I'm troubleshooting an old solid-state bass amplifier (Ampeg B-15). The power supply has a 56 VAC transformer tap into a 4-diode full bridge rectifier circuit.

One side of the bridge goes to ground. The output side (power rail) of the bridge has 3 filter capacitors before any other circuitry: two 2500 uF electrolytic capacitors in parallel to ground and a 0.1 uF non-electrolytic in parallel going to ground.

I know the large caps are filtering caps to reduce ripple. What is the function of the 0.1 uF since it theoretically doesn't add any significant capacitance?

  • 3
    \$\begingroup\$ Please see Jorgen's answer here: electronics.stackexchange.com/questions/298798/… \$\endgroup\$
    – winny
    Commented Dec 25, 2017 at 21:41
  • 5
    \$\begingroup\$ The larger caps are not as good at shunting high frequencies, the smaller caps are. \$\endgroup\$
    – Voltage Spike
    Commented Dec 25, 2017 at 21:54
  • \$\begingroup\$ @laptop2d I'm guessing the high frequencies are introduced by the diodes when they "switch" on and off. \$\endgroup\$
    – Bill N
    Commented Dec 25, 2017 at 22:00
  • \$\begingroup\$ @BillN: no, the high frequencies are due to the load. If you had a passive purely resistive load, you'd only need the big cap to stop voltage from drooping much between beaks of the AC you're rectifying. \$\endgroup\$ Commented Dec 26, 2017 at 0:42

3 Answers 3


What you have there isn't two capacitors in parallel. It's more like this:


simulate this circuit – Schematic created using CircuitLab

The small cap is a ceramic type, it has a low series resistance and inductance, so high frequencies can pass it easily. It doesn't need high capacitance because the current in those frequencies is usually low.

The big cap is an electrolytic type, it has a high series inductance and most times a high series resistance, too. For high frequencies, it's similar to an open circuit.

  • 5
    \$\begingroup\$ "It doesn't need high capacitance because the current in those frequencies is usually low." no, it doesn't need high capacitance beause capacitive reactance is inversely proportional to frequency. \$\endgroup\$ Commented Dec 26, 2017 at 0:20
  • 5
    \$\begingroup\$ @Peter Green, if the ceramic had to supply the same ripple current for high frequencies as the electrolytic does for low frequencies, you needed a much bigger ceramic cap. Not 2500µF because the charge/discharge time is much shorter, but still in the µFs. \$\endgroup\$
    – Janka
    Commented Dec 26, 2017 at 0:42

Big electrolytic capacitors have substantial parasitic inductance in series. This makes its operation at high frequencies less than perfect. The amplifier circuit probably does not stand that much of unwanted inductance and becomes unstable. Bypassing the electrolytic with smaller non-electrolytic capacitor helps.


Beware of "coloration" from resonance in that network.

The smaller cap will resonate with the total loop inductance.

0.1uF and 0.1uH (about 4" of inductance/wiring) resonate at 1.6MHz. To dampen, you need sqrt(L /C) = sqrt(1) = 1 ohms of losses at 1.6MHz.

When you have the amplifier working again, examine the VDD with AC-coupled o-scope and look for high-frequency ringing.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.