Induced photocurrent in a photodiode

Discussion in 'General Electronics Chat' started by sdh314, Nov 30, 2011.

  1. sdh314

    Thread Starter New Member

    Jun 2, 2011
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    Hi all,

    I'm looking at a basic photodiode-transimpedance circuit, which is measuring LED light modulated on/off by a 1kHz square wave.

    I'm particularly interested in the waveform of the induced photocurrent, and subsequent output of the op-amp (Vo = Rf * Ipd). I've a photodiode with a particularly large junction capacitance (1200pF) - will this mean that the photocurrent will not have the same square wave form as the incoming signal, but rather be modified and look closer to a sine wave? I've done a bit of modelling in Spice but the photodiode models vary and I wanted to make sure I'm observing a real phenomena and not a non-physical solution that Spice sometimes spits out.

    I know the output of the op-amp will be dependent on the it's bandwidth and the gain resistor and input capacitance (ie of the photodiode), but I'm particularly interested in what the actual process is that is modifying the form of the photocurrent?

    Thanks in advance,

    sdh
     
  2. Adjuster

    Well-Known Member

    Dec 26, 2010
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    The waveform of photo-current generated within the junction will be subject to modification by different delays in the time taken for carriers to move from their points of origin to the external electrodes, with possibly a significant amount of recombination occurring along the way. With a very big device not optimised for speed, like for instance a solar cell, these effects might be significant at quite low frequencies, but some (usually small) devices work well to MHz or even GHz frequencies.

    The junction capacitance effect is however just what it sounds like, a capacitance in parallel with the junction, albeit a diode's capacitance, subject to its value changing with variations in the voltage across it. To a first order, the behaviour will be as if there were a capacitance in parallel with a current source, so that the externally measured bandwidth will be reduced as the load resistance increases, hence the interest of a trans-impedance circuit which presents an artificial low load impedance.

    Perhaps a really high capacitance device such as you describe might have its response modified a bit further by distributed (transmission line) effects of capacitance and resistance over the device area, but I really do not know if this will be significant.

    In the end, I do not know what the relative size of these effects will be for your particular device: you will have to try it. If junction capacitance were the dominant limiting factor, a roughly triangular (RC exponential, more or less) waveform would be expected, while if more than one effect were contributing the result might be more sinusoidal.

    Good luck!
     
  3. joeyd999

    AAC Fanatic!

    Jun 6, 2011
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    If you use a transimpedance configuration that maintains 0 volts across the photodiode, the effect of the capacitance will essentially be eliminated.

    But beware: the pd capacitance then makes the AC gain of the circuit infinite, resulting in instability or lots of high-f noise.

    To fix this, you will need to add a feedback cap to attenuate the AC gain, which will tend to slow your rise times a bit. The size of the feedback cap will be a balance between noise and speed.

    Yet another thing to be aware of: The two caps (the pd cap and the feedback cap) will interact so as to cause "noise peaking" at some frequency, so you will need to experiment a bit to find the best overall solution for your specific problem.

    You should design your optics so that you have the most light possible available to the photodiode, and as little gain as possible on the amp. This will help to alleviate noise issues.

    When all else fails, a lock-in amp (synchronous detector), will help you to maximize reception of the optical signal when lots of noise is present, but this will slow down your overall response depending on your desired sensitivity.
     
  4. jimkeith

    Active Member

    Oct 26, 2011
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    Where did you get such an animal as this? part #?

    Also, per joeyd999, I agree that zero bias is the way to go--thank you for the tutorial.
     
  5. crutschow

    Expert

    Mar 14, 2008
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    Since diode capacitance is inversely proportional to the applied reverse voltage, it's better to operate the photodiode with a reverse bias if you are concerned about capacitance. You can still connect the diode to a transimpedance amp to minimize the effect of the capacitance.

    Here's an ap note from TI on designing photodiode amps that should help.
     
  6. joeyd999

    AAC Fanatic!

    Jun 6, 2011
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    IMHO, the reduced capacitance wrt a reversed biased pd will only be an improvement if the reverse voltage is held constant. Any change in V will require charging/discharging the capacitance. Also, the reverse bias will create dark current issues (albeit DC) resulting in an offset on the output dependent upon gain and temperature.

    I did learn something from your link, though...according to the app note, higher gains are preferable as the thermal noise of the feedback resistor rises slower than the gain, providing noise improvements. When I wrote my above comment, I was referring back to a pd system I designed years ago. The quantity of light was so small that shot noise was the dominant factor, not thermal noise. In that case, even small increases of light (and lower gain) provided dramatically improved S/N.
     
  7. crutschow

    Expert

    Mar 14, 2008
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    Keeping the voltage constant across the diode is not a problem if it goes to the input of a transimpedance amp.

    You point is valid about the increased dark current. That would have to be evaluated based upon the application requirements. It generally can be made insignificant if you are only interested in AC signals.
     
  8. Adjuster

    Well-Known Member

    Dec 26, 2010
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    The huge capacitance makes me suspect that this may be a big photodiode of the kind that is normally used to generate voltage, i.e. a photovoltaic like an old fashioned camera meter cell or even a solar cell.

    This may not work so well in reverse-bias, as its leakage current may be rather big. In addition, it might not have much tolerance to reverse bias (breakdown). The manufacturer may not necessarily specify it for reverse bias.
     
  9. sdh314

    Thread Starter New Member

    Jun 2, 2011
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  10. Adjuster

    Well-Known Member

    Dec 26, 2010
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    Well, there you have it. Five volts of reverse bias will divide the capacitance by three, at the cost of typically some tens of micro-amps of dark current.

    This thing is specified as having about a 3.5μs response into 1kΩ load at zero bias, of which 1.2μs would be explained by the RC delay, so it should be quite practicable to get a decent response at 1kHz. This seems a much more suitable device than the sort of thing I was afraid you were using.
     
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