A little context: This is part of a bigger project about glowing algae, my job is to design a photodiode array to measure the amount of light the algae produce in response to movement.

I have a good idea of which photodiode I'm gonna use, the S2387-16R, and my plan is to connect them to the DDC1128 Evaluation Module. This thing is a beast, it can measure ridiculously small currents, in the order of femto-Amperes, so I don't need to use amplifiers.

What I'm trying to find out is what would be the maximum amount of light I could possibly detect. I know that:

$$ I_{max} = \frac{V_{bias}}{R_{Load}} $$

And here is where I need help. How do I connect the photodiode to the Evaluation Module?.

To help explain my question, here is the equivalent circuit of a photodiode, I took it from here.

enter image description here

Everything but \$R_L\$ are internal variables of the photodiode.

Now, at first I thought that \$R_L\$ would be the internal resistance of the Evaluation Board, which I don't know and it's not in the Data Sheet but then I thought: "this is like an Ammeter right?, and those are made to have negligible resistances compared to what you are measuring, right?, so I can just ignore the internal resistance".

If that's the case, then I can put some resistance I know, like 1 kOhm, in one of the terminals of the photodiode, connect that to the Evaluation Module, then connect the other terminal to ground, and that's it.

But I'm not sure that's the case, and if I say to the team that we can ignore the internal resistance of the Evaluation Module and I'm gonna have to justify it and justify it well.

And if I have to consider its internal resistance, what do I do? What would you do?

  • \$\begingroup\$ That ADC is a current input transimpedance integrator not a typical Voltage ADC so using 0.3pA/pW for green and the sensor area, you can compute the dark current equivalent lux or lumen level. \$\endgroup\$ Jan 12, 2019 at 22:30
  • \$\begingroup\$ Thanks, but I do not understand, are you saying I can't neglect the resistance but that there is something else I can calculate? \$\endgroup\$ Jan 13, 2019 at 0:02
  • \$\begingroup\$ You can measure the current directly without resistance, but if you wish to make it less sensitive, add a shunt resistor to reduce the current.into the ADC I to V converter. Since it integrates, it must be dumped after each measurement so interval rate can rise with illuminance or you can shunt with R. Slew rate depends on C on the ADC as dV/dt=I/C. Read the manual. \$\endgroup\$ Jan 13, 2019 at 3:58
  • \$\begingroup\$ Thanks but now the question is, how do I find I_max? \$\endgroup\$ Jan 14, 2019 at 10:08

2 Answers 2


If that's the case, then I can put some resistance I know, like 1 kOhm, in one of the terminals of the photodiode, connect that to the Evaluation Module, then connect the other terminal to ground, and that's it.

Why put a series resistor in at all? The 1128 converts current to voltage internally, and the electron source in your photodiode is modelled as a current source - so don't bother. Just connect the lower pin in your PD schematic to ground, and the other to the 1128 input. Easy, peasy, pudding and pie.

  • \$\begingroup\$ Thanks, but then, what is I_max?, I need to find out what it is \$\endgroup\$ Jan 14, 2019 at 10:06
  • \$\begingroup\$ @FernandoFrancoFélix - You start by looking up the efficiency of your diode at your wavelength. Then you calculate the optical power which will fall on the PD area (also found in the data sheet). Multiply the two together and you know the current. \$\endgroup\$ Jan 14, 2019 at 16:20
  • \$\begingroup\$ That's why I commented 0.3pA/pW for green and the sensor area. The min is due to dark current and Vbias and you can shunt this with a shunt resistor and make your ADC less sensitive with less input current being integrated. This also speeds up the response time constant for decay time. This card is a current integrator. Your max current depends on photo emission and how fast you choose to measure the integrator by sample measure then dump and repeat. The min and max ranges are determined by R, n bits and conversion rate with other constants \$\endgroup\$ Jan 14, 2019 at 17:35
  • \$\begingroup\$ @SunnyskyguyEE75 - The proposed application uses the PD in photovoltaic mode, not photoconductive, so bias and leakage should not be an issue. \$\endgroup\$ Jan 14, 2019 at 21:59
  • \$\begingroup\$ How does that affect the choice of R using their intended current mode TIA integrating ADC? if the dark current consumes too much dynamic range of ADC \$\endgroup\$ Jan 14, 2019 at 22:02


simulate this circuit – Schematic created using CircuitLab


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