
#1
Aug909, 12:02 PM

P: 2

can anyone help me with this>>>
* If the input is Given by Vin = V0 cosωt, plot the output of the circuit as function of time. assume an ideal diode model. * 



#2
Aug909, 12:50 PM

P: 461

I can't see the schematic yet, but the ideal diode model is here:
http://en.wikipedia.org/wiki/Diode_modelling 



#3
Aug909, 01:38 PM

Mentor
P: 11,984

Often, in introductory courses, "ideal diode" means zero forward voltage and zero reverse current.




#4
Aug909, 04:06 PM

P: 1,295

Output of the circuit as function of time
Write V_out in terms of the DC voltage source 2V and v_in.
When Vin* is more negative than 2, diode is on so replace it with short circuit. Else diode is off, replace it with open circuit. But, does it really matter whether diode is on/off in this circuit? 



#5
Aug909, 05:52 PM

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P: 11,984





#6
Aug909, 05:57 PM

P: 461

Gahh! I didn't look at the graphic because I thought it was one of those "pending admin approval" things. Its a link! This is easy.
Hint 1: An ideal voltage source is unaffected by the current that is being drawn from it. Hint 2: Consider V_in and the battery as your "ideal voltage source" (Sheesh I practically gave you the answer) 



#7
Aug909, 08:07 PM

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P: 11,984





#8
Aug909, 09:00 PM

P: 1,295





#10
Aug1109, 12:52 PM

P: 2

hahaha. thanks for all your replies guys.
I've already figure it out.. but really thanks to all :D <<< mods can close these thread. 


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