Photodiode Module Dynamic Range

Forgive me if I use the wrong terminology still wrapping my head around things.

Looking at how to calculate the voltage output from the TIA it is simply the current x the feedback resister value. So for instance my photodiode module with the internal 1M-ohm resister if the photodiode current is say 5.00566E-06 amps then the output voltage would be about 5 volts? Which is just Ohms law.

How does the power rail voltage play into this? If I am using the 3.3 volts from the MCU I can't ever achieve 5 volts, is this what is causing my clipping?

Max output voltage of the opamp is "(VS) – 1.3" worst case, according to the datasheet.

Which means that the opamp clips at ~2volt on a 3.3volt supply.
That's not a problem if you choose a lower gain, and set the ADS1115 to 2.048volt Aref.

The ADS1115 has a 15-bit single-ended resolution, so you have 32,768 A/D steps available, which will cause low-end clipping more smootly (assuming the opamp won't clip before that).
Leo..

Appreciate it, I have some more studying to do understand this more clearly.

I am still trying to wrap my head around this and appreciate everyone's patience. If this is not the right category to discuss this please point me to the correct one.

For a layman like me this is tricky stuff. After reading through several articles about TIA's things are still fuzzy. I expect I am going to screw this up but here goes:

TIA

This OP amp will output 0 up to a maximum of 1.3 volts (Vs) per the datasheet as a function of the current generated by the photodiode. Which i believe is what Leo was saying above.

So Vs is NOT dependent upon the voltage at terminal #1 in this diagram (V+) as long it is above the minimum 2.7 volts per the spec??

It also appears that Vs is referenced to V+ (terminal #1) somehow but that doesn't make sense to me yet. Is it just subtracting? I am trying to understand Leo's statement:

Max output voltage of the opamp is "(VS) – 1.3" worst case, according to the datasheet.

Which means that the opamp clips at ~2volt on a 3.3volt supply.

Thanks again for the help...

The circuit is called a transimpedance amplifier, which converts the current produced by the photodiode into a proportional voltage. Ideally, the output voltage is equal to the feedback resistance in Ohms times by the photodiode current in Amperes.

The output voltage can be limited by the sensor power supply voltage Vs. According to the sensor data sheet, the output voltage cannot exceed Vs - 1.3V.

You can use a higher Vs if that is a problem, but make sure that the output voltage does not exceed the maximum allowed ADC input voltage.

I appreciate the reply... This is a minus sign correct?

cannot exceed Vs - 1.3V

I wasn't understanding that part but I think it makes sense now. So say if my Vs is 10 volts just to play with a number then the maximum output from the TIA will be 10 - 1.3 = 8.7 volts?? And at that point any more current from the photodiode that signal is clipped?

I do understand not to exceed the input voltage of my ADC.

Thanks so much for helping me

Correct.

I don't feel so ignorant now, I was just not understanding what you all were trying to say. That is a big help and explains some things for me. My next question would be how do I determine the minimum voltage out? I understand a photodiode will have dark current flowing at all times even without light striking it.

Is this going to determine the minimum voltage out of the TIA?

Put the sensor in complete darkness and measure the output voltage.

Gotcha... this is starting to make a lot more sense to me now. Thanks again.

The datasheet says 7.5mV typical, which is a value of about 120 if you use the ADS1115. I hope you bought a genuine one (Adafruit, Sparkfun) if you're serious about measuring. Consider all others fake.
Leo..

I did see all manner of ADS1115 for sale on Ebay but like you say one has to be cautious of what they buy and my goal is to improve on what the MCU does so it is a small price to pay. I did get one from Adafruit.

Hi All,

This is a continuation of my photodiode project. Here is my initial thread from a couple of weeks ago where everyone was very helpful getting me started: https://forum.arduino.cc/t/photodiode-module-dynamic-range/1198636

I hope it is appropriate to start a new thread instead up picking back up on an older thread. I apologize if it is not.

My current dilemma is understanding the dark current I am in seeing. I must be missing something fundamental because from my research I should be getting a few millivolts when the sensor is in complete darkness but I am getting much more. About 100 times more.

I have tried (3) modules. (2) OPT101 data sheet link
and (1) sd100-42-22-231 data sheet link

The OPT101 shows a dark voltage of about 7.5 mV. I am getting about 650 mV on both modules with my DMM.

sd100-42-22-231 shows dark current of 10 nA and if my math is correct should be about 10 mV with a 1 MEG feedback resistor. I am getting 990 mV with my DMM.

This is the way I wired all of them:

Can anyone give me some insight to what I am missing?

thanks!
Axis

No, it is not. Cross posting wastes everyone's time. Flagged for moderator.

Have not looked at the earlier thread.

If your intention is to create a light detection circuit, I don’t see any biasing for the diode.

Here is a circuit that has the necessary biasing for a detection circuit.

Biasing is not required for the posted circuit, which is a transimpedance amplifier (current to voltage converter). It very effectively converts the diode short circuit photocurrent to a voltage proportional to the illumination level.

1 Like

Again, I am sorry... i don't do a lot of posting on forums so don't know all the rules... if the moderate can move as required that would be appreciated.

Was going by the OP’s circuit posted here.

Which is correct, and internal to the IC OPT101 from TI.

Have no intention or desire to read a long previous thread or data sheet when an OP posts a schematic in their post.