= Impedance bridging with capacitive input
\b[Impedance bridging with capacitive input] treats a receiver's input as a resistance in parallel with a capacitance. Even when its DC <input impedance> is large, the capacitive path can draw appreciable <electric current> at higher <frequency>, changing both the amplitude and phase of the observed <electric potential difference>.
For source resistance $R_s$, receiving resistance $R_{\rm in}$ and input capacitance $C_{\rm in}$, use sinusoidal steady-state <phasors> and a linear source model. The receiving <electrical admittance> is $1/R_{\rm in}+\mathrm i\omega C_{\rm in}$, so
$$
H(\omega)=\frac{V_L}{V_s}=\frac1{1+g+\mathrm ix},\qquad
g=\frac{R_s}{R_{\rm in}},\quad x=\omega R_sC_{\rm in}.
$$
Here $V_s$ is the open-circuit source <phasor>; $V_L$ is the loaded terminal <phasor>. The <impedance bridging> objective is $H$ near one, rather than merely a large quoted DC input resistance.
The amplitude and phase responses are
$$
|H|=\frac1{\sqrt{(1+g)^2+x^2}},\qquad
\arg H=-\arctan\frac{x}{1+g}.
$$
An amplitude-only tolerance and a tolerance on the full <complex number> express different requirements. The latter measures
$$
|1-H|=\sqrt{\frac{g^2+x^2}{(1+g)^2+x^2}}.
$$
For positive resistances, capacitance and a chosen $0<\eta<1$, the <capacitive bridging complex-error bandwidth> condition $|1-H|\le\eta$ requires
$$
x^2\le\frac{\eta^2(1+g)^2-g^2}{1-\eta^2}.
$$
A negative numerator means even the DC loading error exceeds the tolerance.
In an original example, $R_s=1\,\mathrm{k\Omega}$, $R_{\rm in}=1\,\mathrm{M\Omega}$ and $C_{\rm in}=100\,\mathrm{pF}$ give $g=0.001$. A one-percent complex-error tolerance permits frequencies only up to about $15.9\,\mathrm{kHz}$. At $100\,\mathrm{kHz}$ the amplitude ratio remains about $0.997$, but the phase is about $-3.59^{\circ}$ and the complex error about $0.0627$. A small amplitude error can coexist with a materially larger waveform-phase error.
High-input-impedance buffers are motivated by just this source-loading problem. Texas Instruments' three-amplifier presentation shows how current drawn from a source with finite <output impedance> degrades accuracy and how input buffers reduce loading.<Instrumentation amplifier (IA) topologies: three-amp by Tamara Alani>[Presentation slides 3–5]{ref} The chosen component values above are not specifications of a particular amplifier. A real receiver may require additional input paths, frequency-dependent source impedance and a model of the connected cable or probe.
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