This RC high-pass filter calculator gives the cutoff frequency from R and C, or the resistor or capacitor for the cutoff you want, rounded to standard values. See the Bode plot, the effect of the source and load, and how much a square wave sags through an AC coupling capacitor.
Optional: gain and phase at this frequency.
Optional: how much a 50% duty square wave sags (tilt) through the filter.
Optional: output resistance of whatever drives the capacitor.
Optional: input resistance of the next stage (parallel with R).
Cutoff frequency (−3 dB)
159.2 Hz
Assumes a low-impedance source and a high-impedance load. If the next stage has an input resistance, enter it as the load: it appears in parallel with R.
Use fc = 1 ÷ (2πRC), with the capacitor in series and the resistor to ground. A 10 kΩ resistor with a 100 nF capacitor gives RC = 1 ms, so fc = 159 Hz. At that frequency the output is 3 dB down (70.7 %) and leads the input by 45°; below it the output falls 20 dB for every tenfold drop in frequency, and DC is blocked completely.
A capacitor in series with the signal and a resistor to ground form a divider whose ratio depends on frequency. The capacitor’s reactance rises as frequency falls, so less of the signal reaches the resistor, and none of a DC level does.
Rearranged for the part you are choosing:
The RC high-pass filter transfer function gives the gain and phase at any frequency f, relative to the passband:
With a source resistance Rs and a load RL across R, the time constant becomes τ = (Rs + R ∥ RL) × C, fc = 1 ÷ (2πτ), and the passband gain is (R ∥ RL) ÷ (Rs + R ∥ RL).
During each flat half period T/2 the output decays as e^(−t/τ). In steady state the level V₁ at the start of a half and V₁e^(−x) at its end, with x = T ÷ 2τ, must add up to the step V at each edge. Solving gives the tilt as a fraction of the half-amplitude V/2:
The tanh form is derived here from the exponential decay. The small-tilt rule has a published counterpart for a single pulse: General Radio’s 1968 note models low-frequency droop with this same RC high-pass and, for a droop D under 30 % on a pulse of duration Td, gives f1 ≈ (0.159 ÷ Td) × (D ÷ 100). Since 0.159 is 1 ÷ (2π) and a square wave’s flat top lasts Td = 1 ÷ (2f), that rearranges to D ÷ 100 ≈ π × f1 ÷ f.
τ = RC = 10,000 × 100 × 10⁻⁹ = 1 ms. fc = 1 ÷ (2π × 0.001) = 159 Hz. The output is 20 dB down at 16 Hz and 40 dB down at 1.59 Hz.
TI’s application note for the TLV320ADCx120 audio ADC sizes the coupling capacitor against the ADC’s input impedance. For a 20 Hz cutoff with 2.5 kΩ, C = 1 ÷ (2π × 20 × 2,500) = 3.18 µF. The nearest E12 value is 3.3 µF, giving fc = 19.3 Hz. A 20 Hz tone is then 2.86 dB down, as it must be near the −3 dB point. To hold 20 Hz within 0.5 dB the cutoff has to drop to 6.99 Hz: C = 9.11 µF, so a 10 µF part.
1 µF into 100 kΩ gives τ = 100 ms and fc = 1.59 Hz. A 1 kHz square wave sags by 0.500 % across each half cycle (π·fc/f predicts 0.500 %). At 50 Hz the tilt grows to 9.99 %. Any square wave above 500 Hz stays under 1 %. After a DC step the output takes 461 ms to settle within 1 %, which is why large coupling capacitors are slow to recover from a power-up or overload.
fc = 1 / (2πRC) for standard values, with R to ground and C in series. Pick the cutoff you need and read off a pair.
| R \ C | 10 nF | 100 nF | 1 µF | 10 µF | 100 µF |
|---|---|---|---|---|---|
| 1 kΩ | 15.9 kHz | 1.59 kHz | 159 Hz | 15.9 Hz | 1.59 Hz |
| 4.7 kΩ | 3.39 kHz | 339 Hz | 33.9 Hz | 3.39 Hz | 339 mHz |
| 10 kΩ | 1.59 kHz | 159 Hz | 15.9 Hz | 1.59 Hz | 159 mHz |
| 47 kΩ | 339 Hz | 33.9 Hz | 3.39 Hz | 339 mHz | 33.9 mHz |
| 100 kΩ | 159 Hz | 15.9 Hz | 1.59 Hz | 159 mHz | 15.9 mHz |
Every first-order high-pass has the same shape; only fc moves it. Multiply your cutoff by the ratio in the first column.
| f / fc | Gain | Output amplitude | Phase (lead) |
|---|---|---|---|
| 0.01× | -40.00 dB | 1.0 % | +89.4° |
| 0.1× | -20.04 dB | 10.0 % | +84.3° |
| 0.5× | -6.99 dB | 44.7 % | +63.4° |
| 1× | -3.01 dB | 70.7 % | +45.0° |
| 2× | -0.97 dB | 89.4 % | +26.6° |
| 5× | -0.17 dB | 98.1 % | +11.3° |
| 10× | -0.04 dB | 99.5 % | +5.7° |
Steady-state tilt of a 50 % duty square wave at a multiple of the cutoff, exact and from the π·fc/f rule of thumb.
| f / fc | Tilt (exact) | π·fc/f |
|---|---|---|
| 10× | 31.2 % | 31.4 % |
| 30× | 10.5 % | 10.5 % |
| 100× | 3.14 % | 3.14 % |
| 314× | 1.00 % | 1.00 % |
| 1000× | 0.314 % | 0.314 % |
To use this as an AC coupling capacitor calculator, solve for the capacitor with R set to the input resistance of the next stage, often a bias resistor in parallel with an amplifier input, and read that resistance from the datasheet rather than assuming. The source resistance adds to it in the time constant and divides the signal. Large values bring their own problems: aluminum electrolytics are polarized and leak, and X5R or X7R ceramics can lose a large share of their capacitance under a DC bias, which raises the cutoff; film or C0G parts hold their value. A big time constant also means a long settling time after power-up, and a resistor above about 1 MΩ to ground lets bias currents set the DC level. When you need a steeper edge than 20 dB per decade, use a second-order LC or active high-pass.
The cutoff moves inversely with C and R, so tolerance matters. Aluminum electrolytics are often ±20 %, so the 3.3 µF part in example 2 could put the cutoff anywhere from 16.1 Hz to 24.1 Hz. Anything else at the node that the calculation leaves out (a probe, a pull-up, a second input) changes R the same way; if the cutoff is critical, simulate the whole circuit.
fc = 1 / (2πRC), the same as the low-pass, with R in ohms, C in farads and fc in hertz. At fc the output is 3.01 dB below the passband (70.7 % of the amplitude) and leads the input by 45°.
159 Hz. RC = 10,000 Ω × 100 nF = 1 ms, and 1 ÷ (2π × 1 ms) = 159 Hz. Below that the output falls by 20 dB per decade, reaching −20 dB at 16 Hz.
Put the cutoff well below the lowest frequency you need, using the input resistance the capacitor drives as R: C = 1 / (2π × fc × R). For audio into a 2.5 kΩ ADC input with a 20 Hz cutoff that is 3.18 µF, so a 3.3 µF part; to keep 20 Hz itself within 0.5 dB, use 10 µF. With fc a tenth of the lowest signal frequency the loss there is only 0.043 dB.
Between edges the input is flat, so the capacitor charges and the output decays toward zero with time constant τ = RC. The sag (tilt) is about π × fc / f of the half-amplitude: under 1 % needs the square wave to be at least 314 times the cutoff.
Yes. A load resistance appears in parallel with R and a source resistance in series with C, so τ = (Rs + R ∥ RL) × C. For example 1 µF with 47 kΩ to ground, a 600 Ω source and a 47 kΩ load gives 6.6 Hz instead of 3.39 Hz. Enter both in the calculator.
20 dB per decade (6 dB per octave) below the cutoff, because it has one pole and a zero at DC. A tenth of the cutoff is about −20 dB and a hundredth about −40 dB. For a steeper edge use a second-order filter such as an LC or Sallen-Key high-pass.
The gain is flat above the cutoff, 3 dB down at fc, and falls 20 dB per decade below it. The phase starts near +90° well below fc, passes +45° at fc and settles toward 0° above it.
The same two parts swapped: in a high-pass the capacitor is in series and the resistor goes to ground, so DC is blocked and high frequencies pass. The cutoff formula is identical, but the phase goes from +90° (well below fc) to 0°, instead of 0° to −90°.
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