I have asked question in the past about this circuit, this is a new question based on some information provided in this answer here namely concerning the "throwing away" of the signal.

Op-amp supply rails:

ISL28210: -38V   +2V
AD8676:  -2V    +10V

enter image description here

I've tried to do a basic online circuit simulation and playing around with the resistors, it would seem that if I reduce R8, R9 and R10 by a factor of 2 then I will get the same output voltage.
It also means I wouldn't need the -38V supply rail which currently comes from a bulky charge pump configuration on another board and could use a smaller voltage (I currently have -10V I use elsewhere so could piggyback off of that I assume)

I have looked at other questions and answers concerning the size of the resistors used, this answer in particular. I don't think power dissipation will be a problem as the circuit i working in the uW range, so I can't understand the choice for the 2.2M resistor in the first stage

The circuit was originally designed by a colleague who has since left so I have nobody to consult on this

My questions are:

  1. Would reducing the size of these resistors cause a problem?
  2. In reference to this answer, how does reducing the amount of signal "thrown away" increase speed and reduce noise?

As an extra aside question from this:

  • Why would C15 and C19 be different values? I calculate it that the cutoff frequency in the first stage at ~72kHz and in the second stage at ~125kHz, does it make sense to have different cutoff frequencies for each stage?

I calculated the cutoff frequency using enter image description here which I hope is correct


  • \$\begingroup\$ Start by defining the signal you wish to measure and noise you want to reject and how flat a frequency response you need since each OA is an integrator with a fixed GBW product. The PD has a leakage R and junction pF value you can include in your model. then choose the best R values according to BW and SNR and sensitivity and max value in LUX or similar value. You will not be able to get 450kHz BW if the GBW product is only 10MHz for example with high gain, so lower R increases BW while reducing gain. Also you can invert 1st OA output voltage and supply by reversing the PD then choose 0~5Vo \$\endgroup\$ Commented Dec 6, 2016 at 16:02
  • \$\begingroup\$ @TonyStewart.EEsince'75 Sorry you'll have to help me here, I'm relatively new to electronics. As I reduce R8 the BW of the first OA increases whilst the gain decreases. By reducing R9 and R10 I'm increasing the gain of the second OA but reducing the BW. So I need to make sure that the BW on the second OA is greater than the 850kHz cutoff frequency in order for the circuit to perform the same function? As the fastest frequency the second OA would need to respond to would be less than the cutoff frequency? Does that sound right. \$\endgroup\$
    – user103993
    Commented Dec 6, 2016 at 16:37
  • \$\begingroup\$ If I reduce R9 and R10 by a scale factor of 10, that gives me a gain of 2.82dB, if I compare this to the closed loop gain vs frequency chart in the AD8676 datasheet it gives me a bandwidth of >125kHz which should be ok in theory? \$\endgroup\$
    – user103993
    Commented Dec 6, 2016 at 17:07
  • \$\begingroup\$ sounds ok. but how fast you expect light to change rate and what noise level do you expect? \$\endgroup\$ Commented Dec 6, 2016 at 17:42
  • \$\begingroup\$ @Tony I'm sampling at 2.5kHz so once every 400us. As for noise I'm after the minimal possible noise but there was never a spec for it. I think the original designer was a "try it and see" kind of guy \$\endgroup\$
    – user103993
    Commented Dec 6, 2016 at 20:57

1 Answer 1


You seem to be in a tricky situation. You are trying to make sense of a circuit which makes very little sense, but you haven't the experience to realize it. The things which puzzle you are indeed indications that your predecessor is better off somewhere else.

You can get equivalent performance with a single op amp with a 300 k feedback resistor. Since you are sampling at 2.5 kHz, a 500 pF feedback capacitor will produce a single pole lowpass response down 3 db at 5 kHz. The inversion produced by the second op amp can be handled by reversing the polarity of the photodiode. And, of course, the use of compensating RC at the + input (especially the capacitor value) makes no more sense on the second op amp than it did on the first.

The output filter produced by R28/R29/C38/C39 has a response down 20 dB at 200 kHz, so it's not at all clear that it does you any good. Although it may be recommended by the ADC manufacturer.

The parallel 39k resistors are only useful if you need the option of selecting a gain which can be varied by a factor of 2 by installing only 1 of the resistors. This is potentially useful, but if you're not using that option as part of system tweaking it serves no purpose.

I suspect that the purpose of using the two amp configuration is to provide limiting in case of exceptionally large optical pulses. If the TIA limits at -35 volts the attenuation produced by the second will give an output level of just about 5 volts, which sounds like a reasonable ADC input range. Since the AD8676 is a rail-to-rail op amp, you can do this for a single amp configuration simply by using +5 as the + supply voltage. +5 and -10 would work just fine.

  • \$\begingroup\$ This TIA configuration looks like one designed for photoconductive biasing, but the bias point is fixed at zero volts giving photovoltaic operation. \$\endgroup\$
    – Ben Voigt
    Commented Dec 7, 2016 at 1:20
  • \$\begingroup\$ @BenVoigt - Yup. That's why he can get away with simply reversing the PD to change output polarity. \$\endgroup\$ Commented Dec 7, 2016 at 1:24
  • \$\begingroup\$ @BenVoigt - So, from what I can gather, it was designed as if it was to be used like this picture which is a photoconductive circuit, however as the end of the PD is tied to neither V+ or V- but instead to 0V, I can use it in both forward and reverse bias? Is this a photoconductive circuit or photovoltaic or both/neither? \$\endgroup\$
    – user103993
    Commented Dec 7, 2016 at 9:06
  • \$\begingroup\$ Does this also mean I can simplify the circuit by finding the relationship between the TIA + Inverting amp and my Vo and replace it as follows: Reverse the PD and use a single TIA that gives the same Vo as the original circuit with a parallel cap across the resistor to create a low pass filter at 2x my sampling rate? - Circuit simulation of proposed change \$\endgroup\$
    – user103993
    Commented Dec 7, 2016 at 9:54
  • \$\begingroup\$ @Hayman - The CIRCUIT is neither photovoltaic nor photoconductive. The PHOTODIODE is being used in photovoltaic mode. And yes, your proposed circuit ought to work OK. Although I don't recommend using parallel resistors as gain setting. Just get the closest to 300k you can easily find. After all, you're not doing calibration, you just want to determine if the fiber has failed. If you're thnking of redesigning the production circuit, you need to determine your requirements and error budget. Without knowing your requirements, it's not possible to determine if your frequency response is adequate. \$\endgroup\$ Commented Dec 7, 2016 at 13:37

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