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Hi Sven,
Yahoo went down here so this is a resend with an addition (the last paragraph).
Then what is the inductance of the transformer that the 720V charged capacitors look into? This determines the current. If the transformer goes into saturation its inductance almost vanishes making it look like a short to the capacitors. Also the load on any secondaries determines the inductance as such load sends a current in opposite direction around the legs to cancel out the current from the primary winding. This is real or 'Ohmic' load. Without load the current is reactive storing its energy as a magnetic field through the legs of the transformer (pure inductive load).
The bumps in the transformer could be because of saturation. Saturation is determined by the total current level (its saturation level) around a leg. This also determines the energy level for saturation. Thus how fast saturation happens is determined by the capacitor value. As the capacitor has to store the amount of energy to bring the transformer into saturation its voltage must increase if the capacitor decreases in value for being able to store the same amount of energy. The discharge then happens faster as this is given by the LC constant. Less capacitance C means faster discharge if the inductance L is held constant. L drops off when saturation happens. This means even faster discharge of any remaining energy in the capacitor.
A series inductor (choke) can absorb some of the energy and thus limit the current. Else the voltage has to be decreased and the capacitance of the capacitor increased. This gives a slower discharge though. If there isn't time for that the transformer is too small for the purpose or the energy level is just too big.
The stored energy in the capacitors if both have 720V across them is 0.5 X 60uF X (720V)^2 = 16J. The transformer has to be able to store this energy without saturating unless it is meant to saturate. If knowing the inductance of the transformer without load the current around the leg can be calculated for storing this amount of energy. If the current exceeds the saturation level it will make the transformer go into Ferro resonance. The current around the leg is squareroot(2E/L) where E is the energy (stored in the capacitors being discharged (16J) and L is the inductance of the winding. The current in a single turn is found by dividing the total current by the number of turns making up that winding. If the core is saturated the current will be even bigger than the calculated value because of the decrease in inductance. In a normal transformer design this is unwanted as the inductance almost vanishes making the winding act like a short.
Regards
Ole
---In [email protected], <s.friedrich@...> wrote :
Hi Ole, I have not measured the current that flows when unloading. 2 x 30uF capacitors are discharged with approx. 720 volts into a 6.3 kva transformer. The bumps in the transformer sound like an old diesel engine. Hector also used 180 amp diodes why he does not care. I think the reactors in buckplug should receive the split resonance. The coupling capacitors reflect the energy on the other side or a 180 degree shift. It is difficult to find the right dimensions. Greetings Sven
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-------- Ursprüngliche Nachricht --------
Von: "onielsen@... [EVGRAY]"
Datum:21.01.2018 22:38 (GMT+01:00)
An: [email protected]
Betreff: Re: AW: Re: AW: [EVGRAY] Re: buck plug ...
Hi Sven,
What kind of current and voltage are we talking about? Was it to dampen or smooth out too fast discharges from some capacitors?
Regards
Ole
---In [email protected], <s.friedrich@...> wrote :
Hello Ole, Warren, the big heatsinks with the stronger diodes work wonderfully, the coolers are lukewarm. The 350 amp module does not feel warm at all. now I still need suitable reactor inductors. Someone an idea? Greetings Sven
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