Thanks for your careful explanation: it took me a couple of goes to get this right even after I watched your work-through. Your attention to detail helped me to finally "get it".
You got it wrong there, remember that the battery during Positive half cycle is -5V, where in the input Voltage is greater because 0-20V, so 0 is still greater than -5V. So there should be no output waveform during the Positive half Cycle
A doubt, what is the difference between voltage source of 5V and Vi? and why have we taken them separately. Thanks again for the detailed explanations!
sir... what will happen to the -ve output cycle when the basing DC supply is greater then the input -ve DC voltage in case of positive baised series clipper.......the DC supply in series should be in forward direction.....
Doubt!!!! I think you will get at least 5V, whenever you supply input source or not. For problem no. 1: For the negative half cycle, you should get 5V as output ...
I completely agree with Vivek .But even though the diode immediately switches off the circuit..there is an external 5v DC source which depends on the diode in no way . Also the circuit is complete. So why dont we get 5v as output during the negative half cycle ? please explain
NO..during negative half cycle, current passing through load resistor is 0 amp, So obviously output voltage is zero...no matter how many DC sources you have they all are inactive
Doubt.!! In example 2 the negative peak should be 20v because 3v is a constant dc supply and 20v is an alternating. So i think peak value should not be changed...🙄🙄
Have you studied circuit Analysis 1 or 2 in your life. if (You == studied) { print: (then: Its your default that you did not Understand What is going on in class ); But if (You == studied) { And don't get it Then You are A fu**** A**.H***; else There is a straight path 'Just Go To Hell';
if u apply kirchoffs law in clock wise u get -vo-vi-3 =0 here vi direction is reversed because it is in negative half cycle,take time and see it clearly:}
don't think so much guys just remember its written in a book we always have to take upper part of output voltage + and lower part as - wherever the current direction is in clockwise or in anticlockwise whenever we apply kvl always take V output as +ve thats simply its not a rocket science
Vikas Kale when the diode is Reverse biased we will be replacing it with open circuit so from the input side there is no current enters into the load resistor R. The voltage across the load resistor is the output voltage so the current flowing into the resistor is zero. As V=IR, therefore output voltage is zero
At the time 5.39s you started writing the equation to find the Vo in the Forward biased manner. And my question is, can not we write it down including the 0.7V of the Si diode if we know that value?
In solving some circuits by this methode i dont get correct results,means the clipper does not clip the waveform in any way neither positive clipper nor negative clipper.The waveform comes as it was.It is not being clipped.Please explain this point.
If you would check by doing KVL clockwise, -Vi-3V-Vo=0; Vo=-Vi-3V. And since it is forward biased on the negative half-cycle, Vm was used for the purpose of plotting the waveform produced by the clipper.
As a battery show only potential difference ,how can we say that its negative terminal attached to n side of a pn junction diode ,is forward biasing it when its positive terminal is not connected to p side of pn junction diode ,the negative of battery can be at zero voltage,how we say definitely it is less than 0volt
No Vo =Vi+3 this equation doesn't exist in any of two cycles, for +ve cycle: using kvl, Vi = 3v+Vo Vo=Vi-3v as diode is connected reversely, so for +ve cycle the Vo will generate till Vi=3, the Vo stop generating as now the diode behave as short circuit for -ve cycle: using kvl, as the cycle is negative Vi will be treated as -Vi,so, -Vi = 3v+Vo Vo=-Vi-3v
ankit burnwal don't think so much just remember its written in a book we always have to take upper part of output voltage + and lower part as - wherever the current direction is in clockwise or in anticlockwise whenever we apply kvl always take V output as +ve thats simply its not a rocket science
dear sir, your videos lectures is easy to understand but i cant understand one thing in this lecture. IN second example of sinusoidal waveform at the first positive half cycle where the condition of forward bias ( Vi
Remind Kvl: -->Total Voltage = Sum of All Voltages in the circuit Here: --> Vi = Total Voltage -->3v+Vo = sum of voltages so by that: for +ve cycle: using kvl, Vi = 3v+Vo Vo=Vi-3v as diode is connected reversely, so for +ve cycle the Vo will generate till Vi=3, the Vo stop generating as now the diode behave as short circuit for -ve cycle: using kvl, as the cycle is negative Vi will be treated as -Vi,so, -Vi = 3v+Vo Vo=-Vi-3v
the equation that is written at the time of 5 minutes and 37 seconds is Vo= Vi - 3 volts. but when we apply KVL it should be --> total voltages = -Vi - Vo +3volts --> Vi= - Vo +3volts please reply again to make it clear
in +ve half cycle at 5.37, total voltages = -Vi+Vo+3 u cant take -Vo as its +ve terminal is connected to the +ve terminal of Vi and its -ve terminal is connected to the -ve terminal of Vi so Vo will be taken as +Vo if Vo is connected to Vi in reverse than Vo will be taken as -ve Vo in KVL equation
@@kamimart sir how can we say that the upper terminal of resistor is +ve and lower terminal of resistor is -ve as we are considering current in anticlockwise direction, so current in the resistor should be from lower to upper. In that case the lower part will become +ve and upper will become -ve. (as current flows from higher potential to lower potential). Sir kindly please help me to figure out the fault.
Hlo sir I have a doubt related to direction of current in this question no 1 how you have given the direction of current from negative to positive because in first half cycle the current must flow clockwise please please clear this doubt
In the second example the equation will be Vo=-Vi-3 And you said if there is -Vi then we must take its magnitude in the last lecture so 20-3=17volts .so the answer must be -17 volts not -23 volts. Sir please reply to it
Sir..i have one doubt in 1 st sum..during -ve Half cycle..after 5 volts only ..the diode is open so there is no current flow..but u said ..there is no current flow in the circuit during -ve cycle..thats why sir...can u explain that part...
You explained this leaps and bounds better than my electronics teacher!!! I can't thank you enough!
Thanks for your careful explanation: it took me a couple of goes to get this right even after I watched your work-through. Your attention to detail helped me to finally "get it".
Am an Ethiopian Electrical student, i love your lectures , thanks sir
Last ques is full of doubts ...😔
Wow Neso Academy is Back!!!!
I am very happy!
You got it wrong there, remember that the battery during Positive half cycle is -5V, where in the input Voltage is greater because 0-20V, so 0 is still greater than -5V. So there should be no output waveform during the Positive half Cycle
I am looking for this explanation, thank you for making clear
This exercise bothered me a lot. I couldn't figure out the output voltages. Finally I found that V0=Vin-3 when Vin
Finally my doubt cleared, thank you sir..
Dhaga khol diya sir!!!! Maja hi aagaya
thanks for this. But what about BATTERY connected to Diode in Reverse Bias?
one hr before the exam. great explanation thank u thank u thank u
Sir there output graph is wrong as Vout is 17v .But your assumed kvl expression Vout =-Vin-3 is right
5:05 In 2nd ques, how Vi F.B. and Vi>3V => R.B.
Yes it’s wrong. It should be reverse biased and for whole of positive half cycle it will also be 0
@@mohanmohanty7331 Thanks bro keep growing 💗
Make vido on gate previous year q ECE
A doubt, what is the difference between voltage source of 5V and Vi? and why have we taken them separately. Thanks again for the detailed explanations!
Vi is input voltage as given by the graph and 5V is an additional voltage.
@@shlokashekhar4793
5:05 In 2nd ques, how Vi F.B. and Vi>3V => R.B.
sir... what will happen to the -ve output cycle when the basing DC supply is greater then the input -ve DC voltage in case of positive baised series clipper.......the DC supply in series should be in forward direction.....
Doubt!!!!
I think you will get at least 5V, whenever you supply input source or not.
For problem no. 1: For the negative half cycle, you should get 5V as output ...
actualy 1st problem is correct because the voltage is changing from 10 to -10 instantaneously with no voltages in between.
The input signal is square wave which give +10v or -10v so.. If it would be sine then there would be plot in negative cycle
@@terribleloser.24 then why they shows a value of 14.3 V?if the square and sine are diff
I completely agree with Vivek .But even though the diode immediately switches off the circuit..there is an external 5v DC source which depends on the diode in no way . Also the circuit is complete. So why dont we get 5v as output during the negative half cycle ?
please explain
@@94D33M are you mad... You don't know anything about square wave voltage
why diode is forward bais when Vi
I think in positive cycle, direction of current is anticlockwise by Vi.
for the first problem in the +ve half cycle how vo = vi - 0.7v +5v it should be vo = -vi - 0.7v + 5v right ?
I LOVE THIS CHANNEL !!!
Why vi should be less than 3v for diode to be in forward bias?
Sir why u didn't mention 5V in output waveform in the first problem?During -ve half cycle when Vi
Yes! Exactly
In the first problem the voltage is changing from 10 to -10 instantaneously with no voltages in between.
@@sayanbanerjee2722 oh thanks, completely forgot about that...was thinking it's a mistake
good work seriously
I love this channel
In this vedio why u taken Vm instead of Vi???
Could u plz explain...
In the first example , why the output voltage is zero???
There is a 5V voltage source...so the output voltage could be 5V ....am I correct???
NO..during negative half cycle, current passing through load resistor is 0 amp, So obviously output voltage is zero...no matter how many DC sources you have they all are inactive
Sir in first question when our input is -3v then diode is also forward bias?
In other videos(double diode clippers and clampers)you are saying that when v(i) vdc it is f bias. Here you are saying opposite why??
Doubt.!! In example 2 the negative peak should be 20v because 3v is a constant dc supply and 20v is an alternating. So i think peak value should not be changed...🙄🙄
Then what about the voltage cycle up to -5v ,until this value it will be forward biased ,after this it is reverse biased ........can you explain?
It's variable DC supply and its value from -10 to 10 not from zero to 10
So anode of diode is always connected to minus ten not to zero.
You are correct... He had done it wrong
SIR ,if diode is not ideal then what will be the output graph?
Thank you very much. very good explanation
When we apply KVL in circuit (at 7.45 s of the video) Vo = Vi +3. Can you please explain how did u get Vo=-Vi-3. please
Have you studied circuit Analysis 1 or 2 in your life.
if (You == studied)
{
print: (then: Its your default that you did not Understand What is going on in class );
But if (You == studied)
{ And don't get it Then You are A fu**** A**.H***;
else
There is a straight path 'Just Go To Hell';
bhawani is right..so that straight path is going to be for you dude.idiot.
same doubt. please reply sir.
if u apply kirchoffs law in clock wise u get -vo-vi-3 =0 here vi direction is reversed because it is in negative half cycle,take time and see it clearly:}
don't think so much guys just remember its written in a book we always have to take upper part of output voltage + and lower part as - wherever the current direction is in clockwise or in anticlockwise whenever we apply kvl always take V output as +ve thats simply its not a rocket science
can you pls explain me why Vo is 0 at 3:30 ..when diode is RB, it's replaced by OC. then Vo should be -5 V...pls clarify my doubt ASAP...
apply kvl
all voltage will be across open circuit
Vikas Kale when the diode is Reverse biased we will be replacing it with open circuit so from the input side there is no current enters into the load resistor R. The voltage across the load resistor is the output voltage so the current flowing into the resistor is zero. As V=IR, therefore output voltage is zero
It's variable DC supply and its value from -10 to 10 not from zero to 10
So anode of diode is always connected to minus ten not to zero.
Nice explanation
Output should be 5v for negative half of first example
At the time 5.39s you started writing the equation to find the Vo in the Forward biased manner. And my question is, can not we write it down including the 0.7V of the Si diode if we know that value?
sorry the time should be corrected as at 7.35s
how can you say if vi
But in clipper, the shape of the waveform should not be changed... Then how is this possible
And what if the additional DC source reverse biases the diode
Can you give some examples on that
It will hold the diode in reverse condition till the AC sourc voltage is greater than the DC one...
In regards to the second question why did we solve it for FB and RB? is it because the Vi is a sin wave?
In solving some circuits by this methode i dont get correct results,means the clipper does not clip the waveform in any way neither positive clipper nor negative clipper.The waveform comes as it was.It is not being clipped.Please explain this point.
thank you very much!!
Apnader sob video guli amake onek help kore
Thank u
You are seriously hrpling sir
in the first question why didnt we mark the 0.7 in the graph as we did in the previous video
Why you everytime (during applying KVL) taking Vo +ve? This one confusing me
How u get for negative output, vo=-vi-3v,(I'm asking about the sign) second problem
If you would check by doing KVL clockwise, -Vi-3V-Vo=0; Vo=-Vi-3V. And since it is forward biased on the negative half-cycle, Vm was used for the purpose of plotting the waveform produced by the clipper.
@@christiancastillo3857 in negative cycle drop across RL will be -ve
If we have resistance and si daiod connected with series.. what we should do with that case.? 🤔
In ques 2, i think diode is reverse biased and not forward bias. Pls correct me , if i am wrong
Feeling the same
Will the Time period get changed in question two? As it is taking some extra time
NICE JOB SIR
I think kvl in forward bias is incorrect
it is correct
As a battery show only potential difference ,how can we say that its negative terminal attached to n side of a pn junction diode ,is forward biasing it when its positive terminal is not connected to p side of pn junction diode ,the negative of battery can be at zero voltage,how we say definitely it is less than 0volt
Sir at 7;38, using kirchoff's law, the equation comes as Vi-Vo+3=0.
therefore, Vo=Vi+3.
why have you taken as -Vi-3. Please explain.
No Vo =Vi+3 this equation doesn't exist in any of two cycles,
for +ve cycle:
using kvl,
Vi = 3v+Vo
Vo=Vi-3v
as diode is connected reversely, so for +ve cycle the Vo will generate till Vi=3, the Vo stop generating as now the diode behave as short circuit
for -ve cycle:
using kvl,
as the cycle is negative Vi will be treated as -Vi,so,
-Vi = 3v+Vo
Vo=-Vi-3v
ankit burnwal don't think so much just remember its written in a book we always have to take upper part of output voltage + and lower part as - wherever the current direction is in clockwise or in anticlockwise whenever we apply kvl always take V output as +ve thats simply its not a rocket science
see the polarity bro
dear sir, your videos lectures is easy to understand but i cant understand one thing in this lecture.
IN second example of sinusoidal waveform at the first positive half cycle where the condition of forward bias ( Vi
Remind Kvl:
-->Total Voltage = Sum of All Voltages in the circuit
Here:
--> Vi = Total Voltage
-->3v+Vo = sum of voltages
so by that:
for +ve cycle:
using kvl,
Vi = 3v+Vo
Vo=Vi-3v
as diode is connected reversely, so for +ve cycle the Vo will generate till Vi=3, the Vo stop generating as now the diode behave as short circuit
for -ve cycle:
using kvl,
as the cycle is negative Vi will be treated as -Vi,so,
-Vi = 3v+Vo
Vo=-Vi-3v
the equation that is written at the time of 5 minutes and 37 seconds is Vo= Vi - 3 volts.
but when we apply KVL it should be
--> total voltages = -Vi - Vo +3volts
--> Vi= - Vo +3volts
please reply again to make it clear
in +ve half cycle at 5.37,
total voltages = -Vi+Vo+3
u cant take -Vo as its +ve terminal is connected to the +ve terminal of Vi and its -ve terminal is connected to the -ve terminal of Vi so Vo will be taken as +Vo
if Vo is connected to Vi in reverse than Vo will be taken as -ve Vo in KVL equation
thanks sir,, finally get it
@@kamimart sir how can we say that the upper terminal of resistor is +ve and lower terminal of resistor is -ve as we are considering current in anticlockwise direction, so current in the resistor should be from lower to upper. In that case the lower part will become +ve and upper will become -ve. (as current flows from higher potential to lower potential). Sir kindly please help me to figure out the fault.
Thank You Sir 😊
sir , in the first question, in negative half cycle will the biased voltage( 5 ohms) make the diode forward bias for the Vi
You are right
i am not sure why u did not consider si diode voltage in calculation???
Thankyou
thank you sir
sir ,can you explain me in 1st problem in reverse bias why you have taken 0 V while there is 5V DC ....why it is not there in out put graph
It's variable DC supply and its value from -10 to 10 not from zero to 10
So anode of diode is always connected to minus ten not to zero.
@@ElkhedrAli are you mad...
oopppsss that's great you bro.... i love this channel .......... we all time stay withe you ..
First problem, if input is sinewave then output waveform?
why you don't consider potential barrier of diode in 2 question
Thanks
Hlo sir I have a doubt related to direction of current in this question no 1 how you have given the direction of current from negative to positive because in first half cycle the current must flow clockwise please please clear this doubt
Very nice sir
what if the diode is Silicon will we get a Vo of -22.3 or is it -23.7 hmm? do we add it or subtract it
5:24 please explain how the diode is becoming FB? Confused
Plse reply to it
Sir i think In first question... The negative half cycle should come upto -5 V
It's variable DC supply and its value from -10 to 10 not from zero to 10
So anode of diode is always connected to minus ten not to zero.
Can anyone explain why we are putting -0. 7V for the first sum
because it's a silicon diode, not ideal. it is given in the question
sir will you please explain about Theron's like Norton ,superposition,etc and me analysis and machines and they are much helpful for us sir
Can you make video on clippers and clampers sign convention rule Sir please
In the second example the equation will be Vo=-Vi-3
And you said if there is -Vi then we must take its magnitude in the last lecture so 20-3=17volts .so the answer must be -17 volts not -23 volts. Sir please reply to it
But it is not positive half cycle it is negative half cycle that's why polarity of Vi is changed
Best teacher
Sir can u please upload some more questions
plz do a video on filters if u have already done it plzz tell me the name of the video I'm nt able to find it in neso academy
Sir in question 2 how diode is fb when Vi voltage is less than 3v plse tell me iam confusing
i think you made a mistake at 7:40 where Vo should be -Vi+3 but you wrote Vo = -Vi-3
Fawad Khan --Correct or not that equation..
But it is not positive half cycle it is negative half cycle that's why polarity of Vi is changed
the equation will be -Vi-3 only.
I think it should be V0=Vi-3 ???
يعني اقسم بالله جامعتنا ومنهجنا منتهي ب فهمك دا 😭💔💔💔💔💔💔💔
Thanks bhai....
if we say ideal diode it does not have a value? that's it?
if we replace square wave by sine wave in this case diode will forward bias at 5volt ??
explained good
Explain KVL Sign how u r taking
Vo=vi+3v ...so down side of graph must be -17v
No, it will be -23 only. Apply Kirchoff voltage law to the new polarity of the input.
@@deepanshuchaudhary9759 then polarity of output also changes
Sir..i have one doubt in 1 st sum..during -ve Half cycle..after 5 volts only ..the diode is open so there is no current flow..but u said ..there is no current flow in the circuit during -ve cycle..thats why sir...can u explain that part...
Because the input voltage is a dc voltage so it will always give 10 volts. But the second one is an ac source that's why it keeps changing its values.
hello sir is there take place diode voltage drops ?
Thank u Sir..
In the previous que 0.7 was taken in 2nd que y 0.7 is not consider
because in 2nd question the diode is ideal
where is the voltage drop across diode ?
how to solve the problem when there is a resistor just after the diode
How the graph of dc current(3V) has drawn.How can i understand whether it is draw to the upside or below🙄.pls help me anybody
Sir in 2nd qn it is positive clipper!!
In this what is to be grounded?
sir i have watched ur video on 4 bit even parity generator does odd parity generaror will show same results
kindly post soon i have exams next week
hi tasmiaa
at 5:40 V(output) must be 3-V(input).
Let me know if i am wrong....
No it's V-3 use kvl
Sir please solve this problem when Resistance is Unknown
So you ain't gonna mention Kawhi and PG?