Introduction to Clampers
ฝัง
- เผยแพร่เมื่อ 29 พ.ค. 2016
- Analog Electronics: Introduction to Clampers
Topics Covered:
1. What is clamper circuit .
2. Difference between clippers and clampers.
3. Time constant of capacitor.
4. Time constant constraint in clamper circuit.
5. Clamper example.
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7:17 ....vi=-v
then vo=-vi-v=0.
i feel u did wrong. :-(
Yes..here at the tym of applying kvl negative Vi had taken earlier so not using later at tym of substitutn
Had the same doubt. Got it cleared by watching the linked video.
Thank you so much neso academy.
throught the episode ,while operation u said clipper ckt to clamper cky
Yeah got cleared with same doubt
@@ankitburnwal2569 link bro
so far the best explanation because you showed through KVL or mathematically.Thats why easy to understand.
you make clampers easy to me .. very thankful to you
Thanks
Oh shit I wasted my time reading other books to understand this
This is way simple and so understanding
Your explanation was really helpful.
But I would like to add that practically there should be some voltage drop across the diode. So output signal will change a little bit.
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very nice video! helped me understand clampers :)
thanks a ton! :) :)
Well explained.. thank you very much sir
sir , you done a great job in past . your videos are very helpful till today date .
Thank u. very good explanation in detail
Thank you! great video! Subscribed :)
Actually in the _ve half the capacitor is already charged so it has to discharge in the _ve half cycle right?
sir you are a fabulous teacher
thank u so mush , I hope u have a great day sir .
THANKU SIR, it is very helpful
isnt the direction of current across the resistance in the negative half wrong? shouldn't it be in the direction as indicated by the round arrow in the centre..and thus V0 be negative
what will be PIV of each diode in tripler and quadrupler corcuits?? and why??
very nice, but can you solve problems with numbers? would be great if you did.
can we replace a capacitor charged upto 10 V by an uncharged capacitor with a 10 V battery in series?
Sir, Generally at resistor there is voltage drop (so while using KVL we take - sign)
then why sir you took +ve sign for load resistor ? ( does it mean that across load resistor potential increases?
When we use tha mesh analysis then we first offer polarity to tha resistor according to direction of current in such way that first terminal of resistor taken as positive and second terminal will be negative .
Now while moving through loop its our choice whether we take sign of first terminal of battery and resistor or sign of second terminal of battery and resistor .
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 ?).
Super explaination yaaar,
Sir pls give these slides as pdf
it will be a great source of notes
could you place tell me why did you take Vc as negative in KVL while Capacitor is charging??
Thank you sir
Wow man , where were you all this time ??????????????????????
Chole baat Raha tha
😂
Chote aloo bade aloo alag kar raha tha
sir please give example if time constant is very less than time period as it is asked in gate 2017
What will be output, if we reverse the direction of diode in this circuit with given input....
When input changes it's direction.. wont capacitor will change its polarity as capac oppose change of polarity???
How does the peak value change?
(1)Sir if the capacitor charges within the time t/2, then it should discharge within t/2??
(2) you have assumed that the time constant is much more compared to half of the time period then how the capacitor charges to V volts for the half cycle???
Capacitor gets charged 99% in mean time but till the time reaches T the capacitor is charging betn 99-100 so we have to assume it has charged
at 6:55 why the polarity of the output voltage Vo is not changing
I have the same doubt
Excellent sir
How we have maintain the condition that tau is greater than t/2
are you sure you have applied KVL properly in the 2nd half?
I had a challenging time with that as well. If you redraw the circuit it might make more sense. It seemed to for me at least. I put the capacitor underneath the source:
__________________ +
| | |
| | |
~ V RL
| | |
= | |
|_______|________|__ --
It's basically just an output in parallel with the battery voltage + alternating source. Don't know if that answers your question regarding KVL or not.
great video.☺
You had taken tau (time const) = 0 as R=0 but we've asummed that tau > T/2 then it contradicts our assumption.
How does RMS remain same wen peak value changes?
How did u apply kirchhoff's law for T/2 to T ??
😎😎😎
great video
In the ladt video clipper in short circuit of a diode the output voltage was equal toh the voltage passing through it wherease in this video output voltage is 0. 4:34
Why will current flow in downward direction across load resistance when you have the path anticlockwise with red pencil. I guess it should be in upward direction.
kindly tell. See at 6:30
*Thanks*
sir why is RMS value remains same?
Tq sir..... it's very clear
what will be the output waveform if time const
Spikes waveform.
I have a doubt. The current flows through the circuit during the first half cycle. Then why the output voltage becomes 0 and how.?? Kirchoff's law is not the same for the positive and negative half-cycles.
Thank You Sir 😊
at 5.08 how is Vc negative while the polarity you have made shows positive ?
He's using Kirchoff's Voltage Law (KVL). Vc is negative in his equation (even though it shows positive in his circuit) because at that point, when the diode is forward biased, all of the voltage is on the capacitor, so Vin = Vc and so Vin - Vc = 0. This is because KVL states that the sum of all the voltages in a closed loop circuit will equal 0.
Time constant is zero, in case of zero resistance. So in this circuit, how does the condition, tao>>T/2 hold?
same quest. bro
SAMAPAN BHADURY since tau is 0, the capacitor will be immediately charged to V volts.
sir....if capacitor charges upto vm then while reversing the....supply voltage capacitors polarities must be revesed....then how? 2vm comes....correct me if iam wrng..
pankaj prasad my teacher told you have to keep the capacitor polarity same even if u are giving the negative input
Awesome man
2nd loop.. KVL may be wrong.. if you don't mind clear it
won't the capacitor change it's polarity in the second half cycle?
it will change
Sir how rms will be same since the peak value of signal is changed
Check the o/p of the signal, here we're just shifting the signal, there's no change in the magnitude to the overall signal.
what if tau < T/2 ?
6:14 why Cap polarities not getting reversed?
it has already charged in +ve half cycle, in -ve half cycle it will take more time to discharge because of the resistance(note that we have chosen the resistance and capacitance such that its time constant is much greater than the time period @ 2:10 ).
How to apply kvl
Nice.
As Tau=0 the capacitor will also discharge in 0 time when diode will be reverse biased. Can't get it properly
in reverse bias condition capacitor doesn't get discharged...it is acting as a battery in negative half that is why we get some output.
Vrms=Vo/root(2) if Vo changes, Vrms is also supposed to change right ?
thank you
Sir if the peak value is changing then how the RMS remain same ???
No it need not.....
Function of tou in both cycles
Sir please write in big size b/c it no visible
Peak to peak remain same is it true????
Thanks a million,
why did you open circuit?
Sir why we take T(taw) >t/2
Because during half of time-period of the input signal,capacitor is going to discharge(during which cycle that depends upon the diode orientation though)
In the first half time constant was zero. Then how it can be greater than T/2??
The condition is applied when the diode is nonconducting. According to this video the diode was nonconducting at second cycle with a finite time constant.There is nothing to be worry about first half cycle.
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why is tau>>T/2?
I have one doubt. At the first half cycle, you defined polarity of capacitor as the input voltage, that is fine. But in the next half cycle, the polarity of input voltage got changed. So according to that, the polarity of the capacitor should change. Then why it did not change? Plz clear my doubt.
bcuz time constant is very high,
Hi Sir, thank you so much for your explanation! However, I am still confused why τ is greater than t/2...May I ask why is it so?
Because during half of time-period of the input signal,capacitor is going to discharge(during which cycle that depends upon the diode orientation though)
Capacitor will be very well charged if Tau >> t/2 which would allow the capacitor to have its full effect at the output
It seems like charging time of capacitor is very small as compared to discharging time of capacitor
What if it is a sine wave
when does capacitor discharges?
it discharges slightly during the -ve half cycle.
Thank you
How should take this voltage intially= V
rms should be root 2 times v.......right?..it cant be the same
I think that by clamping a signal RMS value will change
+Vi - Vc =0. Here can you plz explain why it is -Vc, not +Vc....according to KVL
Thanku sir
There is confusion in taking charge on capacitor
7:01..
How do you consider the direction of current in load resistance to be opposite to that of input voltage ?
The direction of current through the load resistor should switch technically speaking, but I think it's so negligible because of the charge already on the capacitor from when the diode is forward biased that he doesn't include it because he's teaching from an "ideal" circuit. The circuit will do the exact same thing without a load resistor connected implying that the current is next to nothing when the input voltage is opposite. Basically when the diode is reverse biased we're almost literally just measuring across a battery in series with an oscillating source because the capacitor technically doesn't discharge. But you're right: in the real world the direction through the load resistor will change directions. If you lower the load resistance significantly in the "ideal" circuit then then the current through the load resistor switching directions becomes much more prominent.
ur lecter is amessing to be countioue i am halofom from ethiopia....
can u understand hindi?
Please detail why time constant is greater than time period..?😅
τ should be greater than T, otherwise the capacitor will discharge sooner than the period and the circuit will not function as intended
TAU must be greater than t/2 but in the (+) half cycle TAU =0 , ????
Agar input waveform sinusoidal hua to , output to st line hona chahiye
Output bhi sinusoidal kyu ho ja Ata hai
Koi Bata do plzz
hi!
e or f section
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The negative half cycle is given wrong ( revered baised)
That's KVL?
Capacitor block the DC if I am wrong crocet me
Vi=-v
So -vi comes +v na ..So how do u did as -v only
Nice 👍 😊😊😊😊👍
Tq
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7:12 colorblind can't see the color change. Colorblind sucks.
You keep on saying clipper instead of clamper !