Output of the circuit as function of time

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atan691988
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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. *
 
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Write V_out in terms of the DC voltage source 2V and v_in.

When V-in* 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?
 
Last edited:
rootX said:
But, does it really matter whether diode is on/off in this circuit?

Now that I can see the circuit ... no, you're right it does not affect Vout here.
 
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)
 
fleem said:
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.

It was a "pending admin approval" thing. Eventually an admin approved it.
 
Redbelly98 said:
Now that I can see the circuit ... no, you're right it does not affect Vout here.

That was question to the OP because I didn't want to provide him a direct answer :smile:
 
hahaha. thanks for all your replies guys.

I've already figure it out..

but really thanks to all :D

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