Here finally in 2nd part we got Vo=Vi and the out part is exactly equal to negative part of Vi. In it's just previous video , we got Vo=-Vi and then too you drew the exact negative waveform of Vi as Vo. How sir? Pls explain
when vi is less than 4 volt , the diode is off but if we connect a load resistance in output then a current flow through 2.2k resistance also then why we won't consider the voltage drop across this R??
Doubt : At time 2: 58 , how is the Vo = Vi when diode is reverse biased. Since the diode is open-circuited , no current will flow through the diode. Hence we have current =0 and Resistance = infinity. V= 0x infinity = zero. Please correct my understanding.
bro bcz in diode modelling we had studies that if there is ideal diode and diode is Reverse biased to input voltage is equal to output voltage because there will be no voltage drop across load resistance rL
4:45 if both circuits are reverse biasing the diode then the diode will act as an open circuit and current will not flow. Hence Vo will be zero. How come Vo is equal to Vi. Please someone explain me
current will not flow through the path containing diode. but Vo is voltage across terminals which can be seen to equal to Vi since both are sharing the same terminal (since vr=0 it can be considered as a short circuit).alternatively by kvl we will get Vo=-Vi. hope I could get to the point
sir, how is this circuit "parallel " ? . Sir i do agree with you that it is baised as an external source is there and clipper as it clipped + cycle but how is it "parallel " ?
Exactly what I am battling with. Conventional current is +ve to -ve and that's what we use for designs unlike electron current. If we take a closer look at his analysis from the AC input, he used +ve to -ve for the loop current so I don't know why he changed it for the DC source
When both of the voltage sources are reverse biasing, there occurs an open circuit, but there will be current flow (if load is added in parallel) through R if load is added in parallel with the diode.Why is R ignored
Thanks for this elaboration. Will you please explain how Vr (2.2k) is zero in case of reverse biased.The parallel blocked elements (where diode is present) will be open circuit. However it forms a closed loop from Vin to Vout through the resistance 2.2k. How there will not be any voltage drop across the 2.2k resistance ? Confused a bit, please explain...
Ignore Vr , Then for ideal diode , resistance offered by the diode in reverse bias is infinity , then whole input voltage will appear across the Vo not at resistance 2.2 kilo ohms , therefore the output voltage is equal to positive of the input voltage. Hope you guys have understood it .
in baised parallel clippers of the first question when diode is reversed bias ,how come Vo=Vi,as if no current flows then vo should be equal to zero right ?
Using KVL and the polarities assigned, you get Vi-Vo=0, making Vo subject of the formula yields; -Vo=-Vi. then cancel the negative signs by dividing both sides of the equation by -1 you then get Vi=Vo
@@nesoacademy You must be shameless to respond to this unimportant message while there are many much more important ones pointing out your errors waiting to be answered!
@@luismais4742 diode is nowhere near the resistor... you are mistaking the voltage source with resistor, and even then your argument is stupidly wrong.
May I know the voltage during the (-)ve half cycle? I found it being mentioned in the comments as -8V, I think it should be -12V (8V+4V). Please clarify me.
When the diode is ON , it is a short circuit. So how can we just take the Battery alone when there is input Vs connected as well to consider. You acre correct if you did not have the V input and only had the battery. Think again !! Again we need to apply KVL to figure out the V output. It is not just a battery standing alone.
4:45 if both circuits are reverse biasing the diode then the diode will act as an open circuit and current will not flow. Hence Vo will be zero. How come Vo is equal to Vi. Please someone explain me
your analysis is true for very high load resistance parallel to diode isn't it, else all the voltage drops across 2.2 k ohm resistor, so the voltage at top terminal of output is -Vi for positive half cycle and +Vi for negative cycle, therefore on negative half cycle Vo is zero until the Vi reaches 4v (because until then the potential at top terminal of output hasn't reach 4v to forward bias the diode). Therefore negative half cycle in output can have maximum of -4v not -8v.
I tried simulating this circuit in proteus and found that I am wrong and your analysis is correct. Yeah indeed, the proper way to anlayze the circuit is by analyzing the voltages in two terminals of diode. I think i got it.
NO ,in the above ckt,though we the increase the biased 4v DC supply to a value greater than input peak value,the diode doesnt get on.and hence output appears same as input.
In reverse bias,two voltage sources,i.e.,Vi as well as 4V-both make the diode to operate in reverse bias,as a result,there is no conflict between 2 voltage sources,so there is no need to consider the effect of 4 volts separately.
4:45 if both circuits are reverse biasing the diode then the diode will act as an open circuit and current will not flow. Hence Vo will be zero. How come Vo is equal to Vi. Please someone explain me
@@emotionaljackfruit because that diode will be open ckt and open ckt have infinte resistance so no current flows through that branch so current flow through only that upper part so that's why Vo=Vi
Across all academies i have ever crossed yours is the best until now in every aspect and subject!
Neso academy is the best platform for education.
Those who are having problem to understand just add a load resistance at the output. It will help you understand the problem.
yaa.. that's very helpful
Can u explain ehy
Why
Becoz first of all to be parallel circuit there must be a load resistance in parallel with diode
For negative half cycle vo=-vi 4:40
Ya
Here finally in 2nd part we got Vo=Vi and the out part is exactly equal to negative part of Vi. In it's just previous video , we got Vo=-Vi and then too you drew the exact negative waveform of Vi as Vo. How sir? Pls explain
Sir thank you for giving all such videos for free of cost.
in 1:57 min
F.B. V(i) > 4v
u neglect the resistance voltage drop
Sudip Roy because in clippers you are assuming that theres no current thats why there will be no voltage drop across in the resistor
@@luismais4742 sir i did not understand why there is no current
@@Amityphysics because that series resistance have less value than load resistance and can be neglected
@@opre6451 may I ask why is the series resistance has less value?
when vi is less than 4 volt , the diode is off but if we connect a load resistance in output then a current flow through 2.2k resistance also then why we won't consider the voltage drop across this R??
Doubt : At time 2: 58 , how is the Vo = Vi when diode is reverse biased. Since the diode is open-circuited , no current will flow through the diode. Hence we have current =0 and Resistance = infinity. V= 0x infinity = zero. Please correct my understanding.
bro bcz in diode modelling we had studies that if there is ideal diode and diode is Reverse biased to input voltage is equal to output voltage because there will be no voltage drop across load resistance rL
@@engineergem but in series clippers question sir showed 0. Output in reverse bias
It actually contains resistor parallel to diode so
You are not answering many doubts , just simply showing a method . Anyways a big fan of your c programming lectures.👍
Itne easily to mko ABCD nhi yaad Hui thi
Bhot easy explanation hai yrr.
Loved it.
4:45 if both circuits are reverse biasing the diode then the diode will act as an open circuit and current will not flow. Hence Vo will be zero. How come Vo is equal to Vi. Please someone explain me
current will not flow through the path containing diode. but Vo is voltage across terminals which can be seen to equal to Vi since both are sharing the same terminal (since vr=0 it can be considered as a short circuit).alternatively by kvl we will get Vo=-Vi. hope I could get to the point
@@shatakshibharti5222 Thank you for your explanation I had the same confusion like him.
Doubts got clarified in a single video.thank u sir
Sir why we use the resister (2.2kohm) in sries withe biased diode.
You are a life saver,, very well explained, thank you! Good job🙏
sir, how is this circuit "parallel " ? . Sir i do agree with you that it is baised as an external source is there and clipper as it clipped + cycle but how is it "parallel " ?
Series
Because the output is taken parallel to the diode
At 3:02 why voltage across resistor is zero ?
Coz we take the r as veryy low quantity. It merely takes any voltage drop
Your way of explanation is really excellent ☺️☺️☺️☺️
🎉😢
at 1:59 when in f.b +ve cycle how Vo = 4v ? either (Vo = V) or because (Vo = Vi - V = 8-4=4 )
sir where is the load resistance which is parllel with the diode making it a parllel clipper
He didn't draw it
sir why did you not consider voltage drop at resistor ??
When Vi>4v then Vo=4v.but when we apply KVL in the loop...the above relations does not satisfied.Please expain.
Thanx sir for clearing my dought and also give short trick to solve this kind of problem
at 1:13,actually the 4v is forward biasing the circuit right as its flows from '+'to' -' tve
the current draws from negative(-) to positive (+)
Exactly what I am battling with.
Conventional current is +ve to -ve and that's what we use for designs unlike electron current.
If we take a closer look at his analysis from the AC input, he used +ve to -ve for the loop current so I don't know why he changed it for the DC source
When both of the voltage sources are reverse biasing, there occurs an open circuit, but there will be current flow (if load is added in parallel) through R if load is added in parallel with the diode.Why is R ignored
Thanks for this elaboration. Will you please explain how Vr (2.2k) is zero in case of reverse biased.The parallel blocked elements (where diode is present) will be open circuit. However it forms a closed loop from Vin to Vout through the resistance 2.2k. How there will not be any voltage drop across the 2.2k resistance ? Confused a bit, please explain...
There will not be any current flow through the 2.2k resistor. Hence voltage drop in 2.2k resistor is zero.
@@bharathvarun then from where current pass through to output load ?.from your home?
@@jb-bt4gq 🤣
Ignore Vr ,
Then for ideal diode , resistance offered by the diode in reverse bias is infinity , then whole input voltage will appear across the Vo not at resistance 2.2 kilo ohms , therefore the output voltage is equal to positive of the input voltage.
Hope you guys have understood it .
in baised parallel clippers of the first question when diode is reversed bias ,how come Vo=Vi,as if no current flows then vo should be equal to zero right ?
Tq so much sir
I love teaching sir ❤️ ❤️......
4:40 output voltage equal to -Vi
Using KVL and the polarities assigned, you get Vi-Vo=0, making Vo subject of the formula yields; -Vo=-Vi. then cancel the negative signs by dividing both sides of the equation by -1 you then get Vi=Vo
Very well explained Sir....🎉😊❤
all confusions are vanished after watching this . thanks Neso Academy
@@nesoacademy You must be shameless to respond to this unimportant message while there are many much more important ones pointing out your errors waiting to be answered!
for negative half cycle there will be 4v also in the input, so is it v0=-(vi+4v) instead of v0=-vi, please reply tomorrow is my exam
thank u so much from egypt sir
Nice explanation 💙💙
Sir why we haven't considered the voltage drop across 2.2k despite having large resistance value?
i=0 implies that voltage drop=i.R=0
sir can you explain me why diode is forward bais when Vi>4 and reverse bais when Vi
When vin =0 then Vo=-4v in +ve cycle then why u start waveform from 0 can u elaborate..i m confused
When input voltage =0 then there is no current flow through circuit bcoz then circuit is open
sir how vo=vi in first cycle in reverse bias, i didn't get it?
in reverse bias, whole of the input voltage appears across output. isn't there some voltage drop across the 2.2kohm resistor?
Yeah. I'm wrong here.
@@luismais4742 diode is nowhere near the resistor... you are mistaking the voltage source with resistor, and even then your argument is stupidly wrong.
why in the negative half vo=vi how you applied kvl
Sir, Why is Voltage across the Resistor 0 when Vi
Because current is zero (open circuit)
@@varshakp3693 no its not open cicuit theload resistance is connected to complete the cicuit
When Vi>4V, don't you think that there will be voltage drop in resistance R, which can make diode Reverse biased.
No because at the instance of transition the current flow is 0
If we change value of resistor, will it affect Output????
how is this parallel biasing if load resistance is not connected in parallel to the diode
How we come to know that diode is ideal or practical? If not given in statement
If it's not given in statement you can assume it to be ideal. It will be given in statement if it's anything other than ideal.
why the voltage across resistor is zero for reverse bias of diode??
Thanks. Understood perfectly
May I know the voltage during the (-)ve half cycle?
I found it being mentioned in the comments as -8V, I think it should be -12V (8V+4V). Please clarify me.
its because the diode is in reverse bias.
Luis Mais Thanks got it!
why in reverse bias current from resistor is zero??
Sir,. at 1.39 during forward biased in + ve half cycle when vi > 4 v...vo should be 0...how it becomes 4 v??
When the diode is ON , it is a short circuit. So how can we just take the Battery alone when there is input Vs connected as well to consider. You acre correct if you did not have the V input and only had the battery. Think again !! Again we need to apply KVL to figure out the V output. It is not just a battery standing alone.
@@robertfaney4148 Vin will be dropped across R
this is very helpful
How come 4v became RB in +ve half cycle?
sir baised circuit Mai load resistant ke across voltage kiyon nahi lete hai please tell me
Sir can you please explain how Vo is equal to Vi in the reverse biased condition of the diode
Shouldn't it be Vo = -Vi in the -ve half cycle? (4:50)
No
great video! I like how you broke it down
U saved me! Thank u
But what will be the -ve pick value of output 🤔🤔
And also if we attach load resistance across output, what will happen to output????
4:45 if both circuits are reverse biasing the diode then the diode will act as an open circuit and current will not flow. Hence Vo will be zero. How come Vo is equal to Vi. Please someone explain me
your analysis is true for very high load resistance parallel to diode isn't it, else all the voltage drops across 2.2 k ohm resistor, so the voltage at top terminal of output is -Vi for positive half cycle and +Vi for negative cycle, therefore on negative half cycle Vo is zero until the Vi reaches 4v (because until then the potential at top terminal of output hasn't reach 4v to forward bias the diode). Therefore negative half cycle in output can have maximum of -4v not -8v.
I tried simulating this circuit in proteus and found that I am wrong and your analysis is correct.
Yeah indeed, the proper way to anlayze the circuit is by analyzing the voltages in two terminals of diode. I think i got it.
I don't understand why its called parallel clippers when there is no load resistance that is parallel to the diode
Yup
Hey Thankyou💚💜💛
sir when diode is reverse biased it acts as open circuit so how Vo becomes equal to Vi shouldn't it be 0 pls correct me if i am wrong
Diode not pass current only in those wire inwhich he connect
See carefully v input is directly connected to v output when diode is reverse biased
How is it a parallel clippers example,sir?
Dude just assume the load across the output voltage..It won't have any effect on the resultant output wave.
can we get
Vo>Vi in clipper circuit???
NO ,in the above ckt,though we the increase the biased 4v DC supply to a value greater than input peak value,the diode doesnt get on.and hence output appears same as input.
How voltage drop is across the resistance is zero ..... Vr =0 ???
How is this a parallel clipper? Pls ANSWER!
sir if the diode voltage is less than the dc voltage we have to change the diode direction , sir please elobrate
Thank you sir
Why in FB vo=4v
And in RB vo=vi
Can you please tell me why there is no current flowing through the resistor 2.2 Kohm when the diode is reversed biased?
Connection nhi h.. Diode open ho rha h
what if R input is not negligible and there is voltage drop in it?
Current does not flow through R since the diode is open circuited. So R has no significance
where is the load resistance?
In negative cycle Vo = -8v.
depends to the biasing source if its positive or negative.
thank you
How come V not = 4v , it needs some elaboration
Thank u so much sir
negative half cycle Vo = -Vi isn't ?
in previous unbiased parallel clipper lecture sir u have applied kvl and get Vo=- Vi in negative half cycle
In unbiased parallel clipper at 3:24 sec
i have the same confusion plz tell me which lecture the point was mentioned.
Vi 4 then output voltage is vi
in reverse biasing why you had not considered the effect of 4 volts????
2:15
In reverse bias,two voltage sources,i.e.,Vi as well as 4V-both make the diode to operate in reverse bias,as a result,there is no conflict between 2 voltage sources,so there is no need to consider the effect of 4 volts separately.
Whenever forward biases explation not clearly
thanks sir
sir parallel clipper load resistance parallel to diode .this is not a parallel clipper
bhai kis nalle college mei hai. Input voltage is parallel to the diode. dekh dhyan se
tu kis college me h???
agree with u.
according to "introduction to clampers" videos
Parallel clippers are those where load resistance is parallel with diode,.
This Is Not An Example Of Parallel . Resistance And Diode Are In Series In This One
Thank U
Isme vi=-vo na hoga for negative cycle
kisi ke pass sir ne jo bhe krvaya hai uske hand written notws hai to plzz de doo yar
Why Vr=0 ........at 3.00min
Sir i am understand parallel clipper please clarify
4:45 if both circuits are reverse biasing the diode then the diode will act as an open circuit and current will not flow. Hence Vo will be zero. How come Vo is equal to Vi. Please someone explain me
@@emotionaljackfruit because that diode will be open ckt and open ckt have infinte resistance so no current flows through that branch so current flow through only that upper part so that's why Vo=Vi
Thanks a lot...sir
The videos are good but tentatively a bit slow
I don't know but this was a series biased clipper.
Sir,there is a mistake...once the diode becomes forward baised... The Vo is 4+diode knee voltage
If it was silicon then Vo becomes 4.7
he used an ideal diode there. which is equivalent to a close circuit.
Tomorrow my exam. So please quickly make a vidoe on previous year gate ECE ques
love you
👏
😍😍
this must be biased series clipper as diode is connected in series with resistor
Please don't say you can clearly see that " this is that " 😑..
baised parallel circuit