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Circuit breakers and kA ratings

Craig

Watts are Watts!
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I am looking at 2 different 16 amp circuit breakers one has a rating of 4.5kA and the other has a rating of 6kA I understand the 6kA breaker will hold back more amps but for solar or from our batteries to our inverter is the higher rating necessary or should they even be higher like 10kA
 
50,000 aic anyhow. Midnite solar battery inverter breakers (CBI)
 
Inrush currents...... on battery circuits require great electro-
mechanical strength.
 
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It's the breaking current capacity of the breaker. It's defined as the maximum peak current the breaker can interrupt safely.

When you have some over-current condition the breaker will only have to break the current of the circuit (for example 60 A if you use 60 A through a 50 A breaker) but in case of a short circuit the current to break will be a lot higher, usually kA to dozens of kA on our systems.

Inrush currents...... on battery circuits require great electro-
mechanical strength.

It doesn't help with inrush current as it's the breaking capacity, not the making capacity.
 
It's the breaking current capacity of the breaker. It's defined as the maximum peak current the breaker can interrupt safely.

When you have some over-current condition the breaker will only have to break the current of the circuit (for example 60 A if you use 60 A through a 50 A breaker) but in case of a short circuit the current to break will be a lot higher, usually kA to dozens of kA on our systems.



It doesn't help with inrush current as it's the breaking capacity, not the making capacity.
Yes, good catch. There is arcing on make contact also. The magnetic hydraulic breakers have sufficient strength to open and the correct distances, materials, arrangement and speed to handle the arc, which is larger on opening. Amps interrupting current.

Breakers are rated for make/ break its the contacts that wear and weld.

When there is a short, current starts to move rapidly has inertia as it is and the aic rating takes this inrush current into account else we dont need 50,000 amps to be interrupted or as a safety margin for the dozens of kA that is present, not when switching, but in protection of a live battery short.

The best demonstration of this are test cell which demonstrate the need to de-burr conduit during assembly. A tiny metal sliver is penetrating wire insulation and when the disconnect switch is closed...... boom! Vaporized refrigerator sized electrical enclosure! Different operation, but in order to interrupt large flows of inrush current to a short, high ratings are required.
 
Ok so can we get an Formula for this. If my battery is capable of pushing 800Amps how many la do I need to break. I understand it's more than 800A due to momentum
 
Ok so can we get an Formula for this. If my battery is capable of pushing 800Amps how many la do I need to break. I understand it's more than 800A due to momentum
That, is a question for breaker design engineers!
 
Sipping coffee made atop an iron box with wood on fire in it!
Very electric and solar at the base.
 
I did not see that one i was more looking at the
MNEPV 80A or 100A

yours are actually a good find as the breakers I have been looking at for my original question had a max of 63 amps
 
I would buy an inverter battery breaker in the size you require. The 10kA should be ok, but the high ratings cover surges and massive inrush and break strength occuring in those circuits.

Additionally the breaker is sometimes feeding a bus and has multiple smaller value breakers on it. I look at it as a dc mains breaker.
 
Ok so can we get an Formula for this. If my battery is capable of pushing 800Amps how many la do I need to break. I understand it's more than 800A due to momentum

It's pretty simple but you need the internal resistance of the battery and the resistance of the circuit; then you do Isc = Vbatt / (Rbatt + Rcircuit) so for example let's say you have a 13.5 V battery with an internal resistance of 5 mOhm and a circuit resistance of 8 mOhm then the short circuit current would be: 13.5 / (0.005 + 0.008) = 1038 A so about 1 kA
 
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