I am a bit confused about the current flowing in the magnetizing inductance L_m when the switch is off.

I am reading a book(Erickson & Maksimovic, Fundamentals of Power Electronics, 2020), explaining the working principle of isolated SEPIC. It is mentioned that the current in the secondary side will be (I_Lm + I_1) / n, however, this will imply I_1 is all flowing into the coupled winding on the primary side and no current is divided by the branch where magnetizing inductance lies. But if so far I've understood properly, why there will be no current flowing in L_m in this phase?

Can someone help explain it to me? Thanks in advance.

Here did I post the topologies of the converter and its current waveform which I referenced from the book.

The topology of an isolated SEPIC converter enter image description here

The current flow when the switch is off (I indicated my understanding with markers) enter image description here

Current waveforms on the primary and secondary side enter image description here

The book text where I am having a question enter image description here

  • \$\begingroup\$ Probably a typo and they meant Ic1, instead of I1. \$\endgroup\$ – a concerned citizen Apr 25 at 15:24
  • \$\begingroup\$ The flyback current from L1C1 = +i1 and the conduction Lm=L2 = i2 perhaps for primary although a transformer is optional for Sepic and could be simply L2 but L2 is much smaller than an ideal transformer \$\endgroup\$ – Tony Stewart EE75 Apr 25 at 16:51
  • \$\begingroup\$ @aconcernedcitizen thanks for your comment! Now I also had this problem with the current why they can simply take as I_1 instead of I_C1 now...Also, considering it's a DC source, why capacitor C_1 is not blocking DC current but allowing current to flow into the primary winding? My confusion adds up xD \$\endgroup\$ – Martineo Pang Apr 25 at 20:24
  • \$\begingroup\$ @TonyStewartEE75 thank you, Tony! I think you were mentioning something important that I overlooked. As you said "L2 is much smaller than an ideal transformer", however, given that L2 is small enough, should it take more current from I_p when it comes to the current divider in a parallel circuit? 😳 Or it's simply neglected in this case, if that's what you meant? \$\endgroup\$ – Martineo Pang Apr 25 at 20:38
  • \$\begingroup\$ @MartineoPang There is no DC source except the source, itself. Everything else is switching, therefore C1 will let current pass through. It's actually part of the topology that makes it a SEPIC converter. \$\endgroup\$ – a concerned citizen Apr 25 at 22:12

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