Excellent, thank you!! I went from barely understanding how they work to full basic understanding of inductors and buck converters, something the papers and other explanations I found online was not able to do. I love the graphical explanation with the graphs, I'm a bit of a visible learner so they helped a lot.
@@sambenyaakov No one in a world can help me with this question 😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏 Why we don't consider Miller effect in Common Gate configuration when we study high frequency response
This is such an underrated video teaching how buck circuit works. Trust me, study this video is better off than watching 10 other videos on buck converters that only touch surface level understanding! 🌟
Amazing explanation and very simple to understand...t Your students must do wonders ! If this sort of explanation... there no enough words to describe your approach and command in the topic
Cannot thank you enough for this video. I pay thousand's of dollars to my university and didn't get a single concept from my professor with 1 week of lectures. And here you are explaining it in 30 mins and now I have crystal clear concepts about buck converter. I m very grateful to you for this video. Thank you
You are really great, and the way you teach is fantastic. thanks a lot, sir. My suggestion is if it is, not compulsory, please add some model or simulation for the practical point of view. well done sir.
Thanks. There are many videos in my TH-cam channel with simulation in connection with a Buck converter. In TH-cam search window look for : sam ben-yaakov Buck
Very clear explanations. Thanks a lot. Before watching this video, I had very little understanding of the Buck converter. But I have a much better understanding now.
Very nicely explained, Prof Ben-Yaakov. A request to please arrange your english videos in the form of a playlist (similar to the hebrew ones) so that we can view more of them! Thanks again, clearly explained.
Thank you Prof. Ben-Yaakov for the excellent lectures. I have been following them and they have been very helpful. Just a question on this lecture: at 3:17 when you showed the real LR circuit. You explained that the decay of current will be exponential. Shouldn't the rise of the current be also exponential? In your diagram however the current has a constant rising slope. Is there anything I am missing here? Thanks
Good point. You have a keen eye. I should have explained it better. When the battery is connected and duration is short you are at the beginning of exponential which is approximately to straight line. The discharge section should have been longer. Thanks for pointing this out.
Just loved the explanation 🙏... Thanks for educating the world 🙏👏 Only the sudden introduction of DCM is little confusing to me... I couldn't understand how for the same circuit both CCM and DCM concept came.. please clarify
So helpful! I'm taking my own notes when going through this lesson I'm sure when explaining that he is waving and circling on his drawing pad and there is no indicator on the screen to show which part he is talking about 😂
Hi professor, first, thank You for sharing very educational video where we can get back to basics analysis of a electrical circuits. Probably one of the best video I could find so far of this subject and since You have many more of them I decide to subscribe on the channel. But I would like to ask some question which troubles me. Maybe this comment section is not quite appropriate for it, but still I will ask since it's troubles me and I would like to have better understanding. It is about constant voltage that is often shown across inductor in fast switching circuits like buck convertor. In my head I have a physical picture that as soon as inductor is connected on constant voltage source it will start to build magnetic filed and the voltage over inductor will exponentially dropped almost to 0 value. Of course, during short period of time this drop won't be huge, but still voltage is not quite constant during that short switching period. My problem is fit this picture into the model of fast switching circuit where the voltage over inductor is presented as constant voltage which imply linear current rise or drop. Could you please help me understand why we can assume constant voltage on inductor in a short periods of switching?
HI Marko, thanks for interest. An inductor does not control the voltage across it when connected to a voltage source. It will comply with any voltage and adjust dI/dt. (similarly, a capacitor will not control the current when connected to a current source). You can write to me at sby@bgu.ac.il
I had also thought about this.Usually why Vl is a constant is prolly because the Vo is treated as constant.The possible explanation I can think is that 1)once the circuit reaches steady state , the capacitor say charges to some voltage.The natural time constants involved like RC,LC etc are very large compared to switching time.This implies that capacitor discharge is very less over a switching cycle , hence Vo is almost constant 2)In terms of frequency domain, the Inductor passes lower order harmonic currents(as its impedance is jWL, it offers more impedance to higher harmonics) and the higher harmonics among these lower harmonics itself are mostly absorbed by The filter capacitor as they absorb higher order harmonics easily.Hence the current which passes to the Resistor load is almost dc current (and hence almost constant) due 'a kind of 2-stage' filtering of higher harmonic currents by inductor followed by capacitor.
Dear sir your lectures are excellent.I would like to listen all lectures regarding converter design,but I couldn’t find lectures in sequence,kindly suggest me Where can found all these lectures in sequence. Thanks a lot sir .
@@sambenyaakov I will definitely subscribe. I just have one question, which is related to converters in the discontinuous conduction mode. Would you ever want to operate a converter in DCM? Does it give any advantage for certain applications? Thanks, best regards
Please confirm that the load in the final example ( 31.23 min) is 5V/10A= 0.5 ohm. Time constant is L/R 1.2microH /0.5 ohm= 2.4 micro Seconds to 63% of peak. So frequency should be about 300kHz ( not 100kHZ) with a duty cycle of 0.5. Only correct if you ignore the capacitor. Where have I gone wrong or am I completely wrong
Dear Professor, you have an amazing talent to teach these topics. It would be great to follow your lectures in an order. Is there such order or link or are these random uploads? I saw your lectures in Hebrew are in order but couldnt figure out the English ones. And one question: in simulations a linear voltage regulator at the output of a dc dc converter or smps attenuates the differential switching noise a lot. Is this way also used in practice? I tried LC cascaded by a linear voltage regulator at the output of a noisy ripple voltage source and was amazed with he result. Would be glad if you have comment on it.
Two reasons: vout is approximately constant voltage and the resistance of L is low so the time constant is very long and we are are just at the beginning of the exponential curve
WHAT IS the effect of entilrely removing the inductor ....the rule of vout = d* vin is still completely valid even if no inductor exist ??? or just the inductor ensures large amounts of current supply ??
@@sambenyaakov dr as if you stating that as long as we run our converter in the continous mode we wont need regulation (ices and feedback networkds) as it can be done by the inductor(with proper design ) ?
Hello Sir, I have question In the Buck convertor , If Inductor value is more than calculated value what will be impact on Vout? For example: calculated value is 68uH and I will use 300uH Inductor.
In general, there is not much change as far as voltage regulation. But: on the pro side: ripple will be lower. On the con: slew rate will be slower (the ability to ramp up the inductor current in case of large load change)
Sir do you have a video about Full bridge DC-DC converter ?, or could you please recommend to me a book with a lot of exercises about Power electronic ( AC-DC, DC-DC, DC-AC) thank you for the videos
@@sambenyaakov its the end of the semester. The kids have other things on their mind and didn't really share my excitement about why the current through the inductor doesn't have an exponential slope. That's ok, I'm excited enough for all of us about V/L forcing dI/dt and vice versa.
Dear Professor, can this be another explanation of negative spike? This is before the buck converter explanation. " since inductor current can't change instantaneously. So current will continue to flow and keep on sucking charge out of the right side of the switch node. So the switch node will see a negative spike."
But sir, as you explained , assume some imaginary resistor, which is actually not present. So I think current can't flow. Then I don't understand it intuitively. Mathematically since di/dt = towards minus infinity so Voltage across the inductor will also be towards minus infinity.
The mathematical explanation is is V= L dI/dt/ The intuitive is indeed the the inductor's " attempt to keep the current flowing. I was somewhat uneasy by "sucking charge" since it has the connotation of just charge moving while we need to look for the current loop.
Hi, Professor I have a question. The waveform of current of an inductor at the second slice, it should be pass through origin. Why doesn't it pass through? In the beginning, iL(0) = 0? It that wrong? Thank you so much!
Thank you for your feedback. ^ ^ I knew that it's the steady -state result. The current camp of an inductor would act like that, when converter goes to steady- state response with a fixed switching frequency. So, if we talk about at very beginning, which means I turn on a dc source at the first time, the current of an inductor should rise from zero amp, and after the system goes to the steady state(after several cycles), We just see the common waveform(general case) in the slide. Is that right? Thank you so much.
Dear Sir when the switch is Open, the charges from output capacitor wont flow through Load,R? If it does then how does the output voltage remain constant as charges stored on capacitor are that from ripple? I am confused about capaitor role
I am not sure I followed you question. The role of the output capacitor is to keep attenuate the ripple so the load sees a DC voltage with some ripple on it. In a CCM Buck the inductor current always flows to the output. Does this answer your question?
Thank you sir for your response. When switch S is opened, the inductor tries to maintain the current and therefore provides current to the load. The same way charges are also stored on capacitor, will the capacitor try to discharge itself through load aswell? what happens to the charges on capacitor? Because I watched a video where it was stated that the capacitor grounds the high ripple current, the term grounding was quite ambiguos to me. Can you please shed light on what happens to capacitor as regarded to the charges on it?
At steady state, the average inductor current is equal to the load current while the inductor ripple current is passing through the low impedance of the capacitor which is "shorting" it. This is why the ripple voltage is low.
I spent only 4 minutes and I can already feel that it is going to be a fantastic lecture
🙂👍
I am a former student of yours, graduated in 2008. It's great to see your lectures on youtube, just as good as they were back in the day.
👍😊
Thank you Prof for explaining so beautifully, mind blowing clarity
😊👍🙏
Your tip on how to resolve for the polarity of the inductor voltage is pure genius.
🙏👍
It is rare to find someone clearly explaining what happens in a somewhat complex circuit.
Thanks for sharing.
Thanks for comment.
Excellent, thank you!!
I went from barely understanding how they work to full basic understanding of inductors and buck converters, something the papers and other explanations I found online was not able to do. I love the graphical explanation with the graphs, I'm a bit of a visible learner so they helped a lot.
Thanks for comment
@@sambenyaakov
No one in a world can help me with this question
😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏
Why we don't consider Miller effect in Common Gate configuration when we study high frequency response
This is such an underrated video teaching how buck circuit works. Trust me, study this video is better off than watching 10 other videos on buck converters that only touch surface level understanding! 🌟
Wow, thanks!
Amazing explanation and very simple to understand...t
Your students must do wonders ! If this sort of explanation... there no enough words to describe your approach and command in the topic
Thanks for warm words. I highly appreciate it.
The best explanation on this subject I have ever seen on youtube. Thank you very much.
Thank you. I took this course with an extremely incompetent "professor" at Cairo university. There's a world of difference here. Thank you again.
😊I missed you in class while I was teaching at the university.
I have been searching for the derived equations for hours... Thanks Sam..
Now it is so simple...
😊🙏
Cannot thank you enough for this video. I pay thousand's of dollars to my university and didn't get a single concept from my professor with 1 week of lectures. And here you are explaining it in 30 mins and now I have crystal clear concepts about buck converter. I m very grateful to you for this video. Thank you
Thanks for kind words. Comments Iike yours keep me going.
You are really great, and the way you teach is fantastic. thanks a lot, sir. My suggestion is if it is, not compulsory, please add some model or simulation for the practical point of view. well done sir.
Thanks. There are many videos in my TH-cam channel with simulation in connection with a Buck converter. In TH-cam search window look for : sam ben-yaakov Buck
What a difference it makes learning from an educator who really understands the subject!!! Thank you Sir.
🙏😊
From undergrad to grad school to working as an fae, your videos have been really helpful. Thank you :)
Wow. Thanks for sharing.🙏😊
Very clear explanations. Thanks a lot. Before watching this video, I had very little understanding of the Buck converter. But I have a much better understanding now.
Thanks. Commits like yours keep me going.
Truly amazing presentation. Thanks.
Thanks
Excellent, easy to understand, Thank you , Professor !
🙏🤔
The best explanation I saw about buck converters. Great lecture!
Thanks.
by far the clearest explanation i have encountered in video or text . it is very much appreciated
Thanks for comment.
This is an excellent lecture. It cleared up some confusion I had. Thank you.
Thanks
i have been going thru power electronics material since 2018. this is best I have seen
Thanks for taking the time to write the comment.
Very clear explanation .. Great job
Thank you!
Well, I've got my M.Sc. In EE already, but still... WOW! The explanation was amazing!
😊🙏
One of the best explanations ever
Thanks.
Excellent lecture. Thank you for your time in clearing out many doubts surrounding inductor voltage/current behavior.
Thanks for note
Very nicely explained, Prof Ben-Yaakov. A request to please arrange your english videos in the form of a playlist (similar to the hebrew ones) so that we can view more of them! Thanks again, clearly explained.
Noted
Excellent analysis for Inductor and Buck Converter. Thank you!
Thanks
Thank you for the content. very insightful and Intuitive.
Thanks.
The world's best teacher thanks sir
Wow! 😊Thanks
Thank you Prof. I have learn a lot from this tutorial video.
Thanks
You are an amazing professor. God bless you
Thanks for comment.
Excellent explanation, you are a very good professor.
Thanks
Thanks you, sir. Detail and systematic explaination.
Thanks
Thank you Prof. Ben-Yaakov for the excellent lectures. I have been following them and they have been very helpful. Just a question on this lecture: at 3:17 when you showed the real LR circuit. You explained that the decay of current will be exponential. Shouldn't the rise of the current be also exponential? In your diagram however the current has a constant rising slope. Is there anything I am missing here? Thanks
Good point. You have a keen eye. I should have explained it better. When the battery is connected and duration is short you are at the beginning of exponential which is approximately to straight line. The discharge section should have been longer. Thanks for pointing this out.
Thank you very much for explaining this point👍🙂.
Thank you Professor Ben-Yaakov! Very clear explanation and video.
Thanks for comment.
Thank you very much professor. This is a great video and has helped me to understand my textbook a lot better.
Thanks.
The Best Explained.Thank you, Sir.
Thanks
Big thank you from Thessaloniki Greece!
Thanks. AS you might know, we Israelis, have a worm spot in our heart for Thessaloniki
Excellent explanation. I love your lectures.
Many thanks!
Dear Sir, Excellent, thank you :)
Thanks
I am an amateur exploring electronics.
I could understand how it worked..
Thanks.
Thnks for taking the time to write the note, I am happy to learn that the video helped.
Just loved the explanation 🙏... Thanks for educating the world 🙏👏
Only the sudden introduction of DCM is little confusing to me... I couldn't understand how for the same circuit both CCM and DCM concept came.. please clarify
Please indicate the relevant minute in video
So helpful! I'm taking my own notes when going through this lesson I'm sure when explaining that he is waving and circling on his drawing pad and there is no indicator on the screen to show which part he is talking about 😂
Sorry, in this particular video synchronization to pointer/pen is corrupted.
Great class, that was a great explanation! Thank you.
Thanks.
Very nice video sir but my doubt is, what should be the critical (max) value of the load resistance for CCM operation?
The answer is here th-cam.com/video/HtwiIIPekfs/w-d-xo.html
Thanks, you are an excelent teacher !!!!
Thank you! 😃
Hi professor, first, thank You for sharing very educational video where we can get back to basics analysis of a electrical circuits.
Probably one of the best video I could find so far of this subject and since You have many more of them I decide to subscribe on the channel.
But I would like to ask some question which troubles me. Maybe this comment section is not quite appropriate for it, but still I will ask since it's troubles me and I would like to have better understanding.
It is about constant voltage that is often shown across inductor in fast switching circuits like buck convertor.
In my head I have a physical picture that as soon as inductor is connected on constant voltage source it will start to build magnetic filed and the voltage over inductor will exponentially dropped almost to 0 value. Of course, during short period of time this drop won't be huge, but still voltage is not quite constant during that short switching period.
My problem is fit this picture into the model of fast switching circuit where the voltage over inductor is presented as constant voltage which imply linear current rise or drop.
Could you please help me understand why we can assume constant voltage on inductor in a short periods of switching?
HI Marko, thanks for interest. An inductor does not control the voltage across it when connected to a voltage source. It will comply with any voltage and adjust dI/dt. (similarly, a capacitor will not control the current when connected to a current source). You can write to me at sby@bgu.ac.il
I had also thought about this.Usually why Vl is a constant is prolly because the Vo is treated as constant.The possible explanation I can think is that
1)once the circuit reaches steady state , the capacitor say charges to some voltage.The natural time constants involved like RC,LC etc are very large compared to switching time.This implies that capacitor discharge is very less over a switching cycle , hence Vo is almost constant
2)In terms of frequency domain, the Inductor passes lower order harmonic currents(as its impedance is jWL, it offers more impedance to higher harmonics) and the higher harmonics among these lower harmonics itself are mostly absorbed by The filter capacitor as they absorb higher order harmonics easily.Hence the current which passes to the Resistor load is almost dc current (and hence almost constant) due 'a kind of 2-stage' filtering of higher harmonic currents by inductor followed by capacitor.
Very basic and important concept with well concise explanation!!
Thanks for comment.
wonderful job, professor
Thanks
thanks again very well explained and well worth the time
Thankls
Best video on buck converters that I have ever see. Thanks
Thanks for comment
Dear sir your lectures are excellent.I would like to listen all lectures regarding converter design,but I couldn’t find lectures in sequence,kindly suggest me Where can found all these lectures in sequence.
Thanks a lot sir .
Sorry, they are not arranged. You can search in the TH-cam search box for a given subject e.g. " Sam ben yaakov control"
Hello Sir, really great video. Thank you very much!!
👍😉
Very clear explanation, thank you.
Thanks for comment
شكرا يا سام بن يعقوب
תודה
amazing lecture. great insight, I learn a lot from you, thank you
Thanks for kind note.
Thanks for the explanation. I wish you would have covered the DCM, just as you did in the buck converter.
Good point. Thanks.
@@sambenyaakov I will definitely subscribe. I just have one question, which is related to converters in the discontinuous conduction mode.
Would you ever want to operate a converter in DCM? Does it give any advantage for certain applications?
Thanks, best regards
There are advantage running Flyback at DCM . See www.linkedin.com/feed/update/urn:li:activity:6671448579661529088
Hlo sir,good explanation.
I have a doubt why we are drawing inductor (i min to max)and capacitor (-ve side onwards)charging waveforms
It a convention; in Germany they do it differently
Extremely good explanations. Thank you.
THANKS
So educative and future life skill
Thanks
Please confirm that the load in the final example ( 31.23 min) is 5V/10A= 0.5 ohm. Time constant is L/R 1.2microH /0.5 ohm= 2.4 micro Seconds to 63% of peak. So frequency should be about 300kHz ( not 100kHZ) with a duty cycle of 0.5. Only correct if you ignore the capacitor. Where have I gone wrong or am I completely wrong
WE assume a large cap so the output is like a battery of 5V.
excellent fucken video! thanks prof!!
awesome tutorial! If at the end you can talk about the ripples at the capacitors and what are the contributing factors that would complete everything
Iדמ,א ןא גןבודדקג שא th-cam.com/video/HtwiIIPekfs/w-d-xo.html
very helpful. thanks
Thanks
Dear Mr Yaakov The video is very clear and helpful thanks a lot , but I have a question do you have more videos ?
Look for Sam Ben-Yaakov channel on TH-cam
Would you recommend the art of electronics? Or do you use some other bibliography
This is a good book more to the advanced engineer. I find application notes and reference designs of of companies useful.
Thank you very much sir
👍🙏
wonderful explanation Liked + subsribed! thx ❤
🙏👍
Dear Professor, you have an amazing talent to teach these topics. It would be great to follow your lectures in an order. Is there such order or link or are these random uploads? I saw your lectures in Hebrew are in order but couldnt figure out the English ones. And one question: in simulations a linear voltage regulator at the output of a dc dc converter or smps attenuates the differential switching noise a lot. Is this way also used in practice? I tried LC cascaded by a linear voltage regulator at the output of a noisy ripple voltage source and was amazed with he result. Would be glad if you have comment on it.
Thanks. See index at www.advicepoweracademy.com/video-tutorials
Yes, LDOs are used successfully in low power applications.
Dear professor,
Have you ever made a video about buck converter in discontious conduction mode?
Splendid
👍🙏
Thank you Sir!
You are welcome!
Thank you for great video, do u have a website where i can quickly refer notes in video?
professor do you mean the the DC current goes to the resistor and the AC goes to the capacitor?
Yes. No DC via capacitor. Resistor will have some AC current depending on ripple voltage.
Why current across inductor increases linearly instead of being exponential with time? Is it because of constant Vout?
Two reasons: vout is approximately constant voltage and the resistance of L is low so the time constant is very long and we are are just at the beginning of the exponential curve
Hi, i have problems understanding which minimum value of capacitor i have to use, thanks and excelent video.
Voltage ripple and loss (due to ESR) are the limiting factor
very nicely explaination
Thanks
Interesting and very well explained.
Thanks!
Thank you sir
Thnks
Great
🙏👍
WHAT IS the effect of entilrely removing the inductor ....the rule of vout = d* vin is still completely valid even if no inductor exist ??? or just the inductor ensures large amounts of current supply ??
Without the inductor there will be no regulation The output cap will be charged to input voltage
@@sambenyaakov dr as if you stating that as long as we run our converter in the continous mode we wont need regulation (ices and feedback networkds) as it can be done by the inductor(with proper design ) ?
Hello Sir, I have question
In the Buck convertor , If Inductor value is more than calculated value what will be impact on Vout? For example: calculated value is 68uH and I will use 300uH Inductor.
In general, there is not much change as far as voltage regulation. But: on the pro side: ripple will be lower. On the con: slew rate will be slower (the ability to ramp up the inductor current in case of large load change)
@@sambenyaakov hello Sir, thank you very much for answer, sorry for the late reply.
thankyou...
Thanks
thank you ~~ greet from China
Thanks.
Thank you for these courses
Thanks. If not a memeber please join www.linkedin.com/groups/13606756/
Thank you so much, sir.please make a video on capacitor behavior as well.
Hi Prince, see my channel for capacitor related videos such as switched capacitor converters and voltage dependent capacitors. .
Sam Ben-Yaakov thank you sir
Excellent lecture
Thanks.
Thank you
You're welcome, Thanks
Sir do you have a video about Full bridge DC-DC converter ?,
or could you please recommend to me a book with a lot of exercises about Power electronic ( AC-DC, DC-DC, DC-AC)
thank you for the videos
Thanks for comment. There are good power electronics books. I am not sure which include exercises.
Power electronics by D. Hart, Fundamentals of Power electronics by R.Erickson, Elements of Power electronics by P.T.Krein
I used the voltage forcing current equation in a presentation. I felt good about it. The audience was a little lost.
Why? Thanks for sharing.
@@sambenyaakov its the end of the semester. The kids have other things on their mind and didn't really share my excitement about why the current through the inductor doesn't have an exponential slope. That's ok, I'm excited enough for all of us about V/L forcing dI/dt and vice versa.
@@intheshell35ify 😊😊👍👍
Dear Professor, can this be another explanation of negative spike? This is before the buck converter explanation. " since inductor current can't change instantaneously. So current will continue to flow and keep on sucking charge out of the right side of the switch node. So the switch node will see a negative spike."
Hi Abhishek, please send me the time coordinate (minute) of the location you are referring to.
Sir, I was referring to time around 9:00 minutes.
"keep on sucking charge out of the right side" Current can flow only exist in a closed loop.
But sir, as you explained , assume some imaginary resistor, which is actually not present. So I think current can't flow. Then I don't understand it intuitively. Mathematically since di/dt = towards minus infinity so Voltage across the inductor will also be towards minus infinity.
The mathematical explanation is is V= L dI/dt/ The intuitive is indeed the the inductor's " attempt to keep the current flowing. I was somewhat uneasy by "sucking charge" since it has the connotation of just charge moving while we need to look for the current loop.
I love you professor
😊 If not a memeber please join www.linkedin.com/groups/13606756/
Very good lecture
😊
Hi, Professor
I have a question.
The waveform of current of an inductor at the second slice, it should be pass through origin.
Why doesn't it pass through?
In the beginning, iL(0) = 0?
It that wrong?
Thank you so much!
THis is a general case (as you will find in a PWM converter) in which the initial inductor current is not zero.
Thank you for your feedback. ^ ^
I knew that it's the steady -state result. The current camp of an inductor would act like that,
when converter goes to steady- state response with a fixed switching frequency.
So, if we talk about at very beginning, which means I turn on a dc source at the first time, the current of an inductor should rise from zero amp, and after the system goes to the steady state(after several cycles),
We just see the common waveform(general case) in the slide.
Is that right?
Thank you so much.
Thank you! That was great.
Thanks for comment.
Can we give input to buck converter through a battery
Sure
@@sambenyaakov thank you sir i want to give input from a 150ah battery, is it safe sir
thanks
👍
Great explanation! Keep on! :-)
Thanks for comment
Dear Sir when the switch is Open, the charges from output capacitor wont flow through Load,R? If it does then how does the output voltage remain constant as charges stored on capacitor are that from ripple? I am confused about capaitor role
I am not sure I followed you question. The role of the output capacitor is to keep attenuate the ripple so the load sees a DC voltage with some ripple on it. In a CCM Buck the inductor current always flows to the output. Does this answer your question?
Thank you sir for your response. When switch S is opened, the inductor tries to maintain the current and therefore provides current to the load. The same way charges are also stored on capacitor, will the capacitor try to discharge itself through load aswell? what happens to the charges on capacitor? Because I watched a video where it was stated that the capacitor grounds the high ripple current, the term grounding was quite ambiguos to me. Can you please shed light on what happens to capacitor as regarded to the charges on it?
At steady state, the average inductor current is equal to the load current while the inductor ripple current is passing through the low impedance of the capacitor which is "shorting" it. This is why the ripple voltage is low.
Thank You sir :)