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ur most welcome..if possible can you share ur reviews about this video in learn gate free group...link is given below facebook.com/groups/learnGATEfree/
what ever sohali sir explain is correct .but if u want more clarification do one thing ...during turn of condition the inductor changes its polarity. yeah ur absolutely correct . change inductor polarity and instead of applying the volt sec balance apply the energy conservation principle. chaging energy equal to discharging energy (assuming inductor is completely discharged what ever energy it charged. as energy =v * i * t (Vs-Vo) * (Imin+Imax)/2 * (DT) = (Vo) * (Imin+Imax)/2 *(1-D)T after solving u may get Vo=D Vs ....this may applicable to other choppers also.
@@smrautomations2168 you are justifying that whatever sir had written is right.. sir had written VL=-Vo during off state and in ur explanation u have written +Vo during energy release i.e. during off state u have considered Vo instead of -Vo( as sir had written -Vo during off) How does it mean ur justifying sir's explanation? Please answer
point no. 9 , while calculating switch current, you hv assumed Io=Il . that is true for average values but not for the instantaneous values. the increasing current flows through switch and decreasing current flows through FWD. so while calculating avg current through switch, the +slope portion of inductor current should be taken in consideration, and not the avg load current.
+rajat saxena u r right..i have taken avg value..not instantaneous value..instantaneous waveform will be different..now if u will take instantaneous value also to find avg switch current..u will get same answer..no prblm..u can approach like that..
23:22 sir I have a question ⁉️ we know ripple is peak to peak. than here ripple voltage should be 2∆V ? because capacitor voltage flactuate between +∆V to -∆V why it is ∆V but not 2∆V??
at 6:23, during switch off condition, inductor discharges stored current. at that instant voltage polarity of inductor get changes but here in the video, you kept the same polarity for inductor during discharge, why? is there any mistake my point?
what ever sohali sir explain is correct .but if u want more clarification do one thing ...during turn of condition the inductor changes its polarity. yeah ur absolutely correct . change inductor polarity and instead of applying the volt sec balance apply the energy conservation principle. chaging energy equal to discharging energy (assuming inductor is completely discharged what ever energy it charged. as energy =v * i * t (Vs-Vo) * (Imin+Imax)/2 * (DT) = (Vo) * (Imin+Imax)/2 *(1-D)T after solving u may get Vo=D Vs ....this may applicable to other choppers also.
yeah sir, When inductor discharges during off condition of switch, It reverses its polarity and then discharges. But in the video, you did not change the polarity. Reason? At 6:23
for all time the polarity of voltage across inductor is assumed to be same...that does not mean inductor voltage is not reversing...it is reversing thats why u are getting negative voltage during off..
In transients we studied that when the supply is connected to inductor then it charges exponentially by equation Is(1-exp(-t/Tau)) but here u mentioned that the inductor will.chsrge linearly? Why is there this contradiction? Plz explain
transient equation is valid only for finite time constant. and when inductor is connected with a ideal current source so time constant= L/R when you will remove ideal current source in order to calculate R you will get R=infinite and hence time constant=0 hence transient equation is not applicable. And this becomes linear charging of inductor case. If you got it then fine otherwise apply laplace transform to the circuit you will get the same result. if then also u cant understand. TAB TOH BHAGWAN HI BACAHYE AAPKO.
For practicle purpose linear approximation is good enough, in reality it not only is exponential but also has noise and ringing due to thermal noise and inductor resistance
If avg current across capacitor is zero then how capacitor is charged during turn on time?? 2. If capacitor current is zero then how we use Ic= I inductor- I load in wave form .
Dear Sir At 1:37 inductor current waveforms are shown linear. why only linear shape? i mean they could have been exponential... also while we are writing KVL eqns then Vo=Vs-VL but in waveform we have shown it to be equal to the Vs. during turn-off Vo= -VL but in waveform it is zero. i mean there is lack of consistency in what kvl is saying and what we are drawing... avg VL is zero but there must be some instantaneous value of it.
exponential graph u will get in case of RL circuit..not purely inductor circuit..in purely L circuit IL= integration of VL...right..so integration of constant will give you ramp fxn..i.e. linear.. now coming to sec query..you have written Vo=Vs-VL..that is correct..but for full cycle we know that average voltage across inductor is zero acc to volt sec balance..ok..so VL avg=0....but my question is why to find Vo like this..i apply the volt sec balance to find Vo...and in Vol sec balance i find the VL(on) and VL(off)..ok
also if you will take RL load then also at the first few second the graph shows linear behaviour..so we are operating inductor in linear region of graph...got it..
actually the KVL eqns (Vo=Vs-VL) we r writing .. are they taking avg values or instantaneous values? in rectifier module we were concerned about taking instantaneous values and based on those we drew waveforms...
when we are applying volt sec balance there we are taking instantaneous value..not average value..also average value of inductor over full cycle is zero..till DT you will get voltage across inductor and from DT to T also you will get avg voltage but overall avg voltage across L ovet full cycle is zero..got it..
during the Inductor discharge, the polarities of inductor should be changed right? ( Initially inductor would charge with polarity +,- ; while discharging current direction will not change but the polarity of the inductor should be changed to -,+ ) can you give the reason why it is not changed?
Dear Sir At 27:04 decreasing value of inductor is shifting the waveform downwards without changing its waveshape. but if L changes then its time-constant shall change thus waveshape should change na...
by default i took polarity of VL like this..that does not mean it is not changing..VL you will get negative with this polarity means the polarity is changing..got it..
switching frequency is 20hz then uss rate pe switch get on and off both i.e. Get on at 10 hz and off at 10hz...because f=1/T..then and off both will be included na.
During T_on the supply charges the inductor and capacitor & during T_off the inductor delivers (discharges) its power to the load so when does the capacitor discharge its charge?
During on time Isw is same as inductor current IL So why are you taking constant current It must be triangular shaped wave Please clear doubt sir Self study student
when you switch on the supply then current will start from zero...after that we adjust the switching frequency such that the inductor current can be continuous..so waveform i drawn here is at steady state...means after 5 to 6 cycle...i will get the waveform like this..got it..
Hello Sir, I have one question. In the buck convertor, If we use Inductor value more than calculated value what will happened? example: calculated value is 100uH, I will go with 500uH value.
Thanks for the video. I have one question.How the capacitor is charging and discharging in the ON period or OFF period. I mean when the capacitor current is positive , it is getting charged and negative means it is discharging. how it is happening ?
Sir I don't understand one thing.. i.e.. in the above circuit when switch is closed then source current and switch currents will be equal to inductor current na.. then their current waveforms should be same as that of inductor current during on state but why are u considering source and switch currents as constant sir?
I have came across a little different circuit. There is another diode in place of the switch and a transformer between the supply and the other elements.
sir i have seen your all video related to PE. The videos are so good. But sir i have seen some more topics for gate 2018- Dc drives, resonant converters and applications of PE, AC drives. could you send video related to this.
where it is shown that delVo= Vs-0...Vc is the voltage across capacitor and is connected parallel to load..so when Vc will change then Vo will also change same as Vc..thats y delVc=delVo
@24.49, your differentiation is not correct even though the final answer was right. You have used the quotient rule, but you should have used the product rule. I know it’s just a slip. Your videos are excellent. Thanks:)
D = Ton / T....... Ton = DT...... T = Ton+Toff....... Toff= T-Ton i.e., Ton.=DT...! So substitute Ton in that Toff equtn....! So that Toff = T-DT..... So that taking T common T(1-D)...
capacitor does not allow dc ...,its fine... But just see the current from inductor ...its triangular DC.if .by using fourier series you get a DC + SINE waves with different frequencies and amplitudes equivalent...so capacitor doesnot allow that dc component ..but it allows all other ac component of current....And other thing here is we kept the capacitor in parallel with load for just maintaining the constant voltage across load as in case of the rectifier....
My dear friend capacitor current is zero only in steady state(when DC source is present).. but look there is continuously switch is being on and off so whenever switch is been on and off in any circuit where inductor or capacitor is present then that particular inductor and Capacitor will have voltage and current respectively till it reach steady state.. pls analyse the circuit properly it's in transient state in this operation and in transient state L and C's will posses voltage and current values
Because inductor is initially charged by il(min), but if u want to consider inductor to be initially relaxed and charging takes place after switching operation then consider that.🙂🙂🙂
Awesome lecture...My questions are two 1. in the Amp sec balance section, how did IL change to IL(avg). IL(avg) is a constant but not IL. 2. Why didn't you use I=cdv/dt for ripple in Capacitor voltage as you did for the boost converter. Thank you
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Sir, could u explain about the voltage across switch and diode ,like u explained there avarage currents
I cleared an interview by watching your videos sir.
Thank you so much sir.
ur most welcome..if possible can you share ur reviews about this video in learn gate free group...link is given below
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Hats off to u Sir. What a great explaination. Never got this much clearity on buck converter before.
Simply awsome. way of explanation is better than dr erickson course era lectures for students
thanks in such a less time i got the topic clearly ..
what ever sohali sir explain is correct .but if u want more clarification do one thing ...during turn of condition the inductor changes its polarity. yeah ur absolutely correct . change inductor polarity and instead of applying the volt sec balance apply the energy conservation principle.
chaging energy equal to discharging energy (assuming inductor is completely discharged what ever energy it charged. as energy =v * i * t
(Vs-Vo) * (Imin+Imax)/2 * (DT) = (Vo) * (Imin+Imax)/2 *(1-D)T
after solving u may get Vo=D Vs ....this may applicable to other choppers also.
Correct explanation
@@GATEMATICEducation thanks sir
@@smrautomations2168 you are justifying that whatever sir had written is right.. sir had written VL=-Vo during off state and in ur explanation u have written +Vo during energy release i.e. during off state u have considered Vo instead of -Vo( as sir had written -Vo during off)
How does it mean ur justifying sir's explanation? Please answer
Thanks for this great lecture. I have just one question, in 30:26 why ∆Vc min and not Vc min ?
Thank you so much...Just in 30 minutes the whole topic with its equations are clearly explained...
sir your teaching technique is excelent.
point no. 9 , while calculating switch current, you hv assumed Io=Il . that is true for average values but not for the instantaneous values. the increasing current flows through switch and decreasing current flows through FWD. so while calculating avg current through switch, the +slope portion of inductor current should be taken in consideration, and not the avg load current.
+rajat saxena u r right..i have taken avg value..not instantaneous value..instantaneous waveform will be different..now if u will take instantaneous value also to find avg switch current..u will get same answer..no prblm..u can approach like that..
Sir but when inductor will discharge then polarity should be reversed hence v0=vl.plz clarify
i also have same doubt
Nice sir...In easy way u explained power electronics
Great video sir,thanks a lot ❤❤❤❤ searching this conclusion for long time finally found here
Your explanation is super sir ..🙏
23:22 sir I have a question ⁉️
we know ripple is peak to peak. than here ripple voltage should be 2∆V ?
because capacitor voltage flactuate between +∆V to -∆V
why it is ∆V but not 2∆V??
Nice video and explanation Sir, Thank you
Really thank you for your helps it really help me again i understand power electronics with and thank you for that
ur most welcome...all the best for future..
sir, you are studying from basic i like it very much . thanks for it
Thank you Mr.Suhail.
thank you very very much sir for teaching this concept in very easy manner.
at 6:23, during switch off condition, inductor discharges stored current. at that instant voltage polarity of inductor get changes but here in the video, you kept the same polarity for inductor during discharge, why? is there any mistake my point?
post ur question in learn gate free facebook group..and tag me there..i will resolve ur query..all the best..
right now I'm not using Facebook.
it's doubt regarding polarity of inductor during charging and discharging.
inductor tries to keep the current in the same direction. Thats why it "supports" the current or "opposes" the change
yes polarity will be changed but we already took one terminal as higher and other is lower . so final expression wrote according to this.
what ever sohali sir explain is correct .but if u want more clarification do one thing ...during turn of condition the inductor changes its polarity. yeah ur absolutely correct . change inductor polarity and instead of applying the volt sec balance apply the energy conservation principle.
chaging energy equal to discharging energy (assuming inductor is completely discharged what ever energy it charged. as energy =v * i * t
(Vs-Vo) * (Imin+Imax)/2 * (DT) = (Vo) * (Imin+Imax)/2 *(1-D)T
after solving u may get Vo=D Vs ....this may applicable to other choppers also.
Excellent sir... Good explanation thank you...
Loved all your videos Sir. Made it very simple. Sir, when you will upload the lectures of PWM?
after my mtech thesis
Thank you very much sir. May God bless you!
amazing concept!
Sir ur lectures are just amazing ....
Plz upload the videos for inverter also
You are great lecture may God bless you
yeah sir, When inductor discharges during off condition of switch, It reverses its polarity and then discharges. But in the video, you did not change the polarity. Reason? At 6:23
for all time the polarity of voltage across inductor is assumed to be same...that does not mean inductor voltage is not reversing...it is reversing thats why u are getting negative voltage during off..
@@GATEMATICEducation yes sir exactly
In transients we studied that when the supply is connected to inductor then it charges exponentially by equation Is(1-exp(-t/Tau)) but here u mentioned that the inductor will.chsrge linearly? Why is there this contradiction? Plz explain
Exponential fxn is also have linear region at starting....so this ckt are working in linear region
.ok
transient equation is valid only for finite time constant. and when inductor is connected with a ideal current source so time constant= L/R when you will remove ideal current source in order to calculate R you will get R=infinite and hence time constant=0 hence transient equation is not applicable. And this becomes linear charging of inductor case. If you got it then fine otherwise apply laplace transform to the circuit you will get the same result. if then also u cant understand. TAB TOH BHAGWAN HI BACAHYE AAPKO.
It’s because inductor has very large value
For practicle purpose linear approximation is good enough, in reality it not only is exponential but also has noise and ringing due to thermal noise and inductor resistance
Age inductor ki across constant voltage source lgya jata hn;tb inductor linearly charge hota hn...i ramp function hoga
Excellent presentation
All doubt clear sir 👌
Very amazing lecture sir
explanations superb. thank you!
If avg current across capacitor is zero then how capacitor is charged during turn on time??
2. If capacitor current is zero then how we use Ic= I inductor- I load in wave form .
Dear Sir
At 1:37 inductor current waveforms are shown linear. why only linear shape? i mean they could have been exponential...
also while we are writing KVL eqns then Vo=Vs-VL but in waveform we have shown it to be equal to the Vs. during turn-off Vo= -VL but in waveform it is zero. i mean there is lack of consistency in what kvl is saying and what we are drawing... avg VL is zero but there must be some instantaneous value of it.
exponential graph u will get in case of RL circuit..not purely inductor circuit..in purely L circuit IL= integration of VL...right..so integration of constant will give you ramp fxn..i.e. linear..
now coming to sec query..you have written Vo=Vs-VL..that is correct..but for full cycle we know that average voltage across inductor is zero acc to volt sec balance..ok..so VL avg=0....but my question is why to find Vo like this..i apply the volt sec balance to find Vo...and in Vol sec balance i find the VL(on) and VL(off)..ok
also if you will take RL load then also at the first few second the graph shows linear behaviour..so we are operating inductor in linear region of graph...got it..
actually the KVL eqns (Vo=Vs-VL) we r writing .. are they taking avg values or instantaneous values? in rectifier module we were concerned about taking instantaneous values and based on those we drew waveforms...
when we are applying volt sec balance there we are taking instantaneous value..not average value..also average value of inductor over full cycle is zero..till DT you will get voltage across inductor and from DT to T also you will get avg voltage but overall avg voltage across L ovet full cycle is zero..got it..
Hello, thanks a lot for this session, do you guys have some challenging practice problem for this lesson ?
A Buck converter operates in Continuous Conduction Mode. Find L and C values.
Vin=100V; fsw=50kHz; Vout= 10V; Δv=1%; 1watt < Pload < 10watt
Very nice explanation
thank you for making this video
சிறப்பு💞
during the Inductor discharge, the polarities of inductor should be changed right? ( Initially inductor would charge with polarity +,- ; while discharging current direction will not change but the polarity of the inductor should be changed to -,+ ) can you give the reason why it is not changed?
Due to high negative release of energy
Thank you sir informative jazakhalahu khair
Dear Sir
At 27:04 decreasing value of inductor is shifting the waveform downwards without changing its waveshape. but if L changes then its time-constant shall change thus waveshape should change na...
yes u r right..the slope will change ..i m drawing the rough waveform..just to make u understand the for particular value of L the Imin=0..got it..
time saver I love your sir only love you are my hero
You are nothing but good
Sir, when we kept the switch open, why is the polarity of L not changed as it starts behaving like a voltage source itself ? Please reply..
by default i took polarity of VL like this..that does not mean it is not changing..VL you will get negative with this polarity means the polarity is changing..got it..
GATEMATIC Education then expression will change
switching frequency is 20hz then uss rate pe switch get on and off both i.e. Get on at 10 hz and off at 10hz...because f=1/T..then and off both will be included na.
During T_on the supply charges the inductor and capacitor & during T_off the inductor delivers (discharges) its power to the load so when does the capacitor discharge its charge?
Sir same doubt.
@@PriyankaSingh-je6ro during Ton capacitor is also charging up to DT, and when the switch turned off capacitor discharging.
During on time Isw is same as inductor current IL
So why are you taking constant current
It must be triangular shaped wave
Please clear doubt sir
Self study student
sir..I m not understanding one thing .why the current is not starting from 0
....
when you switch on the supply then current will start from zero...after that we adjust the switching frequency such that the inductor current can be continuous..so waveform i drawn here is at steady state...means after 5 to 6 cycle...i will get the waveform like this..got it..
a good video indeed
Hello Sir, I have one question.
In the buck convertor, If we use Inductor value more than calculated value what will happened? example: calculated value is 100uH, I will go with 500uH value.
Thanks for the video. I have one question.How the capacitor is charging and discharging in the ON period or OFF period. I mean when the capacitor current is positive , it is getting charged and negative means it is discharging. how it is happening ?
Thank you so much!
i interested in your lecture and i need the another lecture in ac to ac voltage converter i)voltage regulators ii)cycloconvertors lecture
Idc1
Powerdc1
Vdc2- 24V
Idc2 -
Powerdc2
Loss power-
Please help me to find out this values
you are amazing thank you sir
Sir I don't understand one thing.. i.e.. in the above circuit when switch is closed then source current and switch currents will be equal to inductor current na.. then their current waveforms should be same as that of inductor current during on state but why are u considering source and switch currents as constant sir?
Awesome sir
amazing sir.......
nice sir.thank you
I have came across a little different circuit. There is another diode in place of the switch and a transformer between the supply and the other elements.
Thank you 🇱🇰
sir, is the inductor is charged initially??bcoz u hv written IL min??
sir i have seen your all video related to PE. The videos are so good. But sir i have seen some more topics for gate 2018- Dc drives, resonant converters and applications of PE, AC drives. could you send video related to this.
these topic has been removed from gate syllabus..so dont worry
If Ton and Toff are not equal , then how have you taken T/2 during calculation of ripple capacitor Del Vc
Thank you sir
thank you
ye one unlike wala kaun h ......wonderful lecture sir......superb.....
ye one unlike jaise hi mai upload karta hu to aa jata hai..uske baad jo log dekhte hai wo like kar dete hai...
gatecoaching wala honge
Sir,During Turn off time,Why the sign of Vl is won't change?
Thanx Sir
Thnx....:) Bro
Sir,one thing...why current is linear in inductor???
Dear Sir At 24:44 we are calculating del Vc = del Vo . from waveform it is clear that DelVo=Vs-0. but that is not the result why?
where it is shown that delVo= Vs-0...Vc is the voltage across capacitor and is connected parallel to load..so when Vc will change then Vo will also change same as Vc..thats y delVc=delVo
sir till DT Vo=Vs and from DT to T it is zero that means delVo=Vs-0. from output voltage waveform ..
no..till DT , Vo= Vs-VL....and from DT to T Vo=-VL..how you are saying i dont know..apply kvl here..u will this much only...
sir i am saying the above fact by looking directly to the output voltage waveform.. thats my confusion.. ki kvl alag bta rha ...
@24.49, your differentiation is not correct even though the final answer was right. You have used the quotient rule, but you should have used the product rule. I know it’s just a slip. Your videos are excellent. Thanks:)
Sir when output current equal to inductor current then capacitor current be zero.
Cap. Current = inductor current - load current
You can see that average current of Capacity equal to zero
sir this is ccm mode only sir again dcm mode we different formulas sir??
sir off time ic current is confusing.becuase kcl is not allowed
Sir capacitor current waveform mein T/2 kaise aaya? Ye nhi bataya aapne.... Point 11
Geometry lagao..aa jayega..thoda sa dhyan lagana hai.
@@GATEMATICEducation can we say T/2 time mein discharge hoga and T/2 time mein charge, to maintain capacitor charge 0... That's y its taken?
U can say
@8.54 ; could you plz tell how IL= Io arrived , looks like its wrong.
Expand kar lo yaar
Good
sir please a lecture on a to e choppers
sir when will inverter lecture be uploaded?
Sir how this (D-1)T is derived
D = Ton / T....... Ton = DT......
T = Ton+Toff....... Toff= T-Ton i.e., Ton.=DT...! So substitute Ton in that Toff equtn....! So that Toff = T-DT.....
So that taking T common T(1-D)...
Sir pls upload the inverter lecture
why current through inductor is not starting from zero
Sir in buck converter how to find peak current in switch
Isn't it true that capacitor blocks dc? Then how current is flowing through capacitor?
Bcoz current capacitor is not dc it is triangular
capacitor does not allow dc ...,its fine... But just see the current from inductor ...its triangular DC.if .by using fourier series you get a DC + SINE waves with different frequencies and amplitudes equivalent...so capacitor doesnot allow that dc component ..but it allows all other ac component of current....And other thing here is we kept the capacitor in parallel with load for just maintaining the constant voltage across load as in case of the rectifier....
My dear friend capacitor current is zero only in steady state(when DC source is present).. but look there is continuously switch is being on and off so whenever switch is been on and off in any circuit where inductor or capacitor is present then that particular inductor and Capacitor will have voltage and current respectively till it reach steady state.. pls analyse the circuit properly it's in transient state in this operation and in transient state L and C's will posses voltage and current values
🙏🙏🙏🙏
where is the lecture pdf?
What is the expression for rms values of inductor current
It depends on the duty cycle of on/off, if duty cycle is 50% then rms inductor current is average current = (imax - imin)/2
Sir volt second balance in buck converter vo=dvs please explain sir
correction after writing del q in del vc formula .c will also come in denominator
in final formula..its written..that is slip of pen..no worry..
amazed by art of your work and understanding of subject to great depth .MAY ALLAH BLESS U
thnx for the compliment..all the best for gate 18..
pls can you guide that why inductor current is not start from zero.
Because inductor is initially charged by il(min), but if u want to consider inductor to be initially relaxed and charging takes place after switching operation then consider that.🙂🙂🙂
IT Gopeshwar waalo like kro re
Sir but source constant nai hoga na
Please make video on inverter sir please please
How to find out power loss in buck converter
Awesome lecture...My questions are two 1. in the Amp sec balance section, how did IL change to IL(avg). IL(avg) is a constant but not IL. 2. Why didn't you use I=cdv/dt for ripple in Capacitor voltage as you did for the boost converter. Thank you
By Amp-Sec balance, he is saying that average current through the capacitor is zero.But i think this is not correct.
check amp sec balance..its wrong for all 3 calcuations...
SUBHANKAR GHOSH its correct
Why not capacitor is discharging by giving current in case when Switch in OFF?