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Ceramic capacitor is of too low voltage?...som,etimes blows up at switch on.

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Doctorandus_P:
In this case it is not the (stable state) of an RLC circuit which has to be considered, but the inrush current.
Inductance of your main cables is hard to guess / measure and varies upon location.

Go blow a few leds like I did a long time ago to get some respect what 230Vac will do with unprotected electronics.
BigClive (youtube) has a very nice switch box for working relatively savely with Mains voltage & electronics.

Terry01:

--- Quote from: T3sl4co1l on July 06, 2018, 12:38:18 pm ---
--- Quote from: Terry01 on July 05, 2018, 11:35:50 pm ---I thought a wall socket was only good for 32 Amps? I don't know for sure myself, i'm still learning. I read it on the forum somewhere.

--- End quote ---

Some sockets? I guess??

You can easily draw over 2000 amperes from a typical outlet... it's just a question of, for how long.  At that rate, the breaker will trip within a cycle (< 10ms).  Which, mind you, is an electrical eternity -- say a transistor was dropping that voltage (even while drawing far less current -- transistors can't draw fault current, but a fault condition like SMPS shoot-through, or driving a shorted load, will still quickly be fatal), it will die in about 20us.  So the breaker opens after about 500 transistor deaths, by which time the one transistor has thoroughly turned into an expanding plasma ball. :)

Over longer time scales, you want to keep the RMS current below the fuse/breaker rating (whatever that happens to be -- and for UK circuits, I believe that includes a fused plug and cord, versus the US system that assumes any outlet may be asked to deliver full rated current).  RMS is an averaging process, so the peak current can still be much more than rated -- again, it's just a question of, for how long.  :)

In the present case, blowing a ceramic cap will have an equivalent RLC circuit, which will peak at around 1.5A in 3.5us, then 680V at 7us (worst case, switching at the instant of maximum line voltage, and assuming mains inductance ~500uH and capacitance 0.01uF constant).  In practice, the ceramic's capacitance will drop off rapidly as voltage rises, so the peak current will be lower, and the peak voltage will be far higher, probably over 1kV.  At that point, the LR8 is guaranteed to be unhappy; it may well happen that it can bear modest avalanche currents (<1A in this case), but it may also be gradually damaged in the process.  Of course if the capacitor fails internally, you don't get a chance to see if that happened... :)

Tim

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I've read this 4 times now and still don't fully understand it but I will...even if I have to read it another 4 or 5 times! It's very very interesting reading.
Thanks for the info..

james_s:

--- Quote from: T3sl4co1l on July 06, 2018, 12:38:18 pm ---You can easily draw over 2000 amperes from a typical outlet... it's just a question of, for how long.  At that rate, the breaker will trip within a cycle (< 10ms).  Which, mind you, is an electrical eternity -- say a transistor was dropping that voltage (even while drawing far less current -- transistors can't draw fault current, but a fault condition like SMPS shoot-through, or driving a shorted load, will still quickly be fatal), it will die in about 20us.  So the breaker opens after about 500 transistor deaths, by which time the one transistor has thoroughly turned into an expanding plasma ball. :)


--- End quote ---

2,000A seems a bit high for a real world situation. Looking up wire resistance I see 12AWG is 1.588 Ohms per 1,000'. If you figure a 50' run between the panel and receptacle then that's 100' round trip so 0.1588 Ohms. If we assume 170V peak and ignore the source impedance of the distribution feeding the panel, resistance of the breaker, receptacle, power cord, etc then I get 1,070A as the absolute peak you could get. Now that's still a whole lot of current, but it's not 2,000A.  Apologies for a pedantic moment.

T3sl4co1l:
No worries about pedantism, 2kA was merely a rough ballpark figure.  YMMV, depending on many things, of course. :)

Tim

exe:

--- Quote from: james_s on July 08, 2018, 03:46:19 am ---If we assume 170V peak and ignore the source impedance of the distribution feeding the panel, resistance of the breaker, receptacle, power cord, etc then I get 1,070A as the absolute peak you could get. Now that's still a whole lot of current, but it's not 2,000A.  Apologies for a pedantic moment.

--- End quote ---

There are some counties with 230VAC RMS :P

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