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Shortcircuiting a pn diode

  1. Jun 25, 2005 #1
    A thought experiment: (note that this comes from a student with only amateur experience in solid state physics)

    suppose you could shortcircuit a pn diode... ie you just link a wire between both ends, or maybe even bend the pn diode in a ring (so you'd have a n-surface between two p-surfaces, and a p-surface between two n-surfaces).

    What would you get?

    Based on that second idea, I was thinking something among the lines of : you'd get two depletion areas at both ends, with a width that's equal to half of the width you'd get from a single end connection.

    I'm having problems with realizing what the differences in characteristics are between the 'internal potential' of the diode, evoked by the difference in free electron concentration (I hope that's an ok interpretation) , and the 'external potential', evoked by the source. I understand how the external potential can 'amplify' or 'lessen' the diode effect - that's somewhat like simple charge induction - however, when I try to combine the idea of the diode and the external potential as in above experiment, I get stuck. Maybe it would help if someone showed me convincingly why a diode can't be used as a power source.
  2. jcsd
  3. Jun 26, 2005 #2


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    You may want to look into what a solid-state transistor is. These are either npn or pnp junctions.

  4. Jun 26, 2005 #3
    A transistor is not really the same as shortcircuiting a diode I think..
  5. Jun 26, 2005 #4


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    There is little if any free charges in a semiconductor. Short circiuting a diode by itself will do nothing.
  6. Jun 27, 2005 #5
    if you had come up with this idea like 40/50 years back, you would have the same financial capacity like Bill Gates.

    Google for MOSFET or look at my latest entries in my journal. In the very last one i explain (briefly) two of the three stages of MOSFET operation : ie accumulation, depletion and inversion

  7. Jun 27, 2005 #6
    Actually a diode can be used as a power source if you drive it out of equilibrium by an external light source. On the other hand, no power can be generated by any device in equilibrium- it should obey to the first and second law of thermodynamics.
  8. Jul 6, 2005 #7
    Thank you very much - but I have to give the credit to a guy from my class. He came up with the original idea, I simply came up with the ring experiment.

    (And I'll be sure to read your journal as soon as I have the time... my vacation seems to be more hectic than the rest of the year.)
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