Worked examples: sizing an audio coupling cap and a speaker-crossover inductor

Example 1 — Audio coupling cap for a 10 kΩ input

A microphone preamp feeds a downstream stage with input impedance $R_\text{in} = 10$ kΩ. We want to AC-couple the two stages with a series capacitor and ensure the response is flat down to 20 Hz (the low end of audible sound). What value of $C$ do we need?

1. Identify the filter. The series cap plus the downstream $R_\text{in}$ to ground form an RC high-pass. Cutoff: $f_c = 1/(2\pi R C)$.
2. Set f_c well below 20 Hz. At the cutoff we're already down 3 dB (≈30% loss in power). To keep 20 Hz “flat” we want $f_c$ roughly a decade below — target 2 Hz.
3. Solve for C.
   $$C = \frac{1}{2\pi R f_c} = \frac{1}{2\pi \cdot 10^4 \cdot 2} \approx 8.0 \,\mu\text{F}$$
   Round up to 10 µF for a safe margin — standard electrolytic value, easy to find.
4. Sanity check at 20 Hz.
   $$|H(20\text{ Hz})| = \frac{\omega RC}{\sqrt{1 + (\omega RC)^2}}$$
   $\omega RC = 2\pi(20)(10^4)(10^{-5}) \approx 12.6$. So $|H| = 12.6/\sqrt{1 + 158} \approx 0.998$, i.e. “0.02 dB down”. Flat to audible ear.

Example 2 — Two-way speaker crossover at 2 kHz

A simple two-way speaker system runs a woofer (4 Ω) behind a series inductor and a tweeter (8 Ω) behind a series capacitor. The goal is a clean split at 2 kHz. Size both reactive parts.

1. Inductor for the woofer (RL low-pass). Cutoff: $f_c = R/(2\pi L)$. With $R = 4$ Ω and $f_c = 2$ kHz:
   $$L = \frac{R}{2\pi f_c} = \frac{4}{2\pi(2000)} \approx 0.318\,\text{mH}$$
   Large air-cored inductor territory — expect a chunky part with a few hundred turns of copper.
2. Capacitor for the tweeter (RC high-pass). Cutoff: $f_c = 1/(2\pi RC)$. With $R = 8$ Ω:
   $$C = \frac{1}{2\pi R f_c} = \frac{1}{2\pi(8)(2000)} \approx 10\,\mu\text{F}$$
   A bipolar electrolytic or MKP is standard.
3. What happens at 2 kHz. Both branches are at their −3 dB point, so each speaker gets 0.707× the source voltage — i.e. half the power. Summed, the two branches reconstruct the full signal energy, just split between the drivers.

Real crossovers use higher-order filters (second order for a 12 dB/oct slope, fourth order for 24 dB/oct) to reduce overlap, but the cutoff formulas are the same.

Example 3 — Crossover snapshot at 2 kHz

Here's Example 2's two-way speaker crossover as a Bode-plot snapshot. Both branches sit at −3 dB right at the 2 kHz cutoff — each driver receives half the power at that frequency, and the two pass-bands combine to cover the whole audio spectrum.