{title2=$g$}

= EMCCD
{c}
{synonym}

An electron-multiplying CCD adds a high-field multiplication register before the output amplifier. <Impact ionization> generates a large mean gain, reducing input-referred <read noise> but adding <multiplication excess noise>.

Electron-multiplying CCD

Let $N$ be the input photoelectron count and $\mu=\eta Q$ its mean, with <quantum efficiency> $\eta$ and mean incident photon count $Q$. Independent arrivals give a <Poisson distribution>, so $\operatorname{Var}N=\mathbb EN=\mu$. Conditional on $N$, the preceding <gamma distribution> gives $\mathbb E[X\mid N]=gN$ and $\operatorname{Var}(X\mid N)=g^2N$. The <law of total variance> therefore yields
$$
\mathbb EX=g\mu,\qquad
\operatorname{Var}X
=\mathbb E[g^2N]+\operatorname{Var}(gN)
=2g^2\mu.
$$
One contribution is the ordinary <photon shot noise>; the other is <multiplication excess noise>. Thus, neglecting <read noise> and backgrounds,
$$
\mathrm{SNR}=\frac{g\mu}{\sqrt{2g^2\mu}}=\sqrt{\frac{\eta Q}{2}},
\qquad \boxed{\eta_{\rm noise\ equivalent}=\frac\eta2.}
$$
This is an excess-noise factor $\sqrt2$ in analog operation: actual photon conversion efficiency has not been halved. At low occupancy, thresholding each pixel as zero or one event can avoid most multiplication noise, but high arrival rates produce coincident events that cannot be counted separately. That is why the high-rate result concerns charge measurement rather than ideal binary photon counting.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 338 / 3 / a / ii / Solution

{title2=$F$}

Random multiplication increases the noise relative to deterministic gain. If the output from one input electron has mean $g$ and variance $g^2$, a <Poisson distribution> of input electrons with mean $\mu$ produces output variance $2g^2\mu$, by the <law of total variance>. The high-gain analog-mode excess-noise factor is $\sqrt2$.

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 Incoming links: Multiplication excess noise