Confusion with a photodiode circuit

Question

I'm using this photodiode with a simple circuit as follows:

And in my setup the photodiode sees the same light input meaning same average power from a laser source which is less than a few mW estimate. Vcc is always 3V.

I keep Rs and Rload always same as 1k and only vary Rpot. And make three measurements where Rpot is 10k, 5.6k and 0k. For each measurement I measure Vout for no light input(Vout_dark) and with light input(Vout_light).

Here are my results for three Rpot values:

Rpot = 10kOhm: Vout_dark= 25mV Vout_light = 69mV

Rpot = 5.6kOhm: Vout_dark= 33mV Vout_light = 76mV

Rpot = 0kOhm: Vout_dark= 73mV Vout_light = 121mV

As you see, Rpot has effect on Vout it means it has effect on Iout = Vout/1k.

But I thought the photodiode is a current source which produces current linearly proportional to input light power(responsivity is 0.85 in my case). So even though I change Rpot I keep laser light power same, so I was expecting same Iout would flow in the circuit regardless of Rpot.

But apparently something wrong in my logic? Why is current changing if the PD is a current source and the input light power is the same. Can someone help me to draw the simplistic equivalent circuit which I can simulate to see what is going on?

As a random example, in my model(PD is a current source) and the Vout is independent from Rpot and always 2mV(meaning Iout is independent from Rpot):

So for some reason, my model doesn't match reality since Rpot changes Iout hence Vout. I would be glad to have an answer with a netter model whcih can show the effect of Rpot.

Answer 1

Rpot = 10kOhm: Vout_dark= 25mV

Dark current is proportional to the reverse voltage you apply to your photodiode. You have 25 mV across Rload (1 kΩ) when the pot is 10 kΩ and, this means the dark current is 25 μA.

Hence, the voltage at the cathode end of the photodiode is 3 volts minus (25 μA × 11 kΩ) = 2.725 volts. In other words, the reverse photodiode voltage is 2.725 volts minus 25 mV = 2.7 volts.

Rpot = 0kOhm: Vout_dark= 73mV

When the pot is 0 kΩ the dark current is 73 μA.

Hence, the voltage at the cathode end of the photodiode is 3 volts minus (75 μA × 1 kΩ) = 2.925 volts. In other words, the reverse voltage is 2.925 volts minus 25 mV = 2.9 volts.

Hence, the reverse voltage across the photodiode has increased and so has the dark current.

The question then you must ask yourself is whether the change in dark current vs the change in reverse voltage across the photodiode matches the data sheet of the photodiode. Well, the data sheet quotes a dark current of 4 μA for a reverse voltage of 1 volt but, it also says this: -

Application of a reverse bias can greatly improve the response speed and linearity 
of the device. This is due to increase in the depletion region width and, consequently, 
decrease in junction capacitance. However, the dark current and noise will increase.

So, either the Thorlabs diode goes a lot leaky when the reverse voltage rises above 2 volts or, you are not truly measuring dark current.

Answer 2

The photo-diode is not a perfect light dependent current source. A better model includes a resistor (perhaps 4MΩ) in parallel with the current source. The resistance might not be linear with applied reverse voltage. You should be able to tell from your measurements.

It is advised to keep the reverse voltage constant to observe a y=mx+b relationship.