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
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
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...
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 (:
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.
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??
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
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
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....
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
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....
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....
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
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
For anyone confused, just change the polarity of Vi in beginning and solve it as previous.
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
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
how can you consider it ?
yes lot of confusion for -ve half cycles
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...
While taking the positive half cycle the polarity of the capacitor will change or not ?????????
Why arent the capacitor pole charges being reversed when cycle changes.
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
(:
@@To_Engineer bro nice explanation ,thnx bhai
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.
i am counfused in -(-vi)????how two -ve signs are taken???
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??
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
Sir you have taken a polarized capacitor. Then how did you change its polarity?
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
For second half , why the polarities of capacitor are not changed ?
sir,why you changed the input waveform i.e, for positive(0 to t/2=-v) and negative clampers(0 to t/2=v)
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
yeah but here after the half cycle the capacitor gets charged so the polarity remains same as in during the case oof charging
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....
sir y didnt u change the polarity of capacitors in the -ve half cycle?
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
I also
I think the polarity of capacitor is wrong . It should be reversed
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
Sir what happen when in positive clamper when signal start from +v
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....
what is dt = 5 t. what are dt and t in 5:45
Sir i didn't understand, y that capacitor won't get discharge during second half cycle?
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
Before the charge drains down, you flip. It is instantaneous.
what will be the output if capacitor discharge is considered?
do you know now?
please repy.
here capacitor discharge is considered
it is considered here ,tats why polarity is unchanged in the next half cycle
hi sir,i cant view the lectures for the op amp please may i know that it is uploaded or not
Sir please explain +ve -ve at time 3:22 again!
In 2nd cycle how you apply kvl
does capacitor want to be charged the highest?
Sir, what is the exact definition of positive and negative clamper circuits. Please explain it clearly.
Please explain for sinusoidal signal
i have a confusion. what will happen after (5*tau) times.? please explain anyone.. :-(
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?
it will be similar, i mean clamped similarly, but if we use non ideal diode there will be clipping
😊nyc teaching
what would be output for further cycle
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.
what if the square wave started from +V instead of -V,..?
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....
@@AnoNymous-po5sx good explanation. from the very first half the capacitor wont get stabilized.
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
Sir I can't understand how -(-Vi) comes
samw here
huzaifa hassan vi=-(-vi) as -ve into -ve is positive.
Same
Same pblm
Same
this video is really helpful, thanks a. lot
Why you are not using some wave as a input
Sine wave
may be because it is simpler to understand ,if we understand this properly, the sine wave one is self explanatory
OPAMP videos plz
sir you give soooo beautiful leacture
does the peak value changes in clippere
The capacitor direction is wrong from the beginning. the curved part should be on the left and straight on the right
Please make a video for additional dc supply in clampers
Sir Vo and Vc ki value samajh nhi aayi
the polarity u put on capacitor is wrong .
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
if vi=-v so it should be vc=-v how vc =v???
at 2:41
Why T>>T/2
Can't understand -vi & v
sir please ZENER LIMITER, ZENER COMPRESSOR AND LIMITER-COMPRESSOR IS MOST IMPORTANT TOPIC
these clamper circuits are biased or unbiased?
Nidhi Suhag these are unbiased clamper circuits as there is no extra source of voltage other than the input voltage
I cant understand anything😢
Real diode please :)
hey bro,,,,,,,,just try to complete all ece syllabus asap...i will be thankfull
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
confused
Clamper is not explain properly
Sir what happen when in positive clamper when signal start from +v