2:50 VL equiation is wrong. When the transistor is off, the voltage of the output is inductor voltage plus input voltage(VL+Vin). Besides that, when there is a diode -VD is also in the equation. That means Vo = VL+Vin-VD. However, it is VL=Vin-Vo for the off state. This is not possible.
@@sambenyaakov Dear proffesor, OK, I defintely see the diode voltage is neglected. This is fine. However, you wrote Vin-Vo in your presentation. I still don't understand this. What I mean is this is a boost converter. Vo is larger than Vin. Thus, VL must be negative. If you want to equalize Vin-Vo, it must be -VL without discussion. Otherwise, the equation will not match. Your explanation or graphics are for sure right but only the equations creates the confusion. Basic KVL loop will show you what I mean.
In this lecture at 6:23 polarity of the inductor is reversed ? In the initial case it was stated as same ? Is it in the initial case the polarity was + on side closest to V source and negative on the other side ? and then in second case when switch is open the polarity is reversed with negative on V supply side and positive on the switch side? I just want to clarify this sir - great lectures and thank you for these lectures.
Indeed. When the switch disconnects the inductor from ground, current indictor continues and it reacts as a source with a flipped polarity. Thanks for interest.
Thank you very much for the video, quite illustrating but got one question. During the off state, would not be a contradiction to say that Vo is greater than Vin if the diode is conducting? how can that happen simultaneously? Thank you very much in advance
@@sambenyaakov So does that mean that the current will still flow through the diode but the excess current will create a voltage drop in the load that will exceed the input voltage (due to the injected current released from the inductor) Thank you so much for your reply and again great videos! big fan!
@@CesarAngeles28 Yes. Think of the inductor as a current source which will adjust to any voltage across it and still inject the current which (as you said) will build up a high voltage at the output.
When an inductor in connected to a voltage source, the current rises as t*V/L so for a given t-on there will be some current increase. Only if the connection is sustained (t is very large) the current will be large i.e a short.
@@niteshsharma6015 Sorry I read wrongly your original question. The diode does not conduct because the voltage drop on Rds(on) is lower. This is why a transistor is used in the first place.
So i want to make a little battery charger. I need to boost from 5V to higher battery voltage and just force 200mA or so into it for a while, and terminate the charge when the battery stops gaining more voltage. Dicing around figures for the off the shelf boost converter was not satisfactory, a lot of dissipation on current limiting resistor depending on the battery charging voltage range. Nor could i come up with a compact and elegant solution to limit output voltage to somewhat above battery's, i didn't exactly want to add a comparator, too much complexity. But looking at this vid now, it seems to be the solution, that i can build just from parts i have at home, and very minimal. It appears the voltage ratio as shown is true for ideal coil and diode, and purely consuming load. But if a battery is present on the output instead, the voltage should get superimposed. Because surely the current off the energised coil has to go somewhere, and there isn't much anywhere else for it to go. So if i drive a little 2306 n-channel MOSFET with a fixed duty cycle from my microcontroller, i should get a vague semblance of a constant current source to charge the battery, and i can stop the operation at will too, so this is ideal. Am i thinking about right? Is there something i may be overlooking, a hazard of some kind? Like it's a steady state calculation, but what if the battery falls off the terminals, should i guard against the voltage shooting up sky high, and is using the highest value coil i can find detrimental in this regard?
@@sambenyaakov Well what i need it for is typical "9V" NiMH battery, the size like used in multimeters, which should usually be 7 cell series, but if i can use it across a larger voltage range from about 6 to 12 NiMH cells in series for various upcoming projects, that would be a nice bonus. The "9V" NiMH should bottom out at about 6.3V when empty, and terminate at about 10.5-11V when fully charged, so i shouldn't need a buck-boost. Funny thing, apparently you can't just buy a nice cheap fast charger for these little batteries, so i was inspired to investigate building one.
Thanks, I was trying to figure out why the input and output of these dc-dc convertors were equal, and it turns out as you said that the average voltage across the inductor must be equal to 0. If this wasn't the case then the inductor would saturate is it? and then it would equate to a short? Need to make some convertors for my thesis on DC microgrids and this was a very useful lecture. I hope you have videos on pid control of dc/dc convertors. I really appreciate this video thanks :)
sir, why the diode could conduction? the Vx always below then Vo I still can't understand why you said the imaginary resister why you can there is a resister
When the current s interrupted V=L dI/dt which generates a high voltage. The purpose of the imaginary inductor is to determine the polarity of voltage. In this case the polarity is such that the voltage is positive. It will increase until clamped by diode.
In general over voltage end over current. If over current is ruled out check layout as you may have high stray inductance the produces high voltages at switching. Try using a clamp to protect transistor.
Then where exactly the DC/DC converter is utilised in the EHV? especially since most of the MOTORs used are AC motors. perhaps, Boost the battery voltage before its rectified?
I don’t understand why people think this is a good video. His equations are wrong. @ 6:24 he has the equation VL = vin - vo then shows his plot to the right that shows VL is below the origin. VL is negative! His equation is wrong. This did nothing but confuse people. The equation should be -VL = vin - vo and then it would at least match your plot
As an engineer, and if I'm trying to teach someone something, my primary goal as a teacher is to SIMPLIFY the concept as best as possible and try my best to use practical examples, to make it easy to understand the actual subject. Sorry to say, that you are sounding too technical and going into unwanted territory unnecessarily, to explain such a simple device.
Hi Tony, Thank you for the input. It would help if you can indicate when it is "too technical and going into unwanted territory unnecessarily" by pointing to the minutes in the video. This is rather an old video, posted 6 years ago. It was watched by 88K viewers and got over 1K likes without a single dislike. So it seems that you in a minority of 1/88K. It would appear that we have different view of what constitutes a good technical explanations. What do think of ruclips.net/video/HtwiIIPekfs/видео.html and the more recent ruclips.net/video/-08rwDt5Wp8/видео.html ?
@@sambenyaakov For me, it was not difficult at all, but my son and his friends couldn't understand anything much, because they are IT guys. This is why I said that. But when I took a paper and explained it to them, using practical, everyday examples, to they got it. Teaching is an art and everybody doesn't get it easily. When I reach, I come down to the level of the student and begin explaining things in the simplest of methods, to bring the person up to a level, where he can grasp the more technical parts with clarity. I'm too busy to give you the time points from your video, so please excuse me. Thank you, anyway for the reply. Cheers.
Hi Tony, I got your point. My videos are inteded for people with basic EE background from the junior to the expert. My objectives are 1. Intuitive explanations with minimum equation. 2. Highlighting issues that many atime even expert miss them. I am getting many comments which seem to indicate that I am achieving these objective.
@@sambenyaakov I don't know what to say at this point. If my comment has touched a raw nerve, I'm very sorry. I'm a blunt person, when it comes to technical issues, usually. Maybe, the others who saw this particular video, looked at it from a casual angle. In my case, I got to look at in detail because these youngsters came back and told me they couldn't understand it well, although they do come from a science (physics) background, after which they have specialised in IT. But when I explained, with paper and then with the device on the table, they got it easily. That's all. Nothing serious and no malice intended. Thank you. Cheers.
You are explaining without using a pointer. Saying "here", "here" many times, but "here" means what? What is that position? How are we going to get what it means if you don't point at it. So annoying
You are a legend professor. Love your intuitive explanation behind things
Thanks. Comments like yours keep me going.
Very impressive lecture. Prof. Ben-Yaakov's lecture helps understanding some fundamental concepts better.
Thank you professor for the efforts. Your channel helps me a lot in understanding the fundamentals of electronics. All the best, Paul
Thanks
I'm an electronics engineer, learning from here a lot ... I learned about Buck now starting Boost
,Thank you
Hi Rami,
I appreciate your taking the time to write this comment.
Again great explanation. Detailed and clear.
2:50 VL equiation is wrong. When the transistor is off, the voltage of the output is inductor voltage plus input voltage(VL+Vin). Besides that, when there is a diode -VD is also in the equation. That means Vo = VL+Vin-VD. However, it is VL=Vin-Vo for the off state. This is not possible.
The inductor voltage (lower left circuit) in not Vin-Vout, diode voltage drop neglected and VL polarity as in the circuit on the upper left??????.
@@sambenyaakov Dear proffesor, OK, I defintely see the diode voltage is neglected. This is fine. However, you wrote Vin-Vo in your presentation. I still don't understand this. What I mean is this is a boost converter. Vo is larger than Vin. Thus, VL must be negative. If you want to equalize Vin-Vo, it must be -VL without discussion. Otherwise, the equation will not match. Your explanation or graphics are for sure right but only the equations creates the confusion. Basic KVL loop will show you what I mean.
Thanks Prof.
Thank you
Thanks for making this video. I learnt something from it.
Thanks for comment
In this lecture at 6:23 polarity of the inductor is reversed ? In the initial case it was stated as same ? Is it in the initial case the polarity was + on side closest to V source and negative on the other side ? and then in second case when switch is open the polarity is reversed with negative on V supply side and positive on the switch side? I just want to clarify this sir - great lectures and thank you for these lectures.
Indeed. When the switch disconnects the inductor from ground, current indictor continues and it reacts as a source with a flipped polarity. Thanks for interest.
Thank you very much for the video, quite illustrating but got one question. During the off state, would not be a contradiction to say that Vo is greater than Vin if the diode is conducting? how can that happen simultaneously?
Thank you very much in advance
The inductor had a positive voltage drop which is added to the input voltage
@@sambenyaakov So does that mean that the current will still flow through the diode but the excess current will create a voltage drop in the load that will exceed the input voltage (due to the injected current released from the inductor)
Thank you so much for your reply and again great videos! big fan!
@@CesarAngeles28 Yes. Think of the inductor as a current source which will adjust to any voltage across it and still inject the current which (as you said) will build up a high voltage at the output.
Nicely explaination, AC and DC just difference in polarity or direction
if source is AC and the output is DC the sistematic will difference
during ON condition, the inductor is directly connected to the DC supply. it will be short circuited.
When an inductor in connected to a voltage source, the current rises as t*V/L so for a given t-on there will be some current increase. Only if the connection is sustained (t is very large) the current will be large i.e a short.
that means, the switch must be open or closed during the transient period before the current reaches its steady state value.
That is correct
Fantastic, thank you again! :)
Thanks
Why is Vi-Vo constant during the off state? And therfore why is the discharge current linear ?
The output voltage ripple in neglected.
Hi, Thanks for your great explanation. May I get your notes?
Sorry, unavailable
You might get some screen shots
My question is : why diode Don't conduct..when switch is on ?
Because it a reverse voltage across it.
@@sambenyaakov you mean to say that diode is reversed biased ?
@@niteshsharma6015 Sorry I read wrongly your original question. The diode does not conduct because the voltage drop on Rds(on) is lower. This is why a transistor is used in the first place.
great lecture. thank you very much
😊
Merci monsieur
Mais comment je peut réaliser un circuit de convertisseur boost multi canaux à base des MOSFET
Sorry for not having the time go into details.
So i want to make a little battery charger. I need to boost from 5V to higher battery voltage and just force 200mA or so into it for a while, and terminate the charge when the battery stops gaining more voltage. Dicing around figures for the off the shelf boost converter was not satisfactory, a lot of dissipation on current limiting resistor depending on the battery charging voltage range. Nor could i come up with a compact and elegant solution to limit output voltage to somewhat above battery's, i didn't exactly want to add a comparator, too much complexity.
But looking at this vid now, it seems to be the solution, that i can build just from parts i have at home, and very minimal. It appears the voltage ratio as shown is true for ideal coil and diode, and purely consuming load. But if a battery is present on the output instead, the voltage should get superimposed. Because surely the current off the energised coil has to go somewhere, and there isn't much anywhere else for it to go. So if i drive a little 2306 n-channel MOSFET with a fixed duty cycle from my microcontroller, i should get a vague semblance of a constant current source to charge the battery, and i can stop the operation at will too, so this is ideal. Am i thinking about right? Is there something i may be overlooking, a hazard of some kind? Like it's a steady state calculation, but what if the battery falls off the terminals, should i guard against the voltage shooting up sky high, and is using the highest value coil i can find detrimental in this regard?
What is the range of battery voltage? You may need a Buck-Boost converter.
@@sambenyaakov Well what i need it for is typical "9V" NiMH battery, the size like used in multimeters, which should usually be 7 cell series, but if i can use it across a larger voltage range from about 6 to 12 NiMH cells in series for various upcoming projects, that would be a nice bonus. The "9V" NiMH should bottom out at about 6.3V when empty, and terminate at about 10.5-11V when fully charged, so i shouldn't need a buck-boost. Funny thing, apparently you can't just buy a nice cheap fast charger for these little batteries, so i was inspired to investigate building one.
Nicely Explained
#PowerElectronicsAPracticalApproach
Thanks, I was trying to figure out why the input and output of these dc-dc convertors were equal, and it turns out as you said that the average voltage across the inductor must be equal to 0. If this wasn't the case then the inductor would saturate is it? and then it would equate to a short?
Need to make some convertors for my thesis on DC microgrids and this was a very useful lecture. I hope you have videos on pid control of dc/dc convertors. I really appreciate this video thanks :)
👍
wonderful professor, thank you
😊
Thank you
👍
What is effect with higher inductance ?
Effect on what?
Thank you..amazing lectures...
Thanks for comment.
Thanks sir
👍🙏
very like it, thank a lot
Thanks
sir, why the diode could conduction? the Vx always below then Vo
I still can't understand why you said the imaginary resister
why you can there is a resister
When the current s interrupted V=L dI/dt which generates a high voltage. The purpose of the imaginary inductor is to determine the polarity of voltage. In this case the polarity is such that the voltage is positive. It will increase until clamped by diode.
nice effort sir super....thnk u.......
2:36 why in parallel? In series?
Look again. This is a parallel connection no matter how you draw it. The input voltage is imposed on the inductor's terminals.
@@sambenyaakov now i see it could be considered as series and parallel both. Thank you a lot for answering!
great lecture sir!
Thanks for comment.
Sir, May i ask about the calculations for Inductor, capacitor and Resistance and the input and output calculations???
This is too broad a question.
Have to 12 v to 24 v dc to DC boosting converter please give me circuit
Hello sir,
I would like to ask reasons of switching mosfet failure in boost converter. What can be probable reasons for mosfet burned out?
In general over voltage end over current. If over current is ruled out check layout as you may have high stray inductance the produces high voltages at switching. Try using a clamp to protect transistor.
Do Hybrid cars use this to boost the battery DC level enabling Toyota to use a smaller battery pack?
I am not sure, but I doubt it. If you mean hybrid cars, they can get by with smaller batteries since they are recharging them.
Then where exactly the DC/DC converter is utilised in the EHV? especially since most of the MOTORs used are AC motors. perhaps, Boost the battery voltage before its rectified?
Hybrid cars mostly used ip>op isolation
Bidirectional ↔ type converter
Plz tell about cuk and sepic or combined cuk and sepic
Some are already in my RUclips channel.
Where can I find RUclips video on this
Rapport cyclique it is faux vo/vin'=1/1-d
sir,,plz upload dc booster in high volt..input500 vdc to 700vdc..3000watts
Too specific
I don’t understand why people think this is a good video. His equations are wrong. @ 6:24 he has the equation VL = vin - vo then shows his plot to the right that shows VL is below the origin. VL is negative! His equation is wrong. This did nothing but confuse people. The equation should be -VL = vin - vo and then it would at least match your plot
Average powe on an inductor is zero so part of the cycle it is positive and part of the cycle is negative . Where is the problem?
Thankyou very much
As an engineer, and if I'm trying to teach someone something, my primary goal as a teacher is to SIMPLIFY the concept as best as possible and try my best to use practical examples, to make it easy to understand the actual subject. Sorry to say, that you are sounding too technical and going into unwanted territory unnecessarily, to explain such a simple device.
Hi Tony, Thank you for the input. It would help if you can indicate when it is "too technical and going into unwanted territory unnecessarily" by pointing to the minutes in the video. This is rather an old video, posted 6 years ago. It was watched by 88K viewers and got over 1K likes without a single dislike. So it seems that you in a minority of 1/88K. It would appear that we have different view of what constitutes a good technical explanations. What do think of ruclips.net/video/HtwiIIPekfs/видео.html and the more recent ruclips.net/video/-08rwDt5Wp8/видео.html ?
@@sambenyaakov For me, it was not difficult at all, but my son and his friends couldn't understand anything much, because they are IT guys. This is why I said that. But when I took a paper and explained it to them, using practical, everyday examples, to they got it. Teaching is an art and everybody doesn't get it easily. When I reach, I come down to the level of the student and begin explaining things in the simplest of methods, to bring the person up to a level, where he can grasp the more technical parts with clarity. I'm too busy to give you the time points from your video, so please excuse me. Thank you, anyway for the reply. Cheers.
Hi Tony, I got your point. My videos are inteded for people with basic EE background from the junior to the expert. My objectives are 1. Intuitive explanations with minimum equation. 2. Highlighting issues that many atime even expert miss them. I am getting many comments which seem to indicate that I am achieving these objective.
@@sambenyaakov I don't know what to say at this point. If my comment has touched a raw nerve, I'm very sorry. I'm a blunt person, when it comes to technical issues, usually. Maybe, the others who saw this particular video, looked at it from a casual angle. In my case, I got to look at in detail because these youngsters came back and told me they couldn't understand it well, although they do come from a science (physics) background, after which they have specialised in IT. But when I explained, with paper and then with the device on the table, they got it easily. That's all. Nothing serious and no malice intended. Thank you. Cheers.
4:05
???
@@sambenyaakov Sorry, I was keeping a reference for myself to come back to
@@ajhussein1347 👍
You are explaining without using a pointer. Saying "here", "here" many times, but "here" means what? What is that position? How are we going to get what it means if you don't point at it. So annoying
This particular video had a problem. Sorry.
Thanks Thanks Thanks 🙏👍
Thanks X 2