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EL7900

Discussion in 'Electronic Design' started by Marco Trapanese, Aug 28, 2012.

  1. Hi,

    has anyone used this photodiode?

    http://www.intersil.com/content/dam/Intersil/documents/fn73/fn7377.pdf

    I'm talking about a single sensor, powered at 3.3V with an Rload of 562 ohm.

    There are several things I can't understand:

    1. the ds states the output may saturate if the illumination is too
    high. Instead the output current will suddenly fall to 0V when I
    increase the light on the sensor too much.

    2. just before that condition, the maximum current read is about 2.7 mA,
    very far from the 6 mA allowed.

    3. the ds also states the output voltage compliance is 300 mV below the
    supply voltage but the figure 7 shows this is true only for very small
    currents (< 400 uA). At 1 mA it raises at 500 mV and there is no
    indication about higher currents.


    I may increase the supply voltage to have the full span I need (2.048 V)
    but I can't deal with the falling of output current when exposed to a
    lot of light.

    I can't understand why it goes down... it shouldn't *saturate* - that is
    the output current won't increase anymore? The figure 9 doesn't show any
    falling edge on the right!

    Thanks in advance
    Marco
     
  2. Bill Sloman

    Bill Sloman Guest

    It's certainly a very odd-looking circuit, if the block diagram on
    page 4 of the datasheet is to be believed.

    The two P-MOSFETs seem to form a current mirror, with what's drawn as
    an N-channel MOSFET as the current source. It would make more sense if
    that MOSFET at the top were another P-channel device. It would then -
    mostly - be a resistor when enabled with more than 2.7V of gate-to-
    source voltage - not less than 2.4V according to the data sheet - and
    guaranteed to not act as a resistor with less than 1.2V of gate-to-
    source voltage

    My guess would be that if the photo-diode draws too much current, the
    MOSFET at the top comes out of saturation (stops looking like a
    resistor and starts looking like a constant current source) and the
    voltage at its drain and the common source connection for the two
    MOSFETs forming the current mirror would then collapse down to
    something very close to the negative rail.

    It could be that the top MOSFET in your particular device comes out of
    saturation at a relatively low current when it's only got 3V of gate-
    to-source voltage.

    See what happens if you boost the device supply voltage closer to 5.5V
    - perhaps with a battery.
     
  3. Bill Sloman

    Bill Sloman Guest

    Impressively insightful analysis. Care to share your insights as to
    how the excess of light breaks it?
     
  4. Il 28/08/2012 18:21, John Larkin ha scritto:

    Because it has a pin to put the output in hiZ. I have several hundreds
    of this IC that share the same output path. With a shift register I
    enable one-by-one without any other components.

    Marco
     
  5. Il 28/08/2012 12:24, Bill Sloman ha scritto:

    I requested a clarification to Intersil's engineers.., I'm waiting for
    their answer.


    I'm pretty sure I'm wrong but to activate the output I need to put the
    enable pin to 0V. So I guess the top MOSFET is a P-channel, despite the
    drawing.


    This is what I measured on the test bench.


    Are you talking about my own specific sensor? I have over 800 sensor on
    my board and all these exhibit this behavior.


    I've already done this test. The output current - before falling down
    close to the negative rail - reaches about 3.5 mA.

    Marco
     
  6. Il 28/08/2012 20:06, John Larkin ha scritto:

    I guess it would.


    Nothing until now.
    I received several emails but with further requests of details.


    Yes, I'm applying too much light because it may happen. It's ok if the
    output saturates but not if falls down.

    Marco
     
  7. Il 28/08/2012 20:52, Phil Hobbs ha scritto:

    But I have to add one mosfet for each sensor in order to enable them
    selectively. On a single board I have 440 sensors.

    Marco
     
  8. Il 28/08/2012 20:57, Marco Trapanese ha scritto:

    I've just read your other message.

    Marco
     
  9. Il 28/08/2012 21:22, John Larkin ha scritto:

    No. The ds says 8 klux to be "linear".

    But there is no an "absolute maximum rating" for illumination.
    Figure 9 for example shows up to 10 klux with a current over 4 mA.

    At page 5 I read: "[...] care must be taken when lux go as high as
    10,000lux because the output current rises above 6mA before reaching the
    device's output compliance".

    Table 6 shows the output current (6000 uA) at 10000 lux.

    The problem is I cannot get a current above 2-3 mA.
    And in the whole ds there is no mention about the "over-light" issue.

    Marco
     
  10. Il 28/08/2012 21:12, Phil Hobbs ha scritto:

    Thanks, very clear now.
    When selecting the photodiode, should I take care of any feature in
    order to work in this way?

    Marco
     
  11. Bill Sloman

    Bill Sloman Guest

    It looks as if the data-sheet data was taken from a sample from a
    rather better batch than the one that provided the parts that got
    shipped to you.

    Batch-to-batch variation is often larger that within-batch variation.
     
  12. Bill Sloman

    Bill Sloman Guest

    So what did you think that your contribution added to the thread? Did
    you feel that your "most prolific poster" status might be be under
    threat, so you needed to make the kind of contribution that Google
    will count, even if everybody will see it as contentless?
    I'm not enough of an ass to add an entirely pointless comment to a
    thread.
    I don't, as I seem to have mentioned before. Technically speaking, I'm
    retired, rather than unemployed, but getting that right would seem to
    require a kind of civility that you seem unlikely to master, but
    complain about failing to get from other people.
    In your depressingly inadequate excuse for a mind.
     
  13. Jamie

    Jamie Guest

    did you put a piece of glass over the top of the sensor?

    Jamie
     
  14. Guest

    datasheet states that; output is not short circuit protect and must
    never exceed 6mA, if you use a load resistor less than 800R you
    may damage the device with strong input light,

    -Lasse
     
  15. Bill Sloman

    Bill Sloman Guest

    You didn't - in fact - include any explicit reference to my comments.
    As far as your - minimal - content went, it would have made just as
    much sense as a reaction to Marco Trapanese's original post.

    "Yup, looks like a bug. Too much light breaks it."
    You'd like to think so. More objective observers might favour John
    Field's point of view - I certainly do, but I'm scarcely objective.
     
  16. Jamie

    Jamie Guest

    Oh I see, you favor what ever grass is greener at the moment.. I can
    see where this is going.


    Jamie
     
  17. Jamie

    Jamie Guest

    :)

    Jamie
     
  18. Il 29/08/2012 01:40, ha scritto:

    I may damage it because the strong input light leads to an output
    current above 6 mA. But this *does not* happen.

    Marco
     
  19. Il 29/08/2012 01:08, John Fields ha scritto:

    As said to Phil Hobbs I just need to scan an array of several hundred of
    photodiodes in order to get their output.

    Marco
     
  20. Il 29/08/2012 02:56, Phil Hobbs ha scritto:

    It's exactly what I did. I have 4 independent lines with 112 sensors
    each one. I scan them together.

    Marco
     
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