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Photodiode Dark Current Measurement

Discussion in 'Electronic Basics' started by GraemeC, Sep 28, 2007.

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  1. GraemeC

    GraemeC Guest

    I am trying to measure the dark current of a photodiode

    I have Chathode of the PD connected to 1Meg ohm resistor which is
    connected to +2V
    The annode is connected to 0V

    I obtain the following measurements
    Voltage accross Resistor: 1.39V
    Voltage accross Photodiode : 0.82V

    I have assumed that the Dark Current (or indeed any current value) is
    (2 - (1.39+0.82)) / 1Meg

    => I obtain a dark current of 210nA. This figrue however is less by
    more than an order of magnitude than that I am expecting.

    Is my understaning wrong?

    Please don't ask why I have this set up am not using Op-Amp circuity!

    Many Thanks

    Graeme
     
  2. Greg Neill

    Greg Neill Guest

    The photodiode and resistor are in series, so they
    share the same current. Given that you measured a
    1.39V drop across the 1M resistor, what's the current
    in the resistor?
     
  3. GraemeC

    GraemeC Guest

    I don't understand that. Because the the value across the resistor
    varies as you adjust the bias voltage. The bias voltage shouln't
    effect the value of the photocurrent generated.

    G
     
  4. What device are you using to measure these voltages? I am
    concerned that it may be distorting these very high
    impedance source readings.
    I think the dark current is just the current through the 1
    meg resistor, so 1.39/10^6=1.39 uA is the dark current
    (while the diode drops what is left of the 2 volts, or
    2-1.39=.61 volts), assuming the volt meter has an impedance
    many times higher than the 1 meg resistor. For a more
    accurate reading, you need to know the voltmeter's
    resistance and solve that formula for the parallel
    resistance of the 1 meg resistor and the meter's resistance.

    It may be simpler to find out what the resistance of the
    voltmeter is, and just use it as the current sense resistor,
    with a higher applied voltage. It is probably a good idea
    to also take several results at several voltages and learn
    how the dark current varies as the applied voltage varies.
    I think your formula is.
     
  5. The reverse voltage across the photo diode certainly varies
    its dark current, which is a leakage current. Most photo
    diode data sheets show this voltage dependence.
     
  6. redbelly

    redbelly Guest

    Your measured dark current, if we neglect the effect of the
    voltmeter(s?) on your circuit, is 1.39 uA as John P. said. At least
    that's closer to the >2 uA you were expecting (*). Perhaps your >2 uA
    figure is a "max/worst case" value, or is the spec at a considerably
    higher bias voltage than the 0.6 to 0.8V you had.

    You might redo the measurement using a 100k resistor, and see if you
    get close to the same current (i.e., 0.14V across the 100k resistor).

    Also, I have some questions about your setup ...

    1. What is the voltage of your "2V" supply, to the nearest 0.01V? The
    diode and resistor voltages should add up to the supply voltage.

    2. Did you measure the resistor and diode voltages at different times
    with the same meter, or simultaneously with two separate meters?

    Regards,

    Mark
     
  7. Jamie

    Jamie Guest

    I would not rely on measuring the voltage across the diode. the Load
    from the meter it self is going to influence the reading, especially
    in the case of a diode. I would how ever, use the measurements across
    the resistor and deduce the 10 meg shunt your meter is adding to it.
    The current in the 1 meg R will be the same of that in the PD.

    You could use a meter like we have at work for special testing that
    places G/T ohm loads on the circuit. In the math you gave, you're
    actually subtracting more than you started with, which gives you a - number.
    This is what's giving you the problem.
     
  8. GraemeC

    GraemeC Guest

    To answer some of the questions.

    I am using a HP 3401 DMM and a Fluke 115. I have taken the results
    from both sepertatly simulatneously.

    When I measure the value across the resistor with a large amount light
    incident on the PD. I obtain a value in excess of 2V (2.6V is the
    maximum iv measued).

    Graeme
     
  9. Guest

    Hello graeme,


    I am not an engineer, rather a garage hacker since the early 1960's
    when all you needed was a Volkswagen van and you picked up love along
    the side of the road.

    When I worked with "photo-diodes" over the years in the "many"
    mega-ohm range you must consider the effect of the magnetic fields
    that fill your work area, often heard as hum over audio amplifiers.

    "I would shield the photo diode and ground everything."

    If I used a cardboard columnating tube to narrow the field of the
    captured light (pin hole) I could detect passive light shifts/movement
    over a quarter mile away at high noon using a single photo-transistor
    and no lens. This is how we guarded our crops. LOL




    * * *
    Christopher

    Temecula CA.USA
    http://www.oldtemecula.com
     
  10. redbelly

    redbelly Guest

    I would remove the meter that is across the photodiode (if that is how
    you are set up), and do the measurement with just a meter across the
    resistor.

    Also: measure the actual source voltage, so you can infer the diode
    voltage. And, repeat the measurement with a smaller (say 100k)
    resistor to check for some consistency in what the diode current is.

    Regards,

    Mark
     
  11. neon

    neon

    1,325
    0
    Oct 21, 2006
    fisrt of all how do you see dark current in the dark. AFTER THAT YOUR READING DO NOT ADD UP 1.39 +82=2.21V HOW DO YOU GET HIGHER SOURCE FROM A 2 V. Furthermore is the photodiode connected to anything at all? like a base of a transistor or what?Torealy measure the diode current your meter must be in series with the 1mega there is going to be some inacuracy because of meter resistance. better yet add a 1k in series and measure viltage across then deternine the current.
     
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