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I am currently tasked with reverse-engineering a buck converter based on the LM2596 adjustable switch. On the PCB that I have, there is a following filter circuit (redrawn by me):

LC-filter

I did some research, and it seems that the two capacitors are needed for better filtering. However, I cannot find any information on how those capacitor values are related.

Thus, the question - how can we calculate those capacitor values if only the switching frequency (150 kHz) is known? Are they really needed for filtering or did I make a mistake?

Thanks in advance!

EDIT: Here is the full schematic: Full schematic

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3 Answers 3

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Large-value electrolytic capacitors are very good for bulk smoothing to produce DC. However adding a small-value poly or ceramic capacitor in parallel is often used to shunt away ("filter") high-frequency ripple as from a switch-mode power supply (SMPS).

It is rather common that those values are empirically derived rather than theoretically calculated.

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  • \$\begingroup\$ So if I derive their values from say, Multisim testing, and present them in a report like that, I will not be shunned upon? \$\endgroup\$
    – Mu3
    Nov 15, 2016 at 17:04
  • \$\begingroup\$ If your audience only wants simulation, sure, why not? However if you are actually designing and building something, don't be surprised if a different value is more effective vs. the simulated prediction. Much is influenced by the Real World. \$\endgroup\$
    – Richard Crowley
    Nov 15, 2016 at 17:11
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    \$\begingroup\$ If the buck converter is integrated, or just has an external switching element, the suggested caps are usually given in the datasheet. \$\endgroup\$
    – Vladimir Cravero
    Nov 15, 2016 at 17:16
  • \$\begingroup\$ @RichardCrowley I will try simulating it and will see which values I get. \$\endgroup\$
    – Mu3
    Nov 15, 2016 at 18:11
  • \$\begingroup\$ As others have observed, when you are talking about high-frequency power circuits, secondary characteristics of the capacitors, especially ESR at particular frequency become much more important than the simple capacitance value. This may be a more complex question than you realize. \$\endgroup\$
    – Richard Crowley
    Nov 15, 2016 at 18:27
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In order to perform a remotely representative simulation you will need to chase down the part numbers of the capacitors used and find the ESR for each.

You will find that the ESR of the smaller ceramic capacitor is much less than that of the electrolytic capacitor, however the capacitance value will be much lower. So you really have two hidden resistances and two capacitors. You will find that the two caps cooperate in reducing the output ripple.

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  • \$\begingroup\$ Thankfully, the capacitors match those suggested by the datashhet, so I should not have a big problem doing that. \$\endgroup\$
    – Mu3
    Nov 15, 2016 at 18:44
  • \$\begingroup\$ The actual manufacturer and part number is important. You may have to go to the manufacturer's website to find the ceramic cap ESR, it may not be on the datasheet. \$\endgroup\$
    – Spehro Pefhany
    Nov 15, 2016 at 18:46
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    \$\begingroup\$ Yes, the LM datasheet mentioned that too. Even if it is there, I will just double-check. \$\endgroup\$
    – Mu3
    Nov 15, 2016 at 18:51
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The LM2596 recommends this: -

An output capacitor is required to filter the output and provide regulator loop stability. Low impedance or low-ESR electrolytic or solid tantalum capacitors designed for switching regulator applications must be used. When selecting an output capacitor, the important capacitor parameters are the 100-kHz ESR, the RMS ripple current rating, voltage rating, and capacitance value. For the output capacitor, the ESR value is the most important parameter.

So, in your example it looks like they have used a lower quality electrolytic shunted by a ceramic to ensure that at high frequencies the LM2596 operates correctly i.e. the output is stable.

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  • \$\begingroup\$ I have added the full schematic, so it is more understandable. I asked not about the buck output, which contains one capacitor, but rather the filters that contain 2. However, your answer also applies here =) Any way to calculate those capacitor values, though? \$\endgroup\$
    – Mu3
    Nov 15, 2016 at 18:19
  • \$\begingroup\$ To calculate the extra capacitor I would need to understand more about the self-resonance of the inductor that feeds it AND the spec the filter was aiming for. \$\endgroup\$
    – Andy aka
    Nov 15, 2016 at 18:22
  • \$\begingroup\$ The filter is meant to reduce the voltage ripple. The inductor is 100uH, and the output capacitor is 200 uF. \$\endgroup\$
    – Mu3
    Nov 15, 2016 at 18:38
  • \$\begingroup\$ But without knowing the SRF of the inductor or what target specification (in the mind or on paper) then this is pointless. \$\endgroup\$
    – Andy aka
    Nov 15, 2016 at 18:42
  • \$\begingroup\$ Unfortunatley, I do not know the SRF of the inductor, since I have to pick a proper one by myself. How would the SRF influence the output voltage ripple, if you don't mind me asking? \$\endgroup\$
    – Mu3
    Nov 15, 2016 at 18:48

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