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I'm selecting parts to build a 3 to 5V boost DC-DC converter using the LTC3525-5. The board I'm planing to build is based on the typical application shown in page 11 of the aforementioned datasheet, which schematic I reproduce below:

DC-DC boost converter schematic

At the end of page 10, the datasheet says:

The input and output capacitors should be X5R or X7R types, not Y5V.

I'm not familiar to the various types of SMD ceramic capacitors, so my main questions are: Why exactly makes Y5V capacitors not suitable for this application? And why X5R or X7R are suitable?

I've tried to search for the answer elsewhere, but nothing stood out on the sea of numbers and parameters on various datasheets out there. The best hits I got were the following documents:

I suspect that the answer has something to do with ESR and maybe the capacitance varying with voltage on switching applications, but I can't tell exactly what's at play in this particular circuit.

And, the really silly question is: can I use a through-hole aluminum bipolar capacitors instead? If not, why not?

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    \$\begingroup\$ Should does not equal must... \$\endgroup\$
    – Passerby
    Commented Jun 26, 2014 at 21:04

1 Answer 1

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One pretty simple reason is that the tolerance on Y5V tends to be much flakier numerically than X5R or X7R. Typically -20% + 80%

A less obvious and possibly more profound reason is that the change in capacitance versus temperature is very poor on Y5V. Typically at low temperatures and high temperatures the capacitance might halve in value.

DC voltage can change the capacitance of most capacitors a bit but Y5V are really prone to it. Typically at full rated voltage the capacitance might be a fraction of what it is when low voltages are present.

Just open this data sheet and look at the first few graphs. See also this: -

enter image description here

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    \$\begingroup\$ At high or low temperatures they practically disappear from the circuit. They lose capacitance with time too. \$\endgroup\$
    – Spehro Pefhany
    Commented Jun 26, 2014 at 17:25
  • \$\begingroup\$ @SpehroPefhany and Andy - But is this circuit I showed likely to heat up the capacitor enough that it looses that much capacitance? \$\endgroup\$
    – Ricardo
    Commented Jun 26, 2014 at 17:29
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    \$\begingroup\$ You must keep the capacitors close to the chip. The chip will get hot and heat the capacitors. As will the outside world. The capacitor itself will not have much self-heating. There's very little good reason to use Y5V parts- X7R and X5R parts of that value/voltage are very cheap and small. \$\endgroup\$
    – Spehro Pefhany
    Commented Jun 26, 2014 at 17:31
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    \$\begingroup\$ Your circuit output power is about 900mW - assuming the device is 85% efficient means the internal power dissipated is about 131 mW. The SC6 package is rated at about 200 degC per watt so that's a temperature rise of about 26 degrees C. Look at the graphs and convince yourself that it's a silly idea to use Y5V because of the imparted heat from the chip. \$\endgroup\$
    – Andy aka
    Commented Jun 26, 2014 at 17:38
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    \$\begingroup\$ Aluminium bipolar probably won't have low-enough ESR for the job but to be sure you'd have to provide a link to one. \$\endgroup\$
    – Andy aka
    Commented Jun 26, 2014 at 17:39

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