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Built-in voltage of pn junction diode

  1. Nov 11, 2015 #1

    cnh1995

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    Why can't the built-in voltage of diode be measured by a voltmeter when the diode is isolated? A potential difference of almost 0.7V exists at the junction, which is clearly not too small for the voltmeter to detect. What's the reason?
     
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  3. Nov 11, 2015 #2

    anorlunda

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    Because the voltage is there only when a current flows. The voltmeter won't cause a significant current flow.

    It is not a voltage source like a battery.
     
  4. Nov 11, 2015 #3

    cnh1995

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    Why does it not act like a voltage source? There are +ve and -ve charges present. Isn't it like an electric dipole?
     
  5. Nov 11, 2015 #4

    anorlunda

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  6. Nov 11, 2015 #5

    marcusl

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    A more precise answer is because the contact potentials at the p-metal and n-metal junctions exactly counteract the built-in potential.
     
  7. Nov 12, 2015 #6

    cnh1995

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    That's what I read somewhere but didn't understand fully. Could you please elaborate? What exactly is the contact potential?
     
  8. Nov 12, 2015 #7

    LvW

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    Did you hear already about a Schottky diode? This a junction between metal and semiconductor - and exactly such a junction exists (necessarily) caused by the contact material at both ends of the pn diode. And the diffusion voltages at these junctions compensate the diffusion voltage across the internal pn junction.
     
  9. Nov 12, 2015 #8

    marcusl

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    This is a common question and has been the subject of a number of threads at Physics Forums such as
    https://www.physicsforums.com/threads/pn-junction-open-conditions-conservation-of-energy.478671/
    A simple graphical explanation showing why this must be true is on the first two pages here:
    https://www.google.com/url?sa=t&rct...rd0cCNe0g6l2U98TPl1EMg&bvm=bv.107467506,d.cGc
    The important point is that the metal part of the circuit is all at the same (0) potential, so Vbi must appear in reverse at each semiconductor contact.
     
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