The reason you see a straight line on the B channel is you have the oscilloscope set for 500V/DIV. I would hope you saw a straight line.

Is the diagram in your post number 1 the correct diagram from your textbook or exam book? While the members here would jump on the improper polarity of V2, that may be exactly what the instructor wanted.

No it's not from my textbook. İt's from my notes. And I think the diagram it's not correct. I set the oscilloscope to 5V, but the result is different from square wave?

 

I believe your circuit is to CLIP the AC signal at approximately 6.7V. The tops may be flat, but it would be a poor square wave.

Which of the three circuits are you trying to solve or are you building up to the one on the original post?

I believe your last portrayal of the oscilloscope has an offset in the B channel.

If you connect the batteries so the diodes are not conducting when there is no AC signal applied you will get just a 6 Vpk sinewave.

 

I believe your circuit is to CLIP the AC signal at approximately 6.7V. The tops may be flat, but it would be a poor square wave.

Which of the three circuits are you trying to solve or are you building up to the one on the original post?

I believe your last portrayal of the oscilloscope has an offset in the B channel.

If you connect the batteries so the diodes are not conducting when there is no AC signal applied you will get just a 6 Vpk sinewave.

Original question is on my first post but V2 has reverse polarity. I'm preparing to my exam and there's no much time I have to understand how can I draw this graphs. I think I have learned the basics but for these kind of questions I really need a help! I have another question like this question with Zener Diode. And it's graph looks fine :
http://yadi.sk/d/bY2TFMh56lepV

 

Why can't you post the picture in the forum?

 

Thanks for your notices! I'll be more carefully now. I have listened this lecture on the web (especially 30:35) and now something make a sense on me. I think there's a mistake on my scheme. The DC sources are connected series with the diodes.(I have to change the direction off one DC source as I done for one diode) So I guess this circuit is clipper at the beginning and then a clamper(I really do not understand what actually is). I have created another circuit with 2 diode, 2 DC source and one load resistor.

For v(t) = 12V sin(ωt) input, I think the output should be the square wave with the amplitude 6V. But I can't see the output on Multisim, it's just show me a line.

Look at the scope settings. Your Channel B has 500V/div.

Think about this circuit. The combination of V2/D1 clamps the green node to a value that has to be greater than 5.3V (assuming a 0.7V drop for your virtual diode model). But V3/D2 clamp that same node to a voltage less than -5.3V. See the conflict?

Look at those four components. You basically have a 12V battery that is shorted by two series-connected diodes. Does that make sense?

You keep changing the circuit and it's almost as though you are just making random changes. What is going on? Please post the exact problem that you are trying to solve.

 

I have uploaded the original question.

 

What is the behavior of this "Virtual Diode"? Does it behave as an ideal diode with 0V across it when it is forward-biased and 0A through it when it is reverse-biased?

What is the voltage at the top node in the following circuit?

 

What is the behavior of this "Virtual Diode"? Does it behave as an ideal diode with 0V across it when it is forward-biased and 0A through it when it is reverse-biased?

What is the voltage at the top node in the following circuit?

I'm not very sure but the value might be 0.7 Volt, and yes it behave when it's forward-biased. Sources are connected series, the voltage at the top should be 12V ? (when diodes are ideal, 0V)

 

Well, is it 0.7V or 12V?

Le'ts take it to be 0.7V. What is the forward voltage drop across D1? What is it across D2? Do either of these make sense? Do the same for an answer of 12V? Is there ANY voltage for which the voltage drops across D1 and D2 would make sense?

Unless you can come up with an answer for this part of the circuit (or see that there is a fundamental problem in doing so), you don't stand a chance of answering the larger problem.

This might help you see things a bit more clearly. What is the voltage at the anode of D1? What is the voltage at the cathode of D2?

 

It might help us if you gave us some context for your situation. What kind of program are you enrolled in and at what type of institution? What is your math background? What is your previous circuits and physics background?

 

It might help us if you gave us some context for your situation. What kind of program are you enrolled in and at what type of institution? What is your math background? What is your previous circuits and physics background?

I'm a physics student, and I've enrolled the course called "Basic Electronics", I don't know a much about electricity or electronic. I just know the basics of electricity and how diodes works.

 

Frankly Erbil, if your exam is looming large I think you are wasting valuable time with this problem. If the last schematic you posted [post #26] was truly "correct" as delivered to you by your teacher, then there would be no point in simulating it, given the glaring schematic error posed by the D1, D2, V2 & V3 diode battery topology - a point WBahn has been trying to get you to understand.

 

Frankly Erbil, if your exam is looming large I think you are wasting valuable time with this problem. If the last schematic you posted [post #26] was truly "correct" as delivered to you by your teacher, then there would be no point in simulating it, given the glaring schematic error posed by the D1, D2, V2 & V3 diode battery topology - a point WBahn has been trying to get you to understand.

Ok. I would like to thanks all users who tried to help me. And I'm very very sorry for my bad English.