I simulated and built a Class AB amplifier recently, which is powered straight from a +-20V lab power supply. When testing, I noticed that there was a dip in the amplifier's frequency response at around 15kHz.

After some research, I realized this was due to the resonant frequency of the PSU - PSU cable - power capacitor RLC circuit. If I model my 0.5m power cable as a 1uH inductor, and use my 100uF power caps in the equation:

$$f_r = \frac{1}{2\pi\sqrt{LC}}$$

I get a resonant frequency of 15.9kHz, which is mostly spot on. Simulations confirm (with variation of the capacitor value) that this resonance is what I am noticing.

How do professional audio amplifiers solve this problem? By making the power capacitors small, the resonant frequency is increased above the audio range, however there is bad distortion at the output. The capacitors can also be made really big to push this frequency below the audible range, however this requires a massive 100F capacitor for a resonant frequency of 15.9Hz. I don't believe audio amplifiers implement this.

For car amplifiers, pretty long (1m+) cables are run directly from the 12V battery of a car into the amplifier. How are these amplifiers designed to allow a variable length of power cables at their supply?

  • \$\begingroup\$ Feedback gives you a degree of PSRR (power supply rejection ratio). High power car amps have dc/dc converters and local capacitance to negate the effects of the power cable. Nevertheless you can’t escape Ohm’s law. \$\endgroup\$
    – Kartman
    Jun 29, 2022 at 8:01
  • 2
    \$\begingroup\$ 100uF power caps are pretty tiny for any normal power amp. You would typically have 10000uF and often more. And very short and thick cables that would have negligible inductance and resistance. \$\endgroup\$
    – danmcb
    Jun 29, 2022 at 9:29
  • \$\begingroup\$ Why the down-votes? Did I not provide enough information? Was I not clear in what I'm asking? \$\endgroup\$
    – Gary Allen
    Jun 29, 2022 at 10:06
  • \$\begingroup\$ You might think 10000uF would knock the resonant dip down to 1.6 kHz which would be anything but ideal. But re-calculate the Q (as a factor of 100 lower) and decide whether the "dip" would even be observable. \$\endgroup\$
    – user16324
    Jun 29, 2022 at 11:03

1 Answer 1


Your setup has a regulated supply of unknown impedance, followed by inductive wiring and then the onboard caps.

A typical audio amp will either use:

  • Transformer, rectifier, and smoothing caps. These are quite large, and usually on the same PCB as the amp, next to the power devices, making power traces quite short.

  • Switching power supply with smaller output caps.

In both cases you will find several smaller decoupling caps on the board where needed, and the wiring is much shorter and less inductive than in your setup.

Resonance between caps can happen because there is always inductance in the PCB. This usually happens between ceramic caps because the low µF values and low ESR are well suited to make LC resonant tanks with the range of common PCB parasitic inductances. It can also happen between ceramics and electrolytics if the electrolytic ESR is low enough.

Two caps in parallel with an inductor between them form a series RLC circuit. with the following parameters:

  • L = total loop inductance
  • R = total loop resistance
  • C = both capacitors in series

This will ring if its damping factor is too low.

\$ \eta = \frac{R}{2}\sqrt{\frac{C}{L}} \$

So, to avoid ringing, enough resistance is needed. In a high current device like a power amp, it can't be put in series with the power supply because that would drop the voltage. So this resistance is usually the ESR of the capacitor at the end of the PCB trace. Small electrolytics usually have substantial ESR, which does the job very well.

Another reason to avoid large capacitors all over the board is that these will couple power supply ripple and signal-dependent noise in your ground.

Car amps of significant power usually have a regulated boost power supply which increases the voltage, that will somewhat compensate for voltage drop in the wires.


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