During negative half cycle, D2 is reverse biased when Vi > Vb2 and is f/w biased when Vi < Vb2 . But here you have said opposite of this. is this because have you merely considered the value not the sign?
Your video tutorials are very helpful and crystal clear for me. But unfortunately, I have a doubt regarding the combination clipper circuits.. When determining the biasing of diodes we are considering the input voltage and the external voltage near to the respective diode not the external voltage which is in parallel with the diode. Why so ?
sir please give whole topics in one solt like u have given topic of flip flop so please give like that sir and u are doing a best literally i am understanding whole things which u are teaching thank u once again but please give one topic all videos in one solt
At 0:27, can we take output across D1,V1 branch and it will act as input for D2,V2 branch, thus final output will be the same as that of whole combination circuit. YES!!!!😚
please upload some more complicated circuit involving both parallel and series with both +ve and -ve clippers
During negative half cycle, D2 is reverse biased when Vi > Vb2 and is f/w biased when Vi < Vb2 . But here you have said opposite of this. is this because have you merely considered the value not the sign?
Yes crct... what could be the final answer?
Your video tutorials are very helpful and crystal clear for me. But unfortunately, I have a doubt regarding the combination clipper circuits..
When determining the biasing of diodes we are considering the input voltage and the external voltage near to the respective diode not the external voltage which is in parallel with the diode.
Why so ?
Very well explained sir
Thanks bro u saved me... 😊😊
great work it helps much more keep it up
sir please give whole topics in one solt like u have given topic of flip flop so please give like that sir and u are doing a best literally i am understanding whole things which u are teaching thank u once again but please give one topic all videos in one solt
Hats off to you, man!
In 7:23, Isnt Vo = - VB2 ?
Yes, it is correct
Correct
no, because Vb2 is defined as a negative voltage
AWESOME !
All doubts cleared :)
Sur D1 me 2nd battery ka and D2 me 1st battery ja effect kyon consider nhi kr rhe h
At 7:14, why is Vo=Vi? Shouldn't it be Vo=-Vi according to Kvl?
Farishta Rahman same doubt boss..u got any clarity
I think it will be Vo=-Vi.. But the plot will be same..
@neso acamedy same doubt
Watch the first 2 videos of sir on clipper diode your dout will be cleared .
@@farishtarahman3597 Yeah it will be minus.....But it doesn't matter because in the end he put -Vi and also - Vb2 in plot
very good video.
but can you make a video which have values of every thing.whether it is a source or whether it is load resistance.
Thanks for another great lecture.
Super explanation, thanks
crystal clear, thanks
thank you so much sir! you're a great teacher. My prof is good and smart but not good at teaching!
At 0:27, can we take output across D1,V1 branch and it will act as input for D2,V2 branch, thus final output will be the same as that of whole combination circuit.
YES!!!!😚
Please teach linear integrated circuit or about opamp
what happens if both are biased positive clipper type diodes?
is this dealing with ideal diodes
Sir we will convert from transfer characteristics to diode crt,,for combination clipper
how do you see that the diode is reverse bias or forward bias?
In negative half the polarity of Vi is also reversed so how your waveform is correct
For t/2 to t, V0 = -Vi and V0 =-Vb2 is what i am getting which is also justifying the waveform. I think he didn't notice the polarities.
awesome ! make more videos!
So this works in similar way as zener diode clipper.
why the peak point of the voltages are equal to vB1 and vB2. Aren't they must be equal to Vi-Vb1 or Vi-Vb2
can u give an emitter-coupled clipper lecture?
I m confused about the load resistance......why we need this???we can also measure the v0 across the diode........can't we????
what's the use of load resistance?
what if the series resistance R not negligible??
u will have a voltage drop at the output by Vr, it will be (Vb1-Vr). And same for Neg. half cycle
SUPERB!!!!!
at 5:00, how vi=vB1
Did you get ur answer bro
@@techenthusiast07 Vb1 is parallel to the resistor...so definitely this is the voltage that will be across the resistor
When vi>vB1, drop across series resistance VR is vi-vB1, but he said VR=0.
how can u suppose vb1 upward from time axix and vb2 downward from time axis please clear it for me
If u name this video to
clipper and clamper engineering 1st year Basic electronics
what if the diodes are in series?
Debalina Banerjee double is only possible for shunt
Same output as zener diode clipping circuit
what if I use a si diode instead of ideal diode? to the output waveform
then just consider its knee voltage and solve in the same process....
What if the polarity of Bias voltages in series with diodes are reversed in the same circuit?
Uzair Mughal output waveform will be different then
Yes, but I was looking how one can find that different waveform because if I am correct then that does not initiate from level "0".
Here in case 1 u said that D1 is reverse bias when Vi is less than Vbe .I think D1 is forward bias and D2 is reverse bias whence is less than Vbe
Vo = -Vb2???
Awesome 👍
vaiya...plz show me another combinatorial clipper ckt maths..plz plz plz
Eureka✌
Bro if you can understand in Hindi it will be more easier to understand 🙂
🙏🙏🙏🙏
Hubby