= RC low-pass output loading
{c}
{title2=$H_L(\omega)=K_L/(1+\mathrm i\omega\tau_L)$}
A real <electrical load> can draw <electric current> from the <output>. For a load <resistor> $R_L>0$ across the <capacitor>, <Kirchhoff's current law> instead gives
$$
\frac{v_{\rm in}-v_{\rm out}}R=C\frac{dv_{\rm out}}{dt}+\frac{v_{\rm out}}{R_L}.
$$
Therefore the loaded <continuous-time frequency response> is
$$
H_L(\omega)=\frac{K_L}{1+\mathrm i\omega\tau_L},\qquad
K_L=\frac{R_L}{R+R_L},\qquad
\tau_L=\frac{RR_L}{R+R_L}C.
$$
Loading reduces the eventual gain and shortens the <time constant>. With the constructed values and $R_L=10\,\mathrm{k}\Omega$, $K_L\approx0.81967$, $\tau_L\approx84.754\,\mu\mathrm{s}$, and the corner relative to the loaded low-<frequency> gain moves to $1.8778\,\mathrm{kHz}$. A $3\,\mathrm V$ step now approaches $2.4590\,\mathrm V$. A quicker rise alone would conceal the error in the final voltage.
With the ideal input <voltage source> set to zero, the <output impedance> seen by the <electrical load> is
$$
Z_{\rm out}(\omega)=\frac{R}{1+\mathrm i\omega RC}.
$$
This follows from <Thévenin's theorem>: the <resistor> and <capacitor> are then in parallel. A sufficient small-loading condition is $|Z_L(\omega)|\gg|Z_{\rm out}(\omega)|$ over the <frequencies> of interest. An <oscilloscope> is itself an <electrical load>: its input <electrical resistance> and cable or probe <capacitance> can both change the response. Pure extra parallel <capacitance> increases $C$ and lowers the corner, whereas a finite parallel <electrical resistance> gives the different gain-and-corner changes above.
The definition of <input> also matters. If voltage is prescribed inside a source with series <electrical resistance> $R_s$, the unloaded response to that internal source uses $R+R_s$. If the actual voltage at the filter input terminal is measured, the ratio of output to that measured input still uses $R$. Measuring both terminals avoids mistaking a source setting for the voltage delivered to the filter.
Back to article page
