😊

Please indicate the minute of video you are referring to.

@@sambenyaakov Dear professor Yaakov, I am sorry I couldn’t get back to you sooner and I feel a bit ashamed for questioning you on a small item in your fly back video especially with the epic job you have done so far. Any way using S1 and S2 is ok with me when explaining power supplies but on this occasion I would like to replace S2 ,as just a contact switch , with a diode instead in order to make it easier for me to explain. Let’s place a diode in the secondary coil circuit in such a way that the cathode faces the coil in the same direction of the current in your example. This way when primary is energised the magnetic field resulted in the secondary will induce a current flowing in the same direction which will result in a voltage in phase with input voltage, but the way I see it you are using 240v which produce peak voltage of 320v in order to produce 5 or 12 volts the other side (with higher current of course) which is not very efficient method .Where as if you change direction of diode and move the dot on the secondary to opposite side of that of primary , then when the primary is on the diode is reverse biased and energy builds up in the coils and when S1 is open magic happens and the magnetic field of B flips and reacts with secondary coil surface area of A which will cause change of flux per time which in tern causes an EMf of negative voltage. B.A= d(flux)/dt=- EMfv (Faradays law). That aside, may I say that your videos on Switch mode power supplies not only good but head and shoulder above the rest and I thank you for it and wish you long life and happiness. PS, I hope to god my boss wont see my comment admiring a lecturer working for a university which is named after a degenerate mass murderer.

Glad it was helpful!

😊Thanks for comment

Thanks for comment

Thanks😊

Thanks for comment.

👍

Thank you for your comment and for the time to write it. Feedbacks like your keeps me going.

Best of luck! Beat the machine: lnkd.in/efPuimEe

Thanks

Thanks. Glad to hear that.

Thanks

Thank you for comment and for taking the time to write it.

Thanks

👍🙏🙂

You're very welcome!

👍😊

Thnks. I appreciate it.

🙏😊

Thanks for comment

Thanks.

😊🙏

Thanks

Good point. I am not sure how to that effectively. Any suggestion? Or help😊

@@sambenyaakov here is a video on how to create a KZfaq playlist. kzfaq.info/get/bejne/q710mbGqz5ebh2Q.html Thanks again for the great lectures.

One playlist to all video? How will this help? One can just go to my channel

@@sambenyaakov one playlist for power electronics or one playlist for flyback, DCM mode flyback and ccm mode flyback.

@@sameqy All that you need to do: go to KZfaq , in the search box type "sam ben yaakov flyback" Walla

Thanks😊

Thanks.

😊🙏

Thanks.

Thanks for comment 😊

Thanks😊

Thanks India from Israel😊

👍😊🙏

Thanks and greetings to our prime minister now visiting Ukraine

Yes, You are correct. The bottom line is in error. But then, I could have said that I did it on purpose to see who is really paying attention to the contents of the slides (-: Thank you for pointing this out.I hope people will see your comment.

😊

شكرًا. من دواعي سروري

Thanks for taking the time to comment. Keeps me going.

Thanks

Thanks

😊

🙂🙏👍

Will try to prepare one. Thanks for comment.

Thanks

Thanks

This is why the symbol of an inductor is a spring😊Yes, this is one intuitive way of visualization. Thanks for sharing.

Thank you for comment and for taking the time to write it.

🙏🙂

Which minute in video are you referring to?

Yes it does but due to gap the H is lower . Search my KZfaq channel for Flyback and gap. e.g. in KZfaq window " sam ben-yaakov gap"

@@sambenyaakov Thank you Professor.

Thanks,

👍🙏

The total reflected voltage is related to the flux (there only one core flux) and the voltage of each winding ( either forced or induced) is n*d(phi)/dt

You are correct! Thanks for pointing this out. But, it could represent a negative current..Just kidding.

Only just watched the video. His solid arrows are correct. His handwritten arrow at the bottom is in the wrong direction so you are not wrong. After two years I am sure you will have figured it out but it may help someone else.

The circuit that I show(with a diode) is generally referred to as Buck-Boost, The non inverting version requires 4 transistors, also called Buck-Boost. Google buck-boost and see. This could be of interst? kzfaq.info/get/bejne/sN6oeKty1ZzOcpc.html

@@sambenyaakov Thank you sir.

The current IS smaller

Thanks. Please search my KZfaq channel for 'control' or 'feedback' there are a number of videos on that.

What matters is the inductance. You can have different turne ratio with same inductance. But there is another issue: interwinding capacitance that can be reflected from secondary. In this case, high number of secondary turns will result in higher capacitance,

When the switch turns off the current in the secondary flow same direction with respect to dot so the secondary diode is conducting. When the switch turns on again, the voltage of the secondary is same polarity as the primary with respect to dots so the voltage imposed on the secondary diode is in reverse polarity and it will turn off.

@@sambenyaakov thank you so much for the lesson Professor

This is because the voltage on diode is in the reverse polarity.

👍Thanks

Spark gap at output? Please clarify.

WE do need, for the leakage inductance.

Forward is generally better at higher power levels, Flyback has the advantage of multiple outputs without inductors.

In this ideal representation there should be no delay. With a delay there will be a voltage spike. In real circuits as shown later there is a diode that captures the current.

Hi cxc# from Japan. Thank you for comment. Good topic. I will try to prepares a tutorial on that.

Yes. Very effectively

@@sambenyaakov thank you very much

This means that they are not galvanically connected

Thanks for comment. About "flyback dc/ ac in discontinuous mode " perhaps in the future.

The futur could be next mouth ?

I doubt but I may surprise myself:)

You apparently have high parasitic oscillation. Try to rewind the transformer for lower leakage add snubbers. Using a controller that goes into burst mode will help.

currently i am operating it in open loop with 555 timer to switch the mosfet, i have used RCD snubbers already, i think i have to use a controller IC with feedback mechanism.

👍

👍

Thank you for comment. DCM, may be in the future.

Thanks.

A transformer by definition conducts at input and output at the same time and does not store energy.

@@sambenyaakov thank you!

You got me there. I do not remember any document that treats the ill effects of reverse recovery per se. Perhaps it is a good subject to present in KZfaq video :-)

I will appreciate if you also teach that on KZfaq ;-)

Hello Siamak! Chapter 3 of Solid State Pulse Circuits by David Bell has lots of helpful explanations on this subjects.

Yes, and more over when the switch is off adangerous voltage may develop. This should be taken care of by the control. .

See kzfaq.info/get/bejne/rc2pg62fndqyc5s.html

The operatic will depend on the turn ration of coupled inductor which affects the output voltage when the transistor is ON.

😊

Flux=L*nI 😊

How close was your body or hand to speaker?

Here they are😃😃 kzfaq.info/get/bejne/osd4gZeJ05udnH0.html kzfaq.info/get/bejne/r7lzYLan1LfFY4k.html I recommend you scan my KZfaq channel for othe videos See kzfaq.info/get/bejne/pK5lg5pyyuCrh2Q.html

See kzfaq.info/get/bejne/rc2pg62fndqyc5s.html

Can you please point to the slide number (or minute in video ) you are referring to?

it's the back EMF of the coil

Thanks

Duty cycle depends on turns ratio which sets the maximum voltage on the switch. Start with specifying the latter.

Thanks for comment. Will try. Enjoy Israel (again?)

Yes, again. I've been twice and I love Israel. I going to come back, hope soon.

Thanks

But the coupled inductors are connected across the voltage sources.

@@sambenyaakov Even if the voltage sources are constant, the voltage across the inductor is not constant, as the current rises. I don`t understand why the current derivative is considered constant.

@@gearstil Even if the voltage sources are constant, the voltage across the inductor is not constant??? The voltage across the inductors are the input output voltages the tipples.

@@sambenyaakov Yes, that is true, I had the impression that there is a resistor load in series. Thank you!

Okay, I might have figured it. Is it because of magnetizing inductance across the primary winding?

Yes, you can look at this way or, there is a continuity of magnetic field (not current as sometime assumed in error) so this result in secondary current also flowing to dot as in primary.

To which minute in video are you referring to?

@@sambenyaakov I think he/she is referring to minute 13:15 in which you said that there would be a voltage across the switch S due to the turn ratio between the primary and secondary coil

Have you seen: kzfaq.info/get/bejne/nKxdnal3tL23iWg.html There are other relevant videos in my KZfaq channel.

Ok sir

What is the required voltage and power? are you interested in a hand held machine?

About 30kv, 45 miliamp, and yes, its a handhold operated machine.

kzfaq.info/get/bejne/admUpZaJ1Z_Jomg.html

@@parametrikaarquitectos960 30kV is bit high for a flyback stage. You might need a second, perhaps spark or resonant based stage. Look up circuits that are used for igniting Xenon lamps.

Not so. Following the transition, the core is determining the direction of current . If before the switching the current was flowing into the dot so will be after the transition.

Thanks

Will try.

Depends on construction: single or distributed air gap (s in powder core)

HI, not sure that I follow your question. Please clarify. (Is it part of an earlier exchange?)

Hi thanks for your response. I was referring to the beginning of the video, forgot to mention it. There you said that there is no energy stored in the core of a transformer, but the field of the input remains the same as in the coupled inductor with the secondary not connected. Except that the primary field of the transformer is opposed by the field from the secondary according to lenz' law. I think there's still energy stored as 2 repelling fields, but not usable? Was also wondering this. Would it be possible to reroute the opposing field caused by the secondary to a third coil instead to the primary,,, which 3rd coil is wound opposite to the secondary, which has a diode to prohibit a current from the primary flux, but not from the lenz flux from the secondary? Using a small dutycycle current on the primary. I'm new to power electronic, am trying to learn some of it on my own. Many thanks for the video's!

Regarding the first point, this video may clear thinks up to you. kzfaq.info/get/bejne/kMyAf5qFr5jag6c.html If still in doubt, write to sby@bgu.ac.il AS for the second, not sure what you mean.

Hi Andre, thanks for sharing your doubts. At first, when seeing that you brought some references I thought you are bringing in Faraday or Maxwell (at least) , but then it turned out that these are internet postings😊. As it turns out, they are OK but I am afraid that you missed the point that my video is about COUPLED INDUCTORS and not a TRANSFORMERS. There is a major difference between them. In a transformer, at any given instance: Pin=Pout. Not so in a coupled inductors (in general, although it may function as a transformer too). Please watch again the video which explains the workings of a coupled inductor in connection with the Flyback converter. There are also other relevant videos in my KZfaq channel "sam ben yaakov".