I am trying to incorporate a smoothing capacitor into a power supply for a model train set. The power supply puts out 18V pulsating DC with a max of 10 amps. The train set draws an average of 5amps and a peak of around 7 amps. I'm guessing that the rectified DC frequency is 120Hz (since regular AC in the US is 60Hz). The DCC protocol (which runs the model trains) operates between 15 and 18 VDC. I found a handy calculator for determining the smoothing cap's value (https://3roam.com/capacitor-smoothing-calculator/) but I'm not very experienced at this and not sure about my logic or my figures. Can anyone tell me if this is right:

Load Current: 7A Frequency: 120Hz Ripple Voltage: 15V

Which calculates to a capacitance of 3888.8888888888887 uF (I presume I'd need to round it up to 3900uF).

Is that correct for the smoothing capacitor?

  • \$\begingroup\$ If the train power source is pulsing as a deliberate attempt to mimic reciprocating steam piston power, it might not be designed to play nice with a capacitor load. \$\endgroup\$
    – Whit3rd
    Commented Sep 13, 2023 at 0:29
  • \$\begingroup\$ I was curious. For those interested, some useful details about DCC (not an overview!) can be found here. For example, it can be found that DCC requires a train to perform full-wave rectification of the track, since the track itself will be alternating the track polarity in order to transmit commands. \$\endgroup\$ Commented Sep 13, 2023 at 2:48
  • \$\begingroup\$ The first question should really be "How smooth of an output do you need?" Select the capacitor size to meet or exceed that requirement. Also, does your present power supply have an 18VDC average or peak output? If it is an average then recognize that the "pulsating DC" peaks will be over 18V. You listed your DCC system as requiring up to 18 VDC, is that an absolute maximum range? Under a light load a large filtering capacitor will tend to charge to the peak of the pulsating voltage. \$\endgroup\$
    – Nedd
    Commented Sep 13, 2023 at 6:37
  • \$\begingroup\$ Careful with assumtions! DCC operates in the kHz range in order to transmit a digital signal with acceptable speed. That lowers the required cap size by at least an order of magnitude. Its also not sinuzuidal but a rectangle wave form - those calculators assume sinuzudial input wave form. \$\endgroup\$
    – Turbo J
    Commented Sep 13, 2023 at 7:24

2 Answers 2


As a rule of thumb, 1000 uF per amp, so you would want at least 7000 uF.

You could come up with an approximation using constant current discharging a capacitor, i.e. \$ \Delta V = {i \ t \over C} \$ which gives about 5.8 volts of ripple at 10000uF, 7 amps and 60 Hz full-wave rectified. This is a conservative number and will actually have less ripple.

You can do a simulation as shown below with LTspice. This shows a ripple of about 4 volts at 10000uF, 7 amps and 60 Hz full-wave rectified. Personally, I would go with something like 20000 uF which will give you a ripple of about 2 volts.

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Load Current: 7A Frequency: 120Hz Ripple Voltage: 15V

The ripple voltage in the equation from the web calculator you used

$$C = \frac{I_{load}}{f \times V_{ripple}}$$

is "the peak-to-peak voltage ripple".

i.e. the high peak voltage minus the low valley voltage

However, you used the average output voltage in your calculation.

If you want the peak to peak ripple voltage to be, say 1V, then plug that into the formula and/or calculator. Note that the equation used by the calculator is not exact, and gets worse with higher ratios of ripple voltage to average voltage.


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