# Can massive ammounts of 0603 SMD 1uF ceramics displace power caps?

Edit:Warning: there's an error in one of my assumptions in this question, which more or less invalidates it. See Tom Carpenters answer.

I'1m designing a switched mode power supply, and of course need various capacitors in the design.

For instance, to spare the mains from all the noise from the switching action, I'm thinking about building an LC filter with a 3uF capacitor.

So now I've been trying to source 3uF capacitors. Of course it's not difficult at all to source a 3uF 450V capacitor. Polypropylene caps seem to be suitable.

But the cost of a capacitor capable of handling at worst 1A of ripple current, is signficiant (say 5 euro). See for example.

However, 1uF ceramics can be had almost for free, see for instance. 2000 caps for ~13 euro.

If I put the 1uF 100V caps 15 in series, 3 in parallell, I need 45 caps. This will cost approximately 0.3 euro.

Of course I'll need some resistors to balance the caps. Placing the caps at a reasonable distance I calculate that they would all fit within a 12.5mm x 12.5mm box, which means they take up a fraction of the space of the Epcos one.

One possible problem could be that DC bias reduces capacitance of ceramics. But for my application I need approximately 400V, so with 15 in series the 100V 0603 caps are loaded only marginally, and should retain most of their 1uF capacitance.

I'm not really considering doing this, but would it work? Or will the 0.3A+ ripple current kill the little 0603:s?

• If you put capacitors in series, the capacitance goes down - so if you put 15 of the 1uF caps in series you would essentially have a 67nF capacitor. Also, unless the capacitors are identical, the voltage drop of each will not split evenly which may or may not cause issues with the voltage across a given capacitor being higher than its rating. – Tom Carpenter Aug 10 '15 at 13:29
• Darn! Hadn't thought about that. That invalidates the entire idea. Many thanks!! – avl_sweden Aug 10 '15 at 13:31
• Generally the same idea is employed for stuff like tesla coil capacitors (google mmc capacitor) but you need bleeder/equalization resistors. You can design it, do the math, and calculate the costs (not only in components, but also in pcb space and power lost). Then it is very likely the dedicated cap doesn't look so bad anymore. – PlasmaHH Aug 10 '15 at 13:34
• "the cost of a capacitor capable of handling at worst 1A of ripple current is signficiant" -- Then how about paralleling smaller caps? 3x 1µF reduce the ESR to 1/3 and provide 3x the current of a single 1µF; scales to 6x 500nF, 30x 100nF, or whatever provides best cost/benefit ratio for you. – JimmyB Aug 10 '15 at 15:14
• Possibly useful: Temperature and Voltage Variation of Ceramic Capacitors (Maxim IC) and COMPARISON OF MULTILAYER CERAMIC AND TANTALUM CAPACITORS (AVX). Note the AVX doc is significantly more optimistic than the Maxim one (based on Murata data). – The Photon Aug 10 '15 at 16:03

When placing capacitors in series, the overall voltage rating does go up as you have identified. You have to be careful though as if the capacitance is not equal some will end up with a larger voltage drop than others which may push them outside their rated operating voltage.

The trouble is a more fundamental one. Placing capacitors in series causes the capacitance of the chain to be less than each individual one. Essentially the following equation becomes valid:

$$\frac{1}{C_{series}} = \frac{1}{C_1} +\frac{1}{C_2} +\frac{1}{C_3} + \space...$$

If all of the capacitors are the same value, this simplifies down to:

$$C_{series} = \frac{C_{single}}{N}$$

Where in this case $N$ is the number of capacitors in the series chain. So if you were to place $N=15$ of the $1\mathrm{\mu F}$ capacitors in series, this would mean the overall capacitance reduces to $~67\mathrm{nF}$, so you would actually have to place $45$ of these chains in parallel to get the $3\mathrm{\mu F}$ that you require (or to put it another way $675$ capacitors in total).

As PlasmaHH points out, for 450V rating, you only need 5 ideal capacitors in the series chain to get the voltage rating required rather than the 15 you mentioned in the question. The feasibility of this would depend on how the capacitance derates with voltage, and would also be contingent on ensuring the capacitors are correctly balanced with bleed resistors to ensure that none of the chain go out of spec. Adding more would increase the total count, but also increase the resilience of the circuit to voltage fluctuations and capacitor tolerance.

So lets repeat the calculations. If you were to place $N=5$ of the $1\mathrm{\mu F}$ capacitors in series, this would mean the overall capacitance reduces to $~200\mathrm{nF}$, so you would have to place $15$ of these chains in parallel to get the $3\mathrm{\mu F}$ that you require which brings to total requirement down to a minimum of $75$ capacitors in total. You would have to determine if this is cost effective in terms of space and component cost.

• I am not sure where (here and in the question) the 15 comes from, isn't he trying to up from 100V to 450V ? – PlasmaHH Aug 10 '15 at 13:45
• @PlasmaHH Here it comes from it what was asked in the question. You would theoretically only need 5 if they were balanced, so that would actually be considerably less required. – Tom Carpenter Aug 10 '15 at 13:48
• Yes, but ceramic caps lose their capacitance when operated near their rated voltage (!). Also, as the voltage sharing won't perfectly even, it's good to have some margin. But the number 15 was just arbitrary. – avl_sweden Aug 10 '15 at 13:48
• 15 might be a high number. But I think even with 5 in series you would need a huge number in parallell. 5 to offset the series reduction, 3 to get from 1uf to 3 uf and maybe 3 again because of ceramic cap reduction at high voltage. – avl_sweden Aug 10 '15 at 13:52
• And 225 caps would be a bit impractical... :-) – avl_sweden Aug 10 '15 at 13:55