What is the formula for BPF op-amp band-pass filter designer (single-op-amp Sallen-Key)?

What engineering notes should I keep in mind when using BPF op-amp band-pass filter designer (single-op-amp Sallen-Key)?

Two design-condition options: ① give the high-pass / low-pass corners fL, fH directly, then f0 = √(fL·fH) and BW = fH − fL; ② give the center frequency f0 and bandwidth BW, then Q = f0/BW. The tool then auto-selects components and derives every R/C value. C1 is chosen automatically: among E6 standard capacitors it picks the value that makes R1 = 1/(π·f0·C1) closest to 33 kΩ, keeping R1/R2/R3 in a common range; R4 is picked so that R5 = R4/(H−1) is close to 10 kΩ. Design step…

BPF op-amp band-pass filter designer (single-op-amp Sallen-Key)

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Single-op-amp Sallen-Key band-pass filter (BPF) designer: given the high-pass / low-pass corner frequencies (fL, fH) or the center frequency and bandwidth (f0, BW), it automatically picks C1 and R4 and derives every R/C value, with standard-value rounding plus slew-rate and GBW checks and a Bode plot.

Key points

The single-op-amp Sallen-Key band-pass filter uses R1 from the input, R2 feeding back from the output, C1 into the non-inverting input, C2 and R3 to ground, and R4/R5 setting the op-amp gain H. Its transfer function is H·s/(R1C2) / [s² + A·s + B]; the design uses k = 2π·f0·C1, H = (6.5 − 1/Q)/3, C2 = C1/2, R1 = 2/k, R2 = 2/(3k), R3 = 4/k, R5 = R4/(H−1), and the center gain A0 = H·Q is set by Q. The design condition can be the high-pass / low-pass corners fL, fH (f0 = √(fL·fH), BW = fH − fL) or the center frequency f0 and bandwidth BW (Q = f0/BW); the tool then derives every R/C value and reports the actual f0, Q, bandwidth and a Bode plot.

Inputs

Auto C1 (so R1 ≈ 33 kΩ) | k = 2π·f0·C1 | H = (6.5 − 1/Q)/3 | C2 = C1/2 | R1 = 2/k, R2 = 2/(3k), R3 = 4/k, R5 = R4/(H−1) (auto R4 so R5 ≈ 10 kΩ) | Q = f0/BW

Design conditionHigh-pass corner fLHzLow end of the passband, −3 dB (high-pass corner)Low-pass corner fHHzHigh end of the passband, −3 dB (low-pass corner)Observed frequency fHzEnter 0 to skip; if > 0, gives the gain and phase at that frequencyInput waveformResistor seriesOp-amp gain-bandwidth productMHzA rule of thumb is GBW ≥ 10·Q·f0Op-amp slew rate SRV/µsInput signal amplitude (peak)VSine peak; output peak = input peak × center gain A0Positive supply V+VNegative supply V−VSingle supply / ground: 0

Calculation model (updates with parameters)

Bode plot (magnitude / phase, updates with parameters)

Center frequency f0	989.6 Hz	target 995 Hz, error -0.54%
Quality factor Q	4.985	target 4.975, error 0.20%
-3 dB passband fL / fH	895.3 Hz / 1.094 kHz	bandwidth BW = 198.5 Hz
Center gain A0	10.793V/V	A0 = H·Q (set by Q, not independently adjustable)
Phase (at center)	0.0°	Second-order band-pass: +90° at low frequency → 0° at center → −90° at high frequency
Phase @ fL / fH	45.0° / -45.0°	Phase at the two −3 dB corners
Op-amp gain H	2.127	H = 1 + R4/R5; target H = (6.5 − 1/Q)/3 = 2.100
Input resistor R1	31.6 kΩ	R1 = 2/k, k = 2π·f0·C1
Feedback resistor R2	11 kΩ	node A → output; R2 = 2/(3k)
Ground resistor R3	63.4 kΩ	non-inverting input → ground; R3 = 4/k
Gain resistors R4 / R5	11 kΩ / 9.76 kΩ	R5 = R4/(H−1)
Capacitors C1 / C2	10 nF / 5 nF	C2 = C1/2 (the 2:1 ratio must be kept)
Output peak	1.079 V	A0 × input peak 100 mV
Required slew rate SR_req	6.71 mV/µs	2π · f0 · Vpk @ 989.6 Hz
Op-amp GBW rule-of-thumb requirement	49.33 kHz	≥ 10·Q·f0; device 10 MHz
Full-power bandwidth FPBW	147.5 kHz	Highest sine frequency the slew rate allows

Engineering notes

Two design-condition options: ① give the high-pass / low-pass corners fL, fH directly, then f0 = √(fL·fH) and BW = fH − fL; ② give the center frequency f0 and bandwidth BW, then Q = f0/BW. The tool then auto-selects components and derives every R/C value.
C1 is chosen automatically: among E6 standard capacitors it picks the value that makes R1 = 1/(π·f0·C1) closest to 33 kΩ, keeping R1/R2/R3 in a common range; R4 is picked so that R5 = R4/(H−1) is close to 10 kΩ.
Design steps: k = 2π·f0·C1, H = (6.5 − 1/Q)/3, C2 = C1/2, R1 = 2/k, R2 = 2/(3k), R3 = 4/k, R5 = R4/(H−1). It requires Q > 0.2857 (i.e. BW < 3.5·f0), otherwise H < 1 and it is unrealizable.
The component ratios C2 = C1/2 and R1:R2:R3 = 6:2:12 fix the relation between Q and the op-amp gain H; changing any ratio alters both Q and gain, so take the values from the formulas.
The center gain A0 = H·Q is set by Q (i.e. by the bandwidth) and cannot be chosen independently; a narrow band (high Q) also raises the center gain, so watch for output clipping. Higher Q is more sensitive to component tolerance and op-amp phase margin, so use 1% or better resistors/capacitors for narrow-band designs.
Square-wave input: after the band-pass it is approximated by its fundamental sine, with peak (4/π)·A0·Vin, and the required slew rate is that of the fundamental slope, 2π·f0·Vpk. To keep the square edges, the filter must pass N harmonics (9 by default), i.e. fH ≥ N·f0, and the op-amp GBW also needs ≥ max(10·Q·f0, N·f0).

FAQ

What is BPF op-amp band-pass filter designer (single-op-amp Sallen-Key) for?: Single-op-amp Sallen-Key band-pass filter (BPF) designer: given the high-pass / low-pass corner frequencies (fL, fH) or the center frequency and bandwidth (f0, BW), it automatically picks C1 and R4 and derives every R/C value, with standard-value rounding plus slew-rate and GBW checks and a Bode plot.
What is the formula for BPF op-amp band-pass filter designer (single-op-amp Sallen-Key)?: Core formula: Auto C1 (so R1 ≈ 33 kΩ) | k = 2π·f0·C1 | H = (6.5 − 1/Q)/3 | C2 = C1/2 | R1 = 2/k, R2 = 2/(3k), R3 = 4/k, R5 = R4/(H−1) (auto R4 so R5 ≈ 10 kΩ) | Q = f0/BW. The tool returns results instantly using this formula and the selected parameters.
What engineering notes should I keep in mind when using BPF op-amp band-pass filter designer (single-op-amp Sallen-Key)?: Two design-condition options: ① give the high-pass / low-pass corners fL, fH directly, then f0 = √(fL·fH) and BW = fH − fL; ② give the center frequency f0 and bandwidth BW, then Q = f0/BW. The tool then auto-selects components and derives every R/C value. C1 is chosen automatically: among E6 standard capacitors it picks the value that makes R1 = 1/(π·f0·C1) closest to 33 kΩ, keeping R1/R2/R3 in a common range; R4 is picked so that R5 = R4/(H−1) is close to 10 kΩ. Design step…