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I have a photodetector device that outputs charge pulses proportional to incident light. From the device's datasheet (here) and some infor provided by the manufacturer (here), I've understood that this particular device, with 900 detectors, outputs about 3.72*10^-13 Coulombs per incident photon, on top of a dark current of about 0.372-0.743 uA.

I'm reverse biasing it with about 64V, and since I only have single supplies at my disposal I've implemented a non inverting charge amplifier/integrator (here, pag.23). However, since the transfer function of this amplifier only relates input and output voltges, how can I estimate the input voltage at my amplifier? The photodetector already includes quenching resistors, but the datasheet does not specify their value. On the other hand, that amound of charge, for just 1 detector, could lead to a current pulse of about 1.24 mA if one uses the FWHM value, which I'm not sure can be used to compute this.

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  • \$\begingroup\$ A charge amplifier is basically a current to voltage converter so why are you using an integrator? \$\endgroup\$
    – Andy aka
    Mar 5 '15 at 14:14
  • \$\begingroup\$ I did not find any non-inverting topology of a transimpedance amplifier. \$\endgroup\$
    – joaocandre
    Mar 5 '15 at 15:09
  • \$\begingroup\$ Use an inverting TIA followed by a standard inverting op-amp circuit. It can all run from a single supply (say) 10V with a "0V" mid-rail at +5V. Real 0V on your system becomes the negative power to the op-amps. OR supply the photodiode array with -64V and have cathode pointing into the TIA. \$\endgroup\$
    – Andy aka
    Mar 5 '15 at 15:31
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I'm going to assume you're interested in the intensity of light (photons/sec) and not photon counting.

Multiplying the intensity of the light by your photodetector's output gain will get you a current.

I've included a schematic for the noninverting transimpedance amplifier below. I wasn't sure how to hide the values, but the schematic is provided for topology purposes. Assuming an ideal opamp, the output is R1*I1. Using your photodetector as the current source, you should be able to get an output voltage that is proportional to the light intensity.

If you wanted to maintain single supply operation, you could use a different reference for the noninverting input than ground.

If you really were interested in photon counting you could convert this into an integrator by replacing R1 with a capacitor. You'll have to do something about the dark current, but this goes beyond my experience. My first attempt would be to calibrate it out with some sort of DAC current source.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ My photodetector needs to be reverse biased, in this schematic I'd need a -64V which I don't have. \$\endgroup\$
    – joaocandre
    Mar 6 '15 at 12:21
  • \$\begingroup\$ Then put the photodiode in the other direction, but to avoid negative output levels, you'll want to set the noninverting input of the opamp to a positive value instead of ground. \$\endgroup\$
    – anon
    Mar 6 '15 at 15:56
  • \$\begingroup\$ If I do that, then current flows the other way - towards the op-amp. I don't think I can have a reverse biased photodetector AND a virtual ground at the inverting input, if I want a positive voltage at the output. \$\endgroup\$
    – joaocandre
    Mar 9 '15 at 15:34

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