= Reactive-load phase shift in a finite-output-impedance amplifier
{title2=$A_v/A_0=(1+j\omega R_oC_L)^{-1}$}
A \b[reactive-load <phase shift> in a finite-output-impedance amplifier] occurs because the load and <output impedance> form a frequency-dependent divider. In the model of <loaded voltage gain>, assume a positive real output resistance $R_o$, an ideal capacitive load $C_L$ and open-circuit gain $A_0$ approximately constant over the band of interest. With $\omega=2\pi f$,
$$
Z_L=\frac1{j\omega C_L},\qquad
\frac{A_v(\omega)}{A_0}=\frac1{1+j\omega R_oC_L}.
$$
The divider follows from the terminal equivalent described by the university circuit notes.<Thevenin and Norton Equivalents, University of Houston teaching notes>[pp. 1–2, The Idea and Important Notes]{ref}
Putting $\tau=R_oC_L$ gives
$$
\frac{|A_v|}{|A_0|}=\frac1{\sqrt{1+(\omega\tau)^2}},\qquad
\arg A_v-\arg A_0=-\arctan(\omega\tau).
$$
Thus even an ideal <capacitor> can change both amplitude and phase when the source driving it has finite resistance. Its reactive current and energy exchange are described by <complex electrical power>; zero average dissipation in the ideal <capacitor> does not mean zero output-current demand.
For an original example, choose $R_o=100\,\Omega$, $C_L=10\,\mathrm{nF}$ and $A_0=10$. Then $\tau=1\,\mathrm{\mu s}$ and the load pole is $f_p=1/(2\pi\tau)\approx159.2\,\mathrm{kHz}$. At that frequency the gain magnitude is $10/\sqrt2\approx7.071$ and the extra phase lag is $45^{\circ}$. A measured input of $10\,\mathrm{mV}$ root mean square gives about $70.7\,\mathrm{mV}$ root mean square at the <capacitor>, requiring approximately $0.707\,\mathrm{mA}$ root-mean-square capacitive current.
A bench test holds the actual amplifier input fixed while sweeping frequency and compares loaded and unloaded output amplitudes and phases. Include probe <capacitance> in $C_L$. The <group delay introduced by a capacitive amplifier load> also matters when comparing pulse waveforms.
This single-pole calculation assumes the stated constant open-circuit gain and resistive <output impedance>. The complete amplifier may have additional frequency dependence and feedback effects. The manufacturer's ideal feedback-gain examples are defined within their operating assumptions.<Op-Amps, Comparator Circuit by Renesas>[Op-Amp Basics (1) and (2), Figures 2–3]{ref} Check the actual output-current and linear operating range before interpreting a departure from this model as a <capacitor>-only effect.
Back to article page
