I currently have a single string of AGM batteries wired for 48 V. 4x12 200ah. Runs to the Lynx dist in the usual fashion with a Smart Shunt at the batteries before the Lynx. I want to add a second string of identical bats for 400ah. I have an extra connection in the Lynx. Should I run the second string to the Lynx and make my parallel connection for the two strings there? Also, would I need a second Smart Shunt for the second string since its an ESS system? What is best way to do this?
Thanks!
best way to do this? There are several ways that will work, and the physical layout will usually determine which is best. With Lead acid batteries you need to keep the sum of cable resistances the same in both legs, to enable equal current sharing.
The negatives are best paralleled either at the battery, or at the battery terminal of the smart shunt. Likewise the positives should be paralleled either at the battery, or at the positive input to the Lynx. I would add a battery isolator switch / bulk fuse disconnect in the positive of each battery string, making it easy to disconnect one string if you have a faulty battery.
@Mike Dorsett So, I'm following this. Have another disconnect for the new string. So in this example, could I just parallel + and - at the terminals and have a single run to the Lynx? If so, I'd only need one smart shunt?
But: In the above diagram, the pack on the left has longer lead resistance than the pack on the right. Whilst I know this is only indicative, the lead length total from the common point (+) to common point (-) must be the same. so the negative lead off would be from the left side, and the positive from the right in the above diagram.
If there is a disconnect in each leg, then the added resistance is equal - so this does not affect the cable length.
The cables should be paralleled at or before the shunt, so this does not affect the symmetry of the battery.
The problem with lead acid and requiring equal resistance in strings comes from the typically 60% charge efficiency (Discharge energy / charge energy), so you need to stop one string supplying more energy than the other(s) during high loads. This is not a problem with Lithium batteries, but some of those require equal current sharing between strings due to limited discharge / charge capability.
Sorry about the rough drawing. Idea is that currents in the battery strings are equal.
This shows a variation with 2 disconnects. The lengths from the shunt to the load (not shown) and from the right hand disconnect do not affect the balance of currents in the battery strings, though they do contribute to overall battery impedance (resistance). The total length of interconnect in each string must be equal as this does. Lengths L2 will also affect the current balance, as will L1. These should be the same total length. i.e. L1+L2 for string 1 must equal L1+L2 for string 2.
Variation 3, with strings paralleled at shunt. Sting cable length is now L2 + L3, length L1 has to be deducted from the -ve cable to the lower string to keep things equal.
Remember that battery impedance will be of the order of a few mili-ohms, so adding cable and connections will have a significant effect.
What is -ve? I see it only in this forum.
