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Why we cannot measure built in voltage in PN juction?

  1. Jun 8, 2013 #1
    I read this from a lecture slide:
    attachment.php?attachmentid=59371&stc=1&d=1370700424.jpg
    • The electric field across the interface of a pn junction gives rise to a voltage across the interface,called the built-in voltage,V0
    The built-in voltage cannot be measured by externally connecting probes to the device.
    • V0 is due to the difference between the Fermilevels of the joined materials, and can be
    calculated from this

    Can you explain the second point "The built-in voltage cannot be measured by externally connecting probes to the device."?
     

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  3. Jun 8, 2013 #2

    marcusl

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    Please read the How to Ask for Help thread. You must show your attempt at a solution, and where you get stuck, before receiving assistance.
     
  4. Jun 8, 2013 #3
    Hi, but this is not a homework. I am totally lost with this. In PN junction, there is a built in voltage between them V0. It is about 0.7V. I don't understand why this voltage can't measure such as a battery. Please help.
     
  5. Jun 9, 2013 #4

    marcusl

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    The reason is that measuring V_bi requires that metal wires be attached from the device to a meter. It can be shown that the sum of the contact potentials at the interfaces from metal to p and at n to metal exactly cancel V_bi. See this thread for a link with more information
    https://www.physicsforums.com/showthread.php?t=478671

    The following thread has a much more in-depth discussion as well, although it might be a bit confusing to follow:
    www.physicsforums.com/showthread.php?t=200790
     
  6. Jun 9, 2013 #5
    Thanks, smallphi's question is exact with mine. I am reading it and will ask for help later.
     
  7. Jun 10, 2013 #6
    My question is solved. Now I have another one.
    Consider PN junction with metal contacts.
    There are three junctions: metal-p semiconductor, p-n junction, and n semiconductor -metal in series.
    Why only p-n juction prevent electrons flow?
    Why two other junctions (metal-p semiconductor and n semiconductor -metal) don't prevent electrons flow?
    Hope you can help?
     
  8. Jun 10, 2013 #7
    Metal is a conductor. In a conductor, the valence and conduction bands overlap. When this metal-semiconductor junction is formed, there won't be diffusion of charge carriers which causes a depletion region to be formed like in a p-n junction. It is the depletion region which prevents electron flow. Since there is no depletion region formed in the other two junctions, they don't prevent electron flow...

    In simple words, conductors conduct. Conduction is their basic property. Why do they need to prevent electron flow? If there are free electrons (Conduction band electrons), they are free to flow!
     
  9. Jun 10, 2013 #8
    Thanks, but it still has a potential barrier in metal - semiconductor junction, right?
    Let's call the contact voltages as follows:
    metal-p semiconductor junction: V1
    p-n junction: Vbi (built in voltage)
    n semiconductor -metal junction: V2
    As in the old thread mentioned above we have: V1 + Vbi + V2 =0. Therefore, we can't measure Vbi by using a normal voltmeter.
    We need an external electric field (voltage source) to flatten Vbi and make it conduct but why we don't need to cancel out V1 and V2?

    V1 + V2 is till exist and equal to -Vbi.
     
  10. Jun 10, 2013 #9
    Can anyone help? I promise this is the last question.
     
  11. Jun 10, 2013 #10

    marcusl

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    Metal-semiconductor (SC) junctions can have virtually any characteristic depending on the particular metal and SC chosen. To make contact to a pn junction, the materials can be chosen to permit a relatively free flow of electrons. With different choices you can get a rectifying contact. This arises from bending of the SC band so that its chemical potential matches that of the metal, resulting in a parabolic band shape that acts similar to that of a pn junction. Schottky diodes are made this way.
     
  12. Jun 10, 2013 #11
    Thanks, to sum up what I understand:
    Then can I infer that elctrostatic field from the applied voltage source will be zero at these contacts ( metal - n semiconductor and p semiconductor - metal)?
    I mean that the voltage drops( supplied by applied source) at these contacts will be zero?
    The contact potentials at these juctions ( metal - n semiconductor and p semiconductor - metal) always cancel out built -in potential no matter how we make these contacts.
    V1 + Vbi + V2 = 0 always mo matter how we make these junctions easier for electrons flow or prevent electrons from flow?
    Sorry for asking but I am very confused and want to know it.
     
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