At the 3rd step of the PSFB cycle, the current at the secondary is not shared equally by the two windings that it would be if it were a full bridge. Does anyone can explain why it is not the case ? And why only the variation of the two output current is shared by the two windings ? I am not sure that the sign of the following pictures are correct. It should be Iout - deltaIout/2 and - deltaIout/2 as no energy is being transfered. I probably do an error ?
The same appears from this diagram :
This one seems to be better for the same architecture :
Here are the links :
https://www.ti.com/lit/ds/symlink/ucc28950.pdf?ts=1642537671521&ref_url=https%253A%252F%252Fwww.ti.com%252Fproduct%252FUCC28950 https://www.ti.com/lit/ds/symlink/ucc2895.pdf?ts=1642523455070&ref_url=https%253A%252F%252Fwww.ti.com%252Fproduct%252FUCC2895 https://training.ti.com/how-design-multi-kw-dcdc-converters-electric-vehicles-evs-how-psfb-works
Have a nice day ,
------------------------------------E D I T --------------------------------
Well I have not still find the answer but by documenting my self I found something well weird :
Here is the simulation that "A concerned citizen" have done. (Thank you). It is a bit modified according to my needs and we clearly see a negative current :
Then I was looking something else and I found this comment on the video (linked below) ...
It seems to be pretty complicated to avoid a negative current with SR ... And I think it is right that it can do a voltage spike as there is no way for the current to go ... There is nothing to limit the rise of voltage when the current is negative...
I am probably missing something...
Here is the last simulation :
Version 4
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SYMBOL diode 688 288 M180
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TEXT -528 -40 Left 2 !.model sw sw ron=50m roff=0.1g vt=0.5 vh=-0.5 ; level=2 oneway ilimit=1k\n.model d d ron=50m roff=0.1g vfwd=0.45 vrev=1k epsilon=50m revepsilon=1
TEXT 952 872 Left 2 !.tran 3m
TEXT -520 -280 Left 2 ;https://www.ti.com/lit/ds/symlink/ucc28950.pdf?ts=1642537671521&ref_url=https%253A%252F%252Fwww.ti.com%252Fproduct%252FUCC28950\nhttps://www.ti.com/lit/ds/symlink/ucc2895.pdf?ts=1642523455070&ref_url=https%253A%252F%252Fwww.ti.com%252Fproduct%252FUCC2895\nhttps://training.ti.com/how-design-multi-kw-dcdc-converters-electric-vehicles-evs-how-psfb-works
TEXT -768 224 Left 2 ;tau = Output RC time constant
TEXT -776 256 Left 2 ;td = Output propagation delay
TEXT -528 -176 Left 2 !.param Fsw = 100k DELAB = 100n DELCD = 100n D=0.8 Lp = 1 Ns = 1 Np = 21 Ls = Lp*Ns**2/Np**2
TEXT -688 920 Left 2 ;Manque la simulation de DELAF et DELBE, utile pour diminuer les pertes
TEXT 928 768 Left 2 !K L1 L2 L3 1
TEXT 720 888 Left 2 ;2u
TEXT 2352 656 Left 2 !K2 L8 L9 L10 1
TEXT 2128 888 Left 2 ;2u
LINE Normal 192 1072 192 320 2
Have a nice day,