Hi, I have read many posts on this with conflicting answers, hence can I have a definitive answer please from Victron:
If input current is set to 10A and the current boost factor is set to 2, does that mean:
A) Total current available, AC1 + Inverter = 2 x 10A = 20A
or B) Total current available, AC1 + Inverter is = 10A + 2 x 10A = 30A.
Without knowing this I can't calculate the boost factor to use, to prevent the inverter overload!
Thanks
Hi @JohnC so I think what that says (and it's still unclear) is the inverter will be asked to find an additional 2 x input current limit, so in affect it is providing twice as much as the shore power, giving 3x the input current limit? I think you saying B is correct. That's a big ask of the inverter for a default of 2! It depends if the assumed current required is the total load, or the additional required from the inverter!
The boost factor is what you tell the inverter to assume it will need to boost by (it does not know what load you have switched on in the first few seconds). it will immediately try to double the input power.
Input 10A - boost factor 2.0 it will assume you need 2 times the input for power so a 20A load - it can do that up to the max of its inverter power before overload.
The inverter can only add its inverting power to the incoming for power assist.
If your load is 7000W incoming is 10A the 5kva inverter can add the to the 10A in to run the load. This is assuming your incomer is 230v. Lower means derating since the setting is amps based. 190v x 10a is 1900w where 10a x 230v is 2300w. If your load is still 7000W you might see an overload. Since now the addition of the (7000 - 1900w) is higher than the inverter can add with nominal inverting power.
The other catch is provided your DC and batteries can do it. So no volt drop that side.
