Model 1

For model $1$, the probability distribution for the number $x$ of electrons generated by impact ionisation from a gain stage when one electron is input is defined by a Poisson distribution:

$$P \left (x \right )=\frac{\alpha^{x}e^{-\alpha}}{x!}$$ (4.1)

where $P \left (x \right )$ is the probability distribution for the number $x$ of additional charge carriers generated and $\alpha$ is the expectation value for the number of additional charge carriers generated. The variance of the Poisson distribution is equal to the expectation value $\alpha$.

It is perhaps more useful to describe Equation 4.1 in terms of the total number of output electrons $n=x+1$ (i.e. including the single input electron):

$$P_{1e} \left (n \right )=\frac{\left ( \mu-1 \right )^{\left ( n -1 \right )}e^{\left (1-\mu\right )}}{\left ( n-1 \right)!}$$ (4.2)

where $P_{1e} \left (n \right )$ is the probability distribution for the total number of output electrons and $\mu$ is the expectation value for the total number of output electrons (equal to the gain of the stage; $\mu=\alpha+1$). As the charge transfer efficiency is perfect, $\mu$ is always greater than or equal to one. The variance in the total number of output electrons is equal to the gain minus one:

$$\sigma_{out}^{2}=\mu-1$$ (4.3)

As the model is linear and the electrons are treated independently, the variance in the number of output electrons for a fixed number of input electrons $m_{in}$ is just:

$$\sigma_{out}^{2}=m_{in}\left (\mu-1\right )$$ (4.4)

If there is a variance $\sigma_{in}^{2}$ in the number $\left <m_{in} \right >$ of input electrons, the total variance in the number of output electrons $\sigma_{out}^{2}$ is just:

$$\sigma_{out}^{2}=\left <m_{in} \right >\left (\mu-1\right )+\mu^{2}\sigma_{in}^{2}$$ (4.5)