How this calculation works
A passive low-pass filter changes both amplitude and phase with frequency. A resistive output load also reduces its DC gain and changes its time constant, so the unloaded one-over-two-pi-RC shortcut does not describe every connected circuit.
DC gain = load/(R + load); pole Hz = 1/[2π(R||load)C]; magnitude = DC gain/sqrt(1 + (frequency/pole)²)A worked example
For 1000 ohms, 100 nF and a 10000-ohm load, loaded DC gain is ten-elevenths. The resistance setting the pole is about 909.091 ohms, giving a pole near 1750.70 Hz. At 1000 Hz the output magnitude is about 0.789 for a one-volt input, with a negative phase shift. Removing the load mathematically would change both the DC level and the pole.
Good to know
Ideal first-order passive response only; no source impedance, parasitics, tolerance, buffering, component ratings or anti-alias compliance. Frequency response is different from an initial-condition charging transient. Use appropriate component and circuit data for a real implementation.
Reference: Texas Instruments — passive RC frequency response (opens in a new tab) ↗
What this calculation answers
A passive low-pass filter changes both amplitude and phase with frequency. A resistive output load also reduces its DC gain and changes its time constant, so the unloaded one-over-two-pi-RC shortcut does not describe every connected circuit.
This page models an ideal zero-impedance input source, a series resistor and a capacitor in parallel with the load at the output. That topology is stated explicitly so the transfer relation can be checked.
The RC time constant charging and discharge calculator follows voltage over time from an initial state. This page instead examines steady sinusoidal gain and phase with an explicit output load.
Use the right input conventions
Capacitance is entered in nanofarads and converted to farads. The output load is resistive and tied to the same reference as the capacitor. Source impedance is excluded rather than silently included in the series resistor.
Input amplitude can use peak, RMS or another consistent sinusoidal convention, provided the output is interpreted in that same convention. The magnitude ratio itself does not require switching conventions.
Interpret the result
The pole frequency is 3.0103 dB below the loaded DC gain. It is not necessarily the point where output is minus three dB relative to input, because the DC gain may already be below one.
The plot uses log10 frequency on its x axis and gain in dB relative to input. It is a modeled response, not a measured Bode plot. Parasitics, tolerances and interaction with later stages can alter real behavior.
Work through the example
For 1000 ohms, 100 nF and a 10000-ohm load, loaded DC gain is ten-elevenths. The resistance setting the pole is about 909.091 ohms, giving a pole near 1750.70 Hz.
At 1000 Hz the output magnitude is about 0.789 for a one-volt input, with a negative phase shift. Removing the load mathematically would change both the DC level and the pole.
Where the model stops
Ideal first-order passive response only; no source impedance, parasitics, tolerance, buffering, component ratings or anti-alias compliance.
Frequency response is different from an initial-condition charging transient. Use appropriate component and circuit data for a real implementation.
A few questions, answered
Why is the loaded pole higher than the unloaded one?
The capacitor sees R in parallel with the load for the pole time constant, which is smaller than R alone.
Is the pole always minus three dB from input?
It is minus 3.0103 dB from the loaded DC gain, which can itself be below unity.