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LED question Vf rise vs. light output

Discussion in 'Electronic Design' started by Anthony Fremont, May 6, 2007.

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  1. Jim Thompson

    Jim Thompson Guest

    Yep. I'll look for the paper on the junction capacitance when I

    ...Jim Thompson
  2. Paul Mathews

    Paul Mathews Guest

    If the OP will simply lower his load resistance, he will quickly
    understand that his idea that load R only affects one of the
    transitions is incorrect. The impedance that drives the photodiode
    capacitance is essentially the load resistance, regardless of rise or
    fall. The photodiode junction acts like a current source with
    consequently high output impedance. Anything more complex is a waste
    of time.
    Paul Mathews
  3. Ok, I had some nuisance work like things to do today, but I was able to do a
    quick test just now. ;-) With a 1M resistor under the photodiode, I
    adjusted the spacing between the IR LED and the photodiode so that the
    pulses were peaking at 5V indicating that I was fully saturating the
    photodiode, IIUC. Rise time on the scope was about 14uS. Changing the
    resistor to 100K, and readjusting the spacing so that the pulses were
    peaking at 5V, the rise time was still ~14uS. Changing to 10K made the gain
    so low that it was impossible to obtain a 5V peak, but at 500mV (the most I
    could get out of it) the rise time was ~1.4uS meaning that the slope (dV/dT)
    doesn't seem to change regardless of the load resistor I use. What say ye
    now? :) Seriously though, if current has to flow thru the resistor
    regardless, then the capacitance of the reverse biased photodiode junction
    must be quite small (as in 1pF or so?).

    I am going to do the current shunt thing that miso suggested, so that I can
    see if that affects the rise times I'm seeing.

    Thanks for the advice. :)
  4. Ok, I'm stupid. I tried some more resistor values and with more careful
    adjustment of the LED and photodiode spacing, I do see changes that do seem
    to correspond to the resistance changes. How on earth can you get usable
    output from a photodiode without major amplification or using a nuclear
    light source?
  5. Guest

    You feed the diode to a transimpedance amp. If you keep an eye out,
    you can find these off the shelf since you aren't the only person on
    planet earth in need of amplifying a photodiode. ;-) The
    transimpedance on these boxes are switch selectable, often reading
    10^9. I have one of these:
    and also one from EG&G PAR. More gain, less bandwidth. I got the UDT
    for $5 at a swap meet. The EG&G was more like $35 since i bought it
  6. Paul Mathews

    Paul Mathews Guest

    1. You're not 'saturating the photodiode''re using up your
    source compliance. Don't.
    2. If the energy you're trying to sense is extended over an area
    larger than your photodiode, you can use a lens to converge it.
    3. Transimpedance amplifiers are fairly simple to implement.
    Paul Mathews

  7. I think I understand what you mean. I was illuminating it to the point that
    the current it "wished" to source could not flow thru the load resistor with
    only 5V pushing it and actually added to the voltage at the junction. If I
    increased the reverse bias to 12V, I wouldn't automatically get 12V peaks on
    my measurements. IOW, I could get some more dynamic range out of it by
    using a higher reverse bias voltage, without losing sensitivity. Am I close
    to getting it yet?

    That would certainly help. :) I'm really not worried about range right
    now (I can aim the LED straight into the photodiode), I just want the
    fastest response that I can get right now.

    I've been looking at this some, but won't I end up compromising my slew rate
    again? I've got some MCP6024 10MHz op-amps, but the specs claim on 7V/uS
    slew. I understand that is probably better than what I'm looking at with
    the big load resistor right now, but if I understand, it's still not going
    to be able to show me say a 10nS rise time (assuming that's how fast an IR
    LED can ramp up its brightness).

    I'll play with the op-amp when I get a chance later.

    Thanks for your help. :)
  8. Paul Mathews

    Paul Mathews Guest

    1. Why do you need so much amplitude at the output? 100 mV signals
    are easy to see on most fast scopes.
    2. High slew rate opamps are easily found.
    3. The usual way to get fast response is to use a very small
    photodiode. Use a lens if necessary. This makes they system more
    directional, which is often a highly desirable characteristic. For
    example, it can reduce the amount of current from extended (not point)
    ambient sources. a 1mm x 1mm photodiode has 1% of the capacitance of a
    similar 10mm x 10mm photodiode.
    4. Phototransistors are surprisingly linear over a few decades of
    photocurrent and easier to use than transimpedance amps. However, you
    would need to choose one with an accessible base terminal and load the
    base collector junction to get decent speed. This degrades
    sensitivity. In any case, do not approach the saturation region (small
    Vce) of the transistor characteristic, since it will slow down
    drastically if you do.
    Paul Mathews
    Paul Mathews
  9. Paul Mathews

    Paul Mathews Guest

    I should have mentioned this earlier: Do not expect your average IR
    emitter to have anything like 10 ns risetimes, although there are a
    few that can approach that figure. To get into the low nanoseconds,
    you generally need to go to a laser emitter. Some LEDs actually have
    risetimes that are nearer to 1 microsecond.
    Paul Mathews
  10. I do suspect that high instantaneous current helps. One thing is that
    LEDs have more capacitance than many would suspect, so the usual
    30-nanosecond or whatever lag has some part from getting the voltage
    across the capacitance to change. Keep in mind that the junction is quite
    thin, much thinner than the chip as a whole, and dielectric constants of
    LED chip materials can be high. I am under the impression that
    capacitance of LEDs can get well into the double digits of picofarads. I
    have heard of people knocking down the usual 10's-of-nanoseconds figure by
    using high instantaneous current and low source impedance. I also suspect
    active sink of charge from the LED when trying to make it turn off can
    shave off some nanoseconds. Also keep in mind capacitance and inductance
    of wire/cable/whatever leading to the LED since this is occaisionally

    White LEDs are likely slower due to response of the phosphor.

    - Don Klipstein ()
  11. Paul Mathews

    Paul Mathews Guest

    No doubt some LED driver circuits themselves have slow risetimes. All
    my fast LED driver experience is with current sources with lots of
    voltage compliance, and we still see significant rise and fall time
    limitations.. The point is this: many high output LEDs are actually
    not very fast. If you really need to modulate at more than 1 MHz or
    so, you will need either special emitters designed for speed or
    (better yet) diode lasers. We're now working with VCSEL clusters that
    act much like IR LEDs optically, have about 1W/A output, and can be
    modulated at 100 MHz and higher.
    Paul Mathews
  12. Jim Thompson

    Jim Thompson Guest

    I just got around to trying (PSpice) a 1N4001 rectifier diode pulsed
    thru 1K, 5V with 10ns risetime.

    Takes 50ns to get up to 400mV... at 5us the voltage is 621mV and is
    still rising at the rate of 7mV/us

    ...Jim Thompson
  13. RHRRC

    RHRRC Guest

    have a look at
    et seq.
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