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I'm trying to set up a transimpedance amplifier reading the response from an avalanche photodiode (Hamamatsu S12572-100P http://www.hamamatsu.com/resources/pdf/ssd/s12572-025_etc_kapd1043e03.pdf) and a low-noise operational amplifier (TI LMV793 http://www.farnell.com/datasheets/1771351.pdf).

The dark current is around 0.15 uA, and the signals I'm expecting to measure correspond to around 0.8 uA current on the APD. My feedback resistor is 1.27 MOhms. However, after building this circuit on a PCB I just get about 0.09V on the output of the amplifier, in a situation where it should saturate, and no variation at all. I've replaced both the APD and the op amp thinking they might be damaged, but the problem persists.

The biasing is being achieved by setting 70V across a 2 KOhm load and connecting a 1 KOhm resistor in series with the APD, as suggested by the datasheet. The breakdown on the APD is 67V, thus I assume the APD is pulling around ~35 mA from the biasing supply, but I don't get any output.

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  • \$\begingroup\$ Inject 1uA (via a big resistor to Vcc) and see if that generates an output voltage. \$\endgroup\$
    – user16324
    Dec 15, 2014 at 14:29
  • \$\begingroup\$ Is that APD being used in the geiger mode? (Where it's biased above the reverse breakdown voltage?) Is so then I assume the breakdown voltage changes from device to device.... that's something you have to set. (It will aslo change with temperature.) You don't really need a TIA after it. You should be seeing big pulses. \$\endgroup\$ Dec 15, 2014 at 14:35
  • \$\begingroup\$ I suppose you know that the opamp unit is decompensated and, thus, needs some external compensation ? \$\endgroup\$
    – LvW
    Dec 15, 2014 at 15:09
  • \$\begingroup\$ Show your circuit. \$\endgroup\$
    – Andy aka
    Dec 15, 2014 at 15:19
  • \$\begingroup\$ Yes, the APD is operating in geiger mode, as APDs are supposed to work. I'm going to post my schematic \$\endgroup\$
    – joaocandre
    Dec 15, 2014 at 18:31

2 Answers 2

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I'm assuming your circuit looks like

schematic

simulate this circuit – Schematic created using CircuitLab

and your expectation that "it should saturate" is exactly correct. It IS saturating. The op amp is trying to drive the output below zero, but it cannot since it's a single-supply amplifier.

If you're going to use a TIA you have to use a negative bias to get a positive output.

ETA: Oh yes, and just to make your day - you may well have killed your op amps. Maybe not, since designers have gotten pretty good at protecting chips against overload, but this is pretty extreme. Not as extreme as getting hit by lightning, but that kills most chips, too. You need to check them in a less stressful circuit if you want to reuse them.

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  • \$\begingroup\$ My setup is rather similar to this, but the APD is reverse biased. What implications would this have? \$\endgroup\$
    – joaocandre
    Dec 15, 2014 at 18:32
  • \$\begingroup\$ The circuit shown makes no claim as to APD polarity. The arrow just shows which direction current flows. The implication would be that my assumption is correct. A constant output of ~0 volts is what you can expect from the circuit, regardless of light levels. \$\endgroup\$ Dec 15, 2014 at 22:16
  • \$\begingroup\$ I followed the schematic for a simple TIA suggested by the op amp datasheet though: farnell.com/datasheets/1771351.pdf \$\endgroup\$
    – joaocandre
    Dec 16, 2014 at 11:49
  • \$\begingroup\$ I assume you're referring to Fig. 1. Please note that they derive Vout/Iin = -Rf. So with a positive Vb, and a positive Iin, Vout is negative. Since you are using a single-supply op amp with a positive supply, you cannot produce a negative output. So what you need is a negative bias and a negative input current. \$\endgroup\$ Dec 16, 2014 at 16:22
  • \$\begingroup\$ That is quite the problem though, as I don't have any negative voltage supply. Is there any other way to offset the op amp? \$\endgroup\$
    – joaocandre
    Dec 16, 2014 at 17:02
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So the simplest model for an Spad looks like this,

schematic

simulate this circuit – Schematic created using CircuitLab

Where the switch is the breakdown of the diode, (mostly from photons)

You're pretty much stuck with R and C of the diode. But you can choose R quench (RQ).
It seems like you'd like to make RQ small (fast recharge time of C diode after a breakdown), but if it's too small then the avalanche breakdown never stops. So you've got to adjust it. Start big.

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