= Square-root collapse of capacitive bridging bandwidth near the DC error floor
{title2=$f_\eta\propto\sqrt{\eta-\eta_0}$}
The <square-root collapse of capacitive bridging bandwidth near the DC error floor> describes how the usable <frequency> interval disappears when a resistive divider has almost exhausted a complex-error allowance. Use the model and exact formula in <capacitive bridging complex-error bandwidth>: $g=R_s/R_{\rm in}>0$, $\tau_s=R_sC_{\rm in}$, and the DC floor $\eta_0=g/(1+g)$. Write the available extra tolerance as $\Delta=\eta-\eta_0>0$. The exact bandwidth becomes
$$
f_\eta=\frac{1+g}{2\pi\tau_s}
\sqrt{\frac{2\eta_0\Delta+\Delta^2}{1-(\eta_0+\Delta)^2}}.
$$
At fixed positive $g$, the leading numerator is linear in $\Delta$, so $f_\eta$ is proportional to $\sqrt\Delta$. Thus demanding four times less extra tolerance approximately halves the bandwidth; it does not reduce it by a factor of four.
For an original example, choose $R_s=20\,\mathrm{k\Omega}$, $R_{\rm in}=1\,\mathrm{M\Omega}$, and $C_{\rm in}=50\,\mathrm{pF}$. Then $g=0.02$, $\tau_s=1\,\mathrm{\mu s}$, and $\eta_0\approx0.019608$. Taking $\Delta=0.00020$ gives about $456\,\mathrm{Hz}$ from the exact formula. Taking $\Delta=0.00005$ gives about $228\,\mathrm{Hz}$. The small deviation from an exact two-to-one ratio comes from the retained quadratic term and denominator.
The qualification \i[fixed positive $g$] matters. If the receiving resistance is infinite, $g=\eta_0=0$ and
$$
f_\eta=\frac{\eta}{2\pi\tau_s\sqrt{1-\eta^2}},
$$
which is linear in small $\eta$. For a small but nonzero floor, the square-root approximation needs $\Delta$ small compared with $\eta_0$ as well as with the remaining distance to unit error. The <crossover from square-root to linear capacitive bridging bandwidth scaling> prevents using the fixed-floor asymptotic expression uniformly as $g$ also tends to zero.
This is a limit of the chosen passive <input impedance> model, not an empirical limit for every amplifier. The manufacturer's shunt-capacitance model motivates the resistive/capacitive paths; an actual bench comparison must include both <amplitude> and <phase shift> and the probe's added <capacitance>.<MT-040: Op Amp Input Impedance by Analog Devices>[Printed pp. 1–2, Figures 1–2 and Op Amp Input Capacitance]{ref}
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