this man teaches things more simpler than my course teacher. My course teacher literally came one day and started these maths with biasing without explaining this simple matters earlier.. thanks a lot maannn!!!
Bro, I am a teacher of electronics. You are doing a great job which is very useful to students all over the world. I was watching this video just to refresh the things after a long time and I found that at 8:50 the question you solved has wrong solution. Please correct it if possible. The question has potential divider and for a max input of 10 volt the potential drop across RL will be 5 volt only i.e. the diode remain off during the whole cycle and no portion of input will be clipped. Output will maintain sinusoidal waveform with a max voltage of 5 volt. Thanks
It would be voltage divider only during the negative wave cycle when two resistors are in a series, so voltage across RL and RS are equal. But during the positive wave cycle RL is in parallel and we need to consider equivalent resistance of the circuit, not individual resistance of RL.
You have great way of explaining things. I learned about these circuit. @9:24 the maximum voltage is 5V, because the voltage divider divides it by two (10/2=5) and never gets to 6.7V, However, if your input voltage was say 20V then we get to 6.7V. Thanks again for good clip.
You have always saved my ass. I have no words to thank you! I am a senior student, I have been following your content since freshman year. Thank you so much
ill be honest, u been uploading videos so in sync with my college sem and modules I have taken you to be my main teacher. im doing analogue electronics this sem and I really needed this. Thank you for all the work you put into this.
@@PunmasterSTP it went well. Got my diploma since then and now on the way to get my degree in mechatronics. I still come back here to refresh everything I learned.
@@alirizvy8229 That's great! Though I got my engineering degree awhile ago, I still tutor related subjects and I definitely come here to refresh myself. I also didn't quite know about mechatronics, but I googled it and it looks awesome!
My college (indianatech) just send us here to for our lessons sometimes. Other times I just watch the professor to know what to look up as they do a horrible job explaining anything. Then I come here to learn. Thank you so much.
@@intheharness I tutor online and (mostly before the pandemic) in person and I think this kind of behavior on the part of professors has gotten worse since the quarantine. What were your observations?
Hey man. You saved me so many times. I really appreciate your videos. I just wanted to thank you. I'm really grateful to you. Your teaching skills are amazing.
Really stretched my mind with this lecture. I never heard of clipper circuits. Definitely learned a lot though probably I'll need to rewatch it one more time.
Remember when solving using kvl, the diode is parallel to RL, and therefore set the "Vout" at node between diode and RL. Solve for Vout and you'll get the (+) or (-) voltage. Also if diode is ideal, no need for 0.7 drop. If not ideal it is given in problems.
For the circuit at around 10:07...The output voltage will never go higher than 5V, because the peak input voltage is 10V. Vout=Vin/2, according to voltage divider rule, right? (Because RL is equal to RS in this particular circuit) So the peak output voltage is 5V. How can the output voltage be clipped at 6.7V if it never goes higher than 5V?
In 10:00 the diode will not be in on state for the given sine wave of amplitude 10 alteast 13.4V i.e 6.7*2 is required for diode to be ON.So the circuit will never clip.
I don't understand the circuit starts at 10:00 of this video.Is the anode peak voltge 5volt?If the peak voltage of athod is 5v then the diode will be always off.
How is it that the diode's value function changes from a 0.7 drop to a limiter of 0.7 when it comes to the shunt clipper circuit? also, how is it the biased positive shunt circuit sees an increase of the 6 volts when it's positive charge runs directly to the block of D1? Thanks in advance for any answer, loving these videos!
Is the output signal at 11:24 possible? In my opinion, the diode would be all the way open cuz the voltage is distributed to two resistor. Thus, the potential across diode is not high enough to make the current go through.
I want to note that in the example were Vz=12volt in the negative half of the single the second diode is fowrward basie not revers based …and thanks for the vedios
At 10:30 seconds you introduced battery in the clipper circuit. And the output is 6.7V at RL. Since input is 10V i cant figure out how the voltage is 6.7V. Is it because the current is flowing back to source(AC)??
I love all of your videos, but at 5:00 you said the diode is parallel to the source. If the definition of two parallel dipoles is a couple of dipoles that are connected to the same nodes (and thus having the same voltage across them), then the diode in this example isn't parallel to the source. As far as I know, if two dipoles aren't in series, that doesn't mean they're parallel. What do you think?
before clipping the width of one of those bumps is lambda/2 right? after clipping does it remain the same. since the signal is vertically shifted by 0.7v, i feel like it should be slightly smaller???? 🤔
So am I understanding correctly that if you have a resistor and diode in parallel that if the diode has the lower voltage of the two elements then that diode element will be the one current flows through and the resistor will share the same voltage?
so for the negative shunt clipper circuit what happens between the parallel resistance and the diode? Is there any voltage going through the resistance, or is it all just going through the diode since it has close to 0 resistance?
How cannot the current flow from the negative terminal of the battery though the cathode for the negative half cycle? The diode should be forward biased for both sides because you have negative voltage flowing through the negative side of the diode and positive voltage flowing through the anode for the positive half cycle?
Late but, it's because of the diode regulating the voltage hence the "clipping". Whilst on the negative cycle, since it has no diode regulating it, its voltage can vary up until it hits the -5 V peak then head towards the "transition" stage. Tho this is based on observation. I suggest studying up more examples.
before clipping one of these bumps has width of lambda/2. after clipping does it change. when clipping output signal is shifted vertically by 0.7v right? so i feel like it should be slightly smaller??? e🤔
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this man teaches things more simpler than my course teacher. My course teacher literally came one day and started these maths with biasing without explaining this simple matters earlier.. thanks a lot maannn!!!
I'm currently finishing my education as an electronics technician and these videos have been gold for studying. Please keep it up man.
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How much do they pay in your job very curious
The man, The myth, The legend!!!!!!!!
Gotta save this semester with this video!
Bro, I am a teacher of electronics. You are doing a great job which is very useful to students all over the world. I was watching this video just to refresh the things after a long time and I found that at 8:50 the question you solved has wrong solution. Please correct it if possible. The question has potential divider and for a max input of 10 volt the potential drop across RL will be 5 volt only i.e. the diode remain off during the whole cycle and no portion of input will be clipped. Output will maintain sinusoidal waveform with a max voltage of 5 volt. Thanks
yeah i was thinking about the same. The input voltage max should be at least 13,5V to see some clipping
It would be voltage divider only during the negative wave cycle when two resistors are in a series, so voltage across RL and RS are equal. But during the positive wave cycle RL is in parallel and we need to consider equivalent resistance of the circuit, not individual resistance of RL.
hmm.. but they are in parallel doesn't mean they have the same voltage?
You have great way of explaining things. I learned about these circuit.
@9:24 the maximum voltage is 5V, because the voltage divider divides it by two (10/2=5) and never gets to 6.7V, However, if your input voltage was say 20V then we get to 6.7V.
Thanks again for good clip.
You have always saved my ass. I have no words to thank you! I am a senior student, I have been following your content since freshman year. Thank you so much
ill be honest, u been uploading videos so in sync with my college sem and modules I have taken you to be my main teacher. im doing analogue electronics this sem and I really needed this. Thank you for all the work you put into this.
I feel that he’s been my teacher too! How did circuits and the rest of your semester go?
@@PunmasterSTP it went well. Got my diploma since then and now on the way to get my degree in mechatronics. I still come back here to refresh everything I learned.
@@alirizvy8229 That's great! Though I got my engineering degree awhile ago, I still tutor related subjects and I definitely come here to refresh myself.
I also didn't quite know about mechatronics, but I googled it and it looks awesome!
My college (indianatech) just send us here to for our lessons sometimes. Other times I just watch the professor to know what to look up as they do a horrible job explaining anything. Then I come here to learn. Thank you so much.
@@intheharness I tutor online and (mostly before the pandemic) in person and I think this kind of behavior on the part of professors has gotten worse since the quarantine. What were your observations?
I love this voice ♥️
Hey man. You saved me so many times. I really appreciate your videos. I just wanted to thank you. I'm really grateful to you. Your teaching skills are amazing.
Really stretched my mind with this lecture. I never heard of clipper circuits. Definitely learned a lot though probably I'll need to rewatch it one more time.
These videos are very helpful in my sem exams.... Keep going 😇
And while scrooling and in tension when I saw this video I said "salute to you sir".
thank you so much it clears my all doubt lots of love from india
my favorite teacher
Mine too!
Me too
So best a teacher we like your explanations world's best
Remember when solving using kvl, the diode is parallel to RL, and therefore set the "Vout" at node between diode and RL. Solve for Vout and you'll get the (+) or (-) voltage.
Also if diode is ideal, no need for 0.7 drop. If not ideal it is given in problems.
Thank u sooooooo much
U r a hero
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For the circuit at around 10:07...The output voltage will never go higher than 5V, because the peak input voltage is 10V. Vout=Vin/2, according to voltage divider rule, right? (Because RL is equal to RS in this particular circuit) So the peak output voltage is 5V. How can the output voltage be clipped at 6.7V if it never goes higher than 5V?
This all rests upon the assumption that the diode will be off until the voltage across RL is greater than 0.7+6. Is this right to say?
I have the same idea as yours
Thank you sooo much. This is very useful for my studies 💚💚💚 keep it up!!!!!
Am kmathaddict a math tutor on RUclips and I like this video 🥰🥰🥰
I love this man a lot!
In 10:00 the diode will not be in on state for the given sine wave of amplitude 10 alteast 13.4V i.e 6.7*2 is required for diode to be ON.So the circuit will never clip.
At 10:46, why did the negative portion of the wave maintained its form when the input voltage is 10V? Shouldn't it be clipped too?
Best teacher
I don't understand the circuit starts at 10:00 of this video.Is the anode peak voltge 5volt?If the peak voltage of athod is 5v then the diode will be always off.
You are the best !!!!!!!
Much love from 😊🇿🇲🇿🇲🇿🇲 Zambia
bruh finally a video with proper english
you're great man! thank you 🙏
thank you for making this. It is super useful.
11:44 There is a mistake in this video: The positive peak is not 6.7V but instead 5V. The diode is reverse biased here.
How is it that the diode's value function changes from a 0.7 drop to a limiter of 0.7 when it comes to the shunt clipper circuit? also, how is it the biased positive shunt circuit sees an increase of the 6 volts when it's positive charge runs directly to the block of D1? Thanks in advance for any answer, loving these videos!
how do you know all things?? :o
Is the output signal at 11:24 possible? In my opinion, the diode would be all the way open cuz the voltage is distributed to two resistor. Thus, the potential across diode is not high enough to make the current go through.
Thanks for another truly amazing video, and all the knowledge you have shared!
Thank you!!
Oh, so that's how it works thank you :)
Thaaank you, you saved me
this show is very good
In 16:23, why he did that? Like the sine wave when it reaches 12.7 volts, it become flat. Pls answer huhu
what is the significance of load resistance ?? what happens if we don't connect RL in parallel circuit?
I want to note that in the example were Vz=12volt in the negative half of the single the second diode is fowrward basie not revers based …and thanks for the vedios
😮thank You legend 👏😀
At 10:30 seconds you introduced battery in the clipper circuit. And the output is 6.7V at RL. Since input is 10V i cant figure out how the voltage is 6.7V. Is it because the current is flowing back to source(AC)??
I don't understand either, I would like note on this too
Great vid bro!
13:51 why clip at 0.7 volt when two 10k resistor are there so there must be 0.35 volt across RL
I hope someday a quantum physical playlist comes up.
Sbinalla
I love all of your videos, but at 5:00 you said the diode is parallel to the source.
If the definition of two parallel dipoles is a couple of dipoles that are connected to the same nodes (and thus having the same voltage across them), then the diode in this example isn't parallel to the source.
As far as I know, if two dipoles aren't in series, that doesn't mean they're parallel.
What do you think?
If they aren't in series they are in parallel. Simple as that.
@@CharlieScarver nope =)
Thank you! It helped a lot.
before clipping the width of one of those bumps is lambda/2 right? after clipping does it remain the same. since the signal is vertically shifted by 0.7v, i feel like it should be slightly smaller???? 🤔
So am I understanding correctly that if you have a resistor and diode in parallel that if the diode has the lower voltage of the two elements then that diode element will be the one current flows through and the resistor will share the same voltage?
At 11:18 how did the output voltage become 6.7v and not 5.3v?
Umukamba uyu 😹 since fresher 3rd year pa EEE achili aletu papa
Wa last JG as a monk wapa Unza natasha
Is this half-wave rectification? Or what's the difference between clipper circuits and rectification?
Half wave rectification and clipper diodes are the same
Wow, thanks for the info!
so for the negative shunt clipper circuit what happens between the parallel resistance and the diode?
Is there any voltage going through the resistance, or is it all just going through the diode since it has close to 0 resistance?
How cannot the current flow from the negative terminal of the battery though the cathode for the negative half cycle? The diode should be forward biased for both sides because you have negative voltage flowing through the negative side of the diode and positive voltage flowing through the anode for the positive half cycle?
awesome explaination thank u
How do you solve for voltage through the diode. Not the forward.
Awesome job!
what about offset in output wave form???
At 13:33 min the out put is 10v then at the graph it should be 10v not 5v
BEAUTIFUL!!
for 10:54 why only the positive part will be clipped but no the negative part too?
Late but, it's because of the diode regulating the voltage hence the "clipping". Whilst on the negative cycle, since it has no diode regulating it, its voltage can vary up until it hits the -5 V peak then head towards the "transition" stage.
Tho this is based on observation. I suggest studying up more examples.
So nice
before clipping one of these bumps has width of lambda/2. after clipping does it change. when clipping output signal is shifted vertically by 0.7v right? so i feel like it should be slightly smaller??? e🤔
Is it series clippers circuit a half wave rectifier?
Is it implied that the AC source is producing 10 volts?
How can I tell when it's parallel and series pls?😢
do you know how?
Perhaps 10v rather than 5v at 14:04
thanks!
Have u tried any of these theoretical exercises at 1mhz ?
How to tell when its "clipped"(squarewave) vs just a regular sine wave
Amazing
perfect
Thanks our prelim is tom.
makasih bang doain aku besok berhasil ya bang
it would be better if you focus more on Diode limiter circuit
thaaank uuuuuuu
Nice
tysm
thank uuuuuuuuuu
6.42 why is is negative .7
11:52
Diode
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Hi everyone 👋
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Gran ya place
everyone sees but no one likes or subscribes :(
Jabe see this?
Exemplary
Understood what you said, just didn't understand the reasoning
Here is a quick revison video
ruclips.net/video/hhovTZCUCjw/видео.html
.
5:20