OpenSpecEngineering

Works out the corner frequency of a first-order RC low-pass or high-pass filter, or the capacitor needed for a target corner, with the nearest standard value. For a band-pass made of a high-pass stage followed by a low-pass stage, it gives the true −3 dB band, center frequency, Q and peak gain. Use it for anti-alias filters, audio tone shaping, sensor smoothing and coupling networks.

Equations

f_c = 1/(2πRC)     C = 1/(2πR·f_c)

R is the resistance, C the capacitance and f_c the cutoff (corner) frequency in hertz. At f_c the output is 3 dB down, about 70.7 % of the input, and beyond it the response rolls off at 20 dB per decade [1]. The same formula holds for the low-pass (R in series, C to ground) and the high-pass (C in series, R to ground). When solving for C, the page also gives the nearest value in the chosen IEC 60063 series [2].

f_L = 1/(2πR₁C₁)     f_H = 1/(2πR₂C₂)

f₀ = √(f_L·f_H)     Q = f₀/(f_L + f_H)

For the band-pass, f_L and f_H are the corners of the high-pass and low-pass stages on their own. f₀ is the center frequency, where the gain peaks. Together the stages form one second-order filter [1]. Its −3 dB edges are f₀·(√(1 + 1/4Q²) ∓ 1/2Q), so the bandwidth is f_L + f_H. The peak gain is 20·log₁₀(f_H/(f_L + f_H)) dB, a little below 0 dB.

Limits and common mistakes

• The band-pass edges are not f_L and f_H. The stages overlap, so the real band is wider, and Q can never exceed 0.5. For a narrow, sharp band use an active second-order filter.
• The results assume the source has low impedance and the load high impedance. In the band-pass, buffer the stages (for example with an op-amp follower) or the second stage loads the first and both corners shift.
• A first-order filter only cuts 20 dB per decade. A signal at twice f_c is reduced by only about 7 dB.
• Ceramic capacitors can lose much of their value with DC bias and temperature. Use C0G/NP0 or film parts where the corner matters.

Worked example

A voice channel needs roughly 300 Hz to 3.4 kHz. The parts on hand give a buffered high-pass stage of 10 kΩ and 47 nF and a low-pass stage of 4.7 kΩ and 10 nF. What band does it really pass?

1. Choose Filter Type Band-pass (two-stage RC).
2. Enter High-pass stage R1 (sets low cutoff) 10 kΩ and High-pass stage C1 47 nF.
3. Enter Low-pass stage R2 (sets high cutoff) 4.7 kΩ and Low-pass stage C2 10 nF, and press Solve. The Solution shows: Lower −3 dB frequency: 285.899 Hz
   Upper −3 dB frequency: 4,010.8 Hz
   Center frequency (f₀): 1,070.83 Hz
   Bandwidth (BW): 3,724.9 Hz
   Quality factor (Q): 0.28748
   Peak gain at f₀: -0.827854 dB
   High-pass stage corner 1/(2πR1C1): 338.628 Hz
   Low-pass stage corner 1/(2πR2C2): 3,386.28 Hz

Check by hand: 1/(2π×10 kΩ×47 nF) = 338.6 Hz and 1/(2π×4.7 kΩ×10 nF) = 3386 Hz. Their sum, 3725 Hz, is the bandwidth, and √(338.6 × 3386) = 1071 Hz is the center. The real band, 286 Hz to 4.01 kHz, is wider than the two corners suggest, as expected for overlapping stages.