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Diode circuit analysis

  1. May 8, 2014 #1
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    I have just started studying this topic and I am confused about several things.

    1) Why does the assumption that D1 is reverse biased mean it is modeled as a switch? Are reverse biased diodes always modeled as a switch?
    2) Also, why does the assumption that D2 is forward biased means it can be modelled as above?
    3) Where did they get V = 7V from?

    Thanks
     
  2. jcsd
  3. May 8, 2014 #2

    phinds

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    1) A reverse biased diode is an open circuit and a forward biased diode is a short circuit. Doesn't that sound like a switch to you?
    2) I don't even understand the question. A forward biased diode is a short circuit and they have shown it as a short circuit. What is it you don't understand?
    3) Uh ... what's 10 minus 3?
     
  4. May 8, 2014 #3

    donpacino

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    2. When a diode is in forward conduction mode it has a voltage drop that is relatively independent from the current through the device. This voltage drop is sometimes approximated as zero (a short circuit).

    3. typically for solving multiple diode problem you make a guess as to whether the diodes are conducting or not. you then evaluate your circuit model to figure out if your guess was correct. In this case with 7 volts across diode 1, it would have to be conducting. Therefore you know that your initial guess was wrong, you will have to make another guess and evaluate the circuit again.
     
  5. May 8, 2014 #4
    I didn't know that a forward biased diode is a short circuit and that a reverse biased diode is an open circuit. Thanks :)

    I'm still confused why its 7V across diode 1 though? Why do we do 10-3 ?

    Also, how do you know whether to assume whether a particular diode is forward biased or reverse biased?
     
  6. May 8, 2014 #5

    phinds

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    This is only true of ideal diodes, of course. Real diodes have non-ideal characteristics

    We do 10 - 3 because it has 10v on one side a 3v on the other side. Just look at the circuit.

    You don't. That's one of the points of your original problem. You make an assumption and see if the circuit can support the assumption. If it can then you have a solution. If it can't, as in your original problem's assumption, then you don't and you have to try another assumption.

    Since the original assumption didn't work, what assumption would you make next?
     
  7. May 8, 2014 #6
    Assume the opposite for each diode. Cool.

    Maybe I am missing something basic but I am failing to see how how that yields the voltage across D1?
     
  8. May 8, 2014 #7

    phinds

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    I am at a loss as to how to explain that. There is 10V on one side of the diode and 3V on the other side. Draw all the voltages in the circuit.
     
  9. May 10, 2014 #8

    CWatters

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    If D2 is conducting the 3V source appears across the 6K resistor. So there you have 4K and D1 in series, and there is 10V on one side and 3V on the other. Note the polarity and work out if D1 is conducting.
     
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