Battery to inverter ampacity calculation

[luckylerado]

Is this an accurate assumption disregarding all derations

10 amp load at 120 volts is 1200 watts. Powered by a 12 volt battery supplying an inverter. The battery must supply the same 1200 watts of power so at 12 volts the load is 100 amps. This would mean that the wires connecting the battery to the inverter must have an ampacity of at leas 100 amps. Is this accurate. I have an engineer type telling me no this is not accurate, but the math seems simple to me. What am I missing. Thanks

 

[hurk27]

I would believe you need to size the wires to the capacity of the inverter not the load when sizing the battery conductors, as for one the fault current will far exceed the load amps, and loads can some times cause the inverter to pull it ratting from the battery, just not allot of stuff in the NEC about protecting battery conductors, as a bolted fault even on a small battery can result in very high amperage.

 

[broadgage]

Assuming a perfect inverter without loses, then the battery current would be as you calculate.
But in practice it will be more, in the absence of more detailed information, for a 12 volt system I would allow one amp of battery current for every 10 watts of line voltage load.
Therefore for a 1,200 watt load I would allow for 120 amps battery current.

If the inverter output is hardwired to a fixed load, then it might be acceptable to size the battery wires for the actual load.
It would however be better practice to size the wires for the maximum continous input current of the inverter, in case the load increases up to the inverter rating.

If the battery cables are smaller than the maximum input of the inverter then they should be protected against overload.
If the battery cables are sized for the maximum input current of the inverter, then it could be argued that overload protection is not required, since this is provided by the load limiting feature of the inverter.
Short circuit protection is advisable in any case.

Unless the battery cables are very short, they will probably have to be upsized to reduce voltage drop in any event.

Ampacity may be based on steady running current, as with line voltage circuits, not normaly any need to upsize for brief starting currents.
Voltage drop should however be calculated at the maximum short term current, since otherwise the voltage at the inverter input terminals may drop below the lower limit when starting a motor.

As an example consider a 2,000 watt inverter, that can supply 4,500 watts briefly for motor starting.
The ampacity of the battery cables should be at least 200 amps.
The voltage drop should be calculated at 450 amps, to ensure reliable motor starting.
No overload protection would be needed since the inverter can not draw more than 200 amps continually.
Short circuit protection could be by a 400 amp fuse.

 

[tallgirl]

Don't forget our dear friend "coup de fouet" or "Crack of the Whip". The battery is going to sag phat the instant you drop a large load on it, and the current is going to rise to make up for the voltage drop from the battery.

If the vendor doesn't provide a maximum current (may not for a small inverter), calculate Imax at Vmin and Pmax, plus a factor for conversion losses.

 

[K8MHZ]

Don't forget our dear friend "coup de fouet" or "Crack of the Whip". The battery is going to sag phat the instant you drop a large load on it, and the current is going to rise to make up for the voltage drop from the battery.

If the vendor doesn't provide a maximum current (may not for a small inverter), calculate Imax at Vmin and Pmax, plus a factor for conversion losses.

Take a look at 690.51 and 52. 'Modules' include the inverter, so the inverter must be labeled with the info you are asking about.

 

[tallgirl]

Take a look at 690.51 and 52. 'Modules' include the inverter, so the inverter must be labeled with the info you are asking about.

I think you're replying to a different thread -- there is no "module" mentioned, only an inverter. And for many inverters in the 12 volt and small power range, there is no manufacturer provided maximum DC current.

 

[luckylerado]

Thank you for your replies

 

[K8MHZ]

I think you're replying to a different thread -- there is no "module" mentioned, only an inverter. And for many inverters in the 12 volt and small power range, there is no manufacturer provided maximum DC current.

Duh....

I incorrectly assumed this was a question about inverters for PV systems and got Article 690 stuck in my head.

FWIW, I use 400 watt inverters at home that had the internal batteries (about 17Ah) removed and I have them connected to 100Ah SLA pulls from the phone company. The DC current from the battery is a function of the AC output current. Since the AC output current is limited by a built in OCPD, the DC current will not exceed the design limits, even with a huge battery connected to it.

However.....

Motor loads can destroy an inverter. I know for a fact that a fractional motor can toast an inverter not capable of handling the inrush current and reactive load before the built in breaker will trip. I don't suggest trying to run a pump or a furnace blower motor with a small inverter.

Inverters can also destroy motors. The voltage can drop to the point of overheating the motor to the point of destruction.