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pn junction diode

Discussion in 'Electronic Design' started by Lax, Jan 10, 2008.

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  1. Lax

    Lax Guest

    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. D from BC

    D from BC Guest

    I sometimes just think of diode as a wacky current dependant resistor.

    D from BC
    British Columbia
  3. Guest

    My 2 pesos:
    A diode is not a battery. The diode obeys i=isubs*exp(v/vt). Say you
    curve trace a diode. Well, you view it linearly, so looking on the
    curve tracer, you say the potential is 0.7V, but really that is only
    true at some current level. if you were the type to work the diode
    harder, you might say a diode drop is 0.75 V,

    Think of the so-called junk buffer circuit. [PNP emitter follower
    followed by NPN emitter follower.] It doesn't have zero offset, yet it
    is a better example of attempting cancellation than the one you
  4. whit3rd

    whit3rd Guest

    Yes, but it can't just be ANY diode. It needs to be open to light.
    There is a threshold effect in photoelectric output that scales with
    the built-in potential.

    Interestingly, ANY two metals take on a potential difference on
    and that's why thermocouples work. But to measure such potentials
    inside a circuit one needs to control ALL the circuit's materials and
    temperatures. Doing the measurement by photoelectric emission is
    possible, there have been experiments with vacuum lathes (because
    a fresh metal surface is required, not a dirty exposed-to-air crust).
  5. whit3rd wrote:

    You might want to look into this belief a bit, since it is
    completely wrong.
  6. whit3rd

    whit3rd Guest

    No, not wrong, just a bit perverse. The little electrons
    quickly into charged layers after contact, but at the moment of
    contact, there IS a field at the interface.
  7. Have it your way.

    Since you brought it up, please tell me more about how
    thermocouples work.
  8. Tim Williams

    Tim Williams Guest

    Contact potentials are well known in condensed state physics. They can be
    quite high and have essentially unlimited current, but the naughty thing is
    getting that voltage out without cancelling it with another junction or
    series of junctions. A difference in temperature generates a low order
    difference, which is how thermocouples work.

  9. Show me the phrase "contact potential" or "low order
    difference" in that article.

    Here is the sentence that tells what makes thermocouples
    work, "The thermopower, or thermoelectric power, or Seebeck
    coefficient of a material is a measure of the magnitude of
    an induced thermoelectric voltage in response to a
    temperature difference across that material." Thermocouples
    are made of two materials that produce different induced
    thermoelectric voltages in response to the same temperature
    difference across those different materials.

    The thermocouple output is the difference of those two
    thermal gradient induced potentials. It has nothing to do
    with anything produced at their contact surface. They don't
    even have to contact each other, They can be connected
    through any number of seriesed contact surfaces with
    intermediate conductive materials, as long as all those
    contact surfaces are at the same temperature. The contact
    surfaces produces no voltage.
  10. whit3rd

    whit3rd Guest

    Geez, there must be a lot of engineer-types on this
    blog. All the discussion is about externals, just
    phenomenology, not about the innards.

    The density of allowed states in two dissimilar metals causes, on
    contact, a diffusion of electrons from the material with the lower
    density of allowed states, into the other (and in the case of PN
    junctions, two separate bands, the holes and electrons, BOTH
    undergo such diffusion). The charge transfer, of course,
    builds up a charge layer and causes an electric field (which
    eventually leads to zero net diffusion current).

    The amount of that field minimizes the Helmholtz free energy
    (see Wikipedia - it's got a temperature term, and entropy
    which relates it directly to the density-of-states effect). The
    temperature dependence of the field, and the voltage drop
    that measures the field, makes it useful for thermometry.

    All the discussion has been on Seebeck effect, tabulated
    thermocouple voltages, etc. That isn't an explanation at all,
    just a description of the apparatus. And, no simple apparatus
    is gonna tell you about the thermocouple voltage of a SINGLE
    junction of two metals. All your meters work only on complete
    circuits (which will have TWO junctions or more).

    The main way this relates to electronic design is in the
    microvolt range, at DC. No feasible amplifier technology
    will ever give you zero errors at low frequency, because
    neither the factory calibration fixture nor the normal
    circuit environment is so uniform in temperature or
    controlled in wiring composition that the thermocouple
    voltages won't randomize (or worse, couple) to make
    input 'offset error', usually in the few-microvolts range.
    The best you can do, is to make the circuits very small,
    with minimum power dissipation, and wrap thermal
    blankets around 'em.
  11. With a Usenet title of "", ya think?
    Keep digging. That hole isn't quite deep enough. We can
    still see you.
    Provably not true by simple experiment with a thermocouple
    and a millivolt meter. The only point in having the second
    thermocouple is to remove the temperature dependence of the
    meter connections on that measured voltage.
    Damn, you can dig two separate holes, simultaneously.
    At this rate, you could disappear from view, permanently, in
    just a few days, in two holes, at the same time. Of course,
    you could go back and actually learn about how thermocouples
    work and fill these holes back in. But I am not betting
    that will happen any time soon. Shame, really.
  12. whit3rd

    whit3rd Guest

    ...   And, no simple apparatus
    When I think of this experiment, I see metal A, joined to metal B,
    with copper leads from A to a meter movement, to B.
    So there is an A-B thermocouple, a B-copper couple,
    and a copper-A couple. At thermal equilibrium, A, B, and the
    copper all take on different potentials, but no current flows in
    the loop. Current only flows if the couples aren't at
    thermal equilibrium.
  13. neon


    Oct 21, 2006
    see boltzman constant for diodes drop. the forward voltage drop of any diodes is never constant varies with current and tenperature, it is an exponantional inpedance. it is assumed a drop of .6 to ,7v fully conducting but it could be much less it depends on the current flowing,
  14. Phil Hobbs

    Phil Hobbs Guest

    Let me come to the guy's defense a bit...the physics of the Seebeck
    effect really is more or less the same as that of the depletion region
    in semiconductors--a balance between diffusion and drift.

    Consider two pieces of metal, e.g. calcium (work function of about 2 eV)
    and platinum (work function about 6 eV), both of which are electrically
    neutral (same number of electrons and protons). Electrostatics is
    conservative force, i.e. in a static field, if you take an electron
    around a closed loop, its energy doesn't change. If you rip an electron
    out of the calcium and take it out to a large distance, it'll cost you 2
    eV. If you now let it fall onto the platinum, it'll release 6 eV. (A
    fine point: the energy released is actually the electron affinity of the
    material, not the work function, but the two are essentially identical
    for partly-filled bands, i.e. conduction electrons in metals.)

    Sooo, when you touch the Ca to the Pt, electrons start flowing across
    the boundary until the two Fermi levels line up. But that results in a
    big boundary layer of free charge inside the Pt, enough so that the
    voltage drop across the charge layer equals the difference in the Fermi
    levels of the metals. (This is not quite the same as the difference in
    the work functions because the electrostatic image potential, which is
    part of the work function, depends on the surface charge density of the

    In a metal, these layers are very thin because the density of free
    charge is so high, which makes the Debye shielding length very short.

    The Seebeck effect arises from the different temperature coefficients of
    the two processes, diffusion and drift. As you increase the
    temperature, more electrons drift into the Pt, which changes the potential.

    If you have two junctions back to back, the common mode temperature
    cancels out by symmetry, but in the presence of a temperature
    difference, the cancellation isn't complete, and a thermocouple voltage
    can be measured via an external circuit.

    The chemical potential and free energy business is just a physicist's or
    chemist's way of describing the result--there's a variational principle
    in thermodynamics that systems seek the state of minimum free energy
    (This is either the Helmholtz or Gibbs free energy, depending on whether
    you're keeping the number of particles constant or not.) It's much
    easier to calculate that way than having to keep the kinetics in view
    all the time.


    Phil Hobbs
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