Deciphering multiphase interleaved PWM converters with coupled inductors

  Рет қаралды 7,964

Sam Ben-Yaakov

Sam Ben-Yaakov

Күн бұрын

Пікірлер: 59
@Txepetxcc
@Txepetxcc 8 күн бұрын
in 14:40 I believe there is a mistake . The M
@ats89117
@ats89117 2 жыл бұрын
Interesting video. I've seen this done with a toroidal coupled inductor with out of phase components, and the main benefit was touted to be the improved performance of the magnetics because of the lack of DC bias. That seemed like a reasonable explanation but would be interested in hearing your thoughts...
@sambenyaakov
@sambenyaakov 2 жыл бұрын
At high frequency, core losses are increasing and they are associated to the AC current ripple component. Thanks for Note.
@SR-ml4dn
@SR-ml4dn 2 жыл бұрын
Thanks for the video. No one can explain it so intuitive as prof. Ben-yaakov. All most every buck converter circuits I've seen utilize the load is on "0 V". Is it possible to operate with symmetrical supply voltage and push-pull configuration so the load is kind of floating. I was thinking of a use case where the rail supply is under +/- 50 V measured to ground and measured over the load it will be 100 V max. For safety reasons where people could some how touch only one of the output at the time and therefore exposed to max 50 V.
@sambenyaakov
@sambenyaakov 2 жыл бұрын
Thanks for note. Yes I guess it can be used in a double bridge configuration.
@benm4784
@benm4784 2 жыл бұрын
Professor, thank you for another great video! I'm not sure I follow the takeaway that opposite polarity is preferred. If low output ripple is a desirable trait, what is the cost of same-polarity coupling? Size to reduce copper losses? Would either approach be preferable to reduce AC core losses?
@sambenyaakov
@sambenyaakov 2 жыл бұрын
Thanks for note. High frequency current ripple in winding will cause excessive conduction losses. Core losses will probably be lower with the revers coupling configuration.
@isacdaimary9908
@isacdaimary9908 Жыл бұрын
Thank you Professor, But at 05:31, I see two variables Ksimpe and K: which is confusing for me. Can you please look into this or just share the LTspice simulation file?
@sambenyaakov
@sambenyaakov Жыл бұрын
Ksimple is 1
@isacdaimary9908
@isacdaimary9908 Жыл бұрын
​@@sambenyaakovand the value for "k"? Am in confusion how a single coupled Inductor can have two "k" (coupling coefficient)
@sambenyaakov
@sambenyaakov Жыл бұрын
@@isacdaimary9908 OK. The circuit represents a coupled inductor with coupling coefficient K which is different from 1. The equivalent circuit includes a transformer with K=1 and inductors. In this circuit K is a parameter (.PARM is not shown) so the equivalent circuit can be adjusted to any original (total) K.
@小鹿666
@小鹿666 4 ай бұрын
2 windings in 1 core @@isacdaimary9908
@solenskinerable
@solenskinerable 2 жыл бұрын
thanks for the video! correct me if im wrong, but i think the lack of dc bias would allow the use of powder cores with a very high permeability per turn for their size, as long as the linkage is v high - while uncoupled or loosely coupled cores would have to use a larger distributed air gap to not lose permeability by the ampereturns of dc flux. ltspice doesnt seem able to couple inductors with nonlinear flux expressions, but maybe it could be done with dependant current sources? 🤔
@sambenyaakov
@sambenyaakov 2 жыл бұрын
The lack of DC bias is only in some of the legs. Powder cores have a higher core losses at high frequency.
@solenskinerable
@solenskinerable 2 жыл бұрын
@@sambenyaakov thank you, i didnt know that! ive mostly done old-timey linear stuff as of yet
@chaiyonglim
@chaiyonglim 2 жыл бұрын
Hi professor, can you comment what is the benefits of using coupled inductors for multiphase buck? most published papers claimed it is good for power density (size), able to use smaller L and C, better transient performance. But as you showed ripple current is actually higher and we would need larger L and Cout to have the same inductor current ripple per phase and output capacitor rms current compared to non-coupled inductors case?
@sambenyaakov
@sambenyaakov 2 жыл бұрын
I think the saving, if exsists, is due the fact that you have one magnetic element, albeit larger.
@RaedMohsen
@RaedMohsen 8 ай бұрын
Great video. I tried to duplicate the simuation model of the two phase coupled inductor buck and I always have an average difference between the current of the two coupled inductors. In other words, they are not sharing equally. Yours seem to be sharing nicely. I am not sure if there is something that I missed in the simulation.
@sambenyaakov
@sambenyaakov 8 ай бұрын
Try to reduce the minimum step size. If it does not work
@sambenyaakov
@sambenyaakov 8 ай бұрын
Send me the .asc to sam.benyaakov@gmail.com
@RaedMohsen
@RaedMohsen 8 ай бұрын
@@sambenyaakov Thanks, reducing the step size certainly helped. I thought I have to try some sort of initital conditions for the inductor currents or the cap voltage to avoid initial difference between the two inductors current.
@AhmedAli-op6ng
@AhmedAli-op6ng 2 жыл бұрын
What is the difference between coupled inductor and transformer? Both have windings around the same core so what makes them different?
@sambenyaakov
@sambenyaakov 2 жыл бұрын
See kzbin.info/www/bejne/p33SiYKip9aNebc
@madaboy121
@madaboy121 2 жыл бұрын
Hello, thank you for the very good explanation. Could you please tell me what are the values of {L} and {R} used in the simulation?
@sambenyaakov
@sambenyaakov 2 жыл бұрын
Thanks Robert
@rahulpj32
@rahulpj32 2 жыл бұрын
Sir , In the video 5:49 how L/K^2 is coming actually it is Lm = KL.
@sambenyaakov
@sambenyaakov 2 жыл бұрын
See slide 22 in kzbin.info/www/bejne/iobJkoFqZblkj8k
@rahulpj32
@rahulpj32 2 жыл бұрын
@@sambenyaakov thanks sir
@AhmedAli-op6ng
@AhmedAli-op6ng Жыл бұрын
Professor i really like your intuitive way of teaching. I also want to enter the realm of power electronics. But i am not sure how do i go with it. Like how do i realize which field of power electronics will be the topic of research in next decade. I am clueless at the moment. I would really like if you could give a generic input for enthusiastic newbies in this field.
@sambenyaakov
@sambenyaakov Жыл бұрын
"I also want to enter the realm of power electronics." as what? engineer? academician?
@AhmedAli-op6ng
@AhmedAli-op6ng Жыл бұрын
@@sambenyaakov i am electrical engineer. I do have basic knowledge of power electronics and want to go in advance field. I did my masters related to ems in multi-microgrid but i also have interest in power electronics i also took took UCL bolder courses on power electronics by Dr. Erikson but i am blank about my direction from there. I need your advice. Thank you for your time.
@sambenyaakov
@sambenyaakov Жыл бұрын
@@AhmedAli-op6ng I think that power eletronics related to automotive application (power train, chargers, energy storage, relevant magnetics, and the like) are going to be in great demand
@AhmedAli-op6ng
@AhmedAli-op6ng Жыл бұрын
@@sambenyaakov thank you for the reply.
@azharajiel8500
@azharajiel8500 2 жыл бұрын
thanks so much for this explain, I want to design more mutual inductors for multi convrtetrs in one core in planar inductor to reduce the size of pcb. can I make that? and How, please?
@sambenyaakov
@sambenyaakov 2 жыл бұрын
kzbin.info/www/bejne/oKubeKiGa6-opsU
@LydellAaron
@LydellAaron 2 жыл бұрын
Are you familiar with the dynamics associated with synthetic inductors made from op amps? These synthetic inductors do not have magnetic cores. Interested in using these in a project and came across your lectures.
@sambenyaakov
@sambenyaakov 2 жыл бұрын
Are you going to use them in a power circuit?
@LydellAaron
@LydellAaron 2 жыл бұрын
@@sambenyaakov I really appreciate your reply. It is a signal circuit, no power handling. Carrying sinusoidal signals.
@sambenyaakov
@sambenyaakov 2 жыл бұрын
@@LydellAaron OK. This emulation is possible. The question is why do you need it? From outside it looks like an inductor. So why not used an inductor? One place it could be useful is to mimic an inductor for reconstructing the inductor current. But for that you do not need an Op Amp. So how are you going to use tis emulation?
@LydellAaron
@LydellAaron 2 жыл бұрын
@@sambenyaakov thank you for your insight, I am looking to investigate how the synthetic inductor will perform when inserted as L in an LC oscillator at a tuned design frequency. I heard they can achieve high Q factor and high precision. The specific LC oscillator I have identified, would be a Clapp oscillator design at single design frequency.
@小鹿666
@小鹿666 4 ай бұрын
what's different of TLVR
@sambenyaakov
@sambenyaakov 4 ай бұрын
Basically the same . Have you seen : kzbin.info/www/bejne/e2LOqXVsZs6BkJY
@pietroferlita
@pietroferlita Жыл бұрын
Prof. I think the model of the coupled inductors, used as in slide 11, is wrong. You can clearly see that the self-inductances on each side leaving the other open are not L. Same if you want to compute the mutual inductance, you don't get kL. The correct one is with L in place of L/k² and k:1 instead of k²:1
@sambenyaakov
@sambenyaakov Жыл бұрын
Hi, you are absolutely correct . I have no idea how this error occured and how come nobody noticed it since it was published a year ago. I am planning to redo the video. Would it be OK if I add a thank you to you? If you wish you can send me your full name for that. BTW the transformation was supposed to have been based on kzbin.info/www/bejne/iobJkoFqZblkj8k
@RaedMohsen
@RaedMohsen 8 ай бұрын
@@sambenyaakov is there an already corrected version of this video? I was trying to derive this model but in vain until I realized that the self inductance is not the same looking into both terminals. Also, how do we go form the T model you showed to the model with leakage inductance on only one side. Is it a star to delta transformation?
@sambenyaakov
@sambenyaakov 8 ай бұрын
@@RaedMohsen Sorry I donot follow you. Which minute in video or slide number are you referring to?
@RaedMohsen
@RaedMohsen 8 ай бұрын
@@sambenyaakov I am referring to slide 11 in which you had a model for coupled inductors with two inductances only... leakage and magnetizing.
@sambenyaakov
@sambenyaakov 8 ай бұрын
​@@RaedMohsen Sorry, never got around to it. I will try to put it again on top of the todo file. Please let me know when you see this. Afterward I will unlist this video.
@hamidk4772
@hamidk4772 2 жыл бұрын
Outstanding Job.
@sambenyaakov
@sambenyaakov 2 жыл бұрын
Thanks
@petegaslondon
@petegaslondon 2 жыл бұрын
Hi Prof nice to see you back (ah you've been quite prolific, i just didnt subscribe - lets fix that) interesting - I thought that the inverse coupling being better was kinda obvious, but didnt expect the rise in circulating current - interesting set of tradeoffs :) Also, pretty relevant now everybodys squeezing more power phases into high performance PC motherboards!
@sambenyaakov
@sambenyaakov 2 жыл бұрын
Welcome back! You have a lot to catch😊Thanks for sharing your thoughts.
@tamaseduard5145
@tamaseduard5145 2 жыл бұрын
👍🙏💖😎
@sambenyaakov
@sambenyaakov 2 жыл бұрын
👍Thanks Aidan for sharing your thoughts. I have published this approach many years ago and for some reason it did not catch. As the saying goes: You Can Lead a Horse To Water, But You Can't Make It Drink.
@justpaulo
@justpaulo 2 жыл бұрын
I think the relevant video you mention at 5:40 is this one: kzbin.info/www/bejne/iobJkoFqZblkj8k I actually have difficulty in understanding the mathematical transformation of the coupled inductor to the particular model you are using.
@sambenyaakov
@sambenyaakov 2 жыл бұрын
Thanks for correcting the slip. I will add a note on this in the description section. This is just a simple tranformation. Perhas the difficulty is due to not taking into account the fact that the turns ratio of ideal transformer includes K?
@justpaulo
@justpaulo 2 жыл бұрын
@@sambenyaakov Professor, one thing that is confusing me is the following: in the schematic of slide #11 shouldn't the ideal transformer ratio be given by sqrt( L3/L10 ) = k ? If so, why are the drawings showing 1: k² ? And if the transformer ratio is 1:k , then shouldn't the magnetization inductance be L instead of L/k² ? Also, as it is, when I try to calculate the voltages at both sides of the transformer I don't get what I expected: V₁,₂ = L·di₁,₂/dt + M·di₂,₁ /dt where M = k·L Something seems not to be consistent or I'm doing some mistake... In fact, if the transformer ratio is 1:k and the magnetization inductance is L, things become consistent and I do get V₁,₂ = L·di₁,₂/dt + M·di₂,₁ /dt .
@johngreen1060
@johngreen1060 2 жыл бұрын
You can also try a more general "pi circuit" model, which can work with multiple coupled inductors: en.wikipedia.org/wiki/Inductance#%CF%80-circuit This doesn't change your findings, I just think you will find the model itself interesting.
@sambenyaakov
@sambenyaakov 2 жыл бұрын
I am not sure I follow. Can you please send me a sketch of the configuration you have in mind to sby@bgu.ac.il ?
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