The Timestamps for the different Topics covered in the video: 0:19 What is Clamper Circuit? Types of Clamper Circuits 1:32 Positive Camper Circuit (with circuit analysis) 6:40 Negative Clamper Circuit (with circuit analysis) 10:27 Negative Clamper Circuit with Biasing Voltage (with circuit analysis) 12:44 Positive Clamper Circuit with Biasing Voltage (with circuit analysis)
Literally you're the only channel which explained the positive clamper with such clarity. Rest of the tutors are considering the negative half cycle first. They're not talking what happens if we have positive half cycle first. Thank you! Thanks a lot!!!!🙏
Thank you very much, I have been very confused about clamper circuits before watching this video, even after going through multiple articles. Though it didn't strike at first to me and I have re-watched it a couple of times, but finally I got it and it felt really great. You guys have explained it pretty well.
I figured out my own question. I am, by no means, an LTSpice expert. But I managed to model my negative clamping circuit. Under normal conditions, this circuit would clamp the output to -2Vin. However, my output is across a capacitor (low pass filter) that is effectively shunting the negative half cycle of Vin to ground. Under that condition, the circuit clamps the output to -Vin. LTSpice clearly shows the output being clamped to -Vin minus the voltage drop across the diode. It does take quite a few cycles for the output voltage to reach a steady state value. The higher the frequency, the longer it takes to reach steady state. At 10MHz it takes about 800ms. In order to smooth out the ripple in the output voltage, I added an additional capacitor to the output. This pretty much removed all the ripple, but it did increase the amount of time it takes to reach a steady state value of -Vin. If it weren't for your excellent video, I might never have figured it out!
Great explanation of clamping circuits! I have a question regarding a negative clamping circuit where the output is across a capacitor rather than a resistor. In my application the resistor and capacitor are basically a low pass filter that effectively grounds the negative half cycle of all frequencies above 20Hz. It would also ground the positive half cycle but the diode is virtually a short circuit which allows the series capacitor to charge to Vin, but the output is zero. Not sure how to model what happens on the negative half of the cycle.
If you apply the KVL, then voltage at the cathode of the diode is Vin + VM. While the voltage at the anode is 0. Therefore, from top to bottom, the voltage across the diode is Vin + VM. I hope, it will clear your doubt.
hi, i have sold TVS528 diode to my customer , he is saying:- During our application test Indexing test failed , Clamping Voltage found unsufficient and indexing angle not detected . i would like to ask you can we check the diode thru multimeter?
Great video mate Just one doubt In the 1nd PART of negative half cycle the capacitor charged to Vm And got discharged in 2nd PART Of negative half cycle So the capacitor drained out r8ght So in the 2nd positive half cycle the capacitor should no longer have Vm voltage with it right
The capacitor won't get discharged completely after the time 3T/4. (during the second half of the first negative half cycle) Because, as I mentioned in the starting of the video, for the clamper circuit, the RC time constant is much greater than the time period. RC >>T. that means there is an only slight reduction in the capacitor voltage. At the end of the video, I have also shown the simulation result. Please go through it. You will get a better idea of the circuit. For the exact timing, you can check the timestamps in the description of the video.
The Clamper Circuit is used when you want to shift the DC level of the signal. While clipper circuit is used for the protection. In the circuit when you don't want your signal to go beyond the certain voltage in any circumstances then the clipper circuit can be used. It will clip the waveform or the signal if it tries to go beyond certain range. I hope it will help you.
Because after 3T/2 time, the input voltage Vin is greater than -Vm, That means the voltage observed at the cathode of the diode (Vin + Vm) is greater than zero. At 3T/2 it is zero, as Vin = -Vm. That's why cathode is more positive than the anode. (Anode is ground potential) For more info, please check the simulation graph at the end of the video.
How is the capacitor getting fully charged to vm ? We know that capacitor takes 5RC time to get fully charged , but here the input is reaching it's peak value in T/4 time , where T=0.1RC approx , that means input is reaching it's peak value in RC/40 time , so how is the capacitor getting fully charged up , as we know it takes atleast 5RC time to get fully charged .
If you notice, the capacitor charges through the diode (when a diode is conducting). The forward resistance of the diode is much smaller than the load resistor. And to make the analysis simple, it has been assumed that the diode acts as a short circuit. So, capacitor charges in no time. The capacitor discharges through the load resistor. But for the clamper circuit, the necessary condition is RC >> T. So, in one cycle, it can be assumed that the charge across the capacitor is almost constant. I hope it will clear your doubt. If you still have any doubt, let me know here.
Actually, it is because of the voltage across the capacitor. If you see the voltage across the capacitor at that time, it is equal to Vm - V. And supply voltage is -Vm. So, the total output voltage is = -2Vm +V. I hope it will clear your doubt.
If you see, the voltage at the anode of the diode is 0V, and the voltage at the cathode is Vin + Vm. At the most, the voltage at the cathode will go up to zero. (when Vin = -Vm). But it will not turn on the diode. Therefore, the diode will remain in the reverse bias condition during the negative half cycle. I hope, it will clear your doubt.
During the second negative half cycle, the voltage across the diode ( between cathode and anode ) is Vin + Vm. The minimum value of Vin is - Vm. So, at the most the voltage across the diode will go to 0. That means the cathode will be always more positive than anode during second negative half cycle. And therefore, it will remain off. I hope, it will clear your doubt.
Here the positive and the negative half cycle refers to the positive and negative half cycle of the input sinusoidal signal. When the input sinusoidal signal is positive that refers to the positive half cycle and for the remaining half period when the input is less than 0, it refers to the negative half cycle. I hope, it will clear your doubt.
Because after 3T/2 time, the input voltage Vin is greater than -Vm, That means the voltage observed at the cathode of the diode (Vin + Vm) is greater than zero. At 3T/2 it is zero, as Vin = -Vm. That's why cathode is more positive than the anode. (Anode is ground potential) For more info, please check the simulation graph at the end of the video.
It is the time required by the capacitor (RC circuit) to charge to 63 % of the applied voltage. I have already covered that in the earlier videos of transient analysis. For more info, you can watch those videos on transient analysis.
During the positive half cycle, the capacitor will get charged to the peak voltage Vm. And here it has been assumed that the RC time constant of the circuit is much larger than the time period of the signal. That means from T/2 to T, when the input signal is reducing, the capacitor will hold its charge and hold the voltage across it to Vm. That means at the beginning of the negative half cycle, the voltage across the capacitor is Vm and hence the voltage across the diode is Vin + Vm.
For a moment, just don't see, how the Vin is changing. (Whether it is negative or positive). It can be expressed as Vm sin (wt). So, just using the KVL, it will be Vin + Vm. I hope it will clear your doubt. If you still have any doubt then do let me know here.
@@ALLABOUTELECTRONICS sir according to convention if we take first sign then as negative half cycle if we move from down to above then +sign of Vin amd then as we move upwards then there is first negative sign appears so we should take vin-vm according to me so plz clear my doubt plz sir
That's what I told you initially. Don't consider the negative half cycle of the input. In the sin(wt), when you put t>T/2, it is already considered. Just assume it is Vin with positive terminal on top and negative on bottom. I hope now you will get it.
Starting from basic, there are two types of sources. AC and DC. In the AC source, voltage continuously changes with time. while in case of DC voltage source, the voltage level remains fixed irrespective of the time. In many practical circuits, you will find both AC and DC sources.
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The Timestamps for the different Topics covered in the video:
0:19 What is Clamper Circuit? Types of Clamper Circuits
1:32 Positive Camper Circuit (with circuit analysis)
6:40 Negative Clamper Circuit (with circuit analysis)
10:27 Negative Clamper Circuit with Biasing Voltage (with circuit analysis)
12:44 Positive Clamper Circuit with Biasing Voltage (with circuit analysis)
Literally you're the only channel which explained the positive clamper with such clarity. Rest of the tutors are considering the negative half cycle first. They're not talking what happens if we have positive half cycle first. Thank you! Thanks a lot!!!!🙏
Your literally the only person on RUclips with comprehensive videos on the subject of electronics. Thank you for what you do.
Fr fr
Thank you very much, I have been very confused about clamper circuits before watching this video, even after going through multiple articles. Though it didn't strike at first to me and I have re-watched it a couple of times, but finally I got it and it felt really great. You guys have explained it pretty well.
The explanation of clamper circuit is the best one. I never seen this explanation any book or lectures. Thank u Man.
I figured out my own question. I am, by no means, an LTSpice expert. But I managed to model my negative clamping circuit. Under normal conditions, this circuit would clamp the output to -2Vin. However, my output is across a capacitor (low pass filter) that is effectively shunting the negative half cycle of Vin to ground. Under that condition, the circuit clamps the output to -Vin. LTSpice clearly shows the output being clamped to -Vin minus the voltage drop across the diode. It does take quite a few cycles for the output voltage to reach a steady state value. The higher the frequency, the longer it takes to reach steady state. At 10MHz it takes about 800ms. In order to smooth out the ripple in the output voltage, I added an additional capacitor to the output. This pretty much removed all the ripple, but it did increase the amount of time it takes to reach a steady state value of -Vin. If it weren't for your excellent video, I might never have figured it out!
Every video about electronics is found on your channel. Thanks for your great effort.
best education channel ever ..!!
Thanks to you, we electronics student found a good place for understand our concept
Thank You So Much Sir. Very Great and Beautiful Explanation Sir. Love From Pakistan. ❤
Indian engineering education is amazing! Very few American universities are at the IIT level.
I have microelectronics midterm tomorrow! This is a lifesaver
You are the greatest of all time bruh
Perfect explanation.. Strongly recommended 👍🏿👍🏿👍🏿
Very helpful video, super explanation, so good efforts, very good teaching
Grand salute to you sir, thanku!! 💜
Thank you broh! U made my day!
How the output voltage for the negative half cycle is - 2Vm+V for the Negative Clamper circuit with Biasing Voltage ? Please explain
Really useful video, thank you All About Electronics!
Thanks a lot. Very nice way of clarification.
Thanks from Bangladesh
Some points are not explained clearly just feels straight Outta text book
Result saver video😎😎
Much much better than neso academy
very very clear explanation,i appreciate it
Thanks for your video! I Appreciate it!
Thank you for this vdo sir ... good doubt clearing explanation & concept.
very nice explanation...great job.
Very helpful ,,thank u
SIR - U R EXCELLENT
Great video sir 👍👌💚
Are you a college professor and what degrees do you have - You are very knowledgeable.
he completed masters from IIT DELHI one of the most prestigious engineering institutes of india..
Hiee! Sir ur teaching was very understandable....Y can't uh add pdf link in description! It will help us more!
I have added the links in few videos. Also, I am providing the notes and Quiz in pdf form under membership.
Great explanation of clamping circuits! I have a question regarding a negative clamping circuit where the output is across a capacitor rather than a resistor. In my application the resistor and capacitor are basically a low pass filter that effectively grounds the negative half cycle of all frequencies above 20Hz. It would also ground the positive half cycle but the diode is virtually a short circuit which allows the series capacitor to charge to Vin, but the output is zero. Not sure how to model what happens on the negative half of the cycle.
Thanks .It was very helpful for me
Thank you sir. Your videos are really helpful.
are there videos on clamping circuits with zener diode?
Yes, please check the third example in this video. Here is the link : ruclips.net/video/u9IP_cKn5io/видео.htmlsi=_CKV9ePjc7IFyDVi
@@ALLABOUTELECTRONICSThank you so much!
Thank you for all
Excellent analysis
5:13 ... Why volatge across diode is Vin + Vm ???
If you apply the KVL, then voltage at the cathode of the diode is Vin + VM. While the voltage at the anode is 0. Therefore, from top to bottom, the voltage across the diode is Vin + VM. I hope, it will clear your doubt.
Thank you to the mooon and back!
hi, i have sold TVS528 diode to my customer , he is saying:- During our application test Indexing test failed , Clamping Voltage found unsufficient and indexing angle not detected .
i would like to ask you can we check the diode thru multimeter?
Thank you so muuuuuuuchhhhh🙏🏻🙏🏻
Great video mate
Just one doubt
In the 1nd PART of negative half cycle the capacitor charged to Vm
And got discharged in 2nd PART Of negative half cycle
So the capacitor drained out r8ght
So in the 2nd positive half cycle the capacitor should no longer have Vm voltage with it right
The capacitor won't get discharged completely after the time 3T/4. (during the second half of the first negative half cycle)
Because, as I mentioned in the starting of the video, for the clamper circuit, the RC time constant is much greater than the time period. RC >>T. that means there is an only slight reduction in the capacitor voltage.
At the end of the video, I have also shown the simulation result.
Please go through it. You will get a better idea of the circuit.
For the exact timing, you can check the timestamps in the description of the video.
@@ALLABOUTELECTRONICS oh
I was careless
Thank you
Thank sir aapki vajase I can also do this
I love u 3000
U made me pass
When a Marvel fan studies electronics
Good material, thx for upload.
You can also explain clipper and clamper circuits using opamps ..in later videos ...😊😊
Yes, definitely.
I m a mechatronics student plz difference b/w and application of clipper and clamper video
The Clamper Circuit is used when you want to shift the DC level of the signal. While clipper circuit is used for the protection. In the circuit when you don't want your signal to go beyond the certain voltage in any circumstances then the clipper circuit can be used. It will clip the waveform or the signal if it tries to go beyond certain range.
I hope it will help you.
Thank you!
At 11:20 ...in negative clamper circuit with biased voltage , how the output voltage becomes V volt in positive half cycle ...it should be zero ?
Which software is used in this video for drawing circuits??
At 04:15 _ how the cathode is more positively charged than the anode?
Because after 3T/2 time, the input voltage Vin is greater than -Vm, That means the voltage observed at the cathode of the diode (Vin + Vm) is greater than zero. At 3T/2 it is zero, as Vin = -Vm. That's why cathode is more positive than the anode. (Anode is ground potential)
For more info, please check the simulation graph at the end of the video.
what is the program you use for your presentation?
also which program you used for simulation?
Multisim Live Online Circuit Simulator.
Thank you for the video bro
How is the capacitor getting fully charged to vm ? We know that capacitor takes 5RC time to get fully charged , but here the input is reaching it's peak value in T/4 time , where T=0.1RC approx , that means input is reaching it's peak value in RC/40 time , so how is the capacitor getting fully charged up , as we know it takes atleast 5RC time to get fully charged .
If you notice, the capacitor charges through the diode (when a diode is conducting). The forward resistance of the diode is much smaller than the load resistor. And to make the analysis simple, it has been assumed that the diode acts as a short circuit. So, capacitor charges in no time. The capacitor discharges through the load resistor. But for the clamper circuit, the necessary condition is RC >> T. So, in one cycle, it can be assumed that the charge across the capacitor is almost constant.
I hope it will clear your doubt.
If you still have any doubt, let me know here.
@@ALLABOUTELECTRONICS excellent
U r excellent bro!!
Very well explained. Thanks a lot!
thank you very much
helpful. thank you
4:44 does the capacitor voltage reach Vm only when the input voltage is -Vm ? Or it can reach Vm before that input voltage
Yes, capacitor voltage reaches Vm during first negative half cycle when input voltage is -Vm.
Super sir
Thank you
@12:00 the Vout is -2Vm, not -2Vm+V. Since V is open ciruited, it will not have any influence on the o/p voltage . Right?
Actually, it is because of the voltage across the capacitor. If you see the voltage across the capacitor at that time, it is equal to Vm - V. And supply voltage is -Vm.
So, the total output voltage is = -2Vm +V.
I hope it will clear your doubt.
Very nice 👌👌
does capacitor at charged condition 13v, gets less charged after some cycles.....???
At 5:41 , why does the diode remain reverse biased? Isnt it forward biased as current flow thru the diode from anode to cathode?
If you see, the voltage at the anode of the diode is 0V, and the voltage at the cathode is Vin + Vm. At the most, the voltage at the cathode will go up to zero. (when Vin = -Vm). But it will not turn on the diode. Therefore, the diode will remain in the reverse bias condition during the negative half cycle. I hope, it will clear your doubt.
Thanku so much
What happens if we introduce a series resistor in input side of the circuit how it behaves ?
thank you sir
❤
sir how in 2nd negative cycle of positive clamper circuit the diode is still reverse biased , the diode should be forward biased right?
During the second negative half cycle, the voltage across the diode ( between cathode and anode ) is Vin + Vm. The minimum value of Vin is - Vm. So, at the most the voltage across the diode will go to 0. That means the cathode will be always more positive than anode during second negative half cycle. And therefore, it will remain off. I hope, it will clear your doubt.
Very nice
At 4:00 how voltage acroos the diode is Vin + Vm ... Vin is still in negative cycle.
why does the Vm (the voltage across the capacitor) not change in the right-hand side of the negative half circle at the first cycle?
Thanku sir
Sir we can't use normal capacitor_?
While connecting the polarized capacitors, you need to consider the polarity.
What software you used to simulate ?
Multisim
why do we ignore the first open circuit region and start from the portion of input wave where the diode is conducting?
at 3:36 can someone explain how we are determining what is positive half cycle etc
Here the positive and the negative half cycle refers to the positive and negative half cycle of the input sinusoidal signal. When the input sinusoidal signal is positive that refers to the positive half cycle and for the remaining half period when the input is less than 0, it refers to the negative half cycle. I hope, it will clear your doubt.
well explained
Some points are'nt explained clearly....at 4:15
Because after 3T/2 time, the input voltage Vin is greater than -Vm, That means the voltage observed at the cathode of the diode (Vin + Vm) is greater than zero. At 3T/2 it is zero, as Vin = -Vm. That's why cathode is more positive than the anode. (Anode is ground potential)
For more info, please check the simulation graph at the end of the video.
what is RC time ? plz reply im 1st year ec student
It is the time required by the capacitor (RC circuit) to charge to 63 % of the applied voltage. I have already covered that in the earlier videos of transient analysis. For more info, you can watch those videos on transient analysis.
How it will be vin+vm at 3:57
During the positive half cycle, the capacitor will get charged to the peak voltage Vm.
And here it has been assumed that the RC time constant of the circuit is much larger than the time period of the signal.
That means from T/2 to T, when the input signal is reducing, the capacitor will hold its charge and hold the voltage across it to Vm.
That means at the beginning of the negative half cycle, the voltage across the capacitor is Vm and hence the voltage across the diode is Vin + Vm.
17:32 why it is still reverse biased ?? Im confused sir
At 3:59min, how Vin+Vm ,why not Vin-Vm
For a moment, just don't see, how the Vin is changing. (Whether it is negative or positive). It can be expressed as Vm sin (wt).
So, just using the KVL, it will be Vin + Vm.
I hope it will clear your doubt.
If you still have any doubt then do let me know here.
@@ALLABOUTELECTRONICS sir according to convention if we take first sign then as negative half cycle if we move from down to above then +sign of Vin amd then as we move upwards then there is first negative sign appears so we should take vin-vm according to me so plz clear my doubt plz sir
That's what I told you initially. Don't consider the negative half cycle of the input. In the sin(wt), when you put t>T/2, it is already considered. Just assume it is Vin with positive terminal on top and negative on bottom.
I hope now you will get it.
Sir do you have anything related to latches and flip flops??
What happens to the Vout if theres a change at the resistor value?
The output voltage may change slightly. But as far as RC is much greater than T, it will not affect much.
I want to go down to the Quicky mart for a squishy
which software did you use to simulate the results?
Multisim
which simulation software is being used in this video?
Multisim
Which circuit simulator do you use
Multisim
im from non electronics background, bro what does this dc level means?
Starting from basic, there are two types of sources. AC and DC. In the AC source, voltage continuously changes with time. while in case of DC voltage source, the voltage level remains fixed irrespective of the time.
In many practical circuits, you will find both AC and DC sources.
@@ALLABOUTELECTRONICS thanks bro
how do you simulate those circuits and get those graphs
Using Multisim
Since capacitor is getting discharged but slowly, we can't take voltage across capacitor as Vm for ever right? Can someone clear this doubt?
Video is nor working 😵😔
Where you do simulation
Multisim Live
Thanks
which software do you use for simulation?
he used multisim
where did you do the simukation part ?? in which software ??
Multisim Live