The Timestamps for the Different Topics Covered in the Video. 0:20 What is Clipper Circuit? Applications of Clipper Circuit 1:14 Types of Clipper Circuit 2:00 Parallel Clipper Circuit 4:06 Parallel Clipper Circuit with Biasing Voltage 10:23 Series Clipper Circuit 11:21 Series Clipper Circuit with Biasing Voltage 13:16 Example 1 15:19 Example 2
Its 8:53AM and My test is at 9:30 AM and I am done with video!! This topic has High weightage..I really studied max topics from ur channel after getting result I will update this comment.!! AMAZING CONTENT!
The input is triangular wave and it is gradually increasing to 12V. But during that rise in the voltage, when the input is less than 5V, the voltage at the cathode of the diode will be less than 5V. While at the anode voltage is 5V. So, for that duration when input is less than 5V, the diode will get forward biased (considering ideal diode). And for that time it will conduct. That means when Vin < 5V, Vout = 5V. I hope it will clear your doubt.
Sir, if it is possible for you then please playlist for all the topics. It is hard to search all topics separately. But if you arrange them in a playlist then it would get easy for us access your content.
No, there won't be any effect during the positive half cycle. Because if you see, during the positive half cycle, the diode remains in the reverse bias condition. I hope it will clear your doubt.
The series clipper circuit clips the certain portion of the waveform (in time domain), much like a rectifier. Similar to the rectifier circuit, here the diode will conduct when the voltage across the anode of the diode is more than cathode. And during the ON time of the diode, the output voltage is Vin - V (assuming ideal diode for a moment). That is why you are seeing the reduced voltage during the conduction of the diode. I hope, it will clear your doubt.
At 14:08, Should the sign of zener diode not be be opposite? I think you have made a mistake declaring the sign of the zener diode. Please check it and tell me if your example 1 is correct or not
When the Zener diode operates in the breakdown region, then the voltage across it is equal to the Zener voltage. The polarity is correct. For more info, please check this video on Zener Diode: th-cam.com/video/MZPeRlst8rQ/w-d-xo.htmlsi=f3Nb-aHyhqGYlhw_
It restricts the flow of current through diode. When the diode is conducting, then remaining voltage will drop across this resistor. (e.g Vin - V will drop across the resistor, when diode is conducting.) I hope, it will clear your doubt.
You can tell me , why In the parallel clipper ckt, when +ve half cycle is applying in the ckt the ckt work as a short ckt ? What is the working of resister her ?
The R is the current limiting resistor. And here during the analysis, it has been assumed that the diode is ideal. That means when it conducts it acts as a short circuit. But actually in the output when the diode is conducting, the voltage would be around 0.7V. I hope it will clear your doubt.
When Vin is between V1 and -V2 both diodes are reversed biased. If you see, the diode D1 will conduct, when the voltage at the anode is greater than V1. Similarly, diode D2 will conduct when the voltage at the cathode is less than -V2. That means when the input voltage is in between V1 and -V2, both diodes will be reversed biased. I hope it will clear your doubt.
Since diode acts as open circuit, so there is no current through the diode. ( Considering ideal diode). And since, diode is open, so circuit will not get completed and hence no current through the circuit. So even though there is a resistor in series with voltage source, but there will not be any voltage drop across it. And hence, Vo = Vin during that negative half cycle. I hope it will clear your doubt.
hi, for example 2, i got the reverse version of your graph ie. my pulse was on the negative side, is this fine? I assumed this was because the direction I chose to be my positive and negative cycles were opposite to yours. Does this matter?
In example 2 when the Vin less than 5V the Vout is 5V ..I have some doubt in that part .Is there is no effect for Vin in Vout when the Vin less than 5V .I think Vout =5V -Vin when Vin is less than 5V..? Pliz clarify this part..
The input signal plays role in turning on or off the diode. (Here it has been assumed that the diode is ideal diode). Depending on the conduction of the diode, the output will change. Vout is the output across the one terminal of the diode and reference node. You can simply find that voltage by applying the KVL. You can also simulate the same circuit and check the result by yourself.
Hi there, thanks for the informative video! Can I check why the Vin need not be larger than (0.7 + 0.5)V? To account for ideal diode voltage + bias voltage. Why is it only accounting for bias voltage (0.5) ?
During positive half cycle one zener will work as normal diode while the second one will work as zener diode. So, it will clip the waveform at vz+ 0.7 v. Similarly on the negative side, the voltage will be clipped at (-Vz + 0.7)V. By selecting the proper value of Vz, the circuit can be used for the clipping the waveform at the desired level.
As its name suggest, it clips the input signal. When diode conducts, the voltage drop across diode is 0.7V. And in this way, it restricts the input voltage.
6:00 why output should be constant after V ? I mean as the current through the diode increases the voltage should also increase following it's I-V characteristic curve
During the entire discussion, the diode is considered as the ideal diode. Yes, what you are saying is true in the case of the actual diode, where the diode also has forward resistance. I hope it will clear your doubt.
If you see, the anode of the diode is connected to the 5V. And through the resistor, the cathode is connected to the input voltage. When Vin < 5V, then the anode (which is a5 5V) will be more positive than the cathode (Forward Biased) and hence the diode will conduct for that time. I hope it will clear your doubt. If you still have any doubt let me know here.
Example 4.23 time. I saw your explanation here. Is ANODE conected to AC input source and CATHODE to the V=5 v DC source. But i still dont understand why DC source allow current to flow in reverse? For DC source, current doesent exit from positive borne and enter back in source in negative borne? But ofcourse it will be opposite to the diode direction. So i dont understand. Or we have a special rule here.
Guys, in parallel clipper circuit in reverse bias the voltage is not zero..... Whereas in series clipper circuit in reverse bias the voltage is not zero na .....why? Like this can anyone explain me
Your video contents are good sir, but beginners will have many questions and doubts while watching your videos. Instead of explaining simply what is given in the textbook like a robot, explain more clearly by adding why this is clipping at this voltage etc. Explain in detail sir.
You mean to say, using Zener diode as a reference voltage, you want to conduct another diode, right ? If so, then yes you can do that. But it depends, how you cannect them in circuit ?
The output resistor will be the load resistor across which we want to get the clipper output. It could be a simple load resistor or it could be and input impedance of the some another circuit. ( In case, if the output of clipper is applied to some another circuit)
@@ToolsAandLogic What I mean to say was V volt in the video. But instead, I said, 5V in the video. That's why the same correction was mentioned there. I hope, it will clear your doubt.
Unbiased series diode negative clipper to remove the negative half of the wave form diode is connected the reverse biased no current passing through the diode How to form out put singnal
just want to ask your last example with triangular input, need some clarification. during positive half cycle the diode is "OFF" since it is reversed bias and will be "ON" during the negative half cycle then the signal will draw when Vin is greater than biasing Voltage but it turns out different from my expectation :) I thought it is negative clipper connected with biasing voltage . may you please check and help me clarify this. By the way thank you for this video, i want also to make one using our own language.
It depends, whether it is a series clipper or a parallel clipper. In a parallel, clipper circuit, the output is measured across the diode. When the diode is short-circuited, then output will also get short-circuited. And hence the output is 0 in case of a parallel clipper circuit. But in a series clipper circuit , when the diode conducts, there is a short circuit between input and output. And that's why the output follows the input.
The Timestamps for the Different Topics Covered in the Video.
0:20 What is Clipper Circuit? Applications of Clipper Circuit
1:14 Types of Clipper Circuit
2:00 Parallel Clipper Circuit
4:06 Parallel Clipper Circuit with Biasing Voltage
10:23 Series Clipper Circuit
11:21 Series Clipper Circuit with Biasing Voltage
13:16 Example 1
15:19 Example 2
Btw how u know where to put.... The horizontal line V of bias
studying this one night before the exam lol
omg same
Same 🤦🏻♂️
😑I am here to complete my assignment
I am here to complete my mini project.. Lol😆
And i am here to prepare shortly only for online class 😁
Well this guy certainly knows about clipping. You can hear it in the audio.
Lmao 😂
i have exam tom and i am laughing everytime he says "it will turn on"
perfect video better then my teacher..
thanks alot
Its 8:53AM and My test is at 9:30 AM and I am done with video!! This topic has High weightage..I really studied max topics from ur channel after getting result I will update this comment.!! AMAZING CONTENT!
When will u upadate this comment ?!
@@atom8509 🤣🤣🤣🤣
@@atom8509 😂😂😂
Bhai kitne number aye ?
@@saitamapreetsingh3057 i don't think he will reply now🙄😂
Who is watching this the night before the exam! 😃
Can someone please explain the last example? I did not get it :((
Seeing this video on November 10 th the day before my semester examination
for the first example i dont understand why the zeiner diode turn to an short circuit ( instead of open circuit ) when Vin < 0v ???
Mein bas itna kahunga
Charan kahaan hai aapkeeeee ........
I may have learned something
But I have learned "during this time" a lot more
15:50
Why will it get forward biased?????
It should be reverse biased and should act like an open circuit !!
Please someone explain!!
The input is triangular wave and it is gradually increasing to 12V. But during that rise in the voltage, when the input is less than 5V, the voltage at the cathode of the diode will be less than 5V. While at the anode voltage is 5V. So, for that duration when input is less than 5V, the diode will get forward biased (considering ideal diode). And for that time it will conduct. That means when Vin < 5V, Vout = 5V.
I hope it will clear your doubt.
Sir, if it is possible for you then please playlist for all the topics. It is hard to search all topics separately. But if you arrange them in a playlist then it would get easy for us access your content.
It is already added in diode and diode circuit playlist. You can check it on the playlist page.
why is the condition Vin
But consfused regarding zener voltage problem
3:40 sir no effects on positive half cycle will be seen for non ideal diode?
No, there won't be any effect during the positive half cycle. Because if you see, during the positive half cycle, the diode remains in the reverse bias condition.
I hope it will clear your doubt.
@@ALLABOUTELECTRONICS Thank a lot sir
12:43 series clipper with biasing , the signal not get clipped , it get smaller form .why?
The series clipper circuit clips the certain portion of the waveform (in time domain), much like a rectifier. Similar to the rectifier circuit, here the diode will conduct when the voltage across the anode of the diode is more than cathode. And during the ON time of the diode, the output voltage is Vin - V (assuming ideal diode for a moment). That is why you are seeing the reduced voltage during the conduction of the diode. I hope, it will clear your doubt.
At 14:08, Should the sign of zener diode not be be opposite? I think you have made a mistake declaring the sign of the zener diode. Please check it and tell me if your example 1 is correct or not
When the Zener diode operates in the breakdown region, then the voltage across it is equal to the Zener voltage. The polarity is correct. For more info, please check this video on Zener Diode: th-cam.com/video/MZPeRlst8rQ/w-d-xo.htmlsi=f3Nb-aHyhqGYlhw_
Studying this one night before an interview 😂
Does all the clipper clamper circuit have zener diode? Coz i needs input voltage more than diode activation voltage
You can also use additional voltage source (with diode) to clip or clamp the waveform at the desired level.
at 4:30 wont R affect the voltage at all? how can we directly compare V and Vin
It restricts the flow of current through diode. When the diode is conducting, then remaining voltage will drop across this resistor. (e.g Vin - V will drop across the resistor, when diode is conducting.) I hope, it will clear your doubt.
You can tell me , why In the parallel clipper ckt, when +ve half cycle is applying in the ckt the ckt work as a short ckt ? What is the working of resister her ?
The R is the current limiting resistor. And here during the analysis, it has been assumed that the diode is ideal. That means when it conducts it acts as a short circuit. But actually in the output when the diode is conducting, the voltage would be around 0.7V.
I hope it will clear your doubt.
At 7:53 when V in
When Vin is between V1 and -V2 both diodes are reversed biased.
If you see, the diode D1 will conduct, when the voltage at the anode is greater than V1.
Similarly, diode D2 will conduct when the voltage at the cathode is less than -V2.
That means when the input voltage is in between V1 and -V2, both diodes will be reversed biased.
I hope it will clear your doubt.
Brilliant way of teaching...hats off
11:33
It shouldn't be (Vin)-V>0.7?
Please explain Sir!!
Here the diode is assumed as an ideal diode. If it's not, then yes, Vin - V should be greater than 0.7V.
Informative and very straightforward. Keep up the good work!
++
0.7volt is shown in time axis?
2:47 why Vo is not 0 here?? There is resistance but no voltage drop
Since diode acts as open circuit, so there is no current through the diode. ( Considering ideal diode). And since, diode is open, so circuit will not get completed and hence no current through the circuit. So even though there is a resistor in series with voltage source, but there will not be any voltage drop across it. And hence, Vo = Vin during that negative half cycle. I hope it will clear your doubt.
I watched many videos of yours, they are awesome. I'll recommend these forever.
Could not Example 2 😢
hi, for example 2, i got the reverse version of your graph ie. my pulse was on the negative side, is this fine? I assumed this was because the direction I chose to be my positive and negative cycles were opposite to yours. Does this matter?
bro I am also facing the same problem...so do help me out if u got the reason to it
In example 2 when the Vin less than 5V the Vout is 5V ..I have some doubt in that part .Is there is no effect for Vin in Vout when the Vin less than 5V .I think Vout =5V -Vin when Vin is less than 5V..? Pliz clarify this part..
The input signal plays role in turning on or off the diode. (Here it has been assumed that the diode is ideal diode). Depending on the conduction of the diode, the output will change.
Vout is the output across the one terminal of the diode and reference node. You can simply find that voltage by applying the KVL.
You can also simulate the same circuit and check the result by yourself.
you know you are very underated
today i found your channel and yes subscribed it too.
helped me a lot
how you edit your videos?
Eg 2 is contradictory to what u explain earlier.when Vi5V then Vout=Vin-5V. Please explain this.
In example 2, when Vin
Now I’ll understand. Thank u 😊
sir last k dono example paralllel k lg rhe the to series k concepts kyu lga h
Me who studying it before half hour of exam....... Engineer wali feeling aa rhi h 🤣😂🤣🤣🤣
I m here after learning that this is in AIIMS mbbs entrance syllabus
Now it won't , cuz it's combined with neet
Hi there, thanks for the informative video! Can I check why the Vin need not be larger than (0.7 + 0.5)V? To account for ideal diode voltage + bias voltage. Why is it only accounting for bias voltage (0.5) ?
If you've found the answer, pls do share with me
How to find the Current across resistor
In 9:19 circuit
The current through resistor R is (Vin - Vo )/ R.
At 9:19, Vo = -V2.
So, accordingly, you can calculate the current.
@@ALLABOUTELECTRONICS
Thank you soo much for your reply
Sir in circuit 6:47 what graph when we reverse V attach to diode
The Vout = -V, when Vin > (-V)
And Vo = Vin, when Vin < -V
Sir, at 10:22 how will the zeners work the same way?
During positive half cycle one zener will work as normal diode while the second one will work as zener diode. So, it will clip the waveform at vz+ 0.7 v. Similarly on the negative side, the voltage will be clipped at (-Vz + 0.7)V. By selecting the proper value of Vz, the circuit can be used for the clipping the waveform at the desired level.
@@ALLABOUTELECTRONICS thanks Sir/Mam, i understood that by browsing other sources as well, but thanks for your reply, it solidified my understanding.👍
thanks a lot sir definitely solved a lot of doubts and cleared a lot of misconceptions . keep up the work.
about the last one....how does it V less than 5 V diode able to conduct?
When the input is less than 5V, then the cathode of the diode is at less than 5V, while the anode is at 5V, so the diode will conduct.
sir you hàve taken àll diodes ideal in series circuit?
Yes.
Good explanation
Watching Before going to exam 😅
If diode is silicon diode..out voltage how
Why clippers are used in tv receivers
Nice explanation
Hlo sir
Does the diode in all types of clippers acts like a switch?
As its name suggest, it clips the input signal. When diode conducts, the voltage drop across diode is 0.7V. And in this way, it restricts the input voltage.
@@ALLABOUTELECTRONICS sir I didn't get it
Draw circuit diagram of positive clipper with VR = 3V and Vi=5Sinwt can you drawn this sir
I am too looking answers for question like this. Where do I get them?
I'm getting used to THE.... cadenceoftheseprentations.
lmfaoooooooo
Very good , calm and full of concepts vedio....excellent work
Nice presentation
today i m just watching this video , tommarow will my end sem exam ......
Studying an hour before exam.
6:00 why output should be constant after V ? I mean as the current through the diode increases the voltage should also increase following it's I-V characteristic curve
During the entire discussion, the diode is considered as the ideal diode. Yes, what you are saying is true in the case of the actual diode, where the diode also has forward resistance.
I hope it will clear your doubt.
@@ALLABOUTELECTRONICS got it...thank you very much sir...
this video very helping for understanding clippers circuit.. many thanks
You are doing well kuch teachers hain even they are highly graduated after that they can't explain simply as like you thanx a lot 🌷
love how the audio in this video is also clipping
lmao fr
Studying one hour before exam🫠
In eg 2, when Vin
If you see, the anode of the diode is connected to the 5V. And through the resistor, the cathode is connected to the input voltage.
When Vin < 5V, then the anode (which is a5 5V) will be more positive than the cathode (Forward Biased) and hence the diode will conduct for that time.
I hope it will clear your doubt.
If you still have any doubt let me know here.
ALL ABOUT ELECTRONICS got it, thanks😃
Example 4.23 time.
I saw your explanation here.
Is ANODE conected to AC input source and CATHODE to the V=5 v DC source.
But i still dont understand why DC source allow current to flow in reverse?
For DC source, current doesent exit from positive borne and enter back in source in negative borne?
But ofcourse it will be opposite to the diode direction.
So i dont understand.
Or we have a special rule here.
Studying in day of exam😌🤣
Guys, in parallel clipper circuit in reverse bias the voltage is not zero..... Whereas in series clipper circuit in reverse bias the voltage is not zero na .....why? Like this can anyone explain me
11 hours student like here
Next day you will be passed best of luck
Thanks sir nice and simple
Thanks sir it help so much
Studying on the day of exam 😅😅😂
Your video contents are good sir, but beginners will have many questions and doubts while watching your videos. Instead of explaining simply what is given in the textbook like a robot, explain more clearly by adding why this is clipping at this voltage etc.
Explain in detail sir.
Studying this on the exam day morning 😂
I saw the vedio before my exam 🤧
😅nov 14th, Tomorrow's exam for me
is it me or his vioce sounds different the whole time
Watching this at night before exam and I got to know that much thing which I couldn't during whole semester..
hello sir, can we also use a Zener diode as a threshold voltage reference to conduct the diode?
You mean to say, using Zener diode as a reference voltage, you want to conduct another diode, right ? If so, then yes you can do that. But it depends, how you cannect them in circuit ?
Thanks a lot... Your the best.
Hi, can I ask what is the output response of a clipper circuit. Hope I can get a respone thank you
Do you mean the transfer function, or the in general output of the clipper circuit.
@@ALLABOUTELECTRONICSthan you for the response. What I mean is, how to get the output respone or voltage of a clipper circuit
what would be the output at resistor in clipping circuit??
The output resistor will be the load resistor across which we want to get the clipper output. It could be a simple load resistor or it could be and input impedance of the some another circuit. ( In case, if the output of clipper is applied to some another circuit)
No..the resistor before diode...what would be output there..
Ok, I thought you are asking about the output resistor.
Are you asking about the output across the resistor?
@@ALLABOUTELECTRONICS yes
Why did you say I voltage? Did not understand..
Would you please mention the timestamp where you are referring to ?
@@ALLABOUTELECTRONICS at 4.30 min from beginning
@@ToolsAandLogic What I mean to say was V volt in the video. But instead, I said, 5V in the video. That's why the same correction was mentioned there. I hope, it will clear your doubt.
I have been looking for this for so long thanks
Positive side of graph is why first taken in side of forward side why it is not taken in reverse side
Unbiased series diode negative clipper to remove the negative half of the wave form diode is connected the reverse biased no current passing through the diode
How to form out put singnal
1.3:08
2.3:15
3.3:45
4.5:35,6:05
5.7:09
6.10:04
7.10:28
8.11:09
9&10.11:21
Could you please explain breafly
Please make a video on communication for radio frequency...
Apply a triangular input signal using function generator to a biased series
positive clipper and draw the output waveforms.
ممكن حله
I wish native english speakers made more videos on this cuz mi lass idu a word hes saying
very nice
nice simple
Sir pls explain in more easy terms coz u are unable to clear some doubts
If you have any doubt please let me know here. I will try to resolve it.
Good lecture sir keep on upload the videos
Thanks a lot sir,u r explanation with examples help me a lot
just want to ask your last example with triangular input, need some clarification. during positive half cycle the diode is "OFF" since it is reversed bias and will be "ON" during the negative half cycle then the signal will draw when Vin is greater than biasing Voltage but it turns out different from my expectation :) I thought it is negative clipper connected with biasing voltage . may you please check and help me clarify this. By the way thank you for this video, i want also to make one using our own language.
Yes it is negative clipper as the negative portion is clipped but due to 5V biasing voltage only the Vin above the 5V mark is obtained in the output
Sir thoda hindi se bhi explain kr liya kijiye......
Omg.. feeling sad im not able to understand this topic by watching this video also 😢
Then ur not a human🤣🤣🤣💁🏻
@@Talksofarjuun2405 🥺😤
Can typical zener diodes handle higher ac frequency sine inputs in the 100kHz range?
THANKS FOR THE GREAT VIDEOS A+++
Sir kuch v nahi samajh aaya
kyuki theoretically ideal diode dekh sakte ho lekin practically ideal diode kon dega tere haath mein?
bro only at short circuit the input follows the output. you re saying that every time at open circuit outrput follows the input
It depends, whether it is a series clipper or a parallel clipper. In a parallel, clipper circuit, the output is measured across the diode. When the diode is short-circuited, then output will also get short-circuited.
And hence the output is 0 in case of a parallel clipper circuit. But in a series clipper circuit , when the diode conducts, there is a short circuit between input and output. And that's why the output follows the input.