Quote:
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Originally Posted by Gary Dillard
Better to keep mouth closed and be thought a fool.....
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<I need a bigger mouth or a smaller foot; arguing electricity with Al Skierkiewicz - Dillard's a moron (please don't make that a spotlight quote>
OK Al, I've been thinking about this and I disagree with you - I'll have to draw up a circuit diagram tomorrow if I didn't explain it well here - although your point about current spikes may be correct (remember M.E.'s think in steady state).
First item is a nit - If you look at the IFI panel, AND all of their documentation, they refer to the return side of the battery as Ground. It's marked ground 2 places on the panel, and each terminal is G-something.
Breaker Panel Reference Guide Section 3, paragraph 2:
Connect Negative(-) side of the battery or Ground to the center post contact labeled GND.
Section 6, paragraph 1:
The return path or Ground for these outputs are located at the bottom of the unit and are labeled G1 through G22.
Second, your point about the resistance of the panel being one tenth that of the wiring is exactly what I was referring to in my first question. If my load (motor/controller) is located on the ground side of the board (bottom left), here are the 2 extreme cases for routing:
(1) Big honking wire from battery terminal negative to ground lug, 1 inch of board from ground lug to ground blade#1, 6 inches of wire from ground blade #1 to Victor, N feet of wire to load, N feet of wire back to Victor, 4 inches of wire from Victor to Output blade #21, 6 inches of board from blade #21 to positive lug, big honking wire from positive lug to battery terminal positive.
(2) Big honking wire from battery terminal negative to ground lug, 1 inch of board from ground lug to ground blade#1, 6 inches of wire from ground blade #1 to Victor, N feet of wire to load, N feet of wire back to Victor, 8 inches of wire from Victor to Output blade #1, 2 inches of board from blade #1 to positive lug, big honking wire from positive lug to battery terminal positive.
Both are series circuits, so the total circuit resistance is simple addition. The only difference in the circuits between the 2 is trading 4 inches of wire in (2) for 4 inches of board in (1). By your numbers that's a difference of .3 milliohms, with circuit (1) - my circuit - having lower resistance. So, for a fixed voltage source (which we have) I will have more current (more power) to my load, or looking at it another way I will have an additional .027 mV in available voltage for a 30A circuit to drive my load that is taken up in by loss in your circuit.
Third - you said there are 400 Amps of current running through the board, same as the battery? What is the 120 Amp breaker for then?
Or should I just take my multimeter and go home?