Shunts

scheek

Thanks for all the help from the forum.

I am about to make my cables for my 3 banks. Should I use shunts to tie all the banks together. I see a couple on the online store here at Wind & Sun. I will be using in series 4 /6 volt batteries (300ah) in each bank.

The shunts seem to be expensive if I needs four but I can do that if necessary. Can I make shunt out of copper? If so, where I can find one to see?

Thanks

vtmaps

Re: Shunts

scheek wrote: »

Thanks for all the help from the forum.

I am about to make my cables for my 3 banks. Should I use shunts to tie all the banks together. I see a couple on the online store here at Wind & Sun. I will be using in series 4 /6 volt batteries (300ah) in each bank.

The shunts seem to be expensive if I needs four but I can do that if necessary. Can I make shunt out of copper? If so, where I can find one to see?

To me a shunt is a precision resistor. Are you asking about buss bars? --vtMaps

scheek

Re: Shunts

I think that is what I see on Wind and Sun. http://www.solar-electric.com/mka-100-100.html... But yes a buss bar. Will you look and see if this is the same?

Thanks

vtmaps

Re: Shunts

scheek wrote: »

I think that is what I see on Wind and Sun. http://www.solar-electric.com/mka-100-100.html yes a buss bar. Will you look and see if this is the same?

That link is for a shunt. It is a precision resistor that allows a battery monitor to calculate the current through the shunt.

It is not a bus bar. What is it that you want to do with this shunt?

--vtMaps

scheek

Re: Shunts

I need a buss bar. I have 3 banks of 4 batteries tied in series. As I get the positive and negative off each bank end, I need to connect the 3 banks together to run to my inverter. I thought these were buss bars. Got any ideas?

Thanks

BB.

Re: Shunts

There are not a lot of off the shelf bus bar setups... Quite a few make their own. Blue Sea makes some up to 600 amp rating... See if this link works to West Marine:

http://www.westmarine.com/webapp/wcs/stores/servlet/ProductDisplay?productId=17458&catalogId=10001&langId=-1&storeId=11151&storeNum=50523&subdeptNum=50549&classNum=50553

The bigger ones are $100 to $135 dollars each (4 or 8 stud mounts).

You can see what "2manytoyz" did with DIY:

http://2manytoyz.com/battbank.html

-Bill

SolInvictus

Re: Shunts

Or connect the ends of the series strings together using 2 ga copper wires with terminal lugs and do not worry about trivial things. The cables between the battery array and inverter should be attached to the center battery string.

Cariboocoot

Re: Shunts

SolInvictus wrote: »

Or connect the ends of the series strings together using 2 ga copper wires with terminal lugs and do not worry about trivial things. The cables between the battery array and inverter should be attached to the center battery string.

Nope.
Although it's less critical on higher Voltage systems, keeping the wiring even length on paralleled battery circuits helps reduce problems with current sharing.

http://www.smartgauge.co.uk/batt_con.html

mtdoc

Re: Shunts

Copper bus bar kits available HERE. They call them "ground bars" but they are the same as bus bars. You likely will want fewer holes though in which case they also sell lengths of copper bar for DIY.

SolInvictus

Re: Shunts

Cariboocoot wrote: »

http://www.smartgauge.co.uk/batt_con.html

Thanks for the link.

"35mm copper cable has a resistance of around 0.0006 Ohms per metre." 35 mm diameter cable is larger than 4/0 gauge copper cable with a diameter of 11.684 mm and a resistance of 164.0 microohms/meter at 25 C. 3 gauge copper wire has a diameter of 5.827 mm and a resistance of 659.1 microohms/meter whereas 2 gauge copper wire has a resistance of 522.6 microohms/meter. Smart Gauge appears to be using 3 gauge wire.

Smart Gauge wrote:

... the 20cm length between each battery will have a resistance of 0.00012 Ohms. This, admittedly, is close to nothing. But add onto this the 0.0002 Ohms for each connection interface (i.e. cable to crimp, crimp to battery post etc) and we find that the resistance between each battery post is around 0.0015 Ohms.

Using his numbers, one wire is 20 cm long with 2 terminal lugs making the combined resistance:

200 microohm (post to first lug) + 200 microohm (first lug to wire) + 120 microohm (wire) + 200 microohm (wire to second lug) + 200 microohm (second lug to post) = 920 microohm

a bit less than his 1,500 microohm. The resistance of a lug is so low I can not measure it (< 20 microohm). How did he calculate it? His assumed resistance of 1.5 milliohm/link is equal to a 10 gauge copper wire 20 cm long (668 microohms) with the terminal lugs adding 832 microohms. Thus the example is equivalent to drawing 16 A to 75 A through woefully undersized 10 gauge copper wire causing the unsurprising imbalances in current.

If I do not round up the resistance of my 20 cm. long 2 gauge copper wire (105 microohms) to one digit of precision and skip the fuzzy math when assessing the resistance of my clean and tight contacts, then my interconnecting links should have a resistance around 200 microohms. Today I checked my estimate against the actual resistance of the 6 interconnecting links in my battery array and got the following results:

2 links: 240 microohm
3 links: 280 microohm
1 link: 140 microohm

There is about 1 digit of precision in these values because I measured voltage drops that ranged from 6 mV to 12 mV with 25 A to 50 A flowing through the conductors. I'll be generous and select the resistance at 300 microohms, five times less than Smart Gauge Electronics assumes, which greatly reduces the variations in current from the batteries. The simplifying assumption that all interconnecting links have the same resistance and that allows the problem to be solved more easily, is incorrect. The resistances will vary. The best way to prevent this from being a problem with parallel connections is to keep the resistances of the links small, about 100 times less than the internal resistances of the batteries.

There is another way to wire the battery array besides the ones shown in the first two diagrams. The main connections can be attached at the center of the battery array to minimize the current flowing through the links as shown in the following diagram:



Because this is how my battery array is currently wired, I will analyze this circuit which is similar to Smart Gauge's Method 2. Skipping to the answer:

Ib2 = Ib3 = (Io/4) * (Rb + 2Rw) / (Rb + Rw)
Ib1 = Ib4 = Io/2 - Ib3

Ib1 = current from battery 1
Ib2 = current from battery 2
Ib3 = current from battery 3
Ib4 = current from battery 4
Io = current output from the battery array
Rb = internal resistance of the batteries, all batteries identical
Rw = resistance of links = resistance of wire, 2 lugs and the contacts, all identical

With Rb = 20 milliohms, Rw = 300 microohms and Io = 100 A, my configuration yields .7 A difference:

top battery B4: .02 ohm + 3 * 300 uohm = 20.9 milliohm, 24.631 A
next battery B3: .02 ohm + 300 uohm = 20.3 milliohm, 25.369 A
next battery B2: .02 ohm + 300 uohm = 20.3 milliohm, 25.369 A
bottom battery B1: .02 ohm + 3 * 300 uohm = 20.9 milliohm, 24.631 A

For comparison and using Smart Gauge's link resistance of 1.5 milliohms, the battery currents are:

B4 23.256 A
B3 26.744 A
B2 26.744 A, Vb2 = 535 mV, Vw4 = 75.0 mV
B1 23.256 A, Vb1 = 465 mV, Vw1 = Vw2 = 34.9 mV

which is slightly better than his Method 2 because it presents less wire resistance to the inverter. Vb is the voltage across the internal resistance of the batteries, not the voltage on the battery terminals. Because the absolute value of the difference between the internal voltage drops of two identical batteries is the same as the difference between the voltages on the terminals when the batteries are under load, these calculations can be compared to easily measured values. For example:

|Vb1 - Vb2| = |465 mV - 535 mV| = |-70 mV| = 70 mV

If battery 2 has an output voltage of 12.000 V under the 26.744 A load, then battery 1 should read 12.070 V with the lesser load of 23.256 A. When Rw = 300 microohms, the difference is 15 mV. When I measured the voltages on my batteries while the array was outputting about 100 A, I got the following readings which were difficult to measure accurately because the voltages were slowly decreasing as I measured them:

B4 12.15 V , 12.10 V
B3 12.14 V , 12.09 V
B2 12.14 V , 12.09 V
B1 12.15 V , 12.11 V

The difference between the voltages of battery 1 and battery 2 range between 10 mV and 20 mV which is consistent with Rw = 300 microohms but not Rw = 1,500 microohms. If your batteries are outputting about 25 A each and the voltage difference between adjacent, parallel connected batteries is larger than about 20 mV, then you should clean the terminals, tighten the terminals or get larger diameter wire with better lugs.

For those of you considering a bus bar to equalize the load on the batteries (Smart Gauge's Method 3) consider the ease of cutting a standard wire to length to get a specific resistance compared to crimping the lugs onto the wire and keeping the connections clean and tight. The connections add 2 to 3 times the resistance of the wires when the wires are ~20 cm long. The bus bar forces one to use longer wires which increases resistance to lessen the variations caused by the contacts, but you still need good, low resistance contacts. If you make those contacts properly in the first place and periodically check them, then you do not need the bus bar. Save $.

vtmaps

Re: Shunts

SolInvictus wrote: »

Rb = internal resistance of the batteries, all batteries identical
<snip>
With Rb = 20 milliohms

Interesting analysis... I never thought much about it because I try to avoid parallel batteries.

For the purpose of designing the battery interconnects, it makes sense to assume (as you did) that the internal resistance of the batteries is identical. In the real world they are not.

I think an even more interesting analysis would be to measure the internal resistance of individual batteries and see if the variation is on the same order of magnitude as the resistance in the wiring. As you point out, the resistance of the wiring is much less than the internal resistance of the batteries. Less than a 2% variation in the internal resistance of the batteries would be as much resistance as the wiring to the battery.

It may be that balanced wiring helps slow down the divergence of the parallel battery strings, but parallel batteries in equilibrium are in an unstable equilibrium and will diverge in their internal resistance. Good wiring cannot stop that.

--vtMaps

NorthGuy

Re: Shunts

SolInvictus wrote: »

With Rb = 20 milliohms, Rw = 300 microohms and Io = 100 A, my configuration yields .7 A difference.

What kind of battery may have 20mOhm resistance? I have 673AH@48V bank and it has 20mOhm resisane (most of the time). If I had 12V bank, the resistance would be 20*12/48 = 5mOhm. To get to 20mOhm, I would beed to decrease the bank soze 4 times, which gives 673/4 = 168AH@12V. I may be out of wack here bexause I didn't measure any batteries except mine.

100A is completely unrealistic current for 168AH@12V bank. The highest current experienced by this bank in a normal RE application would be about 15-20A.

Cariboocoot

Re: Shunts

Ooh lots of numbers and calculations that don't actually help.

Resistance being neither fixed nor predictable nor constant, you take the prescribed measures to reduce the possibility of it going high on one string and causing trouble. Or you don't and gamble that it won't. Your choice.

And for heavens sake NO ONE go around thinking about resistance of batteries and getting any ideas about measuring it with an Ohm meter!

SolInvictus

Re: Shunts

NorthGuy wrote:

What kind of battery may have 20mOhm resistance?

My battery array is composed of 8 Crown CR-395's (6 V batteries) for a total of 1580 Ah at the 20 hour discharge rate and 12 V. The specification sheet states their internal resistance is 8.1 milliohms at 27 C. Because the batteries are in series pairs in my array, their combined internal resistance is ~16.7 milliohms (including ~500 microohm for 4 gauge wire link) which is fairly close to the 20 milliohms used in the calculations. When I measured mine, their temperature was around 15 C which might have altered their internal resistance somewhat. It just occurred to me that I could use 2 gauge wire in the series links that make batteries 1 and 4 to reduce their effective internal resistances to balance the array better.

BB.

Re: Shunts

If I understand you correctly... Interestingly, using longer/thinner cable/wring actually can balance the current flow among the paralleled strings better.

The wiring resistance if fairly consistent and provides negative feedback to the system (as the wire gets hot, its resistance goes up--So, if you have a string that is drawings too much current, the increased resistance will reduce current flow).

Battery "resistance" is not consistent, and in some cases, actually operates in positive feedback mode (when charging, if a single string starts getting hot, the battery voltage falls, which increases current flow, battery gets hotter, etc...).

From my limited experience in working with parallel conductors carrying current, it was the "low resistance" path that actually caused problems. In various computers/test gear that used parallel current paths (very common when trying to bus enough 5 volt power to various circuit packs), the low resistance path would pass the most current, and I²R heating would overheat and cook the connection, which would fail (it was almost always the physical connector that failed). Then the next lowest current path would overheat and fail.

One way to prevent the "unzipping" failures was to use smaller gauge/longer cables where the wiring series resistance overwhelmed the variability in contact resistance. Which then pretty much eliminated the point I²R heat source.

It was also at this time where I pretty much decided (for my designs) where "generic" paralleled wiring connections should be derated by 1/n -- And using that derating, that more than about 3 parallel circuit paths simply did not carry enough current to outweigh the chance of failure with I²R contact heating.

1/1 + 1/2 + 1/3 = 1.83 rated capacity of three parallel wires for short cables with no "ballast" for current control

Adding ore 1/n simply did not add very much more "useful" current carrying capacity.

-Bill

NorthGuy

Re: Shunts

SolInvictus wrote: »

My battery array is composed of 8 Crown CR-395's (6 V batteries) for a total of 1580 Ah at the 20 hour discharge rate and 12 V. The specification sheet states their internal resistance is 8.1 milliohms at 27 C. Because the batteries are in series pairs in my array, their combined internal resistance is ~16.7 milliohms (including ~500 microohm for 4 gauge wire link) which is fairly close to the 20 milliohms used in the calculations. When I measured mine, their temperature was around 15 C which might have altered their internal resistance somewhat. It just occurred to me that I could use 2 gauge wire in the series links that make batteries 1 and 4 to reduce their effective internal resistances to balance the array better.

That's quite a difference. If mine are 20mOhm, and I have 8 of them then a single battery would be 2.5mOhm. But mine are about twice the capacity, same as two of yours in parallel, or 4.1mOhm. Looks like my batteries have internal resistance about half of yours. Would be easier to parallel.

zoneblue

Re: Shunts

SolInvictus wrote: »

My battery array is composed of 8 Crown CR-395's (6 V batteries) for a total of 1580 Ah at the 20 hour discharge rate and 12 V. The specification sheet states their internal resistance is 8.1 milliohms at 27 C. Because the batteries are in series pairs in my array, their combined internal resistance is ~16.7 milliohms

Im a bit lost. You have 4 parallel pairs with each 3 cell battery having 8.1Mohm? Isnt each pair 16.2 and the overall bank 4.05?

SolInvictus

Re: Shunts

When connecting two 6 V batteries in series to form a single 12 V battery, the effective internal resistance of the 12 V battery includes the resistance of the wire link between the two batteries. I estimate the resistance of the 4 gauge wire + 2 terminal lugs + contacts at 500 microohm which results in (8.1 + .5 + 8.1) = 16.7 milliohms for the effective internal resistance.

SolInvictus

Re: Shunts

NorthGuy wrote: »

If mine are 20mOhm, and I have 8 of them then a single battery would be 2.5mOhm. But mine are about twice the capacity, same as two of yours in parallel, or 4.1mOhm. Looks like my batteries have internal resistance about half of yours. Would be easier to parallel.

I looked at Trojan's specification sheet for IND13-6V's and did not see an internal resistance listed. Usually the higher the battery capacity, the lower its internal resistance. Yours might be around 4 or 5 mOhm each, so, yes, yours would need lower resistance parallel links than mine.

I wish my system was 24 V because it would make sourcing 1.5 kW easier, but I would always have at least 2 parallel strings in my battery array. If one battery fails, then I would at least have a functioning half-sized array. A single series string provides no backup, and one needs backup in remote places.

vtmaps

Re: Shunts

SolInvictus wrote: »

A single series string provides no backup, and one needs backup in remote places.

Folks with 48 volt battery banks have been known to operate with only 23 of their 2 volt cells. Many inverters and chargers can be adjusted to work on a 46 volt battery bank.

--vtMaps

NorthGuy

Re: Shunts

SolInvictus wrote: »

I looked at Trojan's specification sheet for IND13-6V's and did not see an internal resistance listed. Usually the higher the battery capacity, the lower its internal resistance. Yours might be around 4 or 5 mOhm each, so, yes, yours would need lower resistance parallel links than mine.

Mine are 2.5mOhm/battery, although it varies. Sometimes may get as high 8mOhm, but for only brief periods.

SolInvictus wrote: »

I wish my system was 24 V because it would make sourcing 1.5 kW easier, but I would always have at least 2 parallel strings in my battery array. If one battery fails, then I would at least have a functioning half-sized array. A single series string provides no backup, and one needs backup in remote places.

When you go higher voltage, there's a different kind of redundancy. You can remove one (or more ?) and go on with lower voltage. With 48V system, I definitely can survive with 23 cells. I think I could get by on 22 cells too. With higher voltage banks (e.g. 200-250V) a removal of a single cell will hardly be noticeable.

SolInvictus

Re: Shunts

I did not know that some of the higher voltage inverters are adjustable.

NorthGuy, because you have 6 V batteries, you would lose 3 cells if one fails. What is your plan?

NorthGuy

Re: Shunts

SolInvictus wrote: »

NorthGuy, because you have 6 V batteries, you would lose 3 cells if one fails. What is your plan?

My plan is to open the battery and by-pass the bad cell. I then may open the bad cell and try to fix it. If this fails, I will start working on buying the new set. I don't think I can buy a replacement battery. They're too unique. I thought I had a warranty, but not with Trojans. With my next set, if I stay with lead-acid, I'll go with Surrettes.

jcheil

Re: Shunts

NorthGuy wrote: »

My plan is to open the battery and by-pass the bad cell.

Now thats a cool idea for an emergency...woulda never thought of that!