# Confusion with a photodiode circuit

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.

• What optical source are you using for the testing? Commented Apr 25 at 21:13

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.

• Are you saying that the dark current causes voltage drop and all three cases reverse bias is different so the output current? But all I know is that the in my case responsivity is 0.85. They dont show any data how it changes with reverse bias voltage. So if I didnt know the input light power and I want to estimate it, which Iout in those three Rpot cases should I use to estimate the light power? To estimate average power in all cases, should I use the relation between light power versus Iout as P = 0.85 * (Iout_light_ - Iout_dark)? This supposed to be equal in all cases correct? Commented Apr 25 at 22:23
• Sorry the formula P = (Iout_light_ - Iout_dark) / 0.85 Can we use this then? Commented Apr 25 at 22:32
• That photodiode is rated for a reverse voltage of 1 volt or less whereupon it draws 4 uA. Applying a larger reverse voltage causes greater dark current and, unfortunately the data sheet tells you nothing about this so, your options are either don't reverse bias it, speak to Thorlabs or, find a device that can have reverse bias and not produce significant dark current. I know nothing about your application so I can't advise. Commented Apr 26 at 10:07
• Hi again, Thanks for feedback. If I dont reverse bias it, do I need a transimpedance amplifier? But I noticed in they require negative supply since I need positive voltage output swing to a 3.3V MCU's ADC and have only positive power supplies:( Is there a way to use transimpedance amplifier without negative supply? I can open a new question if you would have time. Commented Apr 26 at 18:28
• Great thanks seems like what I look for! Commented Apr 26 at 18:39

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.

• if I didnt know the input light power and I want to estimate it from my measuremenrs, which equation in those three Rpot cases should I use to estimate the light power? To estimate average power in all cases, should I use the relation between light power versus Iout as P = (Iout_light_ - Iout_dark) / 0.85? Commented Apr 26 at 7:02