 Shot noise worked example

For an original model, set $q=1.602176634\times 10^{-19}\mathrm{C}$, $I=1\mu \mathrm{A}$, $R_F=100\mathrm{k}\Omega$, $G=10$, 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. ThenDoubling $I$ doubles the voltage variance and multiplies its standard deviation by $\sqrt{2}$. Halving $\tau$ also doubles the variance, provided the detector remains approximately white throughout the enlarged passband. Using $f_c$ in place of $B_{\mathrm{n}}$ underestimates the variance by the factor $2/\pi$. These are calculated predictions, not measured values or settings claimed for the laboratory below.

To read the dimensionless figure, set $a=\lambda \tau$ and $x=\tau i_{\mathrm{filtered}}/q$, 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 hasThe 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 $100\mu \mathrm{s}$.