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Resistance of a diode

  1. Sep 14, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-9-14_20-48-18.png

    2. Relevant equations


    3. The attempt at a solution
    Considering I- V characteristics graph, the slope for D.C. resistance is lower than that of the A.C. resistance. Now, the resistance is inverse of the corresponding slope. So, D.C. resistance should be more than A.C. resistance.
     
  2. jcsd
  3. Sep 14, 2017 #2

    berkeman

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    Staff: Mentor

    Can you show us on the I-V curve? I think I know what the answer is, but I'm not sure.
     
  4. Sep 14, 2017 #3
    upload_2017-9-14_22-5-59.png
    It is clear that the slope for D.C. is lower than that for A.C.
    Hence, the resistance for the D.C.is greater than that for A.C.
    Right?
     

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  5. Sep 14, 2017 #4

    berkeman

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    Yes, correct. With that graph you can show that the tangent to the graph always has a greater slope than the line from the origin to the point on the graph (at least in the 1st quadrant).

    Quiz Question -- is it still true in the 3rd quadrant for reverse bias? Include low-level reverse bias and as the diode starts to go into reverse breakdown... :smile:
     
  6. Sep 14, 2017 #5
    Before the breakdown voltage, the DC resistance is less than AC resistance.
    After the breakdown voltage, the DC resistance is more than the AC resistance.
     
  7. Sep 14, 2017 #6
    Is the definition of linear device following?
    Those devices who follow the rule V = IR, are known as linear devices.
     
  8. Sep 14, 2017 #7

    berkeman

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    Staff: Mentor

    Can you show on the I-V curve why you think this is true? :smile:
     
  9. Sep 14, 2017 #8
    upload_2017-9-14_23-2-40.png
    Sorry, even before breakdown, the slope of AC is more than that of DC.
    So, AC resistance is smaller than the DC resistance.

    So, DC resistance is always more than the AC resistance. Right?
     
  10. Sep 14, 2017 #9

    berkeman

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    Staff: Mentor

    Yep! Good job. Looking at the slope really helps on this type of question. You can also show it mathematically by taking the first derivative of the Diode Equation, but this one is pretty easy to see graphically. :smile:
     
  11. Sep 14, 2017 #10
    :smile::smile::smile::smile:
    Done!
    Thank you, thanks a lot for guiding me.
     
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