2
\$\begingroup\$

I have two identical circuits (see attached schematic for example) residing on one PCB. Each circuit amplifies the signal from a silicon pin photodiode VEMD5080X01 . The diodes are sitting in a black enclousure without the possibility of light entering that enclousure. The whole arrangement is placed inside a temperature chamber for a temperature cylce (almost 24 h) from 25 °C up to 70 °C down to -30 °C and back to 25 °C again (black curve in the diagrams). Both "DIFF" signals (see circuit at Cf8) are sampled and logged by an 12 bit ADC with 2.9 V reference -> q = 700 uV). The red curve is the signal from circuit1 and the green one from circuit2 (both identical as already stated). What seems strange to me is that the circuitry seems to behave strongly different. The red curve is dropping down from 140 mV when temperature goes up and the green one is rising slightly with temperature.

Has anyone have a clue if that is related to some instable operating point of the OPAmps used, or even a condition which is out of specification fot that particular OPAmp AD8692?

I'd appreciate any hint in order to track down the reason of that behaviour.

enter image description here enter image description here

Additional information:

EDIT1: @jonk, thanks for the hint I'll calculate that. According to reverse bias: No It is not reversed biased. I guess for the sake of dynamic range they gave up linearity and decided not to reverse bias. (I am not the initial "designer" of that circuit and have little to none design documentation for that.

@Tony Stewart EE75, thanks first of all for your hints.

  1. "Unbalanced source.." do you mean that I should AC-terminate each differential signal "PD1" and "PD2" to GND?
  2. "Open circuit voltage.." that I have noticed as well. Signal "Diff" at the end of the chain is drifting in the order of same magnitude as the -3 mV/K would suggest. Should I consider the PD as beeing in open circuit conditions with that much of resistance connected (high impedance amplifier inputs)?
  3. "The PD's capacitance.." So it is, but how would it be linked to the behaviour observerd?

Will consider your suggestions (need time foor that) though it means changing much of the circuit. First of all I need to understand them first.

EDIT2: @jonk, the PD is operated with almost 0 V across it, in order to minimize dark current and noise. That's what I was able to investigate. I still need to look for some web ressources to support that statement.

EDIT3: Further thoughts: 200 digits signal correspond to approx. 80 pA current (calculating back through the amplifier chain), which indeed could be dark current. Imagine that the PD is not held at 0 V because the OPamps (TIA) contribute to a Voltage by having certain input offset voltages. Now, if the resulting voltage across the PD is positive I will get current flow in the corresponding direction. But if the resulting voltage is negative the current will also reverse. Is that reasonable?

\$\endgroup\$
7
  • \$\begingroup\$ Before I spend any time, may I assume you have already considered Figure 1 of the datasheet and expect at least that much behavior and that you are applying a reverse bias to the PIN diodes (is that shown in the schematic and I'm just missing it?) \$\endgroup\$
    – jonk
    Apr 30, 2021 at 23:27
  • \$\begingroup\$ Did You have a look at my additional information (see edits)? \$\endgroup\$
    – stowoda
    May 4, 2021 at 8:56
  • \$\begingroup\$ Is Op Amp attached to sensor in oven? It's all noise to me. There is no signal other than heat. The sensor is positive regardless of bias to input wavelengths except for the negative Vo thermal drift . Have you tried what I said? \$\endgroup\$
    – Tony Stewart EE75
    May 4, 2021 at 11:03
  • \$\begingroup\$ Op Amp is together with sensor in oven, they are on a PCB. The are under enclousure to prevent light from coming in and hitting the PD. I have not reverse biased yet because its a major change in circuitry and I need to use the PD in photovoltaic mode in order to get the best senitivity with lowest dark current. How should I balance the stray capacitance? What do you mean by interface cable? The only cable going into the oven is the power supply. You sispect that to catch EM and pass it on as noise to the circuitry? \$\endgroup\$
    – stowoda
    May 4, 2021 at 12:27
  • \$\begingroup\$ @stowoda I've not spent a lot of time thinking about PINs. There was a time when I spent a short time looking into them, though. But as I recall, the whole point of using a PIN instead of a normal PN is that the greater separation between P and N allowed higher reverse voltages, for those applications where that's desired (high speed, for example.) My applications were low-speed (like maybe 100 readings per second, at most) and were used for very high accuracy measurements. So it didn't take long to figure "no PINs here." \$\endgroup\$
    – jonk
    May 4, 2021 at 16:21

1 Answer 1

Reset to default
3
\$\begingroup\$

It seems there are possibly a few interactions going on.

  • Unbalanced source impedance to gnd, causing poor CMRR to noise
  • the Pd has an open circuit voltage Vo of 320mV that is temperature dependent -3mV/‘C.
  • The PD capacitance reduces with Vr negative bias 80pF at V=0 and 50pF at V=-3V

Suggestion

  • Reverse bias the PD to >=5V and use your differential trans-impedance amplifier on a current shunt to provide some gain. Excessive gain will require RC matching feedback to source parasitic and cable capacitance.
  • Using 1nA dark current at room temp, choose your shunt R based on some Noise spec.
  • Try to balance your PD stray capacitance to ground on each pin.
  • Choose your interface cable short for low capacitance but also twisted for CMRR.
\$\endgroup\$
1
  • \$\begingroup\$ Did You have a look at my additional information (see edits)? \$\endgroup\$
    – stowoda
    May 4, 2021 at 8:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.