= Shot noise worked example
For an original model, set $q=1.602176634\times10^{-19}\,\mathrm C$, $I=1\,\mu\mathrm A$, $R_F=100\,\mathrm{k}\Omega$, $G=10$, and the exponential <low-pass filter>'s $\tau=100\,\mu\mathrm s$. Assume an ideal detector without charge multiplication and omit all additive <noise> for this calculation. Then
$$
f_c\approx1591.55\,\mathrm{Hz},\quad B_{\rm n}=2500\,\mathrm{Hz},\quad
S_I^{(1)}=3.204353268\times10^{-25}\,\mathrm{A^2/Hz},
$$
$$
\sqrt{S_I^{(1)}}\approx0.56607\,\mathrm{pA}/\sqrt{\mathrm{Hz}},\quad
\sigma_{v,\rm shot}^2=8.01088317\times10^{-10}\,\mathrm{V^2},\quad
\sigma_{v,\rm shot}\approx28.3035\,\mu\mathrm V.
$$
Doubling $I$ doubles the voltage <variance> and multiplies its <standard deviation> by $\sqrt2$. Halving $\tau$ also doubles the <variance>, provided the detector remains approximately white throughout the enlarged passband. Using $f_c$ in place of $B_{\rm n}$ underestimates the <variance> by the factor $2/\pi$. These are calculated predictions, not measured values or settings claimed for the laboratory below.
\Image[/shot-noise-poisson-counts-filtered-spectrum-and-variance.png]
{title=Discrete counts, a filtered shot-noise spectrum, and two distinct mean–variance laws}
{description=Original exact model calculations, with no experimental trace. Left: <Poisson distribution> for mean count $4$. Middle: normalized one-sided <power spectral densities> after an exponential <low-pass filter>. Right: count <variance> and normalized filtered-current <variance> versus their common normalized mean.}
To read the dimensionless figure, set $a=\lambda\tau$ and $x=\tau i_{\rm filtered}/q$, with <frequency> $\nu=f\tau$ in cycles per unit $t/\tau$. A gate of duration $\tau$ has count mean and <variance> $a$. The continuously filtered $x$ instead has
$$
\mathbb E[x]=a,\qquad \operatorname{Var}(x)=a/2,\qquad
S_x^{(1)}(\nu)=\frac{2a}{1+(2\pi\nu)^2},\qquad
\int_0^\infty S_x^{(1)}(\nu)\,d\nu=a/2.
$$
The count and the filtered <electric current> are different observables. The factor $1/2$ comes from the exponential weighting, not from non-Poisson arrivals. The dimensionless count illustration is a separate choice of rate; it does not assert that the $1\,\mu\mathrm A$ example has only four <Electrons> per $100\,\mu\mathrm s$.
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