How is the DC current consumption in idle mode of Multiplus-II?
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How is the DC current consumption in idle mode of Multiplus-II?
The answer really depends on a few factors, one of them being how it is programmed, another is the systen voltage and which model unit you are referring to.
But there is something on the datasheet referring to power consumption for the models.
See the bottom of the product page here.
Hi Alexandra,
yes, I want to know how big is the DC-current which discharge the battery while no AC-load is connected and no power is generated by the Multiplus II.
I have a MP II 48/3000/35 with one Pylontech US3000C connected to.
It can be that the current have a level of 80mA. This seems not much but with a voltage of 50V we get 4watts.
I can't find anything in the data sheet to this subject.
Steffen
So if you are seeing 4w on the ac from dc what you are seeing is correct.
the idle spec is given in the datasheet is ac but you can convert with the formula. W = VA
11w divide by the voltage of the system (in this case battery voltage) that will be the amperage drawn from the battery.
So if your battery is 56V;
11W ÷ 56V = 0.196A
If your battery is 48V then;
11W ÷ 48V = 0.229 A
Dont forget you may have other DC consumers like a GX and MPPT.
The formula is
Voltage x amperage = wattage.
Or
Wattage ÷ voltage = amperage.
Wattage as a power consumption figure does not care about whether it refers to AC or DC. It is universal to the two.
AC is alternating current grid power)
DC is direct current or battery related power.
The data sheet for that inverter says the zero load power is 11 watts.
@Trevor Bird The 11W is the idle consumption on AC connection. This is in the data sheet and I know this.
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