# Ideal Diode

Discussion in 'Homework Help' started by winzer, Mar 25, 2008.

1. ### winzer Thread Starter New Member

Mar 25, 2008
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0
So I am suppose to consider for the values of V(in)=0,2,6,10 to plot V vs. Vin ranging from -10V to 10V of the circuit. However I am very confused where to begin. IF I take Vin=0. Do I have to start assuming states for each diode and go about trying to find the right combination that will work?

Thanks

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2. ### Pich Senior Member

Mar 11, 2008
119
5
I'm not shure what you are looking for, can you please explain in a differt way?. The drawing has the + - polarity backwards.

3. ### JoeJester AAC Fanatic!

Apr 26, 2005
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Maybe not, for this question. Since it's homework assignment, it could be either way, depending on the instructor's objective in the question.

On edit ... damm, that wasn't the diagram I was thinking of when I responded. I must have got two postings mixed up. CRS was taking it's hold this morn.

4. ### beenthere Retired Moderator

Apr 20, 2004
15,808
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The four diodes are arranged like a bridge rectifier, but the AC input is not correct. Having the one input line connect to ground through a resistor is strange.

5. ### winzer Thread Starter New Member

Mar 25, 2008
7
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So how would I tackle this?

6. ### studiot AAC Fanatic!

Nov 9, 2007
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Can you tell us more of the question as your extract does not make sense.

If V(in) is alternating at 0,2,6,10 volts the peak to peak of 10 volts = 2xsquareroot2x10volts equals 28 volts.

Unless the point of the exercise is to show clipping.

7. ### winzer Thread Starter New Member

Mar 25, 2008
7
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Here are the exact words from the text:
"Consider V(in)=0,2,6,10V. Also for the figure plot V versus V(in) for V(in) ranging from -10V to 10V"

8. ### studiot AAC Fanatic!

Nov 9, 2007
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There must be more to the question than this.

Where did the circuit diagrams come from?

Did you understand my comment about clipping or beenthere's comment about ac input?

9. ### winzer Thread Starter New Member

Mar 25, 2008
7
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The diagram came from from my text(Hambley). Clipping is covered in the section. But it doesn't explicitly say anything about clipping though it may be the case.

10. ### JoeJester AAC Fanatic!

Apr 26, 2005
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It is a clipper circuit.

11. ### winzer Thread Starter New Member

Mar 25, 2008
7
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SO the graph will look like y=x then go flat at 10V on either side of V vs. V(in)

12. ### JoeJester AAC Fanatic!

Apr 26, 2005
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Don't forget the drop across the diode.

KVL is a law; not a theorem.