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Testing a photodiode

Discussion in 'Electronic Repair' started by Robert Wolcott, Jan 4, 2005.

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  1. I have a silicon photodiode and I'm unsure if it is working or not. How
    would I go about testing it? Things I have done so far:

    Measured the resistance - >50 megs in either orientation, regardless of
    lighting condition.
    Measured resistance with a 5mw diode laser shining on it.

    Anything else I should do? How do these typically fail?

    Thanks,
    Bob
     
  2. A diod is a diod, not a resistor.
    See it as a current source, and put it into a opamp to form a current to
    voltage converter...
     
  3. Even simpler: Just hook it to a multimeter on mA and shine a light on it.

    With your laser diode, the output should be somewhere around 0.2 to 0.5 mA/mW.

    For more accuracy, reverse bias it with a few V and measure the current.

    --- sam | Sci.Electronics.Repair FAQ Mirror: http://repairfaq.ece.drexel.edu/
    Repair | Main Table of Contents: http://repairfaq.ece.drexel.edu/REPAIR/
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  4. I'll give that a try. This diode is part of a relative power meter on my
    yag laser. The meter is currently not working and I'm not sure if it is the
    diode or the circuit. The circuit can be seen at:

    http://oregonstate.edu/~wolcottr/Yag_pictures/pwr meter schematic.jpg

    Tests I have run include:

    Measured the voltage supplied to the diode via BNC connector and it is 5.0
    volts
    placed a 1k resistor across the BNC connector and the power meter was
    pegged.
    Hooked up the photodiode to the circuit and shined a 5mw diode laser into it
    and there was no reading (direct illumination).

    Does this sound about right?

    Thanks,
    Bob
     
  5. Is the input 5 VDC or 12 VDC? 5 VDC wouldn't peg a 1 mAFS meter since
    there is a 10K current limiting resistor.
    Sounds consistent with 12 VDC input and the circuit working properly.
    Do the tests above but sounds like it's bad.
    Yep. You can also test the photodiode on the diode check range of your
    DMM. It should test like a silicon diode. I bet it tests open both ways.

    --- sam | Sci.Electronics.Repair FAQ Mirror: http://repairfaq.ece.drexel.edu/
    Repair | Main Table of Contents: http://repairfaq.ece.drexel.edu/REPAIR/
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  6. I measured the voltage across the bnc connector (disconnected from diode)
    again and it was 5.015 volts. The diode was open in both directions with my
    diode tester (Fluke 177). What usually causes these to fail?

    Thanks,
    Bob
     
  7. They can just fail on their own. In that circuit, current is safely
    limited by the 10K resistor. Not much other than excessive current
    or excessive reverse voltage could damage a low speed photodiode.

    Just install any cheap photodiode with a large enough active area and
    see if it works. A $2 one from Digikey would likely work just fine.

    --- sam | Sci.Electronics.Repair FAQ Mirror: http://repairfaq.ece.drexel.edu/
    Repair | Main Table of Contents: http://repairfaq.ece.drexel.edu/REPAIR/
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  8. Bradley1234

    Bradley1234 Guest

    A diode is a variable resistor.

    the circuit reacts around the depletion region to increase or decrease it.
    When the p type and n type silicon meet, the charges mix to form a neutral
    region which is resistive. if the applied electricity is on one side it
    makes that neutral region expand out, if its the other way the neutral
    region is compressed

    But silicon has unique properties where it can convert energy from light or
    heat into current. In the photodiode's case if I recall, it can use light
    or heat to control that neutral region
     
  9. Not really. :)

    --- sam | Sci.Electronics.Repair FAQ Mirror: http://repairfaq.ece.drexel.edu/
    Repair | Main Table of Contents: http://repairfaq.ece.drexel.edu/REPAIR/
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  10. Bradley1234

    Bradley1234 Guest

    Then please to explain me how silicon diodes work? All this time I thought
    it was the depletion region but its something else?

    If my simplified explanation of the solid state physics involved in PN
    junctions is incorrect Im grateful to learn where Im wrong.
     
  11. Please Google "Photodiode principles". There were 33,100 hits but
    the first one is adequate. Thanks.

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    Repair | Main Table of Contents: http://repairfaq.ece.drexel.edu/REPAIR/
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  12. Bradley1234

    Bradley1234 Guest

    I did, and it still explains similarly to what I said. Ive worked on the
    design of semiconductors, lithography, processing, doping, circuits design,
    and enough circuit testing to be content with, and have trained and shared,
    possibly the wrong information with dozens and dozens of people. maybe
    more.

    I enjoy studying, but to say just go do a google search? Id expect any such
    searches will simply verify and validate what I said.
     
  13. Bradley1234

    Bradley1234 Guest

    Since there isnt an edit to these things, what can happen over the years is
    that topics we used to work in daily, if we havent done significant work in
    them for a while we can lose track of certain details. While I strive to be
    accurate I realize its possible that Im mistaken. having a background
    itself does not imply accuracy, only to say that Ive studied something in
    it.

    If a person tells me Im mistaken? Im grateful but will appreciate knowing
    how so we can examine details and I can correct my understanding. Im the
    type who will buy lunch/dinner for that favor, or shave my head in public if
    Im wrong, etc...gotta make it interesting and funny.

    The first thing to go is your eyesight, the second thing is your memory, and
    I cant remember what the third thing was
     
  14. Did you do the search? Here is the first sentence of the first hit:

    "When these diodes are exposed to photons of energy greater than 1.12 eV
    (wavelength less than 1100 nm) electron-hole pairs (carriers) are created.
    These photogenerated carriers are separated by the p-n junction electric
    field and a current proportional to the number of electron-hole pairs
    created flows through an external circuit."

    That is the most simple explanation of how a photodiode works. It's not
    a variable resistor but a a light to current converter.

    A device like a cadmium sulfide photocell is a variable resistor, controlled
    by how much light falls on it. That's totally different than a photodiode.

    I agree that a lot of Web search returns are useless - in fact this one
    goes on to talk about other things that aren't directly relevant, it is
    a start for many thing.

    I'll let others comment on whether your explanation is similar to this one
    but unless I missed something, you didn't even bother to mention anything
    about a photodiode, only a normal PN diode used in an electrical circuit.

    --- sam | Sci.Electronics.Repair FAQ Mirror: http://repairfaq.ece.drexel.edu/
    Repair | Main Table of Contents: http://repairfaq.ece.drexel.edu/REPAIR/
    +Lasers | Sam's Laser FAQ: http://repairfaq.ece.drexel.edu/sam/lasersam.htm
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  15. Bradley1234

    Bradley1234 Guest

    Okay, way back somebody said a diod is a diod, not a resistor, so the theme
    became what is a diode, not what is a photodiode.

    So thats why I mentioned the electrical properties of a diode being used for
    a variable resistance. Of course its not a bidirectional resistance, but by
    any other name becoming a variable resistance to current flow. (yes non
    linear) I described the simple effect of recombination in the depletion
    region at the PN junction that changes based on bias, to which you said this
    was incorrect.

    In photodiodes? If the photodiode is biased and light is applied the amount
    of current flow should change. Net effect its resistance has changed.

    The website mentioned shows a non standard photodiode, and yes I did the
    search, interesting product line
     
  16. In no way, shape, or form, is a diode generally used like a resistor or
    behaves like a resistor.
    I said: "Not really" as an explanation of how a photodiode worked. And while
    it some reasonable information about what happens when voltage is applied
    to a PN junction with respect the electrons and holes and width of the
    depletion region, etc., I doubt anyone who didn't already know how a diode
    worked would have after reading it. There was no attempt made to relate
    the textbook description to circuit behavior.
    The current is proportional to optical power and is more or less independent
    of applied voltage. That's not what is normally considered to be a resistor.
    Yes, but the first sentence was a good summary of how a silicon photodiode
    works.

    --- sam | Sci.Electronics.Repair FAQ Mirror: http://repairfaq.ece.drexel.edu/
    Repair | Main Table of Contents: http://repairfaq.ece.drexel.edu/REPAIR/
    +Lasers | Sam's Laser FAQ: http://repairfaq.ece.drexel.edu/sam/lasersam.htm
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  17. Bradley1234

    Bradley1234 Guest

    AHA, here is a problem, a diode changes its resistance, if you say it
    doesnt, then please explain the effect on the current flow while its in the
    circuit. If you ignore how the process takes place internal to the PN
    junction and observe the diode operates, why would it not simply be a
    component that appears to change resistance?

    Ive never said a diode can used as a replacement for a resistor, the
    resistor is typically linear, a diode is non linear, for a voltage applied
    it is not a constant resistance unlike a common resistor, but I say the
    diodes reaction is resistive.


    Its my observation that many people do not understand WHY a diode works, but
    most everyone knows the diode tends to block current in one direction and
    permit it in the other. Ask them why? I offered a simple model of the
    depletion region which changes based on bias, So maybe your saying not
    really was saying a photodiode doesnt work that way, when I was describing a
    standard diode?

    Ask people why does an LED emit light? its a very interesting and hard to
    explain phenomenon, I would estimate that most people do not know exactly
    where the light comes from or why. and hard to explain meaning some
    foundation of chemistry, minority/majority carriers, etc

    What does a transistor do? its simply a controlled resistor also. Ive
    known "experts" say a transistor generates power, it creates power where
    none existed before, power comes out, nobody knows why, probably the crystal
    creates the energy.


    resistor.

    The current? so we agree that current changes. Im saying its because the
    resistance changes, not because power is created inside a diode that is
    adding to the circuit. Are you saying optical power is transferred right
    out of the diode? It comes in as light and is converted to current?

    Im saying light affects the depletion region that causes a change in
    resistance, its proportional to incident optical power but assuredly not
    linearly proportional

    But Im still wondering how Im wrong, perhaps we were talking apples and
    oranges, I said a pn junctions resistance changes by the variable depletion
    region, you said not really. So then am I still wrong? or do we agree on
    theory?
     
  18. So how do you explain that a photodiode can operate in photovoltaic mode
    with no bias and generate power?
    But the current in a photodiode is very linearly related to optical power
    until the device saturates.

    Each photon generates one or more electron-hole pairs. That is where the
    current comes from.

    --- sam | Sci.Electronics.Repair FAQ Mirror: http://repairfaq.ece.drexel.edu/
    Repair | Main Table of Contents: http://repairfaq.ece.drexel.edu/REPAIR/
    +Lasers | Sam's Laser FAQ: http://repairfaq.ece.drexel.edu/sam/lasersam.htm
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    Note: These links are hopefully temporary until we can sort out the excessive
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    Important: Anything sent to the email address in the message header above is
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  19. Bradley1234

    Bradley1234 Guest

    I dont deny this, its the effect silicon has which is "special" there is
    current flow from the conversion, it doesnt need bias, but if there is bias,
    the converted energy, as I understand it, is typically used to control the
    size of the depletion region, which is the resistance
    Okay, here we are both describing a graph in general terms. If there is a
    point where the device saturates, its no longer linear after that. I call
    this nonlinear, if a typical photodiodes response contains a nearly linear
    region? Then its sometimes very linearly related to optical power

    Well this is the ideal response, but not the actual response. To get to a
    point to be able to get 1 to 1 reaction? This is the photon counter
    problem. To be able count photons takes a very precise, temperature
    controlled, and expensive photodiode, like APD (avalanche photodiode) which
    isnt perfect. But in a general photodiode, its not necessarily efficient
    where each photon gets converted.

    The conversion causes current, but in typical sensor circuits Im guessing
    the application is usually to control a bias current in a known linear
    region, so it would be in the input to an opamp, with some bias and by the
    photodiode varying resistance would provide varying current in response to
    incident light, that output would feed an a/d converter and to be accurate,
    it would require calibration
     
  20. The photodiode response has nothing to do with the size of the depletion
    region. Changing the amount of light on a photodiode doesn't affect the
    depletion region significantly. Light generates charge carriers which
    pass through the depletion region.
    Yes, over several orders of magnitude. Even a resistor is nonlinear once
    it starts smoking. :)

    There is a fundamental difference between a device which is non-linear
    and one which has a linear operating region with some limits where it
    becomes non-linear.
    I can go to Digikey and buy a $2 photodiode which will demonstrate linearity
    over several orders of magnitude. Is it perfect? No, but darn close. Is
    it a photon counter? No. But on average, it's doing a decent job of
    counting large numbers of photons.
    Go do an experiment. Take a photodiode, almost any one will do, Rip it
    out of your PC mouse if you like. :) Put a reverse bias on it of a few V
    and a mA meter in the circuit. Now get your calibrated light source and
    try it at various intensities. The current will be very nearly linear
    with respect to optical power up a point. That point is where secondary
    effects come into play. But for a typical 1x1mm active area, a couple
    mW should be fine. You can try a laser pointer or LED on it.

    Except for a fixed dark current (may be a uA or so), the current will
    be linear with optical power.

    Now remove the bias but leave the mA meter in place. Repeat the experiment.

    In some applications, the photodiode feeds into the summing junction of
    an op-amp, between the + and - inputs, with a feedback resistor to set
    the transimpedance gain. There is no voltage across the photodiode.
    (OK, there is a uV or nV across it depending on the open loop gain of
    the op-amp but an ideal op-amp with infinite gain and exactly 0 V
    across the photodiode would work as well.)

    --- sam | Sci.Electronics.Repair FAQ Mirror: http://repairfaq.ece.drexel.edu/
    Repair | Main Table of Contents: http://repairfaq.ece.drexel.edu/REPAIR/
    +Lasers | Sam's Laser FAQ: http://repairfaq.ece.drexel.edu/sam/lasersam.htm
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    Note: These links are hopefully temporary until we can sort out the excessive
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