A band-pass filter passes signals within a specific frequency band centred on f₀ and attenuates frequencies above and below. The key parameters are centre frequency f₀ (also called center frequency in American usage), quality factor Q (selectivity), and bandwidth BW (the −3 dB width of the passband). They are related by BW = f₀ / Q. Higher Q = narrower passband = sharper transitions; lower Q = wider passband = gentler transitions. This band-pass filter calculator handles all three common topologies so you can move from a target center frequency and bandwidth to a complete bandpass filter design in one step.
Passive series RLC
A resistor, inductor, and capacitor in series make the simplest band-pass: at f₀ = 1/(2π√LC) the L and C impedances cancel (resonance), leaving just R between input and output → maximum transmission. Off-resonance, the LC combination has high impedance and most of the signal voltage drops across it instead of R. Q = (1/R)·√(L/C) — Q grows when R is smaller relative to the LC characteristic impedance. To observe how a band-pass filter shapes a real audio signal in practice, pair it with the frequency range analyzer to see exactly which frequency content is passed or rejected.
Passive parallel RLC
The dual topology: a parallel LC tank in series with the source resistance, output across the tank. At f₀ the tank has infinite impedance (parallel resonance) → all source voltage appears across the tank → maximum output. Off-resonance the tank impedance is low and the source resistance drops most of the voltage. Q = R·√(C/L) — Q grows when R is large relative to characteristic impedance (opposite of series).
Multiple-Feedback (MFB) active band-pass
The standard op-amp BPF topology and the most practical active filter choice for audio work. Three resistors + two capacitors + one op-amp. The op-amp's inverting input is held at virtual ground; positive feedback through the second capacitor + R₂ creates resonance, and R₃ in the feedback path sets the Q. The MFB is well-suited for moderate Q (up to ~10) with practical components, and can have arbitrary mid-band gain G. For higher Q you need state-variable or biquad topologies. Equations for equal-capacitor design:
- R₁ = Q / (G · ω₀ · C)
- R₂ = Q / ((2·Q² − G) · ω₀ · C)
- R₃ = 2·Q / (ω₀ · C)
The constraint G < 2·Q² is needed to keep R₂ positive. At Q = 5 and G = 1, R₂ is comfortably positive; pushing G near the limit makes R₂ very large and noisy.
The step response of a band-pass filter
Unlike LP and HP, the band-pass step response starts at zero (not 1 like HP) — because the DC component of the step is blocked. The response then rings at frequency f₀ with envelope decaying as exp(−t·ω₀/(2Q)). The higher Q, the longer the ringing lasts. A Q = 5 band-pass at f₀ = 1 kHz will ring for about 5/(π·1000) ≈ 1.6 ms before settling. This is exactly why bell-like resonant filters give that characteristic "ringing" sound when struck with a transient.
Group delay and phase distortion
Every 2nd-order band-pass filter introduces group delay — the time delay experienced by different frequency components as they pass through. At the centre frequency f₀, group delay peaks at Q/(π·f₀). For a narrow-band filter (Q = 10 at 1 kHz) that is roughly 3.2 ms — perceptible as smearing of transients. This is why high-Q band-pass stages are avoided in audio signal chains where transient accuracy matters, and why the pole-zero diagram (a pair of complex conjugate poles near the unit circle) is a useful design tool: poles closer to the unit circle mean higher Q and longer group delay. For audio equalisation applications where you want a peak without audible smearing, keep Q below about 5 and use the frequency comparison tool to verify the before-and-after result by ear.
Bandwidth in octaves vs hertz
Engineers often express filter width in octaves or fractions of an octave rather than Hz — especially in audio and acoustics, because human hearing perceives frequency on a logarithmic scale. The conversion is: N octaves = log₂(f_high / f_low). For a symmetric 2nd-order BPF: N ≈ log₂((1 + 1/(2Q)) / (1 − 1/(2Q))) ≈ 1/Q octaves for Q ≥ 2. So a Q = 1.41 filter has a bandwidth of approximately 1 octave; Q = 2.87 gives ⅓-octave; Q = 4.32 gives ⅕-octave. Graphic equalisers traditionally use fixed ⅓-octave bands (Q ≈ 4.3) which is why their controls sound musically natural.
What's the difference between Q and bandwidth?
They're two ways of expressing the same thing: Q = f₀ / BW. A high-Q filter has a narrow bandwidth relative to its centre frequency. Example: a Q=10 filter at 1 kHz has BW = 100 Hz, passing 950–1050 Hz (approximately). A Q=1 filter at 1 kHz has BW = 1 kHz, passing 500–1500 Hz. Q is dimensionless; BW is in Hz.
When should I use series RLC vs parallel RLC vs MFB?
Passive RLC is simple but requires inductors (bulky, non-ideal at audio). Use it for RF where inductors are easy. Series RLC has output voltage rolling off as we approach resonance from outside (band-pass shape with R in series); parallel RLC is the dual — it peaks at resonance with R as the source impedance. MFB active is the practical choice for audio band-pass (no inductors, op-amp does the work, arbitrary gain). For Q > 10, switch to state-variable or biquad topologies (not covered here).
Why is my computed R₂ negative in MFB mode?
You've violated the topology constraint G < 2·Q². At Q = 1 and G = 3, you'd need 2·Q² = 2 to exceed G = 3 — which fails, making R₂ negative (impossible). Solutions: (a) lower the gain G, (b) raise Q so 2·Q² grows past G. For G = 1, the constraint requires Q > 0.707, which is easy. For G = 5, Q must be > 1.58. For G = 10, Q must be > 2.24. The tool flags this with an "N/A" R₂ value when the constraint is broken.
Why does the band-pass step response start at 0 instead of 1?
A unit step contains DC (low-frequency content) plus an instantaneous edge (high-frequency content). The band-pass filter blocks BOTH — only frequencies near f₀ make it through. So at t=0 there's nothing in the passband and output is 0. As the step "settles" into its DC level, the filter still blocks it, so output stays near 0. The transient ringing you see is the BPF's natural response to the brief moment when the step's frequency content overlapped with the passband. Compare: HP step starts at 1 (the edge passes), LP step rises to 1 (DC passes), BPF stays near 0 (neither passes).
What's the relationship between band-pass and cascaded LP+HP?
For a wide band-pass (BW comparable to f₀), cascading a low-pass at f_high with a high-pass at f_low gives almost identical results — passband from f_low to f_high, 6 dB/octave rolloff on each side (for 1st-order pairs). For a narrow band-pass (high Q, BW << f₀), the cascaded approach falls apart: both filters have to be very steep to define a narrow band, and you'd need much higher-order LP+HP than a single 2nd-order BPF. For Q > 3 the dedicated BPF topology is much more efficient.
Why are my computed inductor values so large?
L scales as 1/(ω₀·f) — for low f₀ at audio (say 1 kHz), inductors are in the milli-henry to henry range, which is bulky and expensive. This is why audio band-pass filters almost never use passive RLC — the MFB or other active topologies replace the inductor with op-amp gain + RC components. For RF (MHz range), inductors are tiny (µH or nH) and passive RLC dominates. The same calculator works at any frequency; just expect impractical L values at low frequencies.
How does this compare to the notch filter calculator?
A notch filter is the complement: it REJECTS frequencies in a narrow band around f₀ and passes everything else. Mathematically, |H_BPF|² + |H_notch|² = 1 (for matched topologies). Use BPF when you want to isolate one frequency band; use notch (in the
Notch Filter Calculator) when you want to remove one specific interfering frequency (mains hum at 50/60 Hz is the classic use case). Both have the same Q and f₀ parameters.
What is a 2nd-order band-pass filter and when do I need a higher-order one?
This calculator designs 2nd-order (two-pole) band-pass filters, which roll off at 6 dB/octave on each side of the passband. That gives a gentle, bell-shaped response. If you need steeper skirts — for example, to reject a nearby interferer while passing a narrow band — you need higher-order topologies: 4th-order gives 12 dB/octave per side, 6th-order gives 18 dB/octave, and so on. Higher-order band-pass filters are typically built by cascading multiple 2nd-order stages (e.g., two MFB sections in series). Chebyshev designs trade passband ripple for sharper rolloff; Butterworth maximises flatness in the passband with no ripple.
What practical Q values are achievable with real op-amps in MFB designs?
Real op-amps have finite gain-bandwidth product (GBW). An MFB band-pass stage at f₀ consumes roughly 2·Q² of the op-amp's available gain at that frequency. For Q = 5 at 1 kHz you need at least 50× gain — any general-purpose op-amp (GBW ≥ 1 MHz) easily handles this. At higher frequencies the op-amp's own rolloff starts to shift f₀ and degrade Q. A practical rule: choose an op-amp with GBW ≥ 100·Q·f₀. For very high Q (> 10–15), state-variable or biquad quad-op-amp topologies are more stable because each op-amp only contributes a portion of the total Q, making the design much less sensitive to op-amp gain variations.
How do I convert the calculated component values to standard resistor and capacitor values?
Calculated values rarely match standard E-series (E24, E96) values exactly. The common workflow: (1) fix the capacitor to the nearest standard value — capacitors have fewer standard values, so pick them first; (2) compute the required resistors from the design equations using the chosen capacitor; (3) select the nearest E24 or E96 resistor values; (4) recalculate the actual f₀ and Q from the chosen component values to see how much they shifted. Small deviations in f₀ (<1%) are usually acceptable. For tight Q tolerances, a trimmer potentiometer in place of R₃ (the Q-setting resistor in MFB) allows fine adjustment after assembly.
How is bandwidth expressed in octaves and what Q gives a one-third-octave band?
For a 2nd-order BPF, bandwidth in octaves N ≈ 1/Q for Q ≥ 2. So Q = 1.41 gives roughly 1-octave bandwidth; Q = 2.87 gives one-third octave; Q = 4.32 gives one-fifth octave. Graphic equalisers traditionally use fixed one-third-octave bands (Q ≈ 4.3), which is why their controls feel musically natural — each band spans the same perceptual width regardless of centre frequency. Enter f₀ and Q = 2.87 into the Series or MFB panel here to design a single one-third-octave stage.