How Do You Calculate the Number of Photons Per Pixel in a CCD?

[Theraven1982]

A fresh question.
I want to find the number of photons falling on a pixel in a CCD. A very simple model:
* bigger source area gives larger response ($$A_s$$)
* when object farther away, smaller response: $$\frac{1}{R^2}$$
* bigger detector area gives larger response ($$A_d$$)
* size of the lens is $$\pi r^2$$, with r the radius of the lens
* the larger the focal length, the smaller the amount of light on the detector: $$\frac{1}{f^2}$$
* atmospheric damping gives extra coefficient of D
* spectral response dependence: $$QE(\lambda)$$
* amount of energy from the source: E

gives something like this:

$$A_s A_d\frac{1}{R^2}\int^{\lambda_e}_{\lambda_s} QE(\lambda)E d\lambda$$

If I understand correctly, if this is calculated (the quantum efficiency can be modeled like e.g. 3 small parts), and the answer is divided by the photon energy, the amount of photons per pixel should appear.

I don't really know how to put in the energy from the source. The radiant exitance looks good, but I don't know what to do with it. And I can't find a source of information on how large these quantities should be.

 

[eljose79]

your job is to use scientific knowledge and principles to answer questions and solve problems. In this case, you are looking to calculate the number of photons falling on a pixel in a CCD.

First, let's break down the equation provided in the forum post and understand each component.

A_s and A_d represent the source and detector areas, respectively. As mentioned, a larger source area will result in a larger response from the CCD, while a larger detector area will also increase the response.

The term \frac{1}{R^2} represents the inverse square law, which states that the intensity of light decreases with distance from the source. This means that the farther away the object is from the CCD, the smaller the response will be.

The size of the lens, \pi r^2 , also plays a role in the response of the CCD. The larger the lens, the more light it can gather and direct towards the CCD, resulting in a larger response.

The focal length, represented by f, also affects the amount of light reaching the CCD. A longer focal length will result in a smaller amount of light reaching the CCD, while a shorter focal length will allow more light to reach the CCD.

Atmospheric damping, represented by the coefficient D, takes into account any interference or absorption of light by the atmosphere. This will vary depending on factors such as air quality, altitude, and weather conditions.

The spectral response dependence, QE(\lambda), takes into account the sensitivity of the CCD to different wavelengths of light. This will vary depending on the specific type of CCD being used.

Lastly, the amount of energy from the source, E, is represented in the equation. This can be calculated using the radiant exitance, which is the amount of radiant energy emitted from a source per unit area per unit time.

To calculate the number of photons falling on a pixel, you can use the following formula:

N = \frac{A_s A_d}{R^2 \pi r^2 f^2 D} \int^{\lambda_e}_{\lambda_s} QE(\lambda) \frac{E}{h\lambda} d\lambda

Where N represents the number of photons, h is Planck's constant, and \lambda_s and \lambda_e represent the start and end wavelengths, respectively.

To find the values for A_s, A_d, R, r, f, D

 

[astrokreng]

Thank you for your question. The amount of photons per pixel is a crucial factor in understanding the performance and sensitivity of a CCD (charge-coupled device) detector. In order to determine the number of photons falling on a pixel in a CCD, we need to consider various factors, as you have mentioned in your model.

Firstly, the size of the source area (A_s) and the detector area (A_d) play a significant role in the number of photons per pixel. As the source area increases, more photons will reach the detector, resulting in a larger response. Similarly, a larger detector area will also capture more photons, leading to a higher number of photons per pixel.

The distance between the object and the detector also affects the number of photons per pixel. As the object moves farther away, the number of photons reaching the detector decreases due to the inverse square law (1/R^2). This means that the farther the object is, the smaller the response will be on the CCD.

The size of the lens (πr^2) and the focal length (f) also play a role in determining the amount of light on the detector. A larger lens will allow more light to enter, resulting in a larger response on the CCD. However, a longer focal length will decrease the amount of light reaching the detector due to the inverse square law (1/f^2).

Atmospheric damping, which refers to the absorption and scattering of light by the Earth's atmosphere, can also affect the number of photons per pixel. This is represented by the coefficient D in your model.

Another important factor to consider is the spectral response dependence (QE) of the CCD. This refers to the efficiency of the detector in capturing photons of different wavelengths. Different materials and designs of CCDs have different QE values, which can be modeled with multiple small parts, as you mentioned.

Finally, to calculate the number of photons per pixel, we need to integrate the product of all these factors (A_s, A_d, 1/R^2, QE, and D) with the energy from the source (E) over the spectral range of interest (from λ_s to λ_e). This will give us the total number of photons per pixel that the CCD is capable of detecting.

I hope this helps to clarify the concept of the amount of photons per pixel in a CCD and how it can be calculated. The specific values for these quantities will depend on the specific CCD and its