On 8/7/24 03:41, Viesturs Lācis wrote:
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.
ferrite in such a circuit is generally wrong. If the ferrite see's too
much magnetisim it may saturate, at which point it looses its ability to
resist the change and max current will flow according to the ohmage of
the wire. It also has a magnetic curie point, a $5 word that says if
this happens above its curie point temperature, it become austenitic,
aka non-magnetic, and can only be repaired by returning the core to
Arnold for a factory heat treatment. You might want to remember that for
some of the high performance ferrite alloy's, this "curie" temperature
is below the boiling point of water at 100C. That's not hot enough to
discolor the paint on it. Moral of the story is use silicon steel cores
for such.
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?
T=RC will give an answer for (IIRC) a 67% charge. Where T is time in
seconds, R is ohms and C is farads. 5T will give essentially a full
charge time. Its the low R from the powerline that is clearing your
fuse/breaker. :o)
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.
One of the reasons to use a honking high wattage resistor.
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?
That's the theory. In the case of my go704, 3 seconds seems to be
enough. That does not heat a 100 watt R more than 20F.
Viesturs
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Cheers, Gene Heskett, CET.
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