How this calculator works
A high-pass RC filter takes its output across the resistor; an RL version takes output across the inductor. For the inverting active topology, a series input capacitor and input resistor create the high-pass pole. The non-inverting topology buffers an RC high-pass network and uses feedback resistors to set gain. All response calculations assume negligible source impedance, no extra loading and an ideal op-amp. Real op-amp bandwidth and output swing limit the usable passband.
Formula
RC: fc = 1/(2 pi RC). RL: fc = R/(2 pi L). Inverting gain = -Rf/Ri; non-inverting gain = 1+Rf/Rg. |H| = |A| x/sqrt(1+x²); phase = atan(1/x), x = f/fc. Inversion adds 180 degrees before principal-angle wrapping.
Worked example
A 1000 ohm, 1 microfarad RC filter has a cutoff of 159.154943 Hz. At that frequency its passive magnitude is 0.707107 V/V. With 10 ohms and 0.1 H, the RL cutoff is 15.915494 Hz.
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Frequently asked questions
What does cutoff mean?
Magnitude is 1/sqrt(2) of passband gain at cutoff, corresponding to about -3.0103 dB relative to that passband.
Can the output loading change cutoff?
Yes. The ideal equations assume the next stage does not significantly load the filter. Include the effective source and load impedances in a real design.
Does active mean a steeper filter?
These active circuits are still first order. The op-amp provides gain or buffering; adding a single pole gives a 20 dB-per-decade asymptotic slope.