What is '22μF * 2' in the datasheet?

Question

I am reading this datasheet.

And in the first page, there is 22μF * 2.

I can not find capacitor parts of 44μF so I am afraid I am mistaken.

https://www.lcsc.com/products/Multilayer-Ceramic-Capacitors-MLCC-SMD-SMT_313.html https://www.lcsc.com/products/Aluminum-Electrolytic-Capacitors-SMD_418.html

Is it really meaning 44μF?

EDIT: I think I need to understand Output Capacitor Selection part. In that part, I can see The output voltage ripple can be estimated by ~ but don't know how to calculate right capacitor value from this.

I am trying to convert 12V/1.5A to 3.3V 1.5A.

Answer 1

It means two caps in parallel as explained by Tom.

Besides lower ESR/ESL, the other most likely reason is they're already recommending 22µF on the input, so it will be simpler to use the same value on the output rather than adding a BOM item for 47µF.

These days MLCCs have gotten very good, so quite often, for the same model of cap and same physical size, the various voltage ratings are actually the same cap, just marketed under different references. So it makes no sense to lose the quantity discount and have two BOM lines for two components that are the same.

don't know how to calculate right capacitor value from this.

You have to choose the voltage ripple you want on the output, then the datasheet gives the formula to calculate the capacitor value.

If you use the component values on the datasheet example schematic, you will get the performance specified in the datasheet, which is measured for these values.

Note capacitance of MLCCs depends on electric field, which can be approximated by DC voltage divided by the physical volume of the capacitor:

Now suppose you use 22µF 25V capacitors both on the input and output to avoid buying 22µF 6.3V separately for the output.

The actual value of the 22µF capacitor on the input, when biased at 12V DC, will probably be closer to 10µF. But the actual value of the 22µF caps on the output, when biased at 3.3V DC, will be much closer to 22µF.

Now if you think "OK I'll put 0805 cap on the input, but output voltage is lower, only 3V3, so I can use 0402 22µF 6V3 caps" then the much higher field strength in the physically smaller cap means its capacitance is reduced, and although it has lower ESL due to its smaller size, the result is increased ripple due to lower capacitance. So if you shrink the caps, you may end up having to put more of them to get back to the same µF value, which means the smaller caps doesn't end up cheaper than the larger one!

If this is for a hobby project, just get a strip of 100x 10µF 25V 0805 so you get quantity discount, put two on the input and four on the output.

Answer 2

I would expect this means you should place two 22uF capacitors in parallel.

I've seen several similar cases where they draw a single one to declutter the schematic. Sometimes these are also followed up with notes such as "one per pin" in the case of ICs with lots of power pins.

In applications such as this (DC-DC converter), the parallel capacitors are usually used to reduce parasitic inductance, ESR, and/or to increase current handling. The current is shared between the two capacitors, and the contact area is increased hence the lower inductance and ESR.