Clipper Circuits
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- Опубліковано 22 лют 2020
- This electronics video tutorial provides a basic introduction into clipper circuits. This includes the positive and negative series clipper circuits as well as the parallel or shunt clipper circuits using diodes, resistors, and even zener diodes.
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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.
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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?
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.
Thanks for another truly amazing video, and all the knowledge you have shared!
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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.
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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!
So nice
thanks!
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
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?
perfect
Nice
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?
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 hope someday a quantum physical playlist comes up.
Sbinalla
how do you know all things?? :o
what is the significance of load resistance ?? what happens if we don't connect RL in parallel circuit?
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
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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 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?
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.
thank uuuuuuuuuu
what about offset in output wave form???
Thanks our prelim is tom.
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 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)??
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.
13:51 why clip at 0.7 volt when two 10k resistor are there so there must be 0.35 volt across RL
How do you solve for voltage through the diode. Not the forward.
How to tell when its "clipped"(squarewave) vs just a regular sine wave
Have u tried any of these theoretical exercises at 1mhz ?
makasih bang doain aku besok berhasil ya bang
Is it series clippers circuit a half wave rectifier?
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!
In 16:23, why he did that? Like the sine wave when it reaches 12.7 volts, it become flat. Pls answer huhu
At 13:33 min the out put is 10v then at the graph it should be 10v not 5v
🎉
for 10:54 why only the positive part will be clipped but no the negative part too?
At 11:18 how did the output voltage become 6.7v and not 5.3v?
Is it implied that the AC source is producing 10 volts?
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 =)
Diode
it would be better if you focus more on Diode limiter circuit
5:20
11:52
Gran ya place
How can I tell when it's parallel and series pls?😢
do you know how?
Here is a quick revison video
ua-cam.com/video/hhovTZCUCjw/v-deo.html
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Understood what you said, just didn't understand the reasoning
everyone sees but no one likes or subscribes :(
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