RC Low-Pass Filter Calculator

Introduction to first-order RC low-pass filtering

A series resistor followed by a shunt capacitor, with the output taken across the capacitor, is the most widely deployed filter in electronics. It sits in front of almost every analogue-to-digital converter as an anti-alias filter, it cleans up the pulse-width-modulated output of a microcontroller into a smooth analogue level, it de-glitches sensor lines, and it rolls the high-frequency hiss off an audio buffer. Because the capacitor's impedance falls as frequency rises, the network behaves as a frequency-dependent voltage divider: at low frequencies the capacitor looks like an open circuit and the output tracks the input, while at high frequencies it looks like a short and the output collapses.

Everything that matters about this circuit follows from one complex transfer function, and the whole design job reduces to picking a corner frequency and then choosing an R and C pair that lands on it with parts you can actually buy. This calculator does that end to end. It accepts resistance in ohms, kilohms or megohms and capacitance in picofarads, nanofarads, microfarads or millifarads, because mixing prefixes by hand is where most first-order filter arithmetic goes wrong. It reports the corner both as a frequency in hertz and as an angular frequency in radians per second, so the factor of 2π that separates them can never be lost. It plots magnitude and phase against a logarithmic frequency axis together with the straight-line asymptotes, and it snaps your design onto the IEC 60063 preferred value series so the result is a real bill of materials rather than an unbuildable decimal.

How to use this RC low-pass calculator step by step

1. Enter the series resistance in the R field and pick its unit (Ω, kΩ or MΩ) from the adjacent menu. The value is converted to ohms internally.
2. Enter the shunt capacitance in the C field and pick pF, nF, µF or mF. A 100 nF part and a 0.1 µF part are the same component; either entry gives the same answer.
3. Enter the input signal amplitude in volts. This scales the predicted output amplitude and the time-domain trace but never affects the cutoff, because the network is linear.
4. Enter the test frequency and its unit (Hz, kHz or MHz). This is the single frequency at which gain, phase and output amplitude are reported; the Bode plot always covers four decades either side of the corner.
5. Optionally add a source resistance and a load resistance. Leave the source at 0 Ω and the load blank for the textbook ideal case; fill them in when you want the real, loaded response.
6. Press Calculate. Press Play to animate the time-domain waveform, Reset to restore the defaults, and Download CSV to export the swept response for a spreadsheet or report.

Every field validates independently. Blank, non-numeric, zero and negative entries for R, C and the amplitude are rejected with a specific message rather than being allowed to produce a division by zero, an infinity or a stale number left over from a previous run.

Circuit variables and modelling assumptions

The filter is a series resistor $R$ feeding a shunt capacitor $C$ returned to ground, with the output taken across the capacitor. Units follow NIST SP 811: resistance in ohms (Ω), capacitance in farads (F), frequency in hertz (Hz), and angular frequency in radians per second. The prefix menus apply the SI decimal factors p = 10⁻¹², n = 10⁻⁹, µ = 10⁻⁶, m = 10⁻³, k = 10³ and M = 10⁶ before any arithmetic happens, so the internal computation is always in coherent SI units and a 100 nF part gives exactly the same answer as a 0.1 µF part.

In the time domain the capacitor voltage $v_C$ obeys a first-order linear differential equation driven by the source voltage $v_s$:

Formula: (d v_C) / (d t) = (v_s − v_C) / (R C)

$$\frac{d{v}_C}{dt}=\frac{v_s-v_C}{RC}$$

The product $RC$ carries units of seconds and is the time constant $\tau$. Every other quantity on this page — corner frequency, phase, rise time, roll-off — is a restatement of that one number. The ideal model assumes a source of zero output impedance, a load of infinite input impedance, a linear capacitor with no leakage or dielectric absorption, a resistor that does not drift with temperature, and lumped elements with no lead inductance or stray capacitance. The optional source and load fields relax the first two of those assumptions exactly; the rest are discussed under limitations further down.

Formula set: corner frequency, magnitude and phase

Replace the capacitor by its impedance $Z_C=1/\left(j\omega C\right)$ and the circuit becomes an ordinary voltage divider, so the transfer function is the ratio of the capacitor impedance to the total series impedance:

Formula: H(j ω) = Z_C / (R + Z_C) = 1 / (1 + j ω R C)

$$H\left(j\omega \right)=\frac{Z_C}{R+Z_C}=\frac{1}{1+j\omega RC}$$

The response falls to $1/\sqrt{2}$ of the passband value — half the power — when the imaginary part equals the real part, that is when $\omega RC=1$. That defines the corner, and it is worth writing it twice because the two forms differ by a factor of 2π:

Formula: ω_c = 1 / (R C) rad/s, f_c = ω_c / (2 π) = 1 / (2 π R C) Hz

$${\omega }_c=\frac{1}{RC}\text{ rad/s, }{f}_c=\frac{{\omega }_c}{2\pi }=\frac{1}{2\pi RC}\text{ Hz}$$

This is the single most common mistake in RC filter work. $1/\left(RC\right)$ is the corner in radians per second; $1/\left(2\pi RC\right)$ is the corner in hertz. Quoting the first as though it were the second overstates the cutoff by 6.283 times. The results panel prints both so the distinction stays visible.

Writing $x=f/f_c=\omega /{\omega }_c$ for the normalised frequency, the magnitude and phase of the transfer function are:

Formula: | H(j ω) | = 1 / (sqrt(1 + (f/f_c)^2))

$$\left|H\right(j\omega \left)\right|=\frac{1}{\sqrt{1+{(f/f_c)}^2}}$$

Formula: φ = − arctan(f /f_c)

$$\phi =-\arctan (f/f_c)$$

Both expressions are exact for the ideal network; there is no approximation anywhere in them. The gain in decibels is a voltage ratio, so it uses the factor 20 rather than 10:

Formula: A_dB = 20 log_10 | H | = − 10 log_10 (1 + (f/f_c)^2)

$$A_{dB}=20{\log }_{10}\left|H\right|=-10{\log }_{10}(1+{(f/f_c)}^2)$$

Decibels, roll-off and the −3 dB point versus the asymptote knee

Substituting $f=f_c$ gives $\left|H\right|=1/\sqrt{2}=0.707107$, and $20{\log }_{10}\left(0.707107\right)=-3.0103$ dB. The familiar “−3 dB point” is that exact number rounded to two significant figures, and the phase there is exactly −45°. Those two values are a reliable sanity check on any filter calculation: if a tool does not return −3.01 dB and −45° when the test frequency equals the corner, something is wrong.

Far above the corner, $f/f_c\gg 1$, the magnitude simplifies to $\left|H\right|\approx f_c/f$, so a tenfold rise in frequency costs a factor of ten in amplitude:

Formula: 20 log_10(10) = 20 dB/decade = 20 log_10(2) = 6.0206 dB/octave

$$20{\log }_{10}\left(10\right)=20\text{ dB/decade = }20{\log }_{10}\left(2\right)=6.0206\text{ dB/octave}$$

A first-order section therefore rolls off at −20 dB per decade, which is the same slope as −6.02 dB per octave. Pages that claim “−6 dB per decade” or “−20 dB per octave” have swapped the two: an octave is a doubling and a decade is a tenfold change, and the ratio between the slopes is ${\log }_2\left(10\right)=3.3219$.

The Bode plot drawn by this calculator shows the true magnitude curve together with its two straight-line asymptotes: a flat 0 dB line in the passband and a −20 dB/decade line in the stopband. Those asymptotes intersect exactly at $f_c$, so for a single RC section the knee frequency and the −3 dB frequency are the same frequency. They are emphatically not the same gain: the asymptotes predict 0 dB at the corner while the real response is 3.0103 dB lower, and that 3 dB gap is the largest error the straight-line approximation ever makes. Treating the knee as the point where attenuation begins, rather than the point where it has already reached 3 dB, is a classic way to under-specify an anti-alias filter.

Step response, time constant and 10–90% rise time

Drive the same network with a step of height $V_0$ instead of a sinusoid and the solution of the differential equation is the familiar exponential charge:

Formula: v_C(t) = V_0 (1 − e^−t/τ), τ = R C

$$v_C\left(t\right)=V_0(1-e^{-t/\tau }),\tau =RC$$

The output reaches 10% of the final value at $t=\tau \ln (10/9)$ and 90% at $t=\tau \ln \left(10\right)$, so the 10–90% rise time is the difference:

Formula: t_r = τ ln(9) = 2.1972 R C = ln(9) / (2 π f_c) = 0.3497 / f_c

$$t_r=\tau \ln \left(9\right)=2.1972RC=\frac{\ln \left(9\right)}{2\pi f_c}=\frac{0.3497}{f_c}$$

That derivation is the origin of the ubiquitous bench rule $t_r\approx 0.35/BW$. The calculator prints the exact value, not the rounded rule, and it prints the settling times to 1% and 0.1% ($4.6\tau$ and $6.9\tau$) because those are what actually matter when a filter sits in front of a multiplexed ADC that must settle between channel switches.

Worked example: 1 kΩ and 100 nF in front of an ADC

Take $R=1\text{ kΩ}$ and $C=100\text{ nF}$. Convert first: 1 kΩ = 1000 Ω and 100 nF = 100 × 10⁻⁹ F = 1 × 10⁻⁷ F. Then

Formula: τ = R C = 1000 × 10^−7 = 100 µs

$$\tau =RC=1000\times {10}^{-7}=100\text{ µs}$$

Formula: ω_c = 1 / τ = 10000 rad/s, f_c = 10000 / (2 π) = 1591.55 Hz

$${\omega }_c=\frac{1}{\tau }=10000\text{ rad/s},f_c=\frac{10000}{2\pi }=1591.55\text{ Hz}$$

Note the answer: 1591.55 Hz, roughly 1.59 kHz. If you had used $1/\left(RC\right)$ and called it hertz you would have written 10 kHz, a 6.28× overestimate that would let a large amount of out-of-band noise through to the converter.

Now evaluate the response at a 1 kHz test tone of 1 V amplitude. The normalised frequency is $x=1000/1591.55=0.6283$, so

Formula: | H | = 1 / (sqrt(1 + 0.6283^2)) = 0.84673 → − 1.4451 dB

$$\left|H\right|=\frac{1}{\sqrt{1+{0.6283}^2}}=0.84673\to -1.4451\text{ dB}$$

Formula: φ = − arctan(0.6283) = − 32.14 °

$$\phi =-\arctan \left(0.6283\right)=-32.14°$$

So a 1 V amplitude tone at 1 kHz emerges at 0.8467 V, delayed by 32.14° of its own cycle, which is 89.3 µs. The 10–90% rise time is $2.1972\times 100\text{ µs}=219.7\text{ µs}$. At ten times the corner, 15.915 kHz, the gain is −20.04 dB and the phase is −84.29°: the response has essentially joined its asymptote, which predicts −20.00 dB there, an error of only 0.04 dB.

Finally, add realism. Suppose the driving op-amp has 50 Ω of output impedance and the next stage presents 10 kΩ rather than an open circuit. Entering $R_s=50$ Ω and $R_L=10$ kΩ gives an effective resistance $R_{eff}=(50+1000)\parallel 10000=950.23$ Ω, so the corner rises to 1674.9 Hz and the passband gain drops from 1.000 to 0.90498, a DC insertion loss of −0.867 dB. Both effects are invisible in the textbook formula and both are routine in real hardware.

Comparison table: normalised response and how to interpret it

Because the magnitude and phase depend only on the ratio $f/f_c$, one table describes every first-order RC low-pass ever built. The last column gives the straight-line asymptote so the approximation error is visible.

Read the table as a design chart. Signals you want to keep should sit at $f/f_c\le 0.2$, where the amplitude error is under 0.2 dB and the phase lag is under 12°. Noise you want to kill needs to sit at $f/f_c\ge 10$ before a single RC section buys you 20 dB. That narrow transition band is the fundamental weakness of first-order filtering: between 0.2 and 10 there are almost two decades of frequency in which the filter neither passes cleanly nor blocks usefully. When the specification demands a sharper edge, cascade buffered sections or move to an active Sallen-Key or multiple-feedback topology.

Interpreting a single result follows the same logic. A gain near 0 dB with a small negative phase means the frequency you entered is deep in the passband and the filter is essentially invisible to it. A gain near −3 dB with a phase near −45° means you are sitting on the corner, the worst place for a signal you care about. A gain of −20 dB or lower with a phase approaching −90° means the filter has taken over and the output is dominated by the integrated version of the input.

Choosing IEC 60063 standard component values

Solving $f_c=1/\left(2\pi RC\right)$ for a resistance almost never lands on a value you can buy. IEC 60063:2015 defines the preferred number series: E12 (twelve values per decade, associated with ±10% parts), E24 (twenty-four values, ±5%) and E96 (ninety-six three-digit values, ±1%). The E24 decade is 10, 11, 12, 13, 15, 16, 18, 20, 22, 24, 27, 30, 33, 36, 39, 43, 47, 51, 56, 62, 68, 75, 82, 91, and E12 is every second one of those. The calculator snaps your resistance onto each series and reports the corner you would actually get.

For the worked example, targeting $f_c=1$ kHz with a 100 nF capacitor calls for $R=1591.55$ Ω. The nearest E12 value is 1.5 kΩ, which puts the corner at 1061 Hz (+6.1%); the nearest E24 value is 1.6 kΩ, which gives 994.7 Hz (−0.53%). Capacitors deserve more suspicion than resistors here: a Class 2 ceramic such as X7R can be 15% off its marked value, loses capacitance under DC bias, and drifts with temperature and age, so the corner of a real filter often sits further from the nominal than any resistor tolerance would suggest. For a corner you must hold, use a C0G/NP0 ceramic or a film capacitor and let the resistor take the tolerance.

Assumptions and limitations of the ideal RC model

The closed-form results above are exact for an idealised network, and the gap between that idealisation and a soldered circuit is where filter designs fail. The main limitations are:

• Source impedance. The textbook formula assumes the driving source has zero output impedance. Any real source resistance adds directly to R and lowers the corner. Use the source field to model it.
• Loading. The formula also assumes an infinite load impedance. A finite load in parallel with the capacitor lowers the effective resistance, raises the corner and introduces a DC insertion loss of $R_L/(R_s+R+R_L)$. This is why RC sections cannot simply be chained: each stage loads the one before it, and the cascade is not the product of the individual responses unless buffers separate them.
• Component tolerance and drift. The corner inherits the full tolerance stack of both parts. E12 resistors alone are ±10%, and Class 2 ceramic capacitors add temperature coefficient, DC-bias derating and ageing on top of their initial tolerance.
• Parasitics. Above roughly the self-resonant frequency of the capacitor, equivalent series inductance turns the capacitor inductive and attenuation stops improving; lead and trace inductance and stray board capacitance do the same. The −20 dB/decade slope is not maintained indefinitely.
• Non-ideal resistor and capacitor behaviour. Capacitor leakage, dielectric absorption and equivalent series resistance are all ignored, as is the temperature coefficient of the resistor and its Johnson noise, which for a 1 kΩ resistor is about 4 nV/√Hz at room temperature.
• Linearity and small signals. The model is linear and assumes the source can supply the required current without droop and that no active device downstream is driven into clipping or slew limiting.
• Steady state. The magnitude and phase apply to a steady sinusoid. Immediately after a change of input there is a transient lasting several time constants; the time-domain animation on this page settles the state before it plots, which is why its measured amplitude matches the analytical prediction.

Use the numbers as a design starting point, then confirm the built filter on a network analyser or with a swept-sine measurement before committing to a production board.

Frequently asked questions about RC low-pass filters

Sources and further reading for RC filter design

Sources. The transfer function, corner frequency, magnitude, phase and roll-off used by this calculator were checked against the following primary and institutional references.

• Paul Horowitz and Winfield Hill, The Art of Electronics, 3rd edition, Cambridge University Press — §1.7.9 “RC lowpass filters” (p. 50), §1.7.13 “‘Poles’ and decibels per octave” (p. 52) and §1.3.2 “Signal amplitudes and decibels” (p. 13). The canonical treatment of the −3 dB point and the per-octave slope.
• Anant Agarwal and Jeffrey Lang, 6.002 Circuits and Electronics, Lecture 18 “Filters”, MIT OpenCourseWare, Massachusetts Institute of Technology — derives $V_c/V_i=1/(1+j\omega RC)$ from the impedance divider. ocw.mit.edu
• International Electrotechnical Commission, IEC 60063:2015, Preferred number series for resistors and capacitors, Edition 3.0, 2015-03 — Table 1 gives the E3/E6/E12/E24 two-digit series and Table 2 the E48/E96/E192 three-digit series used by the standard-value snapping. webstore.iec.ch
• Barry N. Taylor and Ambler Thompson, NIST Special Publication 811, Guide for the Use of the International System of Units (SI), National Institute of Standards and Technology — Chapter 4 (SI prefixes and the derived units ohm, farad and hertz) and Chapter 6 (rules for unit symbols) govern the unit handling. nist.gov
• Supporting manufacturer technical documentation: Jim Karki, Active Low-Pass Filter Design, Texas Instruments application report SLOA049, which states that “the attenuation is –3 dB at the cutoff frequency” and that above it “the attenuation is –20 dB/decade/order”. ti.com
• Related tools on this site: the RLC impedance calculator, the RLC damping simulator and the RL circuit current simulator.