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Which path will current follow?

  1. May 26, 2015 #1
    1. The problem statement, all variables and given/known data
    using complete diode model explain which path will the current follow? given dynamic resistance is 240 ohm.compute current.
    2. The attempt at a solution
    when we make complete diode model, ideal diode will be replaced by a switch and the given dynamic resistance. the silicon diode will be replaced by a 0.7v battery, a switch and the given dynamic resistance.
    from figure both the diodes are forward biased.
    now as ideal diode has 0v potential drop across itself, it will start conducting the circuit current immediately, preventing the silicon diode to attain 0.7v and thus from turning on. so the current will follow the path through ideal diode.
    now we can compute current as, the whole drop of battery is across dynamic resistance of ideal diode so by applying ohms law I=0.041667A.
    Now problem is only i want to confirm this solution attempt. is this right or any other modification needed??
    thanks Untitled.png Untitled2.png Untitled2.png
     
  2. jcsd
  3. May 27, 2015 #2

    Zondrina

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    Just to clarify the diode models:

    An ideal diode will have a ##0 V## drop when forward biased. There will be no small signal resistance in this case.

    The constant voltage drop model usually assumes the voltage drop is ##0.7 V## and there will again be no resistance.

    The small signal model assumes the diode is replaced by a ##0.7 V## battery and a small signal resistance given by ##r_d = \frac{n V_T}{I_D}##.
     
  4. May 28, 2015 #3

    rude man

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    Picture 1: you just burned out your diode!
     
  5. May 28, 2015 #4
    oka......thanks #zodrina you mean to say that we have no longer needed to attach dynamic resistance in ideal diode. In this case this will be a short circuit and current will follow the short circuit but we can not compute the current. Is it??? or something else???
     
  6. May 29, 2015 #5

    CWatters

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    Can you check the polarity of the battery is correct in your drawing?
     
  7. May 30, 2015 #6

    Zondrina

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    If you pointed the battery the other way with the current set up, there would be two open circuits. Trivially the current would be zero.

    I'll assume the ideal diode model for the sake of simplicity.

    With the current setup, both diodes will behave as short circuits. The left wire would short the current going to the resistor, hence the current flowing in the left-hand diode is the only current you need to worry about. You can calculate the diode current if you assumed a non-ideal model.

    Perhaps if you flipped the left-hand diode the other way, your circuit would work with the current setup, and the given problem would seemingly make more sense.
     
    Last edited: May 30, 2015
  8. May 31, 2015 #7

    CWatters

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    Not necessarily.

    The problem statement says.. "using complete diode model".
     
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