Wiring batteries in parallel

[KP44]

Originally Posted by Uncle Bob

Sketch C is pretty much how I have my three pairs of six volt lead acid wet cell batteries wired up. seven years now and still going strong.

I have 4 pairs of 6v batteries wired similar. My last set of 8 batteries lasted over 10 years. Then I tried some 8D gel batteries that cracked and leaked under the strain of tropical heat. Now I am back to 8 dependable 6v flooded cell batteries.

If you have the room, extra capacity means less discharge and longer life.

[Montanan]

Meme time:

[beirutsailor]

why can't you artificially lengthen the cables to the middle battery under C to match that of the lateral batteries?

[s/v Jedi]

Originally Posted by SweareDeep

The Victron batteries require the use of a single Victron BMS and, of course, that's what I'll be using (I've already acquired it). And I'll be going with "C" as that's what Victron recommends. Those decisions were made some time ago.

The question, which seems to have been lost up thread somewhere, is whether within "C" each of the positive connections and each of the negative connections needs to be the same length. Electrically, I can see no reason why the condition shouldn't be a little less restrictive, namely that the sum of the positive and negative connections for each battery should be the same. So for a given battery, for example, a longer positive connection should be balanced by a shorter negative connection.

Okay, to answer this: diagram “C” is the best method, no doubt about it. The positive battery cables can be a different length than the negative battery cables but all positive cables must be the same length and all negative cables must be the same length.
Often overlooked: all crimp terminals must be the same, high quality type and a torque wrench must be used for bolting them down on both batteries and busbars. The torque specs are listed for batteries as well as the busbars.

The reason for a torque wrench is to eliminate contact resistance without stripping threads.
The reason for correct crimps is that there are many tools readily available for incorrect crimps. You need the big FTZ Correct Crimp tool or better for American wire gauges while for metric sizes you can use the hydraulic crimpers that come with metric dies.
The reson for equal length cables is that they all interconnect at the busbar so the potential must be equal at that specific point in the cable. I know this is BS because of differences in battery internal resistance but just for rule of thumb without over complicating minor detail, equal length is the standard.

[GordMay]

Originally Posted by s/v Jedi

... Does jumper resistance change when more or less current passes through? I have not heard that before, how does that happen? The manufacturers state that resistance is only defined by wire gauge and length, plus contact resistance for each crimp and stud. Current is not mentioned.

Yes:
The Temperature Coefficient of Copper (near room temperature) is +0.393 percent per degree C. This means if the temperature increases 1°C, the resistance will increase 0.393%. Since increased current will cause a temperature rise, it will also increase wire resistance.

The generic forumla for temperature affects on resistance is as follows:

[s/v Jedi]

Originally Posted by GordMay

Yes:
The Temperature Coefficient of Copper (near room temperature) is +0.393 percent per degree C. This means if the temperature increases 1°C, the resistance will increase 0.393%. Since increased current will cause a temperature rise, it will also increase wire resistance.

The generic forumla for temperature affects on resistance is as follows:

True, but that would mean that the middle battery gets less resistance instead of the observed more resistance

Observations are pretty solid evidence but I have not seen any correct explanation of it yet for diagram B.

[manitu]

No, the middle battery has to deliver current through both of the hottest links, the batteries at each side deliver current trough one.

I'm still trying to wrap my head around this, but it seem to me that whith SLA and AGM batteries there is not much of a problem. Once you pull big amps , the battery voltage will drop ( because of internal resistance in the battery ) , and the load will be distributed across the bank.

But Lithium batteries has so little internal resistance , that the voltage will not drop (much anyway) and any difference in resistance/voltage between batteries will result in an uneven load.

.manitu

[Scubaseas]

Originally Posted by s/v Jedi

Okay, to answer this: diagram “C” is the best method, no doubt about it. The positive battery cables can be a different length than the negative battery cables but all positive cables must be the same length and all negative cables must be the same length.
Often overlooked: all crimp terminals must be the same, high quality type and a torque wrench must be used for bolting them down on both batteries and busbars. The torque specs are listed for batteries as well as the busbars.

The reason for a torque wrench is to eliminate contact resistance without stripping threads.
The reason for correct crimps is that there are many tools readily available for incorrect crimps. You need the big FTZ Correct Crimp tool or better for American wire gauges while for metric sizes you can use the hydraulic crimpers that come with metric dies.
The reson for equal length cables is that they all interconnect at the busbar so the potential must be equal at that specific point in the cable. I know this is BS because of differences in battery internal resistance but just for rule of thumb without over complicating minor detail, equal length is the standard.

This surely for a "stupid" battery bank like AGM or FLA but add to is that if you link the BMS of each battery together they will monitor the cells to make sure they are all at the same voltage. So me thinks (and I could easily be wrong) that the BMS will account for wiring losses regardless of "B" or "C". Ask Victron what their preferred method is. I doubt you could go wrong with "C" but you may not have to due to the BMSs. Ask Victron.

[s/v Jedi]

Originally Posted by manitu

No, the middle battery has to deliver current through both of the hottest links, the batteries at each side deliver current trough one.

.manitu

Please help me understand this... which are the hottest links and how does the current flow through one or both?

[s/v Jedi]

Originally Posted by Scubaseas

This surely for a "stupid" battery bank like AGM or FLA but add to is that if you link the BMS of each battery together they will monitor the cells to make sure they are all at the same voltage. So me thinks (and I could easily be wrong) that the BMS will account for wiring losses regardless of "B" or "C". Ask Victron what their preferred method is. I doubt you could go wrong with "C" but you may not have to due to the BMSs. Ask Victron.

I’m sorry, I can’t read & understand what you write. Nobody needs to ask Victron because they state diagram C is what they recommend.

[manitu]

Originally Posted by s/v Jedi

Please help me understand this... which are the hottest links and how does the current flow through one or both?

The inner link on positive carries the current of one battery. The outer link carries the current of two batteries. The supply cable to the inverter carries the current of all three batteries. Same with the negative links.

Now look at the drawing. Outer battery om the positive side. Current goes through no positive links + both negative links. Outer battery om the negative side. Current goes through both positive links + no negative links.

The Center battery om the other hand, the current goes through the outer positive link + outer negative link. Alas, the outer links carry twice the current of the inner links, get hotter, and will have a greater voltage drop/ohm.

.manitu

[s/v Jedi]

Originally Posted by manitu

The inner link on positive carries the current of one battery. The outer link carries the current of two batteries. The supply cable to the inverter carries the current of all three batteries. Same with the negative links.

Now look at the drawing. Outer battery om the positive side. Current goes through no positive links + both negative links. Outer battery om the negative side. Current goes through both positive links + no negative links.

The Center battery om the other hand, the current goes through the outer positive link + outer negative link. Alas, the outer links carry twice the current of the inner links, get hotter, and will have a greater voltage drop/ohm.

.manitu

Great! In other words:

- current of the center battery travels through two jumpers that each carry the current of two batteries

- current of the outer batteries travels through two jumpers where one carries the current of two batteries but the other only for one battery.

for each degree Celsius higher temperature, resistance increases by 0.4% and this increase has double the effect on the center battery.

So, to fix that we go to diagram C. But why not double up the two jumpers that are affected? this brings the current equal in all of them, increases jumpers to six which is the same as for diagram C but they are shorter so that's better, plus there is no need for busbars.

I know the answer: when you increase to 4 or more batteries, diagram C becomes more efficient.

In short:

2 batteries: diagram B
3 batteries: diagram B with two jumpers doubled up or bigger gauge wire or diagram C (when future expansion is considered)
4+ batteries: diagram C

Shame that the testers in the linked pages couldn't figure this out or plainly state the wrong reasons.

[manitu]

+ the voltage drop over all cables rise as the amperage goes up.
The amps goes up because the load equals a lower and lower resistive value, and the resistance of the cabling (and internal battery resistance) becomes a bigger and bigger part of the circuit's total resistance.

This is the reason battery voltage drops (and cabling voltage loss goes up) when the load increases.

But Lithium.. the internal resistance, though variable, is so low that the voltage drop is absolutely minimal , compared to SLA or AGM. So any extra resistance to any battery , and the other batteries are glad to pitch in.

I'm mounting 5 AGM batteries in "B" configuration, but as 2 batteries alone can take the full load in my system, I'm fine with a tiny bit of imbalance while running heavy loads. For all I know, the outer batteries maybe lasts longer , for being cycled a little (tiny) extra once in a while.

.manitu

[waltknowles]

Originally Posted by SweareDeep

This diagram shows Victron's recommended practice. I'm simply saying that instead of each positive wire being the same length and each negative wire being the same length, what matters is that the combined length of the positive wire and the negative wire should be the same across the three batteries.

SweareDeep, you and I are in the same boat. There's some details in that diagram you posted that I've been puzzling for myself. Where on the Victron site is the whole diagram?


Thanks,
Walt Knowles
Brraesail

[LargeMarge]

Originally Posted by GordMay

The best parallel connection scheme (Pos from one end & Neg from the other - not both from same batt), shown below, charges/discharges the entire bank equally.

Nope!

That method works the end batteries harder while the center battery loafs.

To fill/empty each battery equally:
* each positive post cables to an off-battery distribution post/panel.
* each negative post cables to an off-battery distribution post/panel.