PD4645 charging amps

[tarp]

I am starting some upgrades from this original with only 10k miles on it.
I removed the wardrobe behind the driver to add a seat. One guy said, What's the fiberglass pan about? I can't tell if you can add a seat easily on these that formerly had the seat on a "box."??

Also, any ideas about the adjacent wall. Needs filling in with an upholstered wall. Any idea where supports go between vertical steel beam and driver door?

Thanks!

[Winston]

Originally Posted by RT-NY

How much loss of charging power would be expected from, say, a six foot run of 8 awg copper cable at 40 amps?

In reading your most recent posts, it seems that you may have solved your problem by replacing the temporary wiring with more robust wiring. The purpose in our suggested voltage readings were, first, to determine that, in fact, the PD is outputting the expected 14.4 volts and, second, to ascertain the amount of "loss" (otherwise known as voltage drop) that is occurring between your 14.4 volt PD output and the actual point of connection at the battery. Remember our starting point: if you were to connect a true 14.4 volt source directly to a discharged lithium battery, a near "explosively" high current would result.

When you asked about 6' of 8 gauge wire . . . we assumed that you really are speaking of 12', 6' for the positive and 6' for the negative. The resistance of 12' of 8 gauge wire is approximately 0.008 ohms. That's a very low number which could be dwarfed by other issues such as the quality of your connections . . . i.e. where the wire lugs connect to the buss bars/studs. But assuming these are secondary concerns . . . 45 amps should only cause a voltage drop of 0.36 volt - - lets round that to 4/10ths of a volt. So your batteries should be 'receiving' a charging voltage of 14.0 volts.

Placing 14 volts across a reasonably discharged lithium battery should result in much more than 45 amperes of current. So, we suspect that, with your improved wiring, the output of your PD is dropping to keep the PD within its current limits. So we're uncertain why you're seeing 37 amps rather than 45, but at least your getting closer. Would still be interested in knowing the voltage at the output of your PD and at your battery when charging your mostly discharged battery . . . (along with the charging current).

Hopefully this hasn't been a completely incoherent ramble . . . :~)

[RT-NY]

Originally Posted by Winston

Would still be interested in knowing the voltage at the output of your PD and at your battery when charging your mostly discharged battery

When I was charging, I did check the voltage at the terminals marked "battery" on the fuse board of the PD4645 ("BATT" in pic below) -- is that what you mean by "voltage at the output of your PD"? It was higher there, around 14.05 volts. It might have been 1/2 volt or so lower at the battery terminals, though I will need to double-check these numbers to verify when I next charge up.

[Winston]

Originally Posted by RT-NY

I did check the voltage at the terminals marked "battery" on the fuse board of the PD4645 ("BATT" in pic below) -- is that what you mean by "voltage at the output of your PD"?

Maybe. Here we're mixing some 'theoretical' concepts with practical realities. On the theoretical side we speak of "the battery", "the charger", and, "the interconnecting wires, buss bars, lugs, studs etc."

On the practical side, it's hard to determine, precisely, were the 'charger stops', where the 'battery starts' - - everything between being the elusive connecting wires, the resistance of, and the voltage drop across, we're trying to measure.

Most power supplies/charges have a point (near the output terminal) that the internal wiring of the charger 'taps into' to measure its own output voltage. A power supply/charger often seeks to maintain a precise, predetermined voltage at this spot. If the loads/demands on the power supply/charger increase causing this voltage to drop, the internal circuitry will take corrective steps to force the voltage to return to that predetermined voltage. Ideally this is where one should measure the voltage output of the charger. We presume that the output terminals of the PD (which are marked "Battery") are as close as you're going to get to this "theoretical" measuring point.

If PD says that this voltage should be 14.4 volts, we would expect 14.4 volts. While, 14.05 volts seems close . . . it is a bit disturbing that you've lost nearly a half volt under a 37 amp load (did we remember that correctly?). One explanation could be that the "Battery" terminals are not the correct spot, that these exterior terminals on the chassis of the PD are actually connected to an internal 'output' through a set of wires that have resistance and are losing the .45 volts.

On the Battery side of things, about the best we can do is go directly to the plus and minus terminals (when making these measurements we should not assume that "minus" is equal to "ground" . . . go directly to the minus/ground terminal of both the PD and Battery when measure these voltages). But remember, even in a well designed battery, the terminal has finite resistance between the post/connection on the outside of the battery and the 'chemistry/plates' on the inside.

In case all this mumbo jumble has confused the issue. So to summarize, we're trying to determine why you're not getting the full 45 amps of charging. The possibilities are basically three: 1) the battery is already charged, it's output voltage is already "too close" to the 14.4 volt output of the PD, or, you could have a bad battery, but this possibility, we've been ignoring; 2) the PD is defective, poorly designed, it cannot output 45 amps; and/or 3) your connections between charger and battery are too high resistance, too much voltage is dropping across these connections. By way of an example, say the PD is properly outputting 14.4 volts, say further, that the battery voltage under charging conditions is 13.8 volts. Under this hypothetical example (but not too far from reality), there will be 0.6 volts (total, summed) across the two connecting wires. By Ohm's Law, this means that the maximum resistance of all your connecting wires, lugs etc can be 0.013 ohms. In view of our earlier discussion that your 12' of 8 gauge wire would have a resistance of 0.008 ohms - - which includes no allowance of the other 'interconnecting hardware' - - it is conceivable that you could exceed the maximum allowable resistance.

And it doesn't help that your PD is already down by a half volt. In measuring your battery and PD voltages again, we'd suggest also measuring the PD output with the battery disconnected.