# Transfer Curve of Diodes

Discussion in 'Homework Help' started by Erbil, Jul 9, 2013.

1. ### Erbil Thread Starter New Member

Mar 19, 2013
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Hi. I'm very newbie on electronics. I know just the basic principles, like how diodes was made it. Or we can use the diodes to convert to signals etc. (reverse bias, forward bias) But to graph the output signal, I really need to help! So please help me to graph the transfer characteristics curve and output signal.

I have uploaded my question. İf anyone explain me how I can graph the output signal step by step I'll be very happy (sorry for my bad English)

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2. ### WBahn Moderator

Mar 31, 2012
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First, this is Homework Help, not Homework Done For You. You need to make a good faith effort at an attempt. That will serve as a starting point for discussion.

Second, your English is definitely going to be an issue, but we'll work through that.

Third, you have the notation Vi-15sin(ωt). Assuming that this is 15V, what is Vi? And what is (Vi-15Vsin(ωt)) equal to?

What are the constraints on the voltage on the node connecting R1 and C1?

What are the constraints on the voltage on the output node?

3. ### Erbil Thread Starter New Member

Mar 19, 2013
15
0
@WBahn;

OK. But I just want to specify that I never want from someone to do my homework. I really want to understand that. So before that question, I have uploaded easier question. I know that if the diode is forward bias we have a half wave rectifier output. But what about when we have a DC power too? (like that picture) Can you explain it to me theorically?

Thanks,

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4. ### Erbil Thread Starter New Member

Mar 19, 2013
15
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İnput signal is 15 V. The are no any constraints. And you can use a different values for capacitors and resistances.

Last edited: Jul 9, 2013
5. ### WBahn Moderator

Mar 31, 2012
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Is the input signal a sinusoid at 15V amplitude, or at 15Vrms? You seem to be indicating both, but it is one or the other. They are NOT the same.

If there are no constraints on the voltage on the node between R1 and C1 (let's call it Node A and that voltage Va), then you are saying that Va can be anything? Can it be 0V? Can it be 10V? The carefully about the other branches connected to that node before you answer.

6. ### WBahn Moderator

Mar 31, 2012
23,401
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If you are trying to explore the impact of have a DC supply in the mix, don't you think it might be a good idea to DC couple the scope channels?

To analyze the circuit, make two circuits, one with the diode removed entirely and one with the diode replaced by a properly-oriented 0.7V battery. Us the first circuit to determine the behavior when the diode is reverse biased (or, more accurately, biased below the turn-on voltage) and the second to determine the behahavior when the diode is forward biased.

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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I suspect source V2 in the schematic in the first post should be of opposite polarity to that indicated.
If you are a noobie you will find it difficult to solve the problem anyway.

8. ### Erbil Thread Starter New Member

Mar 19, 2013
15
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Please let me ask me to you first, if we remove the diode entirely we have one AC and one DC source. So does it mean this circuit has (for example : AC is 12sinWt, DC = 5V) 5V for negative voltage values ?

9. ### Erbil Thread Starter New Member

Mar 19, 2013
15
0
All I know that the amplitude of AC source is 15V. And DC sources is 6V. There are no any information on my homework.

10. ### WBahn Moderator

Mar 31, 2012
23,401
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I suspect you are likely correct about V2.

The presence of the capacitor makes this a pretty challenging problem. I would recommend replacing the capacitor with a resistor to at least get a decent feel for the general behavior first.

11. ### WBahn Moderator

Mar 31, 2012
23,401
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Not at all.

Consider a 5V DC source in series with an AC source that is 12Vsin(ωt) in series with a resistor R. The voltage across the resistor is

v(t) = 5V + 12Vsin(ωt)

It will peak at 17V on the positive side and -7V on the negative side.

12. ### WBahn Moderator

Mar 31, 2012
23,401
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If the amplitude of the AC source is 15V, then your simulation schematic is wrong because it has the AC source with an amplitude of 15Vrms, which is NOT the same thing. The AC source in your schematic has an amplitude of 21.2V.

13. ### Erbil Thread Starter New Member

Mar 19, 2013
15
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Yes you're right. I couldn't find where can I change the amplitude. So I though that RMS is near to that value. Anyway, the AC source is 12V sinewave.

14. ### JoeJester AAC Fanatic!

Apr 26, 2005
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On your schematic, it indicates the required amplitude of the input signal. Its in the lower left hand corner.

15. ### WBahn Moderator

Mar 31, 2012
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You need to STOP using ambiguous language!

What do you mean when you say the AC source is 12V sinewave? Do you mean amplitude? Do you mean RMS? Do you mean peak-peak? Don't make people read your mind or between the lines to interpret what you mean.

For a sinewave, the amplitude is sqrt(2) times the RMS value. So if you want to have a voltage source that outputs

v(t) = Vo sin(ωt), then you need to set the AC source to a magnitude of Vo/sqrt(2) as its RMS value.

16. ### Erbil Thread Starter New Member

Mar 19, 2013
15
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I apologies. It's just v(t) = 12 sin(ωt),

Last edited: Jul 10, 2013
17. ### WBahn Moderator

Mar 31, 2012
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And you need to START using units!

If you want v(t) = 12V sin(ωt), then you need to set the source amplitude to 8.5Vrms.

Make that change and also change your virtual scope so that the input channels are DC coupled. Then rerun your simulation and post the results. It won't make much sense to talk about the results in terms of what is happening in the circuit until you do.

18. ### Erbil Thread Starter New Member

Mar 19, 2013
15
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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.

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19. ### Erbil Thread Starter New Member

Mar 19, 2013
15
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Yes, you're correct. I have tried something for that, can you take a look ?

20. ### JoeJester AAC Fanatic!

Apr 26, 2005
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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.