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# Configuration Help for SolArk 12k Inverter and KiloVault HAB

I have 4 KiloVault HAB 7.5 (v2) connected in parallel to a SolArk 12k inverter.  I am trying to optimize the battery and charge settings on the SolArk but am having trouble translating the recommended settings in the KiloVault user guide to the available options in the SolArk configuration screen.  Any input would be appreciated.

I have attached screenshots of the three screens to be configured and the descriptions of the values.

KiloVault Tech Support and gubbool have reacted to this post.
KiloVault Tech Supportgubbool

Hello,

Just a note to let you know that we've read your question and will post a reply shortly.

Hello,
Please pardon the delay in responding.
Here are the settings you're looking for.

##### Batt Tab
• Batt Capacity = 150 Ah per HAB in the battery bank
• Max A Charge = 100 A per HAB
• Max A Discharge = 100 A per HAB
• TEMPCO = 0mV / °C
##### Charge Tab
• StartV (Gen or Grid) = 51.4 V
• Start% (Gen or Gris) = 20%
• A = 100A per HAB in battery bank
• Float V = 52.8
• Absorption V = 56
• Equalization V = No Equalization
##### Discharge Tab
• Shutdown = 48.2 V
• Low Batt = Sets when the Sol-Ark batt symbol will turn yellow.  Personal preference.
• Restart = 49 V
• Batt Empty V = 48
• Batt Resistance = Use this formula. (1/n) * (1/r) = resistance. n * (1/r) = 1/total resistance
n = Number of HABs, r = HAB internal resistance  = 15mΩ
• Batt Charge Efficiency = 94.5%

Thanks so much for the reply.  I have just two follow-up questions.

For Equalization V = No Equalization, does that mean I set that value to 0?

And for the Batt Resistance formula, in my particular scenario I have 4 HABs.  Based on your formula, does that mean the Batt Resistance would be 0.017mOhms?  That number just doesn't seem right based on the other examples in their user manual, so I wanted to make sure I was doing the calculation correctly.

Thanks again for your thorough help.

Hello,

Set the equalization time to zero hours to disable equalization.

Good Catch!!!!

It is actually n * (1/r) = 1/r total.
In parallel circuits, the total resistance is always less than the resistance of any of the branch resistances.
So, with 2 HABs. Each HAB has a resistance of 15mOhms
1/15 = 0.0666666666666667
2 HABs so n = 2
2 * . 0666666666666667 = 0.1333333333333333 (which is 1/r total)
1/0.1333333333333333 = r total = 7.5mOhms

With 3 HABs
1/15 = 0.0666666666666667
3 * 0.0666666666666667 = 1/r total = 0.2
1/0.2 = r total = 5mOhms

Please accept my apologies for the confusion.