Diode as Clipper question

[enigma^6]

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]

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]

@ 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]

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]

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!!!!!

 

[a_tek7]

No problem.It happens for everyone