Can Someone please explain the math to actual amps being pulled from the battery to an inverter?

[Hedges]

Amps from a battery do not increase as state of charge drops.

Amperage capability of the battery does not drop.
But an inverter delivering constant watts will draw more current as voltage of battery drops.

My 2 cents - your looking at it backwards - fuses are to be 125% of whatever the max possible supplied current - Fuses only protect wires from melting or tools if you manage to drop them across the battery posts.

And I say that is 1.25x RMS current in wire.

The current from a battery looks much more like rectified 60 Hz sine wave than DC.
Power comes from Mean average amps x voltage (voltage is relatively constant, near DC.)

I say multiply by an additional 1.12x for Ripple Factor, when battery is supplying single phase inverter.

And consider cutoff voltage at inverter, which will be drooped below battery (and BMS cutoff) voltage.

 

[robbob2112]

So I can get around max fusing requirements by using lots of tiny fuses?

Where exactly do you think the power in this system is coming from?

A dropped tool across the bus bars would pull 28000 amps and blow all 4 MRBF fuses. A battery going south and dumping current would blow its own fuse. If the SCC went crazy they would pop their breakers. This is off grid so no charging from the inverters.

A short at the inverter wire could pull all the battery amps and the SCC amps. It would pop the class T and the breakers and maybe MRBF.

Wires only vaporize when they receive enough joules of energy to melt them and then convert to gas. Been a long time since chemistry, but that is a lot of energy for 4/0 wire, let alone a length of it. The fuses will all pop long before that happens. If they don't pop you have an entierly different problem. If there is an arc big enough to sustain current and overcome the aic of the fuses it will start by burning the crap out of the loose end and starting a fire.

A tool dropped on battery posts would bypass most fuses.

Not properly installed MRBF with all contact points covered.

 

[Checkthisout]

Amperage capability of the battery does not drop.
But an inverter delivering constant watts will draw more current as voltage of battery drops.

Agreed that resistance in the inverter drops as it's load increases.

Disagree that a battery at a lower state of charge will deliver more amps.

If that was the case, then a 6 volt battery would turn a motor just as fast as a 12 volt.

And I say that is 1.25x RMS current in wire.

The current from a battery looks much more like rectified 60 Hz sine wave than DC.
Power comes from Mean average amps x voltage (voltage is relatively constant, near DC.)

I say multiply by an additional 1.12x for Ripple Factor, when battery is supplying single phase inverter.

And consider cutoff voltage at inverter, which will be drooped below battery (and BMS cutoff) voltage.

Victron says a 400 amp fuse for a 12/3000 multiplus so that's a good starting point I think.

 

[Checkthisout]

Where exactly do you think the power in this system is coming from?

A dropped tool across the bus bars would pull 28000 amps and blow all 4 MRBF fuses. A battery going south and dumping current would blow its own fuse. If the SCC went crazy they would pop their breakers. This is off grid so no charging from the inverters.

I'm not talking about a short at the bus bars, I'm talking about a short on one of the 0000 cables.

Lets say you had all 4 batteries hooked up and then had a short to ground in one of the cables. You now have 800 amps available on an 0000 cable. If you fuse at the bus bar you only have whatever the battery is fused at plus whatever its cable is fused at the battery.

A short at the inverter wire could pull all the battery amps and the SCC amps. It would pop the class T and the breakers and maybe MRBF.

Wires only vaporize when they receive enough joules of energy to melt them and then convert to gas. Been a long time since chemistry, but that is a lot of energy for 4/0 wire, let alone a length of it. The fuses will all pop long before that happens. If they don't pop you have an entierly different problem. If there is an arc big enough to sustain current and overcome the aic of the fuses it will start by burning the crap out of the loose end and starting a fire

So then what's the point of fusing at all?

Shunt snapped and caused a fire🔥

Got a call Sunday from the wife about smoke coming from the garage and fire trucks were already on the way. After they had there fun spraying many cans all over the garage I got to have a look inside. I have 5 rows of shelves with 2 shelves for in each with a battery and then busbar along the...

diysolarforum.com

Your argument works if the total amount of fusing does not exceed what is requires to protect a 0000 cable which I would say is 400 amps, give or take so if each battery was fused at 100 amps then you could forego the second fuse at the bus bar.

Make sense?

 

[wattmatters]

Not properly installed MRBF with all contact points covered.

Sure, I was just responding to what you wrote. If the posts are covered then you can't drop a tool on them.

 

[robbob2112]

I'm not talking about a short at the bus bars, I'm talking about a short on one of the 0000 cables.

Lets say you had all 4 batteries hooked up and then had a short to ground in one of the cables. You now have 800 amps available on an 0000 cable. If you fuse at the bus bar you only have whatever the battery is fused at plus whatever its cable is fused at the battery.

So then what's the point of fusing at all?

Shunt snapped and caused a fire🔥

Got a call Sunday from the wife about smoke coming from the garage and fire trucks were already on the way. After they had there fun spraying many cans all over the garage I got to have a look inside. I have 5 rows of shelves with 2 shelves for in each with a battery and then busbar along the...

diysolarforum.com

I actually followed that entire conversation about the shunt. The final conclusion was still in question, but the most supportable answer was one of the partical board shelves weakened and shorted one of the battery cables before it entered the DC breaker. There are other theories but no solid conclusions. In that case a home built battery. And if it had a fuse at the output terminal verse at the bus bar it would have popped. Instead he was depending on a DC breaker attached after several feet of wire.

How do you get a short at one of the 4/0 wires? They have insulation over them. They are bolted to the MRBF at one end with a rubber boot over it and to the bus bar at the other. A ground to the end at the bus bar is the same as shorting the bus bar, blow all MRBF. A short at the top of the MRBF, how? If you managed it through poor wiring practice, then the battery is gonna fry the internal BMS and maybe cause a fire. The other batteries would blow their MRBF.

 

[Checkthisout]

I actually followed that entire conversation about the shunt. The final conclusion was still in question, but the most supportable answer was one of the partical board shelves weakened and shorted one of the battery cables before it entered the DC breaker. There are other theories but no solid conclusions. In that case a home built battery. And if it had a fuse at the output terminal verse at the bus bar it would have popped. Instead he was depending on a DC breaker attached after several feet of wire.

How do you get a short at one of the 4/0 wires? They have insulation over them. They are bolted to the MRBF at one end with a rubber boot over it and to the bus bar at the other. A ground to the end at the bus bar is the same as shorting the bus bar, blow all MRBF. A short at the top of the MRBF, how? If you managed it through poor wiring practice, then the battery is gonna fry the internal BMS and maybe cause a fire. The other batteries would blow their MRBF.

Like I said, so long as the total amperage of all the fuses does not exceed that which will protect an 0000 cable, then just fusing at the battery is fine.

If that's not the case, then why fuse at all?

Arguments about wires being insulated and terminals being covered are just arguments for not fusing at all.

 

[Hedges]

Agreed that resistance in the inverter drops as it's load increases.

Disagree that a battery at a lower state of charge will deliver more amps.

Not delivers more current that it could have at higher state of charge.
Inverter presents lower impedance on its input, drawing more power from battery even though (or because) input voltage is lower.

Similar case if you had AC voltage sagging, so you boosted it back up with buck-boost transformer or Variac. Or utility changes transformer taps. To Deliver same power, it is set for lower input voltage, higher input current.

If that was the case, then a 6 volt battery would turn a motor just as fast as a 12 volt.

A 12V brush-type motor fed 6 volts will run slower.

A DC/DC converter fed 6V and putting out 12V will draw extra current from the 6V battery to deliver full voltage and current at 12V. Same with DC/AC inverter.

Some SMPS are protected from brownout, others will burn themselves out when input voltage drops too low and they pull too much current.

 

[robbob2112]

Like I said, so long as the total amperage of all the fuses does not exceed that which will protect an 0000 cable, then just fusing at the battery is fine.

If that's not the case, then why fuse at all?

Arguments about wires being insulated and terminals being covered are just arguments for not fusing at all.

At this point you are just being argumentative just for the sake of it.

Nobody proposed not fusing. I pointed out that any short you can do other than a wrench across the positive and negative of the same battery is protected by one or more fuses.

You refuse to accept the reality of that so I am done answering you.

 

[Checkthisout]

At this point you are just being argumentative just for the sake of it.

Nobody proposed not fusing. I pointed out that any short you can do other than a wrench across the positive and negative of the same battery is protected by one or more fuses.

You refuse to accept the reality of that so I am done answering you.

I honestly wasn't. I'm pointing out something many people fail to realize with this type of setup.

The point of fusing is to protect wires from carrying more amps than they are rated for.

Fusing right at the battery with 200 amps would allow any of the 4 cables that run to each battery to be exposed to carrying much more current than they are rated for.

With 4 fuses of 200 amps each you have 800 amps of fault current available which is more than 0000 is rated for.

Fusing so that the sum of the 4 fuses at each battery does not exceed what 0000 is designed to carry is the easiest solution.

However, I would make the sum of the fuses no larger than the fuse feeding the inverter provided its the only load placed on the bus bar.


Your argument against this hinged on the fact that cables and battery terminals are insulated which to me implies there is no point to Fusing in the first place.

 

[Checkthisout]

Not delivers more current that it could have at higher state of charge.
Inverter presents lower impedance on its input, drawing more power from battery even though (or because) input voltage is lower.

Similar case if you had AC voltage sagging, so you boosted it back up with buck-boost transformer or Variac. Or utility changes transformer taps. To Deliver same power, it is set for lower input voltage, higher input current.



A 12V brush-type motor fed 6 volts will run slower.

A DC/DC converter fed 6V and putting out 12V will draw extra current from the 6V battery to deliver full voltage and current at 12V. Same with DC/AC inverter.

Some SMPS are protected from brownout, others will burn themselves out when input voltage drops too low and they pull too much current.

Well, ok. Maybe I'm not thinking this through.

So the more dead my battery gets, the more it's amperage output increases for a given load?

If I were to graph this, a dead battery delivers infinite amps and full battery delivers zero amps.

I can't conceptualize this but I do experience a lot of voltage drop across my neurons due to too much brake cleaner use.

 

[robbob2112]

Ok, last time.

My argument was that if you did short the wire at any place on it all you would achive is 4 blown fuses.

The fuses blow in microseconds when you exceed their rating by that many amps. You can look at the specific details on the fuse spec sheet.

You can lookup the resistance per foot of 0000 wire of any type. Now figure how long to bring 2ft of of wire (approx 0.5kg of copper) from 25c to 105c (where the sheath melts). Do that math using the power in terms of heat and the density of copper along with a few other bits and you get around 4 minutes.

The wire might get a little warm before the fuses blow but that is it.

 

[Hedges]

Well, ok. Maybe I'm not thinking this through.

So the more dead my battery gets, the more it's amperage output increases for a given load?

If I were to graph this, a dead battery delivers infinite amps and full battery delivers zero amps.

I can't conceptualize this but I do experience a lot of voltage drop across my neurons due to too much brake cleaner use.

You should have stuck with good old global warming brake cleaner that doesn't turn mechanics into Stephan Hawking.

I also suggest R12 refrigerant, rather than the stuff that causes testicular cancer* and detonates when mixed with compressed air.
*can't say for sure if the risk depends on biological gender, or gender identity.

5kW at 50V requires 100A. (not zero)
5kW at 40V requires 125A.
5kW at 25V requires 200A (assuming inverter even keeps operating.)
5kW at 0V requires infinite amps, as you say. But V = IR keeps that from happening.

Constant power out, constant voltage x current product drawn from battery.
Neglecting any difference in efficiency.
Half the voltage, twice the current, four times the power dissipated in wires and FETs.

 

[Checkthisout]

You should have stuck with good old global warming brake cleaner that doesn't turn mechanics into Stephan Hawking.

I also suggest R12 refrigerant, rather than the stuff that causes testicular cancer* and detonates when mixed with compressed air.
*can't say for sure if the risk depends on biological gender, or gender identity.

5kW at 50V requires 100A. (not zero)
5kW at 40V requires 125A.
5kW at 25V requires 200A (assuming inverter even keeps operating.)
5kW at 0V requires infinite amps, as you say. But V = IR keeps that from happening.

Constant power out, constant voltage x current product drawn from battery.
Neglecting any difference in efficiency.
Half the voltage, twice the current, four times the power dissipated in wires and FETs.

Ok so lets pretend we have a 12 Volt system. Battery is at 13.2 volts @ 3000 watts 227 Amps.

Ok now battery is at 12.2 volts @ 3000 watts = 240 Amps

Ok now battery is at 11 volts @ 3000 watts = 272 Amps.

Ok now battery is at 10 volts @ 3000 watts = 300 amps.

How low does my battery have to get before I blow my 310 Amp Inverter fuse?

 

[Hedges]

Now divide by inverter efficiency at full load (90%)?
And multiply by 1.12 ripple factor, because 60 Hz ripple comes from batteries not capacitors.
3000W / 10V / 90% x 1.12 = 373A

On a good day, a 350A fuse may hold on for quite a while, probably outlast your battery.

In fact, even 300A class T should hold out way over 1000 seconds according to the LittleFuse chart:

https://www.littelfuse.com/media?resourcetype=datasheets&itemid=2925cc6d-2278-44e0-8fdd-106f087f707e&filename=jlln-fuse-datasheet

But I suggest 373A x 1.25 = 466A minimum, so 500A fuse and sufficient gauge wire.
The objective is for expected load to never blow a fuse.

 

[Checkthisout]

Now divide by inverter efficiency at full load (90%)?
And multiply by 1.12 ripple factor, because 60 Hz ripple comes from batteries not capacitors.
3000W / 10V / 90% x 1.12 = 373A

On a good day, a 350A fuse may hold on for quite a while, probably outlast your battery.

In fact, even 300A class T should hold out way over 1000 seconds according to the LittleFuse chart:

https://www.littelfuse.com/media?resourcetype=datasheets&itemid=2925cc6d-2278-44e0-8fdd-106f087f707e&filename=jlln-fuse-datasheet

But I suggest 373A x 1.25 = 466A minimum, so 500A fuse and sufficient gauge wire.
The objective is for expected load to never blow a fuse.

So I have a 500 watt solar panel loaded by a 500 watt inverter producing 10 amps at 50 volts with an 11 amp fuse.

Will my fuse blow if less sun begins the strike the panel because the amps will increase as panel voltage drops?

 

[Dadoftheturkeykids]

Thank you guys for keep bringing it back around to my initial question amidst the disagreements..
So I'm thinking I have to either upgrade my wire or settle with 350ish amp fuse which
(I think) will Give me around 3000w 120v

To add my 2 cents to the load draw disagreement. It makes sense to me that if a load on an inverter is pulling a steady wattage at 120v AC, as the battery voltage dropped off amperage would raise on the DC side to maintain the output current. V*A=W

 

[MisterSandals]

350ish amp fuse which
(I think) will Give me around 3000w 120v

3000W AC requires about 3000W / .85 conversion loss = 3529W DC

3529W / 350A fuse = 10V
If your battery discharges to 10V while your inverter is delivering 3000W AC, your fuse should blow. This "may" be very close to where your inverter cuts off (check your manual).

In real life, you likely wouldn't be pulling the max your inverter can deliver. 350A fuse probably good to go; if not you blow a fuse.

 

[Hedges]

So I have a 500 watt solar panel loaded by a 500 watt inverter producing 10 amps at 50 volts with an 11 amp fuse.

Will my fuse blow if less sun begins the strike the panel because the amps will increase as panel voltage drops?

I think you're right at the edge of source not having enough current to do it.
But if PV panel produces 10 A Imp, what is its Isc rating? 11A?
Assuming inverter doesn't shut down, it just might pull 11A continuous and that doesn't have the 25% margin we want.

If that was a battery and it dropped to 42V, I think blowing fuse gets more likely.

If the grid has resistive loads, higher voltage means higher current.
Induction motors, lower voltage means higher current. Higher frequency means higher RPM and higher current.

In real life, you likely wouldn't be pulling the max your inverter can deliver. 350A fuse probably good to go; if not you blow a fuse.

My fuses are big enough for sustained current the inverters can draw. 4/0 Wires between batteries aren't but batteries capacity is too small to count as "continuous" current. That's 405 Ah. If I used 1200 Ah, would have pairs of 4/0 wire.

 

[MisterSandals]

My fuses are big enough for sustained current the inverters can draw. 4/0 Wires between batteries aren't but batteries capacity is too small to count as "continuous" current. That's 405 Ah. If I used 1200 Ah, would have pairs of 4/0 wire.

Agreed, me too. I recommended in post #4 the use of 500A fuses. I was merely pointing out what the OP might experience with his 350A fuse choice, assuming his/her wiring were sufficient (more than 4/0).