Hi Prof. Sam, I just want to show my appreciation for all these riddles and challenges you post. I have been learning a lot through it. Thank you very much! For sure my favorite Power Electronics channel. 👏👏👏

@sambenyaakov

24 күн бұрын

Thank or kind note

I did get the right answers using magnetic circuit analysis as well as by power balance when assuming the 3 currents are identical. But, one thing that bothered me at first was that such analysis "contridics" basic three-winding ideal transformer models including those used in many simulation platforms such as PLECS and PSIM (there isn't any contradiction - see further on for an explantion). For a 3-winding transformer, you get V1=V2=V3 and the currents at eact side are determined by the loads with the source current determined by power balance. Transformer model: V1:V2:V3 = N1:N2:N3 and magnetic circuit model: I1:I2:I3 = N1:N2:N3 So why is there a contradiction? The answer is simple: The configurations of the magnetic circuits assumed for this case (all 3 MMF are in parallel therefore must be equal if N1=N2=N3, or, the total MMF is the average of the three). The assumed configuration for the transformer is all 3 windings are on the same leg -> series connection of 3 MMFs. Parallel MMF connection -> electric loads are in series (post reflections). Serries MMF connection -> electric loads are in parallel (post reflections).

@sambenyaakov

Ай бұрын

These models are of the basic coupled inductors assembly, assuming that the same flux is passing via each winding within a defined coupling coefficient K.

@arthurm7846

25 күн бұрын

Hi man, I was confused by the analogy with the 3 winding transformers and reading your comment helped a lot, thanks!

Simply the most smart professor on this planet! 🎉❤

@sambenyaakov

Ай бұрын

Wow. Let's not be carried away. Thanks . Comments like yours keep me going😊

Great Video and explanation. Thank you for taking time.

@sambenyaakov

Ай бұрын

Thanks

👍🙏❤️

@sambenyaakov

Ай бұрын

👍🙏😊

Sir please average model of peak current mode control active clamp forward converter.. please

@sambenyaakov

Ай бұрын

Perhaps😊Thanks for interest.

Thanks for the explanation sir. At 6:53 , If the right side winding is opened instead of shorting, then as the center, left and right are in series, no current shall flow in center and left winding. Is it correct sir? In short, if R1 is opened instead of shorted, what are the answers to the above 3 questions.

@electrowizard2000

Ай бұрын

Let me take a stab before Prof gets to it.. based on the presented logic it initially looks like you are correct. 2. R1=inf I=0 because I=10/(inf+R2) =0A 3. V2 = 0×4 = 0V Like you I don't really like this answer. How could a wire breaking on one side prevent any power input entirely. I think something else happens in this case. Prof neglected reluctance in normal circuits, which was the underlying assumption that I1=I2=I. I think it cannot be neglected in this case. MMF B2=0AT The right arm will flux saturate, producing some voltage V1 depending on the ferrite. Some current

@sambenyaakov

Ай бұрын

Not really. See my answer video.

@sambenyaakov

Ай бұрын

The open winding voltage will be 10V and the loaded leg will pass very little current,

Sir, what does the meaning of your statement in the end of this video, "With the B=1mT, this is really core that can sustain a much high power?" Does that mean that the lower/smaller the B is the higher the voltage we can feed to it?

@sambenyaakov

Ай бұрын

The current is limited by the wire cross section, the voltage by B . 1mT is very low so the voltage can be increased much.

@StrsAmbrg

Ай бұрын

@@sambenyaakov interesting Prof. Btw, do you have video explaining how or what is the limiting flux magnet/density?

@sambenyaakov

Ай бұрын

@@StrsAmbrg See kzread.info/dash/bejne/aIGdmrN9dJqXpqg.html

Sir, if the right side is shorted and left is opened, then V2 will be 10 V. How is this possible? Please explain.

@sambenyaakov

Ай бұрын

The flux does not enter the leg with shorted windings so you just have a regular transformer

@jitendra09616

Ай бұрын

@@sambenyaakov That means the flux will get opposition from the flux generated by shorted winding. Is it correct sir?

@sambenyaakov

Ай бұрын

@@jitendra09616 No flux and no current in the short! amazing😊.

At 7:04 I've got confused. Why is V1+V2=10V? In an ideal transformer with 1:1:1 ratio I would say that V1=V2=10V. Now this ferrite does not make an ideal transformer but is it really that far from it to completely change the spirit of the equation? I would expect a small voltage drop when the winding gets loaded but both should still be close to 10V.

@electrowizard2000

Ай бұрын

My understanding is that based on the magnetic circuit, the three leg MMFs are equal, MMF=nI so currents are equal (I think) If input power is 10I, output power cannot be 2*10I, the power must be divided. Only way to do that is divide voltage, because V1*I + V2*I = 10*I

@ebarbie5016

Ай бұрын

This is not a transformer configuration! See my explanation above...

@sambenyaakov

Ай бұрын

Indeed this is the first stage for answers.

@sambenyaakov

Ай бұрын

Equal MMFs makes equal current and by conservation of energy V1 I+ V2 I= 10 I