The timestamps for the different examples covered in the video: 0:36 Example 1 (Clamper Circuit) 4:02 Example 2 (Clamper Circuit) 5:49 Example 3 (Clamper Circuit) 8:16 Example 4 (Clipper Circuit) 12:22 Example 5 (Clipper Circuit)
During the negative half cycle, the input voltage is -14V. And the voltage across the capacitor is 6.5V ( from left to right). So, both voltages come in series. And if you see the voltage across the load, then that is -6.5 - 14 = - 20.5 V. So, that is the minimum output voltage. I hope, it will clear your doubt.
The case, which we considered (-10V) is the minimum value of the input signal. But actually, Vin is is continuously changing (Sinusoidal signal) That means Vo will also change in a similar fashion when input is less than -4. That's why it is curved. I hope it will clear your doubt.
Sir, 1st numerical answer is contradicting with your clamper tuturial. It should be between 10 V to 30 V according to your past explanation. It is creating doubt. Please clear.
It will be between +10V and -10V. Please watch it again (The video on Clamper Circuit). You can also check the simulation of the same circuit. It is between +10V and -10V. www.multisim.com/content/i8TjgyMHGsHuKUYuQet46c/example-1-clamper-circuit/open/ If you still have any doubt then let me know here.
When Vin is negative, it means that, the polarity of the voltage is reverse and when Vin is less than -0.7 V it means that, the voltage at the anode of the Zener diode is more than cathode and it is more than 0.7V. Therefore, the Zener diode will be forward biased and the voltage between the anode and cathode of the Zener diode will be 0.7V. But we are measuring that voltage from the top to bottom (Between Cathode and anode). That's why that voltage will be -0.7V. I hope, it will clear your doubt.
@@ALLABOUTELECTRONICSbut jab hum vth nikalwnge to vo 10 v se kum hoga Issliye xener diode ko off condition me rehna chahiye Plz correct me if I am wrong??
No, thats alright. With the same circuit, you will get the output between +10V and -10V. For more info, you may also check the simulation of the circuit. Here is the link: www.multisim.com/content/i8TjgyMHGsHuKUYuQet46c/example-1-clamper-circuit/open/
The timestamps for the different examples covered in the video:
0:36 Example 1 (Clamper Circuit)
4:02 Example 2 (Clamper Circuit)
5:49 Example 3 (Clamper Circuit)
8:16 Example 4 (Clipper Circuit)
12:22 Example 5 (Clipper Circuit)
This channel is genuinely superb . I have cleared many doubts ..tq sir
So nice the world's best teacher
Saviour before exam🔥😎😎
OUTSTANDING
Very well explained
Good luck for your exams yall.
Hey pls include some questions with non ideal diodes also!!
Good
Thanks you sir🙏🙏
thank you!!!
How voltage minimum value comes - 20.5 v in q3?
During the negative half cycle, the input voltage is -14V. And the voltage across the capacitor is 6.5V ( from left to right). So, both voltages come in series. And if you see the voltage across the load, then that is -6.5 - 14 = - 20.5 V. So, that is the minimum output voltage. I hope, it will clear your doubt.
in example 4 why isn't waveform clipped at -7v? why is it curved?
The case, which we considered (-10V) is the minimum value of the input signal. But actually, Vin is is continuously changing (Sinusoidal signal)
That means Vo will also change in a similar fashion when input is less than -4.
That's why it is curved.
I hope it will clear your doubt.
Thanks!!
4:43
Sir, in example 2, what happens if there is a current source(triangular) instead of voltage source at the input?
Sir, 1st numerical answer is contradicting with your clamper tuturial. It should be between 10 V to 30 V according to your past explanation. It is creating doubt. Please clear.
The battery must be reversed
It will be between +10V and -10V. Please watch it again (The video on Clamper Circuit). You can also check the simulation of the same circuit. It is between +10V and -10V.
www.multisim.com/content/i8TjgyMHGsHuKUYuQet46c/example-1-clamper-circuit/open/
If you still have any doubt then let me know here.
GOOD
sukran
Very useful
Is it minus 10 Volt
dude u r awesome!
In last example, where you said in last condition voltage is less than -0.7 then zener diode is -0.7 why is that?
When Vin is negative, it means that, the polarity of the voltage is reverse and when Vin is less than -0.7 V it means that, the voltage at the anode of the Zener diode is more than cathode and it is more than 0.7V. Therefore, the Zener diode will be forward biased and the voltage between the anode and cathode of the Zener diode will be 0.7V. But we are measuring that voltage from the top to bottom (Between Cathode and anode). That's why that voltage will be -0.7V. I hope, it will clear your doubt.
@@ALLABOUTELECTRONICSbut jab hum vth nikalwnge to vo 10 v se kum hoga
Issliye xener diode ko off condition me rehna chahiye
Plz correct me if I am wrong??
In Example 1 the battery will be reversed...
No, thats alright. With the same circuit, you will get the output between +10V and -10V.
For more info, you may also check the simulation of the circuit.
Here is the link:
www.multisim.com/content/i8TjgyMHGsHuKUYuQet46c/example-1-clamper-circuit/open/
The confusion started bcoz u wrote -10V which basically means the battery is reversed
Thank u ..😇😇🤘👍
what to do when rl has value
Hi, would you please mention where you are referring to in the video. It would be easy for me to answer your question.
Kuchh samajh nhi aya u unable to point the things u teaching
Thanks sir
I was here
Hi, will you make some videos about op-amp combining with clipping and clamping circuits? I want to diy guitar effect units. Thanks!!
Yes, very soon.
bhai +10Vko -10V q bol rhe??
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@@hunainaghai3342 acha sey bhola
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