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200A breaker destroyed by DC voltage

A

azaxi

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Oct 14, 2023
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I bought this 200A breaker: https://www.aliexpress.com/item/4000506382852.html

Tested a small 48V system with standalone inverter and LiFePo4 48V battery (52.2V) at home. boiled water in 1500W electric kettle, worked no problem.

Went to site and installed 4x 31HDC lead-acid bank with 5.5/11kW AIO. When connected the battery bank, AIO came on for few seconds, then died. Found out AIO only receives 36V from battery bank although at battery terminals it was over 50V. Further investigation revealed voltage is being lost at the breaker. Flipping it on and off did not help.

I drilled off the rivets, opened the breaker and contacts were plasma blackened. Btw. it is nicely done inside, wish I took the pic

This breaker never seen higher current than 15A that one time during initial test and it was flipped on/off maybe 1 dozen times. It was destroyed by DC voltage.

Is there a standard R(L)C circuit that would protect breaker contacts from plasma?

I am thinking a parallel capacitor of 100uF would prevent voltage buildup in first 100us (@ 40A) when the space between contacts is less than required dielectric strength (3kV/mm). 1mF should protect the breaker even after shorts when current is couple hundred amps. Parallel resistor about 10-50ohm will immediatelly after breaking discharge the capacitor so there wont be any charge left when breaker is finally flipped on again. AIO will see some voltage through resistor, but voltage drop on it will be too high so it will gracefully shut down

BreakerProtection.png
 
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That breaker isn't rated for your system voltage.
It's only rated for up to 48V. (I wouldn't trust that, either)
Your system voltage will always be higher than that.
 
The Chinese knock-offs of that style breaker are infamous for being dangerous to use.

Eaton/Bussman makes a quality version of that style (Blue Sea sells the same thing with their branding). I would avoid any other brand. They can create more of a fire hazard than they prevent.

When looking at the Eaton/Bussmann version, you will notice that they top out at around 200A. The reason for this is simple: There is not enough room for the size of copper needed for higher currents. I have seen knock-offs that claim 350A but I would not call them breakers... I would call them fire starters.
 
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BTW: I used to consider fuses 'old school' and wanted breakers instead of fuses for everything. However, quality breakers with large current capabilities get expensive fast and I have changed my point of view.

Poping a 200A DC overcurrent protection device should be considered a catastrophic event. If a system pops that size OCPD during normal operation, the design is fundamentally flawed and should be changed. As such, I will almost always use fuses for anything over ~100A
 
FilterGuy said:
BTW: I used to consider fuses 'old school' and wanted breakers instead of fuses for everything. However, quality breakers with large current capabilities get expensive fast and I have changed my point of view.

Poping a 200A DC overcurrent protection device should be considered a catastrophic event. If a system pops that size OCPD during normal operation, the design is fundamentally flawed and should be changed. As such, I will almost always use fuses for anything over ~100A
Click to expand...

Is your reasoning purely financial, or is there some other reason to prefer fuses over breakers? Your reasoning about replacing breakers after a failure makes sense, but my goal is to not pop either, and if I do, I'm willing to pay the price of either.

Related to that, do you consider it a bad idea to use a breaker as a switch? With a fuse you still need a switch/disconnect somewhere, and having both a switch an fuse adds potential failure point, additional connection resistance, cost, etc.
 
Getting back to your question about the circuit:
1699224176637.png

When the breaker is open there will be ~52V across the 10ohm resistor. That is 520W!!! It better be one monster-sized resister.

In normal operation, the inverter will go into undercurrent protection and limit the current to quite a bit less. However, in a short circuit that the breaker is supposed to protect against the 10 ohm resistor will have to dump a huge amount of energy after the breaker pops open. If the wattage of the resister is not high enough the question is whether it will catch on fire or burn open first.
 
webbbn said:
Is your reasoning purely financial, or is there some other reason to prefer fuses over breakers? Your reasoning about replacing breakers after a failure makes sense, but my goal is to not pop either, and if I do, I'm willing to pay the price of either.
Click to expand...
The other reason I started to re-think breakers is the mounting. ( I am guessing you were drawn to that particular breaker because of the ease of mounting it. ) Once above 200A, the chances are that a quality breaker is going need it's own housing box and mounting. This creates a higher cost (sometimes a lot higher) and the extra space needed can create problems.

Beyond that, it is mostly a financial consideration..... but not that I have gotten over the Breaker-is-better mindset, I really don't see a need for using breakers for high-current situations.

I understand that other people will come to different conclusions and I respect that... I can only give you my point of view.
 
Partimewages said:
Unless a breaker is rated for switch duty it is only an overcurrent device.
Click to expand...
Exactly!!!! Some breakers have a rating for 1000s of manual cycles.... they can be used as switches for isolation and maintenance, but still should not be used for daily on-off functions. Other breakers have a very low cycle count and should never be used as a switch.
 
FilterGuy said:
Getting back to your question about the circuit:

When the breaker is open there will be ~52V across the 10ohm resistor. That is 520W!!! It better be one monster-sized resister.

In normal operation, the inverter will go into undercurrent protection and limit the current to quite a bit less. However, in a short circuit that the breaker is supposed to protect against the 10 ohm resistor will have to dump a huge amount of energy after the breaker pops open. If the wattage of the resister is not high enough the question is whether it will catch on fire or burn open first.
Click to expand...

Thanks, I figured later after posting this that short would burn 10 ohm resistor and defeat purpose of the breaker.

1kOhm would still provide reasonable tau=1sec to discharge the cap and wattage would be only around 2.5W.

Btw. 52V across 10ohm resistor is not 520W but 270W ;) Just saying ...
 
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