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Question regarding the photodiode model

Discussion in 'Electronic Design' started by cuca, Aug 22, 2006.

  1. cuca

    cuca Guest

    Question regarding the photodiode model. Just to clarify in my mind,
    if a photodiode has no biasing voltage across it, i.e. it is operated
    under short circuit condition using a transimpedance amplifier,
    theoretically there is no leakage or dark current. Thus the shunt
    resistance could be considered infinite or at least larger that that
    quoted in datasheets where a 10mv reverse bias voltage. Am I write in
    saying this?
     
  2. cuca wrote...
    No. The leakage current, V/R will be zero if V=0
    (which it rarely does), but R will not change.
     
  3. Phil Hobbs

    Phil Hobbs Guest

    No. If you want to increase the shunt resistance, run it at a small
    reverse bias. The I-V curve (current vs voltage) of a diode without
    photocurrent goes through (0,0) but not with zero slope. The shunt
    resistance is dV/dI, which is large but not zero at the origin.

    If you're worried about the shut resistance of your diode, one of two
    things is probably true:

    1. You're using an IR diode such as Ge or InAs, and it's leaky at room
    temperature.

    2. You're trying to maximize your signal-to-noise ratio, and want to
    minimize the dark current shot noise.

    Reason #1 is a good one, but reason #2 is almost always a bad one,
    unless your signals are in the nanoamps or below. Silicon and InGaAs
    photodiodes are so much better behaved with reverse bias, it'll blow
    your mind. Try 7x lower capacitance (which translates to 17 dB lower
    noise density at high frequencies), 10x better linearity, ... I could go on.

    Cheers,

    Phil Hobbs
     
  4. Phil Hobbs

    Phil Hobbs Guest

    My brain went into reciprocal space there for a moment. Should read,
    "large but not infinite".

    Cheers,
    Phil Hobbs
     
  5. Guest

    The dark current won't be zero at zero bias - charge carriers promoted
    to the conduction band by thermal fluctuations, cosmic rays and local
    radioactivity look just like charge carriers promoted by photon
    absorbtion.

    Leakage current will be zero. As has been mentioned in the other posts,
    the shunt resistance of the diode is not infinite, or even all that
    large, at zero bias - IIRR it is just kT -26mV at room temperature -
    divided by the leakage current at a volt or so of reverse bias
    (anything too low to give any significant avalanche multiplication -
    the reverse current at the rated reverse voltage is inflated by
    avalanching, which is why you don't want to exceed the rated reverse
    voltage).
     
  6. Phil Hobbs

    Phil Hobbs Guest

    So you don't believe that a wire with zero voltage across it carries
    zero current? It has conduction-band electrons all the time.

    Cheers,

    Phil Hobbs
     
  7. Joerg

    Joerg Guest

    Hello Phil,

    I had one that was shorted but beginning to let off a stench. Then the
    insulation melted off and glued it to the table. The stench became
    unbearable.

    Ok, it got helped along by the field of the tank coil of a 750W amp :)))
     
  8. colin

    colin Guest

    Interesting, so if I connect a very sensitive current meter accros a totaly
    dark diode will I see a current ?

    Colin =^.^=
     
  9. Guest

    So you don't believe that you can detect photons with a zero-biassed
    diode?
     
  10. cuca

    cuca Guest

    Ok let me see if I get this write; if I have a photodiode operating
    under a short circuit current regime say for simplicity, there will
    always be a current or dark current generated simply because there will
    always be some mechanism that creates an electron-hole pair which gives
    rise to a current, in much the same way as that achieved via the photon
    radiation? In order to theoretically eliminate the dark current, the
    diode would have to be cooled and shielded from external radiation.
    Not that I am trying to eliminate this component I am just interested
    in clearing this up in my mind.

    Then at this point we could define the shunt resistance will be the
    slope of the iv curve at 0v. Is this so?

    Charles
     
  11. cuca

    cuca Guest

    Ok let me see if I get this write; if I have a photodiode operating
    under a short circuit current regime say for simplicity, there will
    always be a current or dark current generated simply because there will
    always be some mechanism that creates an electron-hole pair which gives
    rise to a current, in much the same way as that achieved via the photon
    radiation? In order to theoretically eliminate the dark current, the
    diode would have to be cooled and shielded from external radiation.
    Not that I am trying to eliminate this component I am just interested
    in clearing this up in my mind.

    Then at this point we could define the shunt resistance will be the
    slope of the iv curve at 0v. Is this so?

    Charles
     
  12. Guest

    That's my expectation. It does depend on where the dark current comes
    from and how energetic the corresponding charge carriers are when they
    get into the conduction band.
     

  13. Dark current from cosmic rays and radioactivity is very low. About one q per
    cm2 and second.
    In the absence of cosmic rays and at thermal equilibrium,
    the 'rectified noise current' will cancel the "charge carriers promoted
    to the conduction band by thermal fluctuations". If not we have a Maxwell's
    demon.

    So in practice we will need a very sensitive current meter, I think.
    Another case if the diode and the current meter have different temperatures.
     
  14. Robert Baer

    Robert Baer Guest

    1) You are *right*, NOT "write".
    2) Furthermore, the theoretical capacitance of the photodiode in that
    case is zero (think about it).
     
  15. cuca

    cuca Guest

    This would seem logical to me but would appear to contradict what other
    people are saying!!

    P.s. sorry for the "write" spelling
     
  16. Phil Hobbs

    Phil Hobbs Guest

    Capacitance is the derivative of charge with respect to voltage. Making
    the voltage zero does not make the derivative zero. It's like a boat
    launching ramp--when your wheels hit the sea water, the altitude is
    zero, but the slope is not.

    For a silicon PIN photodiode, the capacitance will go down by a factor
    of about 7 as you increase the reverse bias from 0 to somewhere near the
    diode's max Vr.

    Cheers,

    Phil Hobbs
     
  17. Phil Hobbs

    Phil Hobbs Guest

    Carrier pairs in the depletion region, regardless of how they're
    generated, will get pulled apart by the junction E field, and appear at
    the terminals. But the quantities involved are sufficiently small (a
    few electrons/s at most) to be completely dwarfed by the diffusion
    currents and the Johnson noise current of the zero-bias resistance.

    My beef is not so much with your physics (which is fine) but rather with
    the confusion this hair splitting is likely to produce in the OP--who
    doesn't seem to realize the difference between current and conductance.

    Cheers,

    Phil Hobbs
     
  18. cuca

    cuca Guest

    Thanks
    I realise that there are a number of other parasitic that will dwarf
    these issues As i am writing up my thesis, i was keen to get a firm
    grasp of the meanings or physics that lead to the model. Thanks for
    the great discussion

    Thanks
    Charles
     
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