Solving the Debate Over Photon Counting With a Scintillation Device

In summary: So the fact that you see a Gaussian means either you are not at the limit of large numbers or the light sources is behaving classically.In summary, the conversation revolved around the disagreement between a professor and a student regarding the use of a photodiode model to measure light intensity from scintillators. The student wanted to use the mean voltage and the properties of the photodiode to calculate the number of photons, while the professor suggested using the quantum efficiency and number of counts to determine the total number of photons. The student raised concerns about the applicability of the professor's method and the presence of dark current. The conversation also touched on the Poissonian distribution of photons and the limitations of using a LED as a single photon
  • #1
Frigorifico
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TL;DR Summary
I have to count the number of photons detected by a photodiode, I want to do it based on voltage/current measurements, but my professor wants me to do some statistical analysis which I think don't apply in this case
Hello, I'm working on a scintillation device to detect protons, I have a disagreement with one professor and I would like your opinion.

There is one photodiode model we want to use to measure the light intensity from the scintillators, and we want to relate the signal of that photodiode with the number of photons it is absorbing.

To do this we took a CAEN DT5751, connected a photodiode to it, and put an LED in front of the photodiode. This DT5751 samples the signal, measures its voltage, and counts how many times it has seen such a voltage, producing a histogram with a gaussian distribution, so far so good.

What I want to do is to get the mean voltage, divide by the resistance the photodiode sees (50 ohms, I measured), get the current produced by the photodiode, and use the photodiode's responsivity to get the mean number of photons. Easy piecy.

BUT my professor says that there's a better way (aka "I want this in your report"). We know that the statistical distribution of the photons is a Poissonian Distribution when N approaches infinity, which is a gaussian where sigma^2 = mean. And he is right, that distribution does fit with our results.

Then he says that we should be able to use the quantum efficiency of the photodiode and the number of counts to know the number of photons. Equation 5.57 here

Here's my problem with that reasoning: Our set up does not count photons, it counts current pulses, and each pulse is produced by the photodiode when it absorbs a bunch of photons, something like 10^10 at least, those formulas is for when you have a device capable of counting individual photons, so those formulas don't apply.

I brought up this point but he just told me to read more, and I did, I read this and this but he just said he had better things to do than to teach me what I should already know.

Maybe he's right, maybe there is a way to use photon statistics to get the total number of photons with this set up, so I come to you, is there?, has ay of you done something similar?, because I am lost.

PD:

The manual of the DT5751 says that "Input dynamic is 1 Vpp" and the values are stored with "16 MSB". I take that to mean that it has 2^16 slots to store all the values from 0 to 1 volts. For example a value of "20,000" would be 20,000/2^16 = 0.35 volts, is that correct?
 
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  • #2
I'm not sure where your methods actually differ. Using the voltage and the properties of the photodiode is sensitive to the number of detected photons only. To get the total number of photons hitting the photodiode you need to consider the quantum efficiency anyway.
 
  • #3
You agree the professor's method gives you the right answer, but you think your method is better, do I have that right? How do you deal with dark current?
 
  • #4
Vanadium 50 said:
You agree the professor's method gives you the right answer, but you think your method is better, do I have that right? How do you deal with dark current?

I agree the distribution is poissonian, the problem is that I don't see how can I go from the ADC units to photons with his method
 
  • #5
mfb said:
I'm not sure where your methods actually differ. Using the voltage and the properties of the photodiode is sensitive to the number of detected photons only. To get the total number of photons hitting the photodiode you need to consider the quantum efficiency anyway.

they differ in the fact that he wants me to use that one equation to get the number of photons, but my understanding is that only works if we were counting individual photons, which we weren't
 
  • #6
Frigorifico said:
To do this we took a CAEN DT5751, connected a photodiode to it, and put an LED in front of the photodiode. This DT5751 samples the signal, measures its voltage, and counts how many times it has seen such a voltage, producing a histogram with a gaussian distribution, so far so good.
If the LED is producing continuous photons, then you are measuring the spectrum of noise on the DC photodiode current while it is continuously illuminated. The gaussian voltage distribution seen with the LED, (without the scintillator), is noise from the hot LED, the photodiode efficiency, electronics, and from sampling and A to D conversion. The LED will have a different wavelength to the scintillator. When compared with the uncalibrated LED efficiency and optics, the spectral response is not really relevant.
 
  • #7
Frigorifico said:
they differ in the fact that he wants me to use that one equation to get the number of photons, but my understanding is that only works if we were counting individual photons, which we weren't
I don't see an equation that would stop being true for more photons. The relative fluctuation will decrease with more photons and other sources can become more important, whether that is relevant or not depends on your setup.
 
  • #8
I don't believe this makes sense. The fundamental problem is that a LED does not produce single photons; the fact that a setup gives you a Poissonian distribution is frequency used to show that something is at least approximately behaving like a single photon source (e.g. a very attenuated laser and/or a sub-threshold diode laser) and this is never the case for a LED.
Another problem is in the limit of large numbers the distribution becomes Gaussian; meaning it would be indistinguishable from random noise.
 
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1. What is a scintillation device?

A scintillation device is a scientific instrument used to detect and measure ionizing radiation. It works by converting the energy from incoming radiation into visible light, which can then be measured and analyzed.

2. How does a scintillation device help solve the debate over photon counting?

A scintillation device is able to accurately count individual photons of radiation, providing a more precise measurement than other traditional methods. This helps to resolve any discrepancies or disagreements in the debate over photon counting.

3. What are the benefits of using a scintillation device for photon counting?

Using a scintillation device for photon counting allows for more accurate and reliable measurements, which is crucial in many scientific fields such as medical imaging and nuclear physics. It also allows for a wider range of radiation to be detected and measured.

4. Are there any limitations to using a scintillation device for photon counting?

While a scintillation device is highly accurate, it does have some limitations. It may have difficulty distinguishing between different types of radiation, and it may have a limited range of detection for certain types of radiation.

5. How is a scintillation device different from other methods of photon counting?

A scintillation device is unique in its ability to directly convert incoming radiation into visible light, allowing for more precise and accurate counting. Other methods, such as Geiger counters, rely on indirect measurements and may be less accurate.

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