Wire gauge for parallel battery jumpers.

AbbyTR23AbbyTR23 Registered Users Posts: 1

Method 1 (typical):

Wiring 12V 100 AH batts in parallel + to + to + to load yields 400 AH @ 12 V. Negatives connected in the same way to ground. Seems to me the first jumper needs to carry 100 A, the second 200 A, the third 300 A, and the connector to load 400 A to supply a 3000W inverter (with T450 fuse within 7 " of the final battery in the string). Camping World installed a 2 battery system and used 2/0 for the jumpers and 4/0 to load to handle maximum static load of 3000w/12V=250 amps and up to 400 start-up amps.

Method 2

I replaced the 2 lead acid batteries with 4 12 V 100 AH lithium iron phosphate batteries. This option connects each positive pole to a bus, then the bus to load. Same for the negatives. Each jumper would then only need to carry 100 A (unless one or more of the jumpers connections fail). This was recommended by the manufacturer of the batteries, since the terminals are tightly packed in these 4 cell batteries and better to run each lead straight off the poles rather than try to use jumpers between poles as in method 1. The jumpers are 2' total length (positive plus negative leads together) to the buses. The manufacturer recommended 6 AWG, which seemed small. I use 2/0 for the jumpers and 4/0 for the buses to inverter.

Would appreciate comment, corrections, criticism(s)

Abby23TR

THOR TR23 RV

Comments

  • EstragonEstragon Registered Users Posts: 3,408 ✭✭✭✭✭
    edited April 8 #2

    In a fault (eg internal short in one battery), you could put max output from the other 3 through the fault (potentially huge current). Accordingly each battery should have a fuse appropriate for wire used.

    With only 2 in parallel, this isn't really needed, because a short could only be fed 1x from its twin vs 3x with 4 parallel.

    Off-grid.  
    Main daytime system ~4kw panels into 2xMNClassic150 370ah 48v bank 2xOutback 3548 inverter 120v + 240v autotransformer
    Night system ~1kw panels into 1xMNClassic150 700ah 12v bank morningstar 300w inverter
  • BB.BB. Super Moderators, Administrators Posts: 28,768 admin

    For a 12 volt battery system, I would suggest that around 1,800-2,000 Watts is about the maximum AC inverter that I would suggest. You can go with larger AC inverters, but you are now working with very large cables (expensive, difficult to wire and terminate) to keep voltage drop low (suggest maximum 0.5 volt drop for 12 volt battery bank). A FLA really works better with a C/2.5 maximum surge current (reliable support of surge current for a few seconds over time/state of charge). For C/1 surge current, (C=20 Hour Discharge Capacity) you would need AGM or Li Ion type batteries.

    And for a 2,000 Watt AC inverter, I would also suggest an ~800 AH @ 12 volt battery bank minimum (for a flooded cell lead acid deep cycle battery bank). Otherwise, for example, a 2 kWatt inverter on a 400 AH @ 24 volt battery bank (same energy storage, just 2x the voltage and 1/2 the current).

    For over ~1,800 Watts, a 24 volt (or higher) battery bank would make wiring much easier and cheaper.

    Lithium Ion batteries (and AGM) can supply much higher surge current than FLA batteries--But now you need to also look at your overall power needs (AH @ XX volts, or XX Watt*Hours per day). Do you have an estimate of your overall daily power needs, as well as your peak power loads (like a refrigerator, microwave, etc.)?

    There are other issues, loke Estragon said... A Fuse or Breaker per battery string on 3+ parallel strings is a very good idea--Although, few people do it.

    And, "oversizing" the series cabling can actually create some issues too with current sharing (current "steering" between strings). A too large AWG cable has very low resistance--This makes current steering more based on the condition of each battery string. Those batteries with lower internal resistance will accept more current during charging and provide more current sourcing loads than a mirror "higher internal resistance string".

    I would suggest that your series cabling be rated for the maximum current that you expect the load (and charging) current to the battery bus.

    If you are trying to pull 400 Amps from a battery bank on 12 volts:

    • 400 amps * 10.5 battery cutoff voltage * 0.85 inverter eff = 3,570 Watts

    And if you derate your battery cabling by 1.25 (NEC continuous current derating):

    • 3,570 Watt * 1/1.25 NEC derate = 2,856 Watt maximum suggested continuous load (assuming 2x rated power for starting surges)

    -Bill

    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
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