Here’s a reference design that includes a soft power up feature.

https://www.infineon.com/dgdl/Infineon-AN2018-35_EVAL-M1-IM818-A-UserManual-v01_04-EN.pdf?fileId=5546d462677d0f460167bb2df0121ab1
Infineon-AN2018-35_EVAL-M1-IM818-A-UserManual-v01_04-EN
PDF Document · 1.7 MB



> On Aug 7, 2024, at 7:30 AM, gene heskett <ghesk...@shentel.net> wrote:
> 
> 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.
> -- 
> "There are four boxes to be used in defense of liberty:
> soap, ballot, jury, and ammo. Please use in that order."
> -Ed Howdershelt (Author, 1940)
> If we desire respect for the law, we must first make the law respectable.
> - Louis D. Brandeis
> 
> 
> 
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