Calls himself the organic chemistry tutor. Has posted more physics and math videos than chemistry videos. Has videos on all topics and domains of science. This guy sure is a legend !
I tried to figure this out by myself ALL day without any luck, my text book only gives very little information. And the circuit looks so simply only with few components, so in some point I thought that I am the stupidest person in the world and get frustrated. You save my LIFE. Thank you !!!!
You seriously rock!!!!! Amazing tutorials. Whenever I see one of your tutorials I always get happy because I think to myself "I'll actually understand this now". Thank you so much.
personal notes, I realized you can kvl capacitors because the supply voltage is AC, capacitor are opened during DC. Since it's AC you can get Vc and VL if it's an inductor. Also, the capacitor charges if current goes from + to -, and when doing kvl, initially, Vc = 0V for positive clamp, so doing kvl would equal load to supply.
Thank youuu. I dont get why the textbook does not go deeper into the time constant and explaining how it affects charging vs discharging (I didnt remember). The book just says the capacitor has a large constant. period. and doesnt give any details on how clamping behaves before steady state. Its so much complex to put everything together without a complete explanation like this.
4:05 To clarify, the diode is in parallel with the branch that contains the input signal and the capacitor, and is also in parallel with the load resistor.
So many new concepts for me in this video! I learned a lot although I needed to review several times. He did clarify one point from the earlier video that troubled me - so the Rl resistor does assume the voltage of the diode then?
Great video Sir,I only had to double check whether the voltage fluctuates around the capacitors and max voltage of both the positive and negative clamper circuit...I still love your videos ,Sir .
I think it can get a bit confusing, since discharging a capacitor with a given polarity would have the same direction of current as charging the capacitor with the *opposite* polarity. In either case, in the ideal positive clamper circuit, the capacitor shouldn't discharge appreciably during the positive half-cycle (or, on the very first positive half-cycle, charge the other way). Then during the negative half-cycle, the capacitor would (quickly) charge with the correct polarity.
So what you're saying is that the capacitor charges up in the negative cycle to offer a positive base around which the voltage fluctuates? Thanks for the explanation!
I see now how that capacitor works, it basically is a line except when the voltage exceeds the positive clamp voltage of the diode in the correct direction. When the voltage is bigger then the stable voltage of the diode on the positive direction, all of the rest of that excess voltage gets dumped into the capacitor, causing it to charge up to the maximum voltage remaining. (i.e the input minus the diode stable voltage)Then when it does, it will remain the same voltage and stay the same(idk why tho). Then what happens is as if the load is connected in series with a DC source, the voltage is the capacitor's. The process starts at the exact moment the peak positive voltage is reached, when the capacitor finished charging. After that it acts like a DC source, and founds the base around which the ac voltage fluctuates.
It seems to me that this circuit would not work well when driving a low impedance stage (e.g. a speaker). I suppose you would have to add buffering if that were the case.
How do we determine the polarity of the non-polarized capacitor? I'm not sure when to add the capacitor voltage or subract it from the input when trying to make my waveform?
in the case when diode is forward biased, the capacitor is charged up (because it connects to the voltage source along a low resistance path), you can determine the polarity of the capacitor by looking at the polarity of V input. in the first problem, the diode is forward biased when the top terminal is -ve and bottom terminal is +ve. so the polarity of the capacitor would be -+ from left to right.
At the negative clamper, why put a connection to the earth? If earth is at 0 V potential, than you have a short circuit between power source and ground no (during first wave)?
That’s a good observation. I think that as long as only one side of the voltage source is at ground, then it isn’t a short circuit. If the more negative side is grounded, then the source would generate a relative positive voltage, and vice versa. From the diagram, the other side of the voltage source has to go through the capacitor and then either the diode or the load resistor to get to ground.
Hi! There is some points I didn’t understand : 1- at point c which is the debut of the negative cycle the diode is forward biased but it also needs 0.3 V to conduct. Correct me if I’m wrong but when the negative cycle comes the diode anode is at O V but I can’t figure out what’s the potential at the cathode. In other words I would like to know how the diode cathode potential varies during the half negative cycle. 2- since during the half negative cycle the capacitor becomes more and more charged until it reaches Vm how the diode could still conduct if the cathode which is connected to the positive side of capacitor is more positive than the anode knowing that the anode is at 0V ? 3- I always used to draw the potential arrow from the positive side of the component to its negative one. From D to E the diode doesn’t conduct anymore. Then the circuit is just Vin C1 and RL. So for me I would draw the potential arrow from right to left for C1, from top to bottom for RL and from bottom to top for Vin since we are still in the negative cycle which means the bottom is most positive than the top. According to Kirchhoff law I will sum the arrow in the same direction and substract the arrow in opposite direction : Vout + Vin = VC1. This is how I would do. In others words I don’t get how you could say that from D to E Vout is the sum of Vc1 and Vin. So these are the 3 points I didn’t really get from the video. Any help would be appreciated. Thank you !
What happens if we introduce a series resistor in the input side and for time varying square wave input signal, how to calculate R and C (if the input signal frequency is around 1MHz how the circuit behaves and what type of diode we have to chose ?).
why does the Vm (the voltage across the capacitor) not change in the right-hand side of the negative half circle at the first circle ? (still remains 9.7V)
7:03 how does the + half cycle current flow through C1? Isn’t C1 supposed to accept current from right to left in this arrangement? I mean since it’s it’s an electrolytic capacitor i thought it could only be used in a DC circuit
i do see but the logics are blowing my mind why for the negative clamper you introduce the ground and the Vc becomes negative but in the positive clamper everything was on point mathematically and scientifically?
Physics tells us that capacitor stores energy through electric field, but electrons cannot pass through dielectric between the plates, but the way he explained, electrons pass through the dielectric but block by the diode in reverse bias, is that so? I hope he would explains it further.
@@ohmslaw6856 I think it depends how you think about it. Electrons will move onto one plate when other electrons move off of the other, so from the outside, it looks like current is moving “through” the capacitor. But in actuality, no electrons are crossing between the plates (at least unless the capacitor starts breaking down).
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Calls himself the organic chemistry tutor. Has posted more physics and math videos than chemistry videos. Has videos on all topics and domains of science.
This guy sure is a legend !
I tried to figure this out by myself ALL day without any luck, my text book only gives very little information. And the circuit looks so simply only with few components, so in some point I thought that I am the stupidest person in the world and get frustrated. You save my LIFE. Thank you !!!!
Same
Absolutely same.
@@PunmasterSTP me too lol i think its because if even a single idea is missing then it gets harder to make sense
You seriously rock!!!!! Amazing tutorials. Whenever I see one of your tutorials I always get happy because I think to myself "I'll actually understand this now". Thank you so much.
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wtf now you have this topic as well? you are a legend mate!
personal notes, I realized you can kvl capacitors because the supply voltage is AC, capacitor are opened during DC. Since it's AC you can get Vc and VL if it's an inductor. Also, the capacitor charges if current goes from + to -, and when doing kvl, initially, Vc = 0V for positive clamp, so doing kvl would equal load to supply.
Bro, can't tell you how much I owe you after watching this video
Thank you for existing
Better than my lect. All credits to u Sir
You are a gift from God thank you so much for the work you do 🙂
Amazing tutorial, please keep doing these! I'm going go through all the other EE tutorials now, thanks!
I love ur voice and ur description tnk u i learn a lot
My teacher took 4 periods to teach this, but I didn't understand :((( Thank you so much
you are the best tutor. thank you so much for what you are doing. God bless you.
You deserve so much more attention my dude. Thanks a bunch
Thank youuu. I dont get why the textbook does not go deeper into the time constant and explaining how it affects charging vs discharging (I didnt remember). The book just says the capacitor has a large constant. period. and doesnt give any details on how clamping behaves before steady state. Its so much complex to put everything together without a complete explanation like this.
Organic Chemistry please, please make a video on timing diagrams. The basic ones, ones with MUX's and LUTs! please I beg of you
i trust u more than my aec tutor
Why he always better than my EE lecturer
4:05 To clarify, the diode is in parallel with the branch that contains the input signal and the capacitor, and is also in parallel with the load resistor.
So many new concepts for me in this video! I learned a lot although I needed to review several times. He did clarify one point from the earlier video that troubled me - so the Rl resistor does assume the voltage of the diode then?
You're a great tutor ... GOD bless you
bestt one till now
Thalaiva vera level..
very great explanation video
Amazing information presented in a wonderful way. Thanks so much!
Great video Sir,I only had to double check whether the voltage fluctuates around the capacitors and max voltage of both the positive and negative clamper circuit...I still love your videos ,Sir .
You are just amazing!
Many thanks for the tutorial! Crystal clear, as usual. I guess you meant "will not discharge completely" @7:24.
no he meant charge because the resistance will take all the voltage then the capacitor's voltage will be approxi. zero.
I think it can get a bit confusing, since discharging a capacitor with a given polarity would have the same direction of current as charging the capacitor with the *opposite* polarity. In either case, in the ideal positive clamper circuit, the capacitor shouldn't discharge appreciably during the positive half-cycle (or, on the very first positive half-cycle, charge the other way). Then during the negative half-cycle, the capacitor would (quickly) charge with the correct polarity.
he meant charge because resistance is low, if it was high it wouldn't discharge completely
sir can you please help us with the laplace transform videos problems and answers
So what you're saying is that the capacitor charges up in the negative cycle to offer a positive base around which the voltage fluctuates? Thanks for the explanation!
I see now how that capacitor works, it basically is a line except when the voltage exceeds the positive clamp voltage of the diode in the correct direction. When the voltage is bigger then the stable voltage of the diode on the positive direction, all of the rest of that excess voltage gets dumped into the capacitor, causing it to charge up to the maximum voltage remaining. (i.e the input minus the diode stable voltage)Then when it does, it will remain the same voltage and stay the same(idk why tho). Then what happens is as if the load is connected in series with a DC source, the voltage is the capacitor's. The process starts at the exact moment the peak positive voltage is reached, when the capacitor finished charging. After that it acts like a DC source, and founds the base around which the ac voltage fluctuates.
I think that is a great way of putting it!
Great video!!! Your videos are amazing!!!!
Tech Sure Couldn’t agree more!!!!
It seems to me that this circuit would not work well when driving a low impedance stage (e.g. a speaker). I suppose you would have to add buffering if that were the case.
thank you
How do we determine the polarity of the non-polarized capacitor? I'm not sure when to add the capacitor voltage or subract it from the input when trying to make my waveform?
in the case when diode is forward biased, the capacitor is charged up (because it connects to the voltage source along a low resistance path), you can determine the polarity of the capacitor by looking at the polarity of V input. in the first problem, the diode is forward biased when the top terminal is -ve and bottom terminal is +ve. so the polarity of the capacitor would be -+ from left to right.
great video :) I had a doubt: is the time period of the clamped signal the same as the original? Because just looking at the diagram, I'm not sure.
At the negative clamper, why put a connection to the earth? If earth is at 0 V potential, than you have a short circuit between power source and ground no (during first wave)?
That’s a good observation. I think that as long as only one side of the voltage source is at ground, then it isn’t a short circuit. If the more negative side is grounded, then the source would generate a relative positive voltage, and vice versa. From the diagram, the other side of the voltage source has to go through the capacitor and then either the diode or the load resistor to get to ground.
you are 100% the best
thank you for the great videos
Hi! There is some points I didn’t understand :
1- at point c which is the debut of the negative cycle the diode is forward biased but it also needs 0.3 V to conduct. Correct me if I’m wrong but when the negative cycle comes the diode anode is at O V but I can’t figure out what’s the potential at the cathode.
In other words I would like to know how the diode cathode potential varies during the half negative cycle.
2- since during the half negative cycle the capacitor becomes more and more charged until it reaches Vm how the diode could still conduct if the cathode which is connected to the positive side of capacitor is more positive than the anode knowing that the anode is at 0V ?
3- I always used to draw the potential arrow from the positive side of the component to its negative one. From D to E the diode doesn’t conduct anymore. Then the circuit is just Vin C1 and RL. So for me I would draw the potential arrow from right to left for C1, from top to bottom for RL and from bottom to top for Vin since we are still in the negative cycle which means the bottom is most positive than the top. According to Kirchhoff law I will sum the arrow in the same direction and substract the arrow in opposite direction : Vout + Vin = VC1. This is how I would do. In others words I don’t get how you could say that from D to E Vout is the sum of Vc1 and Vin.
So these are the 3 points I didn’t really get from the video. Any help would be appreciated.
Thank you !
thanks a lot
you are a guru sir
What happens if we introduce a series resistor in the input side and for time varying square wave input signal, how to calculate R and C (if the input signal frequency is around 1MHz how the circuit behaves and what type of diode we have to chose ?).
You ser a genious man
Does a clamping circuit change anything about the shape of the waveform?
at 14:14 , can you please explain why we need to use a polarized capacitor
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why does the Vm (the voltage across the capacitor) not change in the right-hand side of the negative half circle at the first circle ? (still remains 9.7V)
what to do when negative clamper has, freq, capacitor, battery and rl value?
7:03 how does the + half cycle current flow through C1? Isn’t C1 supposed to accept current from right to left in this arrangement? I mean since it’s it’s an electrolytic capacitor i thought it could only be used in a DC circuit
i do see but the logics are blowing my mind why for the negative clamper you introduce the ground and the Vc becomes negative but in the positive clamper everything was on point mathematically and scientifically?
can we use Laplace Transform to draw the waveform at output if all the values are given and the time period of input waveform.
please explain.
I want to ask a question What happens on the graph if I change the resistor value?
Thank you so much
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Love you boss
Amazinnnnng dude I love it
8:35 can the capacitor charge and/or when the current flows through the capacitor in either direction?
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Physics tells us that capacitor stores energy through electric field, but electrons cannot pass through dielectric between the plates, but the way he explained, electrons pass through the dielectric but block by the diode in reverse bias, is that so? I hope he would explains it further.
Current does flow through capacitor
I may be wrong but I think he was saying that if only a small current goes to the capacitor, then the plates will have very close to the same voltage.
@@ohmslaw6856 I think it depends how you think about it. Electrons will move onto one plate when other electrons move off of the other, so from the outside, it looks like current is moving “through” the capacitor. But in actuality, no electrons are crossing between the plates (at least unless the capacitor starts breaking down).
So nice
Hi! may i ask if the clamper circuit out can be used to resonate rlc circuit? Thanks
Why don't they just use a biasing voltage,we will instantly get the clamping
why the beginning of the graph is missing?
please help
To which graph are you referring? Could you provide a time stamp?
What happens if the resistor is not present??
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this was entertaining to read, thanks!
I’m a bit confused; is this an inside joke or does it have more context?