Voltage drop across diode

Discussion in 'General Electronics Chat' started by jaydnul, Aug 18, 2015.

  1. jaydnul

    Thread Starter Member

    Apr 2, 2015
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    0
    I understand that the ~.7V drop across the diode is caused my the depletion layer counter acting the electric field in the diode from the voltage, but it also takes energy to knock the electrons out of their holes and back into the circuit, so where does that energy come from?

    For example, in an LED, the energy released when the electrons combine with the holes is given off as visible light, but that light is radiating away; where does the energy come from to make up for the light?

    My guess would be that the .7V drop is responsible for both getting rid of the depletion region and knocking the electrons out of the holes and back into the circuit. Is this right?
     
  2. dl324

    Distinguished Member

    Mar 30, 2015
    3,235
    619
    This site has a lot of basic information available. You can read about P-N junctions here.
     
  3. shteii01

    AAC Fanatic!

    Feb 19, 2010
    3,383
    495
    Why guess?
    My textbook has a whole section on the physics of diode operation.

    All anyone has to do is read. No guessing required.
     
  4. crutschow

    Expert

    Mar 14, 2008
    12,991
    3,227
    Energy equals V * I.
    In this case the energy comes from the forward drop of the diode times its current.
     
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