts are on the opposite. So the converter has to go through the breaker to charge the batteries.
The converter is on the same side of the breaker as the DC panel. There may not be any protection there, but if there isn't now, there never was. I don't know there is a large fuse downstream somewhere (I haven't seen one), or maybe they let the branch fuses take care of that. The one exception here is that when I used the "return" wire from the DC panel as a second "feed", the converter can now dump it's current into the DC panel without any circuit protection via the old "return" wire. Now, if the old "feed" wire has no circuit protection, then this is no different and all I did functionally was double the number of feed wires to the DC panel without any protection bypass.
Phrased a different way, here's what's connected to the breaker
"Input" - Batteries (only)
"Output" - Converter ouput, DC panel feeds
Is this in error?
EDIT - If you go to the link earlier in this thread: http://www.smartgauge.co.uk/batt_con.html
I used method 3. I don't see what I did wrong.
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