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I have two adjacent photodiodes in photovoltaic mode, with two transimpedance amplifiers. Unfortunately, my MCU (nRF52840) is one meter away, and I expect a fair amount of noise to enter the system. For my application, the only relevant data is the difference in luminosity between these photodiodes.

It seems the right solution would be to perform true differential signaling for each photodiode, but I can't do it in my case for very specific reasons.

By routing each signal next to each other, I hope that most of the noise would affect both signals roughly equally. My goal is to reject common-mode noise, interference, and maximize the dynamic range at the MCU location.

I am considering the following solutions:

  • directly route each signal to the nRF52840 ADC in differential mode
  • use a differential opamp and use the nRF52840 ADC in single-ended mode
  • use a fully differential amplifier to drive the nRF52840 ADC in differential mode

The first solution would be simple, but I am not sure to get enough dynamic range. I am not familiar with fully differential amplifiers and I am not sure about their usage with non single ended signals.

What solution would be the most appropriate? Can you think of a better alternative?

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    \$\begingroup\$ This kind of thing is done all the time with CD, DVD, and blueray players. They use four diodes, though. The four diodes are in a square with each diode occupying a corner. The basic idea is to sum the cross-diagonal diodes and then take the difference between those sums. This is how they maintain focus. \$\endgroup\$
    – jonk
    Mar 29 '21 at 18:58
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    \$\begingroup\$ Remember to check the sensitivity specifications of the photodiodes. If their sensitivity varies a lot, you may need to include a trimmer to adjust the balance in order to get the best common mode rejection ratio. \$\endgroup\$
    – jpa
    Mar 30 '21 at 5:49
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For my application, the only relevant data is the difference in luminosity between these photodiodes.

If both photodiodes are the same, wire them antiparallel to a single TIA. Wiring them antiparallel means that the only current feeding the TIA will be the difference current. Now, the problem of noise is halved. Use a balanced transmission system like this: -

enter image description here

Replace the sensor on the left with a single ended op-amp driver and use balanced cable such as shielded twisted pair. Ground the shield at the receiver end. Clearly you need to use a differential receiver amplifier too. Schematic from here.

The two resistors marked with a blue X represent equal impedance resistors so that a balanced impedance drive to the cable is facilitated and any interference (red lightning symbol) has to affect both inner transmission wires equally. That is what is meant by "balanced" in this context. Of course, with the spare components from the unneeded TIA you can drive the line differentially and increase SNR by another 6 dB.

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    \$\begingroup\$ Excellent solution. Because the current through the feedback resistor could flow either way, the opamp supply might best be bi-polar. Or the antiparallel photodiodes should be referenced to half-supply in the case where opamp supply is uni-polar. \$\endgroup\$
    – glen_geek
    Mar 29 '21 at 17:03
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    \$\begingroup\$ Nice solution, thanks a lot ! \$\endgroup\$ Mar 29 '21 at 21:46
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schematic

simulate this circuit – Schematic created using CircuitLab

Andy's anti-parallel diode idea works nicely in a transimpedance amplifier. The dynamic range is maximized so long as the opamp doesn't run up against its DC supply rail with too much gain. Choose a feedback resistor value (R1, or R4) to stay within the opamp linear range.

The photodiodes and opamp belong close together, with the opamp driving the long cable to ADC.

Most microcontroller ADCs accept a voltage between ground and a +ve reference voltage - the right circuit (unipolar) would be preferable over the left circuit. When photodiode currents are equal, the opamp output voltage is half the DC supply.

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    \$\begingroup\$ Thanks a lot, both your answer and Andy's are very helpful. \$\endgroup\$ Mar 29 '21 at 21:45
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    \$\begingroup\$ This circuit is usually called a balanced photodetector, and it's the standard solution for this problem. \$\endgroup\$ Mar 29 '21 at 22:15

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