Poisson distribution and Shot Noise

In summary: 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.
  • #1
Frigorifico
32
0
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:

dump_028_eh_3.dat.jpg
dump_037_eh_3.dat.jpg


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?.
 
Engineering news on Phys.org
  • #2
Frigorifico said:
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.
 
  • Like
Likes berkeman
  • #3
Baluncore said:
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?.

Baluncore said:
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?
 
  • #4
Frigorifico said:
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?
 
  • #5
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.
 
  • #6
Baluncore said:
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
 
  • #7
berkeman said:
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
 
  • #8
Frigorifico said:
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.
 
  • Like
Likes berkeman

Related to Poisson distribution and Shot Noise

1. What is the Poisson distribution?

The Poisson distribution is a probability distribution used to model the number of events that occur in a fixed interval of time or space, given the average rate of occurrence and the assumption that events occur independently and at a constant rate.

2. How is the Poisson distribution related to Shot Noise?

Shot noise is a type of random noise that arises from the discrete nature of physical phenomena, such as the flow of electrons in an electrical circuit. The Poisson distribution is used to model shot noise because it describes the probability of a certain number of discrete events occurring in a given interval of time or space.

3. What are the applications of the Poisson distribution and Shot Noise?

The Poisson distribution and Shot Noise are commonly used in various fields such as physics, engineering, biology, and finance. They are used to model phenomena such as radioactive decay, traffic flow, and the fluctuation of stock prices.

4. How is the Poisson distribution different from other probability distributions?

The Poisson distribution differs from other distributions, such as the normal distribution, in that it is a discrete distribution rather than a continuous one. This means that the possible outcomes are countable and cannot take on any value within a range, unlike continuous distributions which can take on any value within a range.

5. How is the Poisson distribution calculated?

The Poisson distribution is calculated using the formula P(x; λ) = (e^-λ * λ^x) / x!, where x is the number of events, and λ is the average rate of occurrence. This formula gives the probability of x events occurring in a given interval, assuming a constant rate of occurrence and independence between events.

Similar threads

Replies
7
Views
2K
Replies
1
Views
947
  • Electrical Engineering
Replies
1
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
10
Views
2K
Replies
9
Views
3K
Replies
9
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
12
Views
2K
  • Electrical Engineering
Replies
22
Views
6K
  • Set Theory, Logic, Probability, Statistics
Replies
4
Views
1K
Back
Top