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Pn junction diode

  1. Jan 14, 2008 #1
    Say we have a silcon pn junction (diode) - i.e., a block of p-type on
    left, attached to a block of n-type semiconductor on right:

    anode ------[ p | n ]------ cathode

    Because of diffusion we get a barrier potention at the junction, which
    makes the n-side/cathode 0.7V higher than the p-side/anode:

    ~~~~~~~~ - 0.7V +

    anode ------[ p | n ]------ cathode (*)

    1) Now is there any way to "measure" this potential difference right
    from the diode using some instrument?

    Now to get ride of the depletion layer (barrier) we need an "opposite"
    external voltage equal in magnitude to the 0.7V shown in (*) :

    ~~~~~~~ (barrier potential)
    ~~~~~~~ - 0.7V +
    anode ------[ p | n ]------ cathode

    ~~~~~~~ + 0.7V -
    ~~~~~~~ (external voltage)

    2) Now, why isn't the resulting voltage of the diode 0V (sum of
    barrier and external)? How come we only measure the external 0.7V
    using a voltmeter when the diode is forward biased?

    3) Is there anything wrong with the thought process I've outlined
  2. jcsd
  3. Jan 14, 2008 #2
    The instrument to measure is a voltmeter, or for more accurate results you can use a scope (only when the diode is on).

    Correction: Because of diffusion the electron-hole paris recombine and a depletion layer is formed. To overcome this depletion layer an external potential (generally 0.7V for Silicon) is needed, hence you measure 0.7V and not 0V.

    Once the depletion layer is formed it prevents the further recombination of holes and electrons.
    Last edited: Jan 14, 2008
  4. Jan 14, 2008 #3


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    Gold Member

    Please note that it is not as simple as placing a voltmeter across the terminals of the diode and measuring the barrier potential, Vb. In fact one would expect a reading of zero volts. Give some thought to contact potentials. Remember that we have metal-metal and metal-semiconductor contacts.
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