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Ohmforge Passive Filter Designer

RC, RL and RLC low/high/band-pass filters: cutoff, Q, bandwidth, Bode plot, E-series parts.

Analyse: the resistor (first resistor of the RC ladder). Engineering notation: 1k, 4k7.

More options

Inductance for RL and RLC filters, e.g. 10m = 10 mH.

Capacitance, e.g. 100n = 100 nF.

Hz

Cutoff (−3 dB) for first-order filters and the RC ladder, centre/resonant frequency f0 for RLC.

0.7071 gives the flattest (Butterworth) low/high-pass; higher Q = narrower band-pass/notch.

Design keeps this value and computes the rest: the capacitor for RC, RC ladder and RLC filters, the resistor for RL filters.

Resistance across the output; changes the −3 dB points. Empty = unloaded.

Resistors are rounded to this series; capacitors and inductors to E12.

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Try an example: RC low-pass 1 kΩ / 100 nF · Design a 1 kHz RLC band-pass, Q = 5, C = 100 nF · 50 Hz notch (series RLC band-stop)

More about this tool

How it works

Analyses or designs passive filters: first-order RC and RL low- and high-pass, the two-stage RC ladder and series RLC low-pass, high-pass, band-pass and notch. Gives the cutoff or centre frequency, Q, bandwidth and exact −3 dB points (also with a load), Bode magnitude and phase from the transfer function, suggested standard parts – and checks the curve against Ohmforge's AC simulator.

  • Each filter is a ladder of series and shunt impedances (R, jωL, 1/(jωC)). The transfer function H(jω) = Vout / Vin is the product of the sections' ABCD matrices; with a load RL, H = 1 / (A + B / RL). Magnitude is 20·log10|H|, phase is unwrapped.
  • Formula values: fc = 1/(2πRC) or R/(2πL) for first order; f0 = 1/(2π√(LC)) and Q = √(L/C)/R for the series RLC filters (band-pass bandwidth f0/Q); f0 and Q of the two-stage RC ladder from its denominator 1 + s(R1C1 + R2C2 + R1C2) + s²R1C1R2C2.
  • The −3 dB points are solved by bisection on |H| relative to the passband gain (DC for low-pass and notch, high-frequency gain for high-pass, the peak for band-pass), so they stay correct with a load or a resonant peak.
  • Design keeps the part you choose and computes the rest (RC: R = 1/(2πfC); RL: L = R/(2πf); RLC: L = 1/((2πf0)²C), R = √(L/C)/Q; equal-value RC ladder: ωRC = 0.3742 for low-pass). Every computed part is tried at the series value below and above, and the combination closest to the target frequency (then Q) is suggested.
  • Cross-check: the equivalent SPICE netlist is solved with Ohmforge's AC analysis over ±2 decades and the largest magnitude difference is reported.

Ideal components: no inductor winding resistance, no capacitor ESR, no source resistance. Real inductors limit the achievable Q, and component tolerances shift the corner (5 % parts move fc by about 5 %).

Sources

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Related tools

Design and analyse passive RC, RL and RLC filters

Get the cutoff or centre frequency, Q, bandwidth and Bode plot of a passive filter, or design one for a target frequency with standard parts.

What it is for

Passive filters are everywhere: an RC low-pass in front of an ADC, an RC high-pass that blocks DC between audio stages, a series RLC band-pass that picks one tone, a notch that removes 50 Hz hum. The Passive Filter Designer handles ten common topologies: first-order RC and RL low- and high-pass, the two-stage RC ladder, and series RLC low-pass, high-pass, band-pass and band-stop.

Pick a Filter and a Task. Analyse given parts takes your R, L and C values. Design for a target frequency keeps the part you choose (the capacitor, or the resistor for RL filters), computes the others and suggests standard values. An optional load resistance across the output is included in the response.

The tool assumes ideal parts. It does not model inductor winding resistance, capacitor ESR or the output impedance of whatever drives the filter.

Worked example

You need a 1 kHz anti-aliasing low-pass and have 10 nF capacitors. Settings: Filter RC low-pass (1st order), Task design, target 1000 Hz, part you choose 10n, series E24.

The exact resistor is 15.92 kΩ. The tool suggests 16 kΩ, and with it the cutoff lands at 994.7 Hz, −0.528 % from the target. The Bode plots show the flat passband and the −20 dB/decade slope above the corner, and the simulator cross-check reports that Ohmforge's AC analysis of the same circuit agrees within 1.4e-7 dB.

Second case, analysing a hum notch: series RLC band-stop with R = 100 Ω, L = 1 H, C = 10.13 µF. The result is f0 = 50.01 Hz, Q = 3.142, −3 dB edges at 42.68 Hz and 58.59 Hz, so the stop band is 15.92 Hz wide.

How to read the result

  • Cutoff fc or f0 (formula) is the textbook value of the unloaded filter. Q appears for second-order filters.
  • Lower/Upper −3 dB are solved on the real response, including the load. For band-pass filters they are measured from the peak, for low-pass and notch from the DC gain, for high-pass from the high-frequency gain.
  • Bandwidth is upper minus lower −3 dB frequency (the stop band for a notch).
  • Components lists the values used; in design mode next to the exact ones, plus the frequency error of the standard set.
  • The netlist and project download open the filter with the Bode plotter in the workbench.

Tips and common mistakes

  • A load changes the corner: an RC low-pass of 1k/100n with a 10k load moves from 1.592 kHz to 1.751 kHz. Add the load so you see it.
  • The equal-value two-stage RC ladder is not a sharp second-order filter: its Q is 1/3. Buffer the stages if you need a steeper corner.
  • For series RLC low- and high-pass, Q = 0.7071 gives the flattest passband. Higher Q makes a resonant peak.
  • Inductor values come from E12 in design mode, so RLC designs often miss the target by a few percent. Check the error line.
  • Real inductors have resistance that limits Q; a 1 H choke with 50 Ω of copper will not reach Q = 10 with a 10 Ω resistor.

FAQ

Why does the cutoff differ from 1/(2πRC)?

With a load, part of the signal current flows into it, so the corner moves. Unloaded first-order filters match the formula exactly.

How is the simulator check done?

The tool writes the equivalent SPICE netlist, runs Ohmforge's AC analysis over ±2 decades and reports the largest difference in dB from the transfer function.

Can I design for a bandwidth instead of Q?

Use Q = f0 / bandwidth. A 1 kHz band-pass with 200 Hz bandwidth needs Q = 5.

What does the phase plot show?

The phase of Vout relative to Vin, unwrapped so it is continuous. A first-order low-pass reaches −45° at the cutoff.

Guide last updated:

For scripts and AI agents

  • JSON API: POST https://ohmforgefilterdesigner.usefulweb.site/api/v1/calculate (same calculation as this page)
  • Interface description: OpenAPI 3.1 · AI guide: llms.txt
  • MCP server (Streamable HTTP): https://ohmforgefilterdesigner.usefulweb.site/mcp · tool name ohmforge_filter_designer
Example request
curl -s -X POST https://ohmforgefilterdesigner.usefulweb.site/api/v1/calculate \
  -H 'content-type: application/json' \
  -d '{"topology":"rc-lowpass","mode":"analyze","r":"1k","c":"100n"}'
Feedback · ideas & bug reports welcome