Poisson distribution and Shot Noise

[Frigorifico]

TL;DR Summary

Sometimes I see a Poisson distribution, sometimes I do not, I don't understand why

My setup:

I have the an LED (LED370E) in front of a photodiode (S12915-16R). The photodiode is connected to an ADC (DT5751) which has a counting functionality. The way it works is that it counts how many times the signal goes above a certain threshold and makes a histogram out of it.

I know that photons often follow Poissonian Distributions, but I am aware that Super and Sub poissonian light also exists.

The mark of a Poissonian distribution is that sigma^2 = mean, so I made Poissonian distributions with the same mean as the measurements from the ADC. Here is how 2 of them look like:

The difference between these two measurements is that the one on the left was taken with the LED at 6.5 volts, and the one on the right at 2.5.

First thing I don't understand: why does one fit the Poissonian distribution and the other doesn't?, what kind of noise is affecting my measurement?.

Second things I don't understand: I have read in https://camera.hamamatsu.com/jp/en/technical_guides/photon_shot_noise/index.html that the Shot Noise is equal to the square root of the signal. Is the Shot Noise the same as the Standard Deviation?, is Shot Noise = Sigma?.

 

[Baluncore]

The difference between these two measurements is that the one on the left was taken with the LED at 6.5 volts, and the one on the right at 2.5.

LED voltage is fixed by semiconductor material bandgap in eV that decides wavelength. E = h * u.
LED light output is proportional to LED current, NOT voltage. You must identify LED current.
LED current must be controlled. Maybe you have a series resistor in the circuit.
You are measuring the spectrum of noise peaks on the continuous photodiode current. That is not shot noise.

 

[Frigorifico]

LED light output is proportional to LED current, NOT voltage. You must identify LED current.

The LED had a resistance of 50 Ohms, so the current would be just Voltage/50. If current is proportional to voltage, the LED output would be proportional too, right?.

You are measuring the spectrum of noise peaks on the continuous photodiode current. That is not shot noise.

I think this is what I'm struggling the most to understand. My understanding is that photodiodes will absorb a bunch of photons, convert them into photoelectrons, and that becomes a signal, a pulse. Then the electrons and wholes have to dissipate and leave the LED as it was at the start, at which point the process repeats.

However if that is not the case, then I have 2 questions:

1.-How can I know what the actual signal of the photodiode was?
2.-If this isn't shot noise, what kind of noise is it?, thermal?

 

[berkeman]

The LED had a resistance of 50 Ohms, so the current would be just Voltage/50. If current is proportional to voltage, the LED output would be proportional too, right?.

No. What do you mean that the LED had a resistance of 50 Ohms? Do you mean that you used an LED that has a built-in series resistor of 50 Ohms? Then the load will look like that resistance in series with the LED semiconductor junction. That combination gives a non-linear V-I characteristic.

Can you link to the datasheet?

 

[Baluncore]

Compare the spectral response of the photodiode with the 370 nm light from the LED.
Notice it is operating at the end of the spectral response curve.

The light centred on λ = 370 nm has thermal broadening.
370 nm light has a photon energy in eV of; 1239.84 / λnm = 1239.84 / 370 = 3.35 eV;
The 1239.84 comes from Plank constant and speed of light.
So it will take a forward current at about 3.35 volt before the LED begins to generate significant light.
Below that voltage there will be some light, but it will only be because of the thermal broadening.

You must read up on LED circuits and how to limit the forward current with a resistor. An LED can easily be destroyed by reverse voltage, so you must take care.

I would suggest you get some cheap visible coloured LEDs, some resistors and 5 volt power supply. Notice how the LED colour and forward voltage are related and how a resistor can limit the current.

If you want to limit LED current to 10 mA you can calculate...
λ = 370nm, needs energy = 3.35 eV = 3.35 volt.
5V - 3.35V = 1.65 volt across the series resistor.
Resistor value = 1.65 volt / 10mA = 165 ohms. So use a standard 150 ohm resistor.

Different value resistors are needed for different colour LEDs.
You can calculate it from λ in nm.

 

[Frigorifico]

Compare the spectral response of the photodiode with the 370 nm light from the LED.
Notice it is operating at the end of the spectral response curve.

The light centred on λ = 370 nm has thermal broadening.
370 nm light has a photon energy in eV of; 1239.84 / λnm = 1239.84 / 370 = 3.35 eV;
The 1239.84 comes from Plank constant and speed of light.
So it will take a forward current at about 3.35 volt before the LED begins to generate significant light.
Below that voltage there will be some light, but it will only be because of the thermal broadening.

I measured the spectrum of the LED, it was 358.6 instead of 370 nm

 

[Frigorifico]

No. What do you mean that the LED had a resistance of 50 Ohms? Do you mean that you used an LED that has a built-in series resistor of 50 Ohms? Then the load will look like that resistance in series with the LED semiconductor junction. That combination gives a non-linear V-I characteristic.

Can you link to the datasheet?

I mean that the power supply has a built in resistance of 50 Ohms for anything you connect to it

 

[Baluncore]

The difference between these two measurements is that the one on the left was taken with the LED at 6.5 volts, and the one on the right at 2.5.

With 6.5 volts and 50 ohms in series, you would have been driving more than the rated current through the LED.
The scales seem similar, but 2.5 volts is insufficient to generate any significant forward current through that LED.
You need to know and understand each module of your equipment before you take measurements.
Those modules are the LED, the optics, the photodiode, the current to votage converter and the sampling system.