= Capacitive bridging complex-error bandwidth
{title2=$f_\eta$}
The \b[<capacitive bridging complex-error bandwidth>] is the largest <frequency> over which a receiving port preserves the source <phasor> within a specified <complex relative gain error>. The criterion includes <phase shift> as well as <amplitude>. A large DC <input impedance> alone cannot establish this bandwidth when <input capacitance> matters. The manufacturer's input models explicitly include resistive and capacitive paths.<MT-040: Op Amp Input Impedance by Analog Devices>[p1, Figure1 and shunt-capacitance description]{ref}
For the scalar <impedance bridging with capacitive input> model, assume a positive real source <electrical resistance> $R_s$, receiver resistance $R_{\mathrm{in}}>0$, and <input capacitance> $C_{\mathrm{in}}>0$ in parallel. Let $V_s$ be the unloaded source <phasor>. A <voltage divider> gives
$$
H=\frac{V_L}{V_s}=\frac1{1+g+ix},\qquad
g=\frac{R_s}{R_{\mathrm{in}}},\quad x=2\pi fR_sC_{\mathrm{in}}.
$$
For tolerance $0<\eta<1$, the <complex relative gain error> is
$$
E(f)=|1-H|=\sqrt{\frac{g^2+x^2}{(1+g)^2+x^2}}.
$$
Its square is strictly increasing in $x^2$, since its <derivative> is $(1+2g)/[(1+g)^2+x^2]^2>0$. Thus satisfying $E(f)\le\eta$ defines a continuous frequency interval starting at DC, whenever it is feasible. Rearranging gives
$$
\boxed{f_\eta=\frac1{2\pi R_sC_{\mathrm{in}}}
\sqrt{\frac{\eta^2(1+g)^2-g^2}{1-\eta^2}}.}
$$
The DC error floor is $\eta_0=g/(1+g)$. If $\eta<\eta_0$, no frequency meets the tolerance; if $\eta=\eta_0$, only DC does. The boxed positive bandwidth applies when $\eta>\eta_0$.
For an original near-limit example, choose $R_s=10\,\mathrm{k\Omega}$, $R_{\mathrm{in}}=1\,\mathrm{M\Omega}$ and $C_{\mathrm{in}}=100\,\mathrm{pF}$. Then $g=0.01$ and the DC error is about 0.9901 percent. A 1 percent full-complex tolerance leaves only
$$
\boxed{f_\eta\approx225.65\,\mathrm{Hz}.}
$$
An amplitude-only 1 percent criterion, $|H|\ge0.99$, instead allows about $2273.47\,\mathrm{Hz}$. The half-power (approximately $-3\,\mathrm{dB}$) <signal bandwidth>, relative to DC, is $(1+g)/(2\pi R_sC_{\mathrm{in}})\approx160.75\,\mathrm{kHz}$. These are three distinct specifications of this same ideal model.
\Image[/capacitive-bridging-error-bandwidth.png]
{title=Complex-error, amplitude-error and half-power bandwidths can differ greatly}
{description=Original 10 kilohm source, 1 megohm receiving resistance and 100 picofarad input-capacitance model. A one-percent complex error permits about 225.65 hertz; one-percent amplitude error permits about 2273.47 hertz. The half-power point relative to DC is about 160.75 kilohertz.}
The <square-root collapse of capacitive bridging bandwidth near the DC error floor> follows by factoring the numerator:
$$
\eta^2(1+g)^2-g^2=(1+g)^2(\eta-\eta_0)(\eta+\eta_0).
$$
For fixed $g>0$ and $\eta\downarrow\eta_0$,
$$
f_\eta\sim\frac{1+g}{2\pi R_sC_{\mathrm{in}}}
\sqrt{\frac{2\eta_0}{1-\eta_0^2}}\sqrt{\eta-\eta_0}.
$$
This <asymptotic equivalence> explains the sharp loss of usable bandwidth when the resistive divider has already consumed nearly all the error allowance. Increasing $R_{\mathrm{in}}$ or reducing $R_s$ changes that floor; reducing only $C_{\mathrm{in}}$ rescales a feasible bandwidth but cannot remove infeasibility at DC.
The criterion also controls a complete waveform. Suppose a source waveform $v_s$ is both a <square-integrable function> and a <bandlimited function>, with spectrum supported where $|H-1|\le\eta$. For the corresponding <zero-state response> $v_L$, the <Fourier transform> of $v_L-v_s$ is $(H-1)\widehat v_s$. The <Plancherel theorem>, or the <norm bound for a Fourier multiplier on L2>, gives
$$
\|v_L-v_s\|_2^2
=\int |H(f)-1|^2|\widehat v_s(f)|^2\,df
\le\eta^2\|v_s\|_2^2.
$$
Here the <Fourier transform> uses the frequency-in-hertz convention, so the displayed <Plancherel theorem> has no extra normalization factor. This bound includes waveform changes caused by phase. It applies to the <zero-state response> of the stated <frequency response>. Any additional <homogeneous solution> from initially stored <energy> contributes its own error.
A bench check therefore needs both amplitude ratio and <phase shift>, with probe and cable <capacitance> included in the port model. Magnitude-only observations determine $|H|$ without determining $|1-H|$. The formula describes this chosen passive input network; additional amplifier dynamics require their own <transfer function>.
Back to article page
