Gain

As stated before, the digital readout data is in units of ADU, not the number of photoelectrons. This means that there must be some conversion factor between the number of photoelectrons and ADU. This factor is called the gain. The gain is defined to be the number of electrons over ADU ( $\frac{\char93 \;\;e^{-}}{ADU}$). Therefore we can extract the number of photoelectrons counted by the CCD from the output digitized data by multiplying the pixel value in ADU by the gain.

We can see where the gain comes from mathematically by simply plugging in the relationship between ADU and number of photoelectrons into the equation for the variance (g$\equiv$gain):

$$x_{e^{-}} = gx_{ADU} \Rightarrow \overline{x}_{e^{-}} = g\overline{x}_{ADU}$$

$$\begin{eqnarray*} s_{ADU}^{2} &=& \frac{1}{N-1} \sum_{i=0}^{N}[x_{i,ADU}-\overli... ...frac{1}{N-1}\sum_{i=1}^{N}[x_{i,e^{-}}-\overline{x}_{e^{-}}]^{2} \end{eqnarray*}$$

Now we can see that the sum on the right side of the equation is just equal to the variance of the number of electrons, so we can plug that in:

$$s_{ADU}^{2} = \frac{1}{g^{2}} s_{e^{-}}^{2}$$ (1)

So now if we can plot the mean of the image pixels in ADU versus the Variance of our image in ADU, let's see what we get:

$$\frac{\overline{x}_{ADU}}{s_{ADU}^{2}} = \frac{g^{2}\overline{x}_{e^{-}}}{gs_{e^{2}}^{2}}\\$$

But if $s_{e^{-}}^2$ and $\overline{x}_{e^{-}}$ obey Poisson statistics, then $s_{e^{-}}^2=\overline{x}_{e^{-}}$, so

$$\frac{g^{2}\overline{x}_{e^{-}}}{gs_{e^{2}}^{2}}= \frac{g^{2}}{g} = g.$$

Thus,

$$gain = \frac{\overline{x}_{ADU}}{s_{ADU}^{2}}.$$ (2)

We will come back to this equation in Section 4.3.