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

 

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

 

The drawing has the + - polarity backwards.

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.

 

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.

 

So how would I tackle this?

 

V(in)=0,2,6,10 to plot V vs. Vin ranging from -10V to 10V

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.

 

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"

 

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?

 

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.

 

It is a clipper circuit.

 

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

 

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

Don't forget the drop across the diode.

KVL is a law; not a theorem.