If [Vo=(-Vi)] for negative half-cycle, then shouldn't the polarity of Vo be opposite to that of Vi? ie. if Vi is -ve, then Vo should be +ve, no? What am I missing?
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
actually there is no need to change. because we have already changed the polarity of the input voltage or just without applying any KVL just look at the circuit when diode is short circuited. The output voltage equals to the input voltage
If R is not small, there will be a substantial current through it causing a voltage drop. How to calculate V(R) then? videos with numerical values can aid to comprehension.
Here output is measured across load resistance which is also in parallel with diode, so for first half cycle diode behalf as short with voltage Vi then why Vo = 0.
No, the graph is correct. Coz during the 2nd half cycle, he has marked the upper terminal as -ve & lower one as +ve which is just the opposite of what we have considered earlier while plotting the input waveform.
Have you tried your theory on a practical circuit. I have tried this with 30volt p-p signal at 1mhz using 4 in4148 but I seeing voltages greater than the forward voltage of the diode.
If someone can answer why we started calculating voltage drop now but didn't in previous clipping circuits? Also How do we know its small enough to be neglected in the current circuit?
Vo =-(Vi-3), and here in the negative half when we are changing the polarity of battery then we have consumed that negative sign of Vi ,so we are using Vi as positive term ..U can use Vi as negative only when u do not change the polarity and let remain as same as in 1st half..
As we got Vo = -Vi, the peak voltage should be -(-Vm) which is +Vm and the wave should be above the time axis. Isn't it?
No when the voltage is in anti clock wise direction we take always net voltage as a negative
If [Vo=(-Vi)] for negative half-cycle, then shouldn't the polarity of Vo be opposite to that of Vi?
ie. if Vi is -ve, then Vo should be +ve, no? What am I missing?
Stuck with the same conceptual doubt.
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
actually there is no need to change. because we have already changed the polarity of the input voltage or just without applying any KVL just look at the circuit when diode is short circuited. The output voltage equals to the input voltage
What if the ideal diode replace with pratical diode (silicon) ?
The Vo during the 1st half cycle will be 0.7 ?
Yes it will work like voltage regulator (like zener diode).
Sir for the negative cycle of the voltage source,the maximum voltage will be would be less than Vm due to the voltage drop across R
I also think that because in -ve cycle the circuit will act as a voltage divider.
But why is the value of Resistance of Vr negligible? Did you just assume that its resistance is small or is there some concept I am missing?
In KVL equation it should be Vi-V0-Vr=0?
No it's correct
Yes
If R is not small, there will be a substantial current through it causing a voltage drop. How to calculate V(R) then?
videos with numerical values can aid to comprehension.
Then output voltage will be in phase but a constant factor(
sir
i have doubt in reverse bias we have to take (+vi), but u taken (-vi) why i dont understand that
First Vi is taken positive then on other side of equal it becomes negative
Here output is measured across load resistance which is also in parallel with diode, so for first half cycle diode behalf as short with voltage Vi then why Vo = 0.
For first half cycle side behaves as open circuit.
sorry for replying after 6 years, but since there is no resistance in ideal diode, no potential difference is created
so V is zero there
if the output is - Vi then should the output while plotting should be shown as 180° phase change
No, the graph is correct. Coz during the 2nd half cycle, he has marked the upper terminal as -ve & lower one as +ve which is just the opposite of what we have considered earlier while plotting the input waveform.
Its very useful for new Students..
When tha diode is forward biase then why the registance does not flow through load registance
Thank you Sirji...🙏
Sir why current choose short path and why current does'nt pass through the load resistance ?
Your video helpful for me, Thank you 😊
sir how did you choose the polarity of Vr?
Zaheen Simin because current is flowing in anticlockwise direction
it always depends upon the flow of the direction of the current ........
whatever be the direction of the flow of the current ...... it will always go with that
How to identify the dc wave that it will be (+) or (-) for different battery n diode connection?
Why resistance is small???
What if your diode is silicon?? How will you graph that? Thankyou!
We can then replace the diode with its voltage i.e. = 0.7 volt and apply kirchoff's voltage law...
Try it.
sir why extra resistor is added here in parallel?
Well done sir
What type of program do you use? Thank you, your videos have helped a lot!
Have you tried your theory on a practical circuit. I have tried this with 30volt p-p signal at 1mhz using 4 in4148 but I seeing voltages greater than the forward voltage of the diode.
Please do make videos on GATE ECE PREVIOUS YEAR Q
If someone can answer why we started calculating voltage drop now but didn't in previous clipping circuits? Also How do we know its small enough to be neglected in the current circuit?
Why?
In positive half cycle. Why Vo = 0, why not Vo = Vi - iR ?
bcoz Current through Load resistance is negligible due to f.b so iR=0
why is the output not positive for negative half beside it's calculated that Vo = - Vi
Vo =-(Vi-3), and here in the negative half when we are changing the polarity of battery then we have consumed that negative sign of Vi ,so we are using Vi as positive term ..U can use Vi as negative only when u do not change the polarity and let remain as same as in 1st half..
what is the output graph?
Vi + Vo - Vr=0 is wrong coz polarity of Vo should be -Vo , so it shuld be Vi - Vo - Vr=0 , please correct me if i am wrong ........!!!!!!!!!!!!!!
yes,polarity of load resistor will also change in -ve half cycle, i am also confused by this video.
same confusion
@@salmanbhuttaaa best of luck for exam though 🖤🌼
Vo is Voltmeter...and Voltmeter polarity will not change both half cycle..
ThankYou😊😅
What if it wasn't ideal?
the solution is wrong because in negative half cycle, the output voltage must be given by the voltage divider rule between R and Rl.
-vi becomes +right
Vo = -Vm, not -Vi
need notes
Your kirchoff concept is too poor