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Diode as Clipper question

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enigma^6

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explain clipper diode

Hello Everyone!

Please help me with this simple parallel clipper to me?

How come at positive cycle ( Vi > 4v) diode is active/shorted? Is it because of the 'transition'? I don't understand this transition thingy

hxxp://i203.photobucket.com/albums/aa196/badenigma/ScannedImage-3.jpg

Again any help will be appreciated much! Thanks in advance
 

a_tek7

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the diode will conduct when V_AK(voltage of anode ≥ voltage of cathode) if so then the voltage that will drop across diode will be 0v in ideal diode.
 

enigma^6

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@ a_tek7
I don't quite understand...

so if VA > VK diode will conduct

E.g. Vi = +16 V

+16V ------kA------ +4V

diode will not conduct right?? how did Vo end up with 16V ?

THanks a_tek7
 

a_tek7

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WHEN diode will not conduct the current will be 0A and KVL:V_O=V_i(notice that v_i is a ramp so v_o will be ramp).ok?
 

    enigma^6

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enigma^6

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ooohhh now i get it.... that was stupid of me..

I was thinking vi is 20 then vo became 16... i get it now.... THANKS!!!!!

:D
 

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