Lec 24 Understand Boost Converter in 25 Minutes
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- เผยแพร่เมื่อ 19 ก.ย. 2024
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Sir in the sum r is not giving how we calculate critical l
1:51 when switch will open, the inductor will change it's polarity. So, the polarity will be reversed (- +).
Sir, I have no words to appreciate Your efforts ...These lectures help us a lot In our Exams...Once again Thank U soo much Sir G.
Thank u soo much sir, i also watched your video on buck converters, your videos helped me understand things that i have been struggling with for the whole of 2022, now i feel like a guru😂. Because of your videos, am about to graduate with an Honors degree in electrical engineering 🤝🤝
Keep learning.. god bless you❤️
RESPECTED SIR, I APPRECIATE THE HARD WORK AND EFFORTS YOU HAVE GIVEN IN TEACHING POWER ELECTRONICS. BUT COULD YOU PLEASE TELL US WHEN THE VIDEO ON INVERTER WILL BE UPLOADED AS WE AS STUDENTS ARE QUITE DEPENDENT ON YOUR VIDEOS FOR POWER ELECTRONICS.
Yes sir we need videos on INVERTERS. Please upload as early as possible
Great Explanation Sir, Thank You
And you can play with this :-)
Good explanation sir. Easy to understand
your teaching way is very helpful for us and I waiting for gate previous year question on boost converter
wait for 15 minutes..its uploading..stay tuned..
Great explanation sir❤
Great teacher ❤
Thanks sir for the videos
When the switch is opened, current will be reduced as the impedance is higher. The magnetic field previously created will be destroyed to maintain the current towards the load. Thus the polarity will be reversed (means left side of inductor will be negative now). As a result, two sources will be in series causing a higher voltage to charge the capacitor through the diode D. This results in Vo = VL + Vs
But then how we are getting in off state
VL= Vs - Vo
It shd be
VL= Vo - Vs
YEs, I also think there is little mistake
Yes, logically you are correct. It is given in PS Bimbhra book also. But in one GATE question, the answer was that inductor average voltage is zero. Now if we use the inductor polarity as you have mentioned, average inductor voltage doesn't come zero. So where is the problem?
sohail sir plz reply to this question, i am also getting confused. you made a little mistake
Lovely work. Sir in which area are you doing research in power electronics?
Thank uhh very much sir
Prof. Can you please tell me how to calculate the voltage of diode in boost converter in on and off state.
In section time 2:24 the polarities of the Inductor will change and hence making the diode is forward biased, as per explanation the diode is reversed biased and can't be shorted.
Please explain.
Even I think inductor voltage polarity should change when it discharge when switch is off. In PS Bimbhra book also, polarity is shown to change. But if we change the polarity, average inductor voltage doesn't come out to be zero. What is the misconception actually?
Awesome awesome awesome...it was just awesome
most welcome..share jarur karna..all the best.
i seen ur video and after that i started watching all ur lectures.Good explaination of concepts and all nice videos.
This will help me to understand and recollect all the concepts.
Ok thank you very much for this useful explanation .....but please ,also I need to understand the efficiency of both BUCK and BOOST🙂🙂🙂🙂
Thank u sir for this lect
Superb Explanation
thank you sir
Whats the difference between chopper(step up/ down) and buck boost converter???
Step up is for increasing voltage above source voltage...step down is for decreasing voltage below source voltage...whereas buck boost is combination of both and can b used as step up as well as step down individually
Sir plzz make vedio lecture on Flyback converter as soon as possible....
Sir while calculating ripple in voltage, at on condition of switch, capacitor is supplying to load.. Is capacitor initially charged? Please reply if possible.
Sohail Sir , already said that all analysis is derived at steady state (timing of video = 4:38) not at just starting... okkk
Respected sir thanks a lot for uploading such wonderful video lecture.
I have a question:If during computing switching current are u neglecting ripples?
yes..i m assuming here that load current is constant and circuit is ideal..ok..
Sir! Why the inductor change it's polarities when switch is off?
To oppose the change in current
Because inductor doesn't allow the abrubt change of current, thus after closing switch it will act as a source, and source polarity is always -,+
At T(off), when inductor is behaving like a source then why it does not change its polarity...please reply sir..
polarity of voltage across is changing..u can see that..thats y VL you are getting negative when switch is off..i m taking convention same for Ton and Toff..got it..
GATEMATIC Education wrong
@@GATEMATICEducation what are you saying? Can't get it.
@@gouranganeog6035 by convention, VL is assumed.. If polarity changes, it is taken care of by mathematics.
@@jayeshpawade3135 then what is crct...I didn't understand it plzzz help me with it
hey sohail, in the transient operation of the circuit you have explained the behavior of inductor current. so the inductor current starts from zero and increases to a value, let's say it I. when switch is off, current decays to a non zero value. now again when the switch is closed, the current should be greater than I, as now we are starting from non zero value. why have you shown that it attains the same peak value I?
see we design any converter such that it can operate in stable region..right..so we adjust the switching frequency for steady state operation..if you will consider ur concept then at t= infinity the inductor currrent will be infifnite..which is the case of unstable circuit..got it..
excellent
During discharging inductor polarity will change...its vo-vs.....not vs-vo
Why have U used two different method to find ripple voltage in capacitor voltage in both buck and boost converter?can I find the same technique in both cases?
when the switch is open, does the inductor reverse its polarity or not?
best
Sir while calculating ripple in inductor current considering off conditon sign changes why?
At 13:28 can we use Source current Is= Il ... Inductor Current
Sir if u do videos on inverter total power electronics will be completed plzzzzzz try to make them...kind request
hi sir tq for wonderful video please help me with below.
when switch is ON cap will gives power to Load so the Ic = Io right ?? why it is Ic = -Io ??
sir, I had doubt that in diode current wave form @ switch off ,
diode current will flow from l(max) to I(min), but here u have took constant IL(avg) while drawing diode current...how we can take sir?..bcuz we are analysing only off condition? and also if u can say that @off Ic(avg) =0,so il(avg) =io(avg)..but if this is then @on (Ic=-io),so on taking avg, io equals to zero will come..which is wrong..so I am confused...pls help sir.
why in calculation of Id(rms)=IL*sqrt((1-D)T/T) .
I mean why outside the sqrt. is IL(avg).we can write that when the Id is having a constant value IL(avg) for (1-D)T time.In that case we can get rms value by this procedure.but Id is the current same as IL(off) which is ramp functn.Don't we need to integrate and divide then take sqrt. ?
sir during off period, why didnt you change the polarity of inductor?
Yes, inductor polarity should change logically. But if we change the polarity average inductor voltage doesn't become zero. Where is the problem actually?
Sir, thank you very much. I think I was not paying listening carefully.
Sir there is a difference in procedure for calculating peak to peak ripple volatge in buck and boost converter why?
Bhai 1 baat bta jb Ton hoga to Vo=0 hoga to us ke according Vo jo h wo Ton me zero show honi chahiye graph pr means wave form pr
Sir, most of the times, while calculating avg thyristor current, avg and rms source current, we are taking Io(conduction time/Total time period). But in boost converter, while calculating switch current, diode current u have taken, Inductor current(Conduction time/time period). Why sir?
nice
sir when will u upload lecture on inverter sir please upload soon ur lecture are awesome
Sir, During Ton, Ic=0. But while calculating Ampere sec balance, u have taken Ic= -IoDT.
during off state the capacitor voltage will be there across the load to prevent ripples. hence capacitor current will also be there.
what is buck , boost and buck- boost converter ?
what's it's use? what are it's applications?
see the introduction to chopper...u will understand..these all three converter are used to get variable DC voltage ...buck converter step down the voltage...boost converter step up the voltage and buck boost can step up or step down voltage depending up the duty cycle..ok
plz upload three input dc-dc converter operation for hybrid electric vehicles
three input dc-dc boost converter
Sir why you are not reverse the terminals of inductor while calculating voltage across inductor in off mode
the polarity u assume should be same for both on and off..
@@GATEMATICEducation why?
You find Ripple current when Turn on time We got a formula... And u said Ripple current find at turn off we will get a same formula... But sign will be getting opposite ... In turn on time u got the positive sign but by using turn off time formula i got begative sign...
Sir switch current and switch average current are same are different
can we expect leatures on inverters?
gud eve sir,sir why cant we change the polarity of inductor during OFF state?
the polarity of voltage across each and every element for full cycle should remain same then only the operation is done in a correct way...we know during off state voltage polarity will be reversed but if we will take same as previous then it will come out to be negative..so that is fyn..negative sugn itself says that the voltage polarity will be reversed..got it..
sir theoritically i got it but sir when it comes to practical analysis is there any possibility to keep the polarity of inductor same during charging and discharging mode?
The input voltage is 12 V, the output voltage is 24 V, the load is 48 W, and the maximum output voltage ripple is ±1%. The inductor Lin current ripple should not exceed ±5%. The inductor Lt current ripple and the capacitor Ct voltage ripple should not exceed ±20%. The switching frequency is 50 kHz. Find the inductor values, the capacitor values, and the transistor duty cycle. Assume that the switches are ideal.and sepic converter Ckt given.. Then how to solve?
Is the calculation of critical inductance right?
It does not match with the book(Power Electronics by Muhammad H.Rashid)
yes all are correct
Sir inverters plssssd cover it
please provide class note link please
Mentioned figure of Output voltage vs time is incorrect .
If correct further explaination is required .
Waiting eagerly for your answer
its correct..go through MH rashid or Bimbhra..you will get ur explanation..
the inducter current should have charged and discharged exponentially ..... but here you have assumed linear variation . why?
Generally the switching fre are in kHz...now u can guess why its linear..
@@GATEMATICEducation thanks sir lecture was really helpfull for me
when D=1 what is op voltage???
also sir I has one confusion in where to take avg and where not?
check it out lec 7(b) there i have explained when to take avg and when rms..
@@GATEMATICEducation when the switch is off the inductor should change its polarity right?...but you didn't
sir ,where is the lecture 25 in which problems have been solved
sohail sir lecture on inverter?
most probably next month..
sir i am dependent on these lectures please upload inverter sir i am requesting you
pls send pdf in PE bcus pdf save lots of tym. I'm passed out student .pls sir
just wait for 3 to 4 days..its under process..i will upload all the pdf..follow learn gate free group for the updates..
Hindi me vdo banate to jyada samajh aata.