I’ve managed to find a fuse that matches these specs in the manual:
The manual does say:
“The inverters are fitted with an internal DC fuse (see table above for rating). If the DC cable length is increased to more than 1,5m, an additional fuse or DC circuit breaker must be inserted close to the battery.
Important note: for UL certified (NEMA GFCI) inverters it is mandatory to install a fuse or DC circuit breaker close to the battery, even if the cable length is less than 1,5m.”
The external fuse or breaker is to protect your wiring, not the device. Overcurrent protection devices on wiring must be sized appropriately to the wire’s ampacity. As long as your wiring is capable of 60A continuous, it’s fine to put a 60A fuse or breaker on it.
I’ve always read this on several forums.
That the fuse is there to protect the wiring from catching fire etc…
The datasheet for the 48v/1200VA inverter says the max. wire input size is 25 mm2 / 4 awg
Based on some ampacity charts on several sites it seems like 60A fuse is sufficient for 4 awg wire.
However, my concern is more of what do I use to prevent the 120 VAC outlet from drawing too much current?
i.e. I use a 1800 W hair dryer.
Will the internal fuse blow out?
The datasheet says it can only output
1000 W of continuous power at 25°C
1200 VA at 25°C (nonlinear load ; crest factor 3:1)
900 W of continuous power at 40°C
Since battery voltage will vary depending on charge level,
Will the amperage output from the battery increase as voltage decreases to output the same level of power?
(56.8V - 40V range) battery max. min./range
1000/56.8 ~ 17.6 amps
1000/48 ~20.83 amps
1000/40 ~ 25 amps
I’m not even taking efficiency losses into account.
So in this case am I better off getting a 60V DC breaker with a 20 amp rating?
Will it activate at the amperage rating without going up to 60V DC?
I’m in the US so I have a NEMA GFCI outlet.
page 4 of the manual also says:
" The inverter does not have a fuse in the AC output. The AC cabling is protected by a fast-acting
current limiter in case of a short circuit and an overload detection mechanism which mimics the
characteristics of a fuse (i.e. faster shutdown with larger overload). It is important to size your
wiring properly based on the inverters’ power rating."
If this the case then will the current limiter be sufficient in preventing me from permanently damaging the inverter?
Yes, as voltage decreases then current increases - but proper 4AWG wire outside of an engine space, unbundled, is good to 160A continuous, so you’re fine with anything above about 40A and below 150A.
An 1800w hair dryer simply won’t run on your 1200VA inverter, so that sort of fixes your issue. The inverter will overload and shut down within a few seconds or less.
In terms of ampacity, 1800w at 120vAC is only 15A on the AC side of things, and you’d be hard-pressed to be using wire that can’t handle 15A… so not much of an issue there even if the inverter could supply that sort of power.
editing to add: Don’t get the DC side and the AC side mixed up. Your DC side should have anything under a 150A breaker or fuse to protect the wire (assuming you’re using proper fine-stranded copper wire as directed in the user and installation manual) but your AC side will never see more than 10A continuous.
Fuses or breakers need to be rated appropriately for the voltage yes, so your DC side needs a fuse or breaker that’s rated to at least 58-60vDC, but that’s just to do with the arc distance. Your 60v-rated fuse or breaker will work just as well at a lower voltage.
I found this DC circuit breaker that has a 60 V DC and 60 amp DC current rating.
My concern is if I put this in line between the battery and the inverter will the fuse inside the inverter the BF1 58V 60A blow out before the breaker will trip. In that case it would be pointless for me to use that?
It looks like it can take awg3 wire as max size and it says cu only so I am assuming we shouldn’t use ferrules?
