trešd., 2024. g. 7. aug., plkst. 00:21 — lietotājs Chris Albertson
(<albertson.ch...@gmail.com>) rakstīja:
>
> The capacitor is charged when the switch is closed.   A huge current will 
> flow into the capacitor until it is “full”.  The usual solutions are to (1) 
> use a slow-blow fuse that can withstand the surge current or place a low 
> value power resister in series with the AC to limit maximum current.   
> Inductors can do this better than resistors.
>

Yes, the inrush current that charges the capacitor is what I am
thinking about now. I would prefer taking some ferrite ring and wrap
the wire around it and be happy but I have no idea if that is how it
works and I was not able to find any place to explain how to
calculate. Initial values are: DC voltage is 340 V, current limit -
hopefully around 10A, fuse is adjustable, originally set at 12 A (can
be increased to 16 A). capacitor is 4700 uF.
How do I calculate the time it takes to [almost] charge the capacitor?
That would determine that transient state for inductor and then I have
no idea how to calculate the inductor from here.
I would prefer inductor because then there is nothing that can break.
Input is 3 phase AC so I see following drawbacks with introducing resistors:
1) putting them after rectifier bridge seems like a violation of "no
switches in DC bus" recommendation from 8i20 manual. Maybe except by
having a resistor permanently connected in line and then having a SSR
in parallel to resistor would be solution. But if SSR fails (I have
experienced that) how will I know that? All load will go through
resistor and burn it.
2) putting them in AC line requires 3 sets of that and the issue of
the relay that bypasses resistor remains or have I missed something?
It seems to me that 340V and approximately 10A would require 30-50 ohm
resistor as suggested by Gene but somehow I am not sure 100W is
enough. Or is it fine because it is very short period of time?

Viesturs


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