For the half cycle because the capacitor has charged and the time constant is greater than T/2 it will remain charged with the polarity from the first half cycle . Therefore we will have Vi+V0+Vc=0 where V0=-(-10)-8=2 Volts
most of the comments are asking that how VO =2 v? for first half . here is simple solution.first calculate vc its 8v.. then apply kirchoffs law +vi-vc-v0=0 v0=vi-vc v0=10-8 v0=2v.
bro one doubt....at 4.23 min of the this video from t/2 to t,vi=-10v,,,but int h kvl equation it is Vo= -Vi-Vc but u substituted -Vi as -10v...shoudnt it be just 10v?
prajval kumar that was a mistake on his part. We're just considering the magnitude of Vi here which is 10V. Since he already changed the polarity in the circuit diagram, he shouldn't have written Vi=-10V
when diode conducts replace it wid short circuit( NORMAL WIRE) now when u try to measure the voltage ITS NOTHING BUT 2V BATTERY IN PARALLEL with o/p,,thts y o/p=2v
Hi Sir, where can i find your lectures for a) Clamping Circuit Theorem, b)Practical Clamping Circuits, c) Effect of Diode Characteristics on Clamping Voltage and d) Synchronized Clamping.
Because diode is forward biased and will be replaced by short circuit. So in the diode branch there is only dc supply having value equal to 2V and if you see the output voltage its positive terminal is connected to positive terminal of 2V dc supply and negative terminal is connected to negative terminal of 2V dc supply. This means output voltage is simply equal to 2V. i hope its help!
during negetive half cycle diode reverse bias m hein,,isliye battery diode part open circuit hojata h..aur ek baaat KVL lagao Vin Vc Vo isme refernce voltage ayga hi ni
from T/2 to T: you stated that Vi=-10V and Vc=8V. and then, you have the equation Vo= -Vi -Vc. if you sub (-10V) into the eq. , isnt it gonna be Vo= 10 - 8 ? the sign of the eq. make me confused even though I think -18V is right
For the half cycle because the capacitor has charged and the time constant is greater than T/2 it will remain charged with the polarity from the first half cycle . Therefore we will have Vi+V0+Vc=0 where V0=-(-10)-8=2 Volts
I think i need to make a correction V0=-Vi-Vc and -Vi is will -10 V because the value of Vi is 10. Therefore V0=-18
@@gamar1226 hey i have the same exact doubt can you please explain ?
U r the reason I bunk classes but still pass my exams
Koi na bro
right now im bunking my classes and sitting in the library watching this lecture
glad the channel is back. you r the best. SIR
most of the comments are asking that how VO =2 v? for first half .
here is simple solution.first calculate vc its 8v..
then apply kirchoffs law
+vi-vc-v0=0
v0=vi-vc
v0=10-8
v0=2v.
In T/2 to T cycle
KVL should be
+Vi -Vo + Vc=0
Isn't it???
+Vi+Vo+VC hoga
Lekin ab to aapne btech kar li hogi bahot let me yha pahocha 😂
@@amankumarsrivastava5591 😂😂
@@amankumarsrivastava5591 😂😂🤣🤣
thank you so much sir , your videos are life saviour!
at 4:19 Vo=-Vi-Vc , Vi=-10 , then Vo=-(-10)-(8)=10-8=2v , you took the voltage as wrong.
Sir you have explained it conceptually
Is the diode in question Ideal diode with turn-on voltage of 0?
From T/2 to T,
Vi = -10V
Vo = -Vi-Vc
isn't Vo = -(-10V)-8V = 2V ?
same doubt
@Neso Academy For +ve cycle would'nt it be like 0v across the resister since capacitor act as open circuit for dc voltage.
For the second half cycle Vi is already -10 na so in tht equation shouldn't it be Vo=-(-10)-8 ?
bro one doubt....at 4.23 min of the this video from t/2 to t,vi=-10v,,,but int h kvl equation it is Vo= -Vi-Vc but u substituted -Vi as -10v...shoudnt it be just 10v?
I said that too , i wish they would answer
prajval kumar that was a mistake on his part. We're just considering the magnitude of Vi here which is 10V. Since he already changed the polarity in the circuit diagram, he shouldn't have written Vi=-10V
Hope this answers your question.
SO answer is Vo=-(-10)-8=2v na?
no he is right, in the +cycle +vi=+10 and in - cycle -vi=-10 cuz the circuit diagram has +vi on its top,-vi on its bottom
You're said for clamping no change waveform but shift the wave but here wave is completely changed
why there is no effect of the 10v in the output voltage
Thanks for the explanation.. now, ✓ subscribed :)
Nice teaching sir very helpful for me
how o/p voltage is 2v when diode in f.b. in positive half cycle?..what about presence of input voltage?
when diode conducts replace it wid short circuit( NORMAL WIRE) now when u try to measure the voltage ITS NOTHING BUT 2V BATTERY IN PARALLEL with o/p,,thts y o/p=2v
can you explain why 2V battery made it R> BIASED In 0-T/2 cycle??
Cool explaination bro...
3:50 me how Vo is +ve ??? it shoiuld be negative n/??
THANKS, THATS WHAT I'M SAYING 😅
Hi Sir, where can i find your lectures for a) Clamping Circuit Theorem, b)Practical Clamping Circuits, c) Effect of Diode Characteristics on Clamping Voltage and d) Synchronized Clamping.
Not there 😂😂
In +ve half cycle for Vi2v,diode is forwardbiased,only then Vo=2v...???
thats when input signal is analog. here the signal is digital. the voltage is either 0 or 10.
why v o/p during +ve cycle is 2v
this is very basic thing ? lol
Sir aap -V ki value kaise le rhe hai samajh nahi aa rha hai... Please bta dijiye
can you please explain how v0 is 2V?
Because diode is forward biased and will be replaced by short circuit. So in the diode branch there is only dc supply having value equal to 2V and if you see the output voltage its positive terminal is connected to positive terminal of 2V dc supply and negative terminal is connected to negative terminal of 2V dc supply. This means output voltage is simply equal to 2V. i hope its help!
@@youtubelover1240 It helps, thanks you sir/mam👍.........
@@ashishkhari5565 most welcome
During the negative have cycle why did he not considered the reference voltage 2 volts while applying kvl
during negetive half cycle diode reverse bias m hein,,isliye battery diode part open circuit hojata h..aur ek baaat KVL lagao Vin Vc Vo isme refernce voltage ayga hi ni
outer loop m kvl lagao battery ayga hi ni aur waise bhi diode battry branch open bhi h
when Vi=-10 volts
how this " Vo=-Vi-Vc " could be -18 volts
please explain , thanks
Hosam A. Fiky I answered above. Please check. It's a mistake on his part.
Thanks 😊
Can u do this with AC source
i still wonder y the i/p ac does not contribute to o/p v .....
thank you..nice job
from t/2 to t v(i) is -10v but why the kvl equation u put -v(i)=-10 v?????
+ - mein itna kya 😂😂
How about a sinewave signal??
Shunya I knew it already
Krissy Limon ya the same..but some noise due to cap
well done!!!
why 2V will make diode reversed?
time :- 1:36
thank you sir
sir.clamper add dc level.but sir in clamper cct capacitors are used,as we know capacitor block dc ,thn how clamper add dc level.plz ans me sir.
why cannot capacitor polarities won't change during different cycles
the basic property of capacitor is that it cant change voltage across its plates instantaneously..
To change the polarity of capicator should get discharged and then get charged and it requires time of 10 time constant which is very larger than t/2
ThankU
where is power Amplifier
u r confusing in all questions by changing the polarity , it can be done simply by KVL and answer will come same as yours
yes exactly
Vi=-10 V, -Vi also = -10V?!
from T/2 to T: you stated that Vi=-10V and Vc=8V. and then, you have the equation Vo= -Vi -Vc.
if you sub (-10V) into the eq. , isnt it gonna be Vo= 10 - 8 ?
the sign of the eq. make me confused even though I think -18V is right
@Neso Academy For +ve cycle would'nt it be like 0v across the resister since capacitor act as open circuit for dc voltage.
Thank you Sir