Positive & Negative Clamper Circuits

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  • เผยแพร่เมื่อ 23 ธ.ค. 2024

ความคิดเห็น • 84

  • @alirihan3374
    @alirihan3374 7 ปีที่แล้ว +14

    For anyone confused, just change the polarity of Vi in beginning and solve it as previous.

  • @surendharwatson7626
    @surendharwatson7626 3 หลายเดือนก่อน +2

    for those who are confused in 1st half cycle equation , In 1st half cycle equation vi=vc , vc=v , vi is considered as v not as -v by the given input signal because the 1 st half cycle is start from -vi and the value is -v it gives -vi=-v tends to vi=v. Hope you all understand

  • @agvlogs5773
    @agvlogs5773 6 ปีที่แล้ว +7

    For those who r confusing. Consider Vin=v
    in -ve half,
    Vo=0 v (since the circuit will be closed)
    then -Vin+Vc=0
    or, Vin=Vc
    For -ve half,
    Vo=Vin+Vc
    or, Vo=Vin+Vin
    or, Vo=2v

    • @vikku_19
      @vikku_19 6 ปีที่แล้ว +1

      how can you consider it ?

  • @harshit8092
    @harshit8092 7 ปีที่แล้ว +34

    yes lot of confusion for -ve half cycles

  • @utkarshtamrakar2437
    @utkarshtamrakar2437 7 ปีที่แล้ว +4

    Sir why we take negetive half cycle of input signal initially and what will happen if we take positive half cycle of input signal initialy please tell...

  • @srikantsahoo5699
    @srikantsahoo5699 3 ปีที่แล้ว +2

    While taking the positive half cycle the polarity of the capacitor will change or not ?????????

  • @taha290500
    @taha290500 5 ปีที่แล้ว +7

    Why arent the capacitor pole charges being reversed when cycle changes.

    • @To_Engineer
      @To_Engineer 5 ปีที่แล้ว +4

      no it will not because physically capacitor is not reversed you just changed the polarity of input
      which implies: if capacitor charging in first half cycle then it will discharge in negative half cycle as it act as dc source which changes the dc level of your output
      (:

    • @dalinsixtus7648
      @dalinsixtus7648 3 ปีที่แล้ว +1

      @@To_Engineer bro nice explanation ,thnx bhai

  • @vikku_19
    @vikku_19 6 ปีที่แล้ว +14

    it is still confusing, we all know that vi=-v (given for the first half cycle) , and after applying we get vi-vc=0,so vi=vc . now you put
    vc = - (-vi) that is okay .now if we replace vi to -v then we get vc=v not -v .hope you wil clarify whole things.

  • @huzaifahassan1935
    @huzaifahassan1935 7 ปีที่แล้ว +3

    i am counfused in -(-vi)????how two -ve signs are taken???

  • @abhishekbora273
    @abhishekbora273 9 หลายเดือนก่อน +1

    In positive clamper -ve side of vi is connected to -ve side of capacitor but in negative clamper negative side of vi is connected to positive side of capacitor. Why??

  • @Esther010
    @Esther010 5 ปีที่แล้ว +2

    I have a question. +Vi -(-Vc )-(+Vo) =0 equal to Vi +Vc =Vo
    Is that right? (T/2 to T ) I consider when in clockwise direction ,start from positive terminal +Vin and minus all the other voltage in one loop for KVL.. And after +Vi, the first polarity I saw in Vc is -ve,
    so -( -Vc ) ,and first polarity I saw for Vo is positive (+) so become -(+Vo) ...,is that correct to do this? Pls help. Thx

  • @niharamzan382
    @niharamzan382 7 ปีที่แล้ว +2

    Sir you have taken a polarized capacitor. Then how did you change its polarity?

  • @mere_illusion
    @mere_illusion 6 ปีที่แล้ว +11

    for those of you who are confused in -(-v)....... just take the lower terminal to be ground(0 voltage) and put + or - Vi at the upper terminal

  • @Toosharp1622
    @Toosharp1622 3 ปีที่แล้ว +1

    For second half , why the polarities of capacitor are not changed ?

  • @venkat2830
    @venkat2830 7 ปีที่แล้ว +3

    sir,why you changed the input waveform i.e, for positive(0 to t/2=-v) and negative clampers(0 to t/2=v)

  • @audiblesunoofficial
    @audiblesunoofficial 2 ปีที่แล้ว +2

    As u mentioned earlier in positive half cycle that polarity of capacitor depends upon on which side it is connected
    But in negative cycle u didn't change the polarity of capacitor

    • @ishandas2674
      @ishandas2674 2 ปีที่แล้ว +1

      yeah but here after the half cycle the capacitor gets charged so the polarity remains same as in during the case oof charging

  • @AnoNymous-po5sx
    @AnoNymous-po5sx 4 ปีที่แล้ว +12

    You can either consider positive half cycle or negative one first. But since he explained negative, I'll explain positive half cycle.
    In the first positive half cycle, the diode will remain turned OFF. The current will flow through R and charges the capacitor. But since τ is greater than T/2, capacitor won't charge fully. So, Vo=Vi=Vm.
    Inthe negative half cycle, the diode will be ON and entire current flows through diode and quickly charges the capacitor upto value Vm. In the nest positive cycle, the voltage through the resistor will be Vi+Vm i.e., 2Vm....and so on....

  • @debjitroy5622
    @debjitroy5622 ปีที่แล้ว

    sir y didnt u change the polarity of capacitors in the -ve half cycle?

  • @prof_as
    @prof_as 6 ปีที่แล้ว

    at 2:44 the value of Vc= Vi and the value of Vi at movement is equal to -V, therefore the value of Vc must be equal to Vc=-V?
    cant understand this plz help

    • @neelkanthpatel873
      @neelkanthpatel873 6 ปีที่แล้ว

      I also

    • @everything483
      @everything483 4 ปีที่แล้ว

      I think the polarity of capacitor is wrong . It should be reversed

  • @mamatha_999_9
    @mamatha_999_9 3 ปีที่แล้ว

    Why can't we assume capacitor after charging it is acting like dc source & try to make diode either FB or RB sed...??
    As u said in clippers

  • @bituphukon6954
    @bituphukon6954 5 ปีที่แล้ว +2

    Sir what happen when in positive clamper when signal start from +v

    • @AnoNymous-po5sx
      @AnoNymous-po5sx 4 ปีที่แล้ว

      You can either consider positive half cycle or negative one first. But since he explained negative, I'll explain positive half cycle.
      In the first positive half cycle, the diode will remain turned OFF. The current will flow through R and charges the capacitor. But since τ is greater than T/2, capacitor won't charge fully. So, Vo=Vi=Vm.
      Inthe negative half cycle, the diode will be ON and entire current flows through diode and quickly charges the capacitor upto value Vm. In the nest positive cycle, the voltage through the resistor will be Vi+Vm i.e., 2Vm....and so on....

  • @mohitswain8441
    @mohitswain8441 5 ปีที่แล้ว

    what is dt = 5 t. what are dt and t in 5:45

  • @charankumar4267
    @charankumar4267 5 ปีที่แล้ว +3

    Sir i didn't understand, y that capacitor won't get discharge during second half cycle?

    • @To_Engineer
      @To_Engineer 5 ปีที่แล้ว +1

      dear when you perform this in lab you will see on output -ve half cycle is extended on time axis due time taken by discharging of capacitor

    • @shivamfaraday1752
      @shivamfaraday1752 4 ปีที่แล้ว +1

      Before the charge drains down, you flip. It is instantaneous.

  • @shagundeepsingh9712
    @shagundeepsingh9712 7 ปีที่แล้ว +1

    what will be the output if capacitor discharge is considered?

    • @Shoron09
      @Shoron09 7 ปีที่แล้ว

      do you know now?
      please repy.

    • @guardofhonour3089
      @guardofhonour3089 5 ปีที่แล้ว

      here capacitor discharge is considered

    • @dalinsixtus7648
      @dalinsixtus7648 3 ปีที่แล้ว

      it is considered here ,tats why polarity is unchanged in the next half cycle

  • @abinayas7821
    @abinayas7821 7 ปีที่แล้ว +3

    hi sir,i cant view the lectures for the op amp please may i know that it is uploaded or not

  • @sudeshnabhowmik5928
    @sudeshnabhowmik5928 6 ปีที่แล้ว +1

    Sir please explain +ve -ve at time 3:22 again!

  • @himanshuagarwal2712
    @himanshuagarwal2712 2 ปีที่แล้ว

    In 2nd cycle how you apply kvl

  • @guardofhonour3089
    @guardofhonour3089 5 ปีที่แล้ว

    does capacitor want to be charged the highest?

  • @bhabanishankardas9800
    @bhabanishankardas9800 5 ปีที่แล้ว +1

    Sir, what is the exact definition of positive and negative clamper circuits. Please explain it clearly.

  • @kothatharonie4236
    @kothatharonie4236 6 ปีที่แล้ว

    Please explain for sinusoidal signal

  • @Shoron09
    @Shoron09 7 ปีที่แล้ว

    i have a confusion. what will happen after (5*tau) times.? please explain anyone.. :-(

  • @hellbark
    @hellbark 8 ปีที่แล้ว

    hey nice video but may i know what if the input voltage is 15sin theta, 60Hz how am i going to draw the output graph?

    • @yasirpunathil7143
      @yasirpunathil7143 6 ปีที่แล้ว +1

      it will be similar, i mean clamped similarly, but if we use non ideal diode there will be clipping

  • @khushigowda9153
    @khushigowda9153 7 ปีที่แล้ว

    😊nyc teaching

  • @princechaudhary8828
    @princechaudhary8828 7 ปีที่แล้ว

    what would be output for further cycle

    • @hrishabhrajput9504
      @hrishabhrajput9504 7 ปีที่แล้ว

      Diode will again become forward biased (replaced by short circuit), and Vc will again be equal to Vi, and the cycle will continue with a DC shift.

  • @ShAmS894
    @ShAmS894 5 ปีที่แล้ว

    what if the square wave started from +V instead of -V,..?

    • @AnoNymous-po5sx
      @AnoNymous-po5sx 4 ปีที่แล้ว +2

      You can either consider positive half cycle or negative one first. But since he explained negative, I'll explain positive half cycle.
      In the first positive half cycle, the diode will remain turned OFF. The current will flow through R and charges the capacitor. But since τ is greater than T/2, capacitor won't charge fully. So, Vo=Vi=Vm.
      Inthe negative half cycle, the diode will be ON and entire current flows through diode and quickly charges the capacitor upto value Vm. In the nest positive cycle, the voltage through the resistor will be Vi+Vm i.e., 2Vm....and so on....

    • @ndkarunarathne
      @ndkarunarathne 8 หลายเดือนก่อน

      @@AnoNymous-po5sx good explanation. from the very first half the capacitor wont get stabilized.

  • @og_justinquisitive
    @og_justinquisitive 7 ปีที่แล้ว

    Is there a way to reverse the effects of the positive clamper?, ie. convert a unipolar signal back to bipolar signal without changing the appearance of the input signal

  • @prasanthsrinivas26
    @prasanthsrinivas26 7 ปีที่แล้ว +7

    Sir I can't understand how -(-Vi) comes

  • @poorvamahto3025
    @poorvamahto3025 8 ปีที่แล้ว

    this video is really helpful, thanks a. lot

  • @rishikeshjha3032
    @rishikeshjha3032 7 ปีที่แล้ว

    Why you are not using some wave as a input

    • @rishikeshjha3032
      @rishikeshjha3032 7 ปีที่แล้ว

      Sine wave

    • @yasirpunathil7143
      @yasirpunathil7143 6 ปีที่แล้ว

      may be because it is simpler to understand ,if we understand this properly, the sine wave one is self explanatory

  • @KirkHammer-fj2of
    @KirkHammer-fj2of 7 ปีที่แล้ว +1

    OPAMP videos plz

  • @dariamalik6791
    @dariamalik6791 6 ปีที่แล้ว

    sir you give soooo beautiful leacture

  • @akshayramteke6265
    @akshayramteke6265 8 ปีที่แล้ว

    does the peak value changes in clippere

  • @yonghuiliew8066
    @yonghuiliew8066 4 ปีที่แล้ว

    The capacitor direction is wrong from the beginning. the curved part should be on the left and straight on the right

  • @YuvrajSingh-we8jc
    @YuvrajSingh-we8jc 7 ปีที่แล้ว

    Please make a video for additional dc supply in clampers

  • @AJDJ108
    @AJDJ108 3 ปีที่แล้ว

    Sir Vo and Vc ki value samajh nhi aayi

  • @FaizanAli-qk1mr
    @FaizanAli-qk1mr 7 ปีที่แล้ว +1

    the polarity u put on capacitor is wrong .

    • @manubansal6686
      @manubansal6686 6 ปีที่แล้ว

      hello sir. i have a confusion. as there is no R i.e. R=0, then time constant should be zero. thus capacitor should discharge quickly

  • @huzaifahassan1935
    @huzaifahassan1935 7 ปีที่แล้ว

    if vi=-v so it should be vc=-v how vc =v???

  • @karapureddyjaswanthreddy6053
    @karapureddyjaswanthreddy6053 3 ปีที่แล้ว

    Why T>>T/2

  • @neelkanthpatel873
    @neelkanthpatel873 6 ปีที่แล้ว

    Can't understand -vi & v

  • @radioartevaporwave6423
    @radioartevaporwave6423 5 ปีที่แล้ว +1

    sir please ZENER LIMITER, ZENER COMPRESSOR AND LIMITER-COMPRESSOR IS MOST IMPORTANT TOPIC

  • @nidhisuhag7604
    @nidhisuhag7604 7 ปีที่แล้ว

    these clamper circuits are biased or unbiased?

    • @atulgupta9084
      @atulgupta9084 7 ปีที่แล้ว

      Nidhi Suhag these are unbiased clamper circuits as there is no extra source of voltage other than the input voltage

  • @VIKAS.R.S
    @VIKAS.R.S 5 หลายเดือนก่อน

    I cant understand anything😢

  • @dol--lod282
    @dol--lod282 6 ปีที่แล้ว

    Real diode please :)

  • @triptirawat1173
    @triptirawat1173 6 ปีที่แล้ว +1

    hey bro,,,,,,,,just try to complete all ece syllabus asap...i will be thankfull

  • @thisissharief7651
    @thisissharief7651 6 ปีที่แล้ว

    neso academy ur videos are very nice but why dont u solve our doubts u are saying that u feel happy to solbe out doubts but i dont think so please respond there a re abouy thousand s of subscribers to u re channel including me

  • @akashtiwari3761
    @akashtiwari3761 7 ปีที่แล้ว +1

    confused

  • @sujansinghshinde7172
    @sujansinghshinde7172 5 ปีที่แล้ว

    Clamper is not explain properly

  • @bituphukon6954
    @bituphukon6954 5 ปีที่แล้ว +2

    Sir what happen when in positive clamper when signal start from +v