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I am designing a PCB layout using the SIC438AED-T1-GE3 buck converter. I am trying to understand what voltage the 'PHASE' pin will have, as it needs a capacitor like this:

bootstrap capacitor

I need to know the voltage rating of the capacitor I will use.

Based on the datasheet:

boostrap return path

Phase pin voltage

It is a return path, but it can also have a voltage up to 35V -aka my input voltage- (what is this AC labeling here?) What will the BOOT pin's voltage be? According to the datasheet it is up to 6V:

boot pin

I guess since I will be powering it to 24V, this capacitor should have at least 24V rating. (I will give the IC 24V input (to Vin) and I will need an output of about 12V (I will be driving a halogen lamp at constant current.)

Of course there is a similar question here and it says I should use a capacitor with a rating of my output voltage, not the input, but I wanted to be sure, since the datasheet says up to 35V, and not up to Vin.

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    \$\begingroup\$ there should be some recommendations on that. Also, you never use voltage rating matching to what voltage you have. If you have 5V line, you go with 10V capacitor (6.3V is usable, but borderline, and for the same price you just go with 10V and never get back to this again). If that line is supposed to be 24V (which I don't know), then the capacitor should be at least 30V. There's voltage fluctuations, there is part ageing and variations and tolerances, and the voltage rating is a "burn" voltage, It's a common thing to take extra quarter to third to voltage you're using there \$\endgroup\$
    – Ilya
    Sep 15, 2021 at 11:05

2 Answers 2

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To answer your question, let's first look at the block diagram:

enter image description here

Initially, when the top NMOS is off and the bottom NMOS is on, since the bottom end of the bootstrap capacitor is connected to the SW node (PH and SW node are the same) thus to the GND, CBOOT charges up to VIN. Therefore the voltage at the BOOT node will be VIN w.r.t. GND.

Then, when the NMOS switches reverse their states, the SW node will see VIN (because the top NMOS is on). With the help of the voltage across CBOOT (= VIN, from the previous state), the voltage at the BOOT node will be 2 x VIN.

So, we can say that that the voltage at the BOOT pin switches between VIN and 2 x VIN. Therefore, the maximum voltage across CBOOT will be VIN.

Please note that you should consider the maximum input voltage plus some margin (e.g. 15-25%) when selecting a bootstrap capacitor as the rated voltage may change with the frequency.

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Depending on what kind of transient current your circuit needs for that bootstrapping...

  • (Under 10mA) A jellybean ceramic capacitor would work fine. The unmarked ones have a 50V rating.

  • (100mA +) you could use a 36V electrolytic (a very common value) as long as the negative transient between the two pins isn't going to exceed 5% of the caps rating. Or you could use a 50V electrolytic and call it a day. 50V at 100uF would probably do fine.

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  • \$\begingroup\$ I am planning for 4A lamp. So 4 amps. \$\endgroup\$ Sep 16, 2021 at 9:24
  • \$\begingroup\$ Yeah a little 50V electrolytic of whatever capacity you have around should be fine (at least 10uF). A jellybean ceramic may not have enough energy to be able to switch the high side properly. Speaking of which, you should look up "high side switching" and "ac power analysis in RC Circuits" for a more complete picture of what's going on here. \$\endgroup\$ Sep 16, 2021 at 9:31
  • \$\begingroup\$ Hm, why a electrolytic will do a better job? (even if the ceramic is of larger capacity?) \$\endgroup\$ Sep 16, 2021 at 9:33
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    \$\begingroup\$ Electrolytic are usually bigger (in capacity and size) than ceramics or other non-polarized types of capacitors. If you can find a ceramic that's bigger than the electrolytics you have around then go for it. \$\endgroup\$ Sep 16, 2021 at 9:34

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