Stuck on this diode problem from the Hayt/Neudeck "ECA&D, 2nd Ed."

Discussion in 'Homework Help' started by grasshopper, Feb 19, 2014.

  1. grasshopper

    Thread Starter Member

    Apr 2, 2009
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    0
    Hello,

    I am doing some self-study, but I don't have the benefit of a solutions manual, so sooner or later I knew I would run into a problem.

    This problem is from the book:

    Electronic Circuit Analysis and Design, 2nd Edition
    William H. Hayt, Gerold W. Neudeck
    ISBN: 978-0-471-12501-3

    (Chapter 1, Problem 7)
    A diode for which I_D = exp(33*V_D) - 1 [nA] is in series with a dc source of 0.5V. Sketch a curve of I_D vs. V_in, the voltage across the series combination, for the two possible battery polarities. Include both positive and negative values of V_in.

    To solve this problem, I let I_D = 1mA, and solved for V_D, which came out to be 0.419.

    The answer in the back of the book shows that that if +0.5V was used for the dc source, then the V_in vs. I_D would be (0.919, 1 mA), and for -0.5V it would be (-0.081, 1mA). (1 point on each curve given as the answer)

    To get these answers: +0.5V + 0.419V = 0.919 and -0.5V + 0.419V = -0.081. I assume that this what was meant by "the voltages across the series combination, V_in".

    But I can't make sense of it when I draw the circuit out. Because it would make more sense if it was +0.5V - 0.419V = 0.081V and -0.5V - 0.419V = -0.919V, with the dc source and the diode in series.

    Can someone please help me make sense of this? :confused:
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    Well, we can't make sense of what you see when you draw it out unless you post a sketch of how you are drawing it out.
     
  3. grasshopper

    Thread Starter Member

    Apr 2, 2009
    11
    0
    Hi WBahn and others,

    Attached is a sketch of how I see the problem. There is no schematic given for this problem. So I drew one out to the best of my understanding.
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    You problem is in your interpretation of the circuit. My initial thought was as yours, but there are some clues that tell you that this is wrong. First, they ask you to sketch Id versus Vin for both positive and negative values of Vin. But your circuit doesn't have a Vin, it just has a fixed 0.5V or -0.5V across the diode. Second, your calculation of the point 0.419V for a current of 1mA is meaningless for your circuit since you don't have 0.419V across the diode.

    So, with those clues telling me that it isn't a 0.5V source applied across the diode, the next obvious choice was that they meant that the source and diode are in series as part of another circuit that is applying a voltage to the series combination of the two.

    With that in mind, see if it makes more sense to you.
     
    anhnha likes this.
  5. grasshopper

    Thread Starter Member

    Apr 2, 2009
    11
    0
    Thanks Wbahn,

    I drew the circuit in the manner you suggested (attached below) and it makes heck of a lot more sense with the solutions in consideration.

    However I still have a question. Why would the 0.5V dc voltage that is placed in series with the diode be called "dc source." Is it a terminology that is used for any independent dc voltage, regardless where they are placed in the circuit (I suspect yes)? Prior to this experience, I always used to think that the "dc source" would be more or less equivalent to "V_in".

    Again, thanks very much for your help and also for making me think through this problem.
     
  6. WBahn

    Moderator

    Mar 31, 2012
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    A battery is a battery no matter where you place it. A source is a source no matter where you place it. Now, at times it might act like a load (otherwise battery chargers would be oxymorons), but it is still proper to refer to it as a "source" component.

    Vin, on the other hand, will usually NOT be a source, but rather just a signal at a certain point is a circuit.
     
    anhnha likes this.
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