How a passive RC band-pass filter works C1 and R1 form a high-pass stage that reduces low frequencies. R2 and C2 form a low-pass stage that reduces high frequencies. The middle band passes with the highest gain.
The two stages connect directly, using only resistors and capacitors. The second stage loads the first, which reduces the peak gain and changes the bandwidth. The calculator includes this interaction and assumes an ideal source and no additional output load.
The center frequency is the geometric mean of the isolated RC corners. The −3 dB band edges are measured relative to the filter’s peak gain and differ from those corners. Changing either R/C pair shifts the response; this simple passive filter has a broad passband and an output smaller than its input.
Inspect the chart, then choose Try in CircuitSim to simulate your values. For a simpler starting point, try the RC low-pass filter calculator or the voltage divider calculator .
Do the math The second RC stage loads the first; the extra loading term accounts for that interaction. Band edges are 3 dB below the peak, not the isolated RC corners.
R1/C1 corner (isolated) R 1 C 1 = 10 k Ω × 100 n F R_1 C_1 = 10\,\mathrm{k\Omega} \times 100\,\mathrm{nF} R 1 C 1 = 10 kΩ × 100 nF f H P = 1 2 π R 1 C 1 f_{HP} = \frac{1}{2\pi R_1 C_1} f H P = 2 π R 1 C 1 1 = 1 2 π × 1 m s = \frac{1}{2\pi \times 1\,\mathrm{ms}} = 2 π × 1 ms 1 ≈ 159.15 Hz R2/C2 corner (isolated) R 2 C 2 = 1 k Ω × 100 n F R_2 C_2 = 1\,\mathrm{k\Omega} \times 100\,\mathrm{nF} R 2 C 2 = 1 kΩ × 100 nF f L P = 1 2 π R 2 C 2 f_{LP} = \frac{1}{2\pi R_2 C_2} f L P = 2 π R 2 C 2 1 = 1 2 π × 100 μ s = \frac{1}{2\pi \times 100\,\mathrm{\mu s}} = 2 π × 100 μ s 1 ≈ 1.59 kHz Center frequency f 0 = f H P f L P f_0 = \sqrt{f_{HP} f_{LP}} f 0 = f H P f L P = 159.15 H z × 1.59 k H z = \sqrt{159.15\,\mathrm{Hz} \times 1.59\,\mathrm{kHz}} = 159.15 Hz × 1.59 kHz ≈ 503.29 Hz Loading contribution R 2 C 1 = 1 k Ω × 100 n F R_2 C_1 = 1\,\mathrm{k\Omega} \times 100\,\mathrm{nF} R 2 C 1 = 1 kΩ × 100 nF f X = 1 2 π R 2 C 1 f_X = \frac{1}{2\pi R_2 C_1} f X = 2 π R 2 C 1 1 = 1 2 π × 100 μ s = \frac{1}{2\pi \times 100\,\mathrm{\mu s}} = 2 π × 100 μ s 1 ≈ 1.59 kHz Bandwidth B = f H P + f L P + f X B = f_{HP} + f_{LP} + f_X B = f H P + f L P + f X = 159.15 H z + 1.59 k H z + 1.59 k H z = 159.15\,\mathrm{Hz} + 1.59\,\mathrm{kHz} + 1.59\,\mathrm{kHz} = 159.15 Hz + 1.59 kHz + 1.59 kHz ≈ 3.34 kHz Upper −3 dB edge f U = B 2 + 4 f 0 2 + B 2 f_U = \frac{\sqrt{B^2 + 4f_0^2}+B}{2} f U = 2 B 2 + 4 f 0 2 + B = 3.49 k H z + 3.34 k H z 2 = \frac{3.49\,\mathrm{kHz} + 3.34\,\mathrm{kHz}}{2} = 2 3.49 kHz + 3.34 kHz ≈ 3.42 kHz Lower −3 dB edge f L = f 0 2 f U f_L = \frac{f_0^2}{f_U} f L = f U f 0 2 = ( 503.29 H z ) 2 3.42 k H z = \frac{(503.29\,\mathrm{Hz})^2}{3.42\,\mathrm{kHz}} = 3.42 kHz ( 503.29 Hz ) 2 ≈ 74.14 Hz Peak voltage gain G p e a k = f L P B G_{peak} = \frac{f_{LP}}{B} G p e ak = B f L P = 1.59 k H z 3.34 k H z = \frac{1.59\,\mathrm{kHz}}{3.34\,\mathrm{kHz}} = 3.34 kHz 1.59 kHz ≈ 0.48 V/V (-6.44 dB) Voltage gain at the selected frequency d = Q ( f 0 f − f f 0 ) , Q = f 0 B d = Q\left(\frac{f_0}{f} - \frac{f}{f_0}\right), \quad Q = \frac{f_0}{B} d = Q ( f f 0 − f 0 f ) , Q = B f 0 ∣ H ( f ) ∣ = G p e a k 1 + d 2 |H(f)| = \frac{G_{peak}}{\sqrt{1+d^2}} ∣ H ( f ) ∣ = 1 + d 2 G p e ak = 0.48 1 = \frac{0.48}{1} = 1 0.48 ≈ 0.48 V/V (-6.44 dB) Output amplitude ∣ V o u t ∣ = ∣ V i n ∣ × ∣ H ( f ) ∣ |V_{out}| = |V_{in}| \times |H(f)| ∣ V o u t ∣ = ∣ V in ∣ × ∣ H ( f ) ∣ = 1 V × 0.48 = 1\,\mathrm{V} \times 0.48 = 1 V × 0.48 ≈ 476.19 mV Explore CircuitSim calculators → 