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p-n junction question

Discussion in 'Electronic Basics' started by vic, Jun 3, 2007.

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  1. vic

    vic Guest

    When various textbooks explain the operation of a p-n junction or a
    diode, they usually do not talk about metal contacts. That is usually
    a separate topic. But let's consider the whole thing together since
    this the only way the diode can operate and let's assume that both
    anode and cathode contacts are made from the same metal and therefore
    have the same energy level of free electrons.

    The energy of an electron flowing through the diode under a forward
    bias condition would have to first be elevated from the metal energy
    level to the energy level of the conduction band of the N-type
    semiconductor, then it would drop when the electron crosses the
    junction an falls into a hole in the P-type semiconductor (and
    releases thermal or light energy) and then elevated again to the
    original metal energy level.

    The voltage drop on a diode will be equal to the band gap and the
    energy spent to get an electron through the diode will be eaual to the
    energy released during the recombination. This energy is needed to
    elevate the energy of an electron before it plunges to a hole and,
    again, to get it out of a hole back to the metal, but any textbook
    would tell you that the voltage drop occurs on the junction itself to
    overcome a barrier created there though carrier diffusion before an
    external voltage was applied.

    As far as I can see there is no barrier to overcome at the junction in
    the forward direction - it's a fall, like in a waterfall. But you do
    need to overcome a barrier between the metal terminal and the N-type
    semiconductor of the cathode to get to that high energy level and
    that's where an external energy is spent.

    Am I missing something basic here? Does everyone else understand the
    story about the depletion region and how a dynamic balance (diffusion
    + drift) existing before an external voltage is applied to a diode
    somehow affects the operation of a diode ever after?
     
  2. Marra

    Marra Guest

    In practice all you need know is its 0v7 forward volts drop and it
    wont conduct reverse polarity.

    I can walk but dont understand how my legs work !
     
  3. Guest

    Yep, this is a weird and fascinating topic. Search for info on
    "nonrectifying junction", also "ohmic contact"

    If manufacturers just stuck some metal contacts on the semiconductor,
    they'd form Schottky diodes in series with the main diode (metal/
    silicon junctions.) The diode as a whole would always turn off
    regardless of polarity.

    So instead they apply heavy doping to the semiconductor surface before
    adding the metal contacts. This converts the semiconductor surface
    into a "metal-like" conductor. In diode diagrams you'll often see a
    layer of p+ next to the metal contact on the p-doped layer, and a
    layer of n- next to the metal touching the n-doped layer. In that
    case there still is an energy shift, but it doesn't present a
    barrier. Don't forget that energy shifts are NOT BARRIERS, they are
    more like ideal batteries. Only a depletion layer can act as a high-
    resistance barrier. ALso don't forget the thermocouple effect: that
    whenever you use copper and aluminum or iron wires in the same
    circuit, there is an energy shift at the metal junctions. These
    mismatched "work functions" don't create a high resistance because the
    energy shift around the circuit as a whole is zero. Any energy gain
    at one metal-metal junction will be cancelled by an energy loss at
    other, opposite metal-metal junctions. It's similar to hooking two
    batteries back to back in series: the voltages cancel out, but the
    batteries still form a low-resistance conductive path.

    But why can we prevent the formation of a metal/semiconductor diode by
    adding extra-heavy doping to the semiconductor? It's because of
    quantum mechanics: the heavier the doping, the thinner the
    insulating depletion layer is formed at any particular reverse
    voltage. To form a diode, light doping is required. With heavy
    enough doping, the depletion layer becomes so thin that electrons can
    "tunnel" quantum-mechanically through this insulating region. The
    metal contact becomes like a tunnel diode, but a tunnel diode that
    turns fully on at all values of applied voltage.


    (((((((((((((((((( ( ( ( ( (O) ) ) ) ) )))))))))))))))))))
    William J. Beaty http://staff.washington.edu/wbeaty/
    beaty chem.washington.edu Research Engineer
    billb eskimo.com UW Chem Dept, Bagley Hall RM74
    206-543-6195 Box 351700, Seattle, WA 98195-1700
     
  4. neon

    neon

    1,325
    0
    Oct 21, 2006
    don't forget if electrons flow this way holes flow opposite.metals are conductors only pn is a semiconductor with controlled inpurity injected to minority holes or electrons
     
    Last edited: Jun 11, 2007
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