Is a capacitor with a greater package number better than a smaller package number capacitor with the same voltage and obviously same capacity?

For example, is it true that a 50 V, 100 nF X7R ±10% 1812 capacitor is better than a 50 V, 100 nF X7R ±10% 0603?


5 Answers 5


Better? That depends what you mean by "better"...

Assuming the same dielectric (X7R)...

The 0603 cap may be better because it's smaller, so it'll fit where you want to, and it will also have lower inductance, also probably cheaper.

The 1812 cap may be better because it's larger, so it'll have more dielectric volume, which could imply a lower loss of capacitance with applied voltage. It could also have lower ESR, which may or may not be "better". However this is handwaving, fortunately some manufacturers give accurate specifications (in this case you have to click the "C-DC Bias" button, it will also show ESR and everything else).

A 1206 cap could be "better" because it's easier to hand solder than a 0402 cap.

If you need high voltage, a longer cap will offer more creepage distance between electrodes.

If there's a lot of vibration, a larger cap may be more prone to cracking.

Also, a component that is in stock is usually "better" than one you can't have.

  • 24
    \$\begingroup\$ a component that is in stock is usually "better" than one you can't have \$\endgroup\$
    – Neil_UK
    Commented Feb 22, 2023 at 16:40
  • \$\begingroup\$ Does heat removal form a benefit for a larger component? More surface area, larger pins means better heat transfer ? \$\endgroup\$
    – Criggie
    Commented Feb 24, 2023 at 3:29
  • 1
    \$\begingroup\$ @Criggie just checked on murata website: 1812 100n 630V has 20mOhm ESR ; 0603 100n 50V has 50mOhm ESR so less heat would be produced too in the larger cap \$\endgroup\$
    – bobflux
    Commented Feb 24, 2023 at 6:58
  • 1
    \$\begingroup\$ A larger footprint is also (obviously) a larger surface area over which to 'pick up' flex in the PCB. i.e. given same strain on the board, two caps in the same position, there'd be more exerted on the larger cap. \$\endgroup\$
    – OJFord
    Commented Feb 24, 2023 at 12:38

Is a capacitor with a greater package number better than a smaller package number capacitor with the same voltage and obviously same capacity?

Let's look at some examples.

Larger capacitors tend to have less sensitivity to DC bias, so the net capacitance you get out is usually more stable with larger packages than smaller:

data from Murata

As you can see in this example taken from Murata, for this vendor, going to 0603 at 5V results in double the loss of capacitance as 1206.

However, as you go to large capacitors, the high frequency performance tends to get worse because of increased inductance:

enter image description here

Again from the same vendor's datasheets, going to 0402 will increase the frequency where the capacitor reaches 1 ohm impedance after resonance by a factor of 3. Thus the smaller package is much better in this regard.

To over generalize: Neither is better. Often you use bigger capacitors for lower frequencies and smaller capacitors for higher frequencies. However, it is also important to check datasheets since especially for DC bias the trend is not universal.


No - that's certainly not universally true.

The effective capacitance of a capacitor is reduced as you increase the DC bias voltage. A capacitor rated for 100nF 50V might only have an effective capacitance of 50nF if you operate it at 25V, for example. Larger package capacitors will often able to better maintain their capacitance at a higher DC bias voltage, although this isn't universally true - you have to check the datasheet and the manufacturer's website.

For decoupling and switching circuits (e.g. switching supplies, microcontrollers, processors, etc.), the physically smaller capacitor may be preferable due to its lower parasitic inductance. This allows for a faster rate of change of current through the capacitor, which is useful in switching applications. This also applies to RF and other high speed applications.

Another thing to consider is the equivalent series resistance (ESR), dissipation factor (DF), and ripple current rating. The ESR and DF will tell you how much power the capacitor will dissipate for a given current flowing through it, either at DC (the ESR) or at a given frequency (the DF). The ripple current rating tells you the upper limit of how much ripple current is allowed to be applied over the capacitor. The more ripple current you apply, hotter the capacitor will get, and its other properties will be derated. If you apply too much, it'll fail (possibly by catching fire).

All of these parameters will vary based on the capacitor's size and cost.

  • \$\begingroup\$ For example for these LCSC components: #C967612 (lcsc.com/product-detail/…) and #C30926 (lcsc.com/product-detail/…) what is relative capacitance in a 5V DC circuit? I can't find this information in the datasheets. \$\endgroup\$
    – Simone
    Commented Feb 22, 2023 at 18:02
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    \$\begingroup\$ DC bias derating curves aren't always shown in the datasheets. You'll often have to go to the manufacturer's website and look for their simulation tools, but the more obscure Chinese vendors on LCSC don't always offer such specs at all, and you'd have to contact them directly to ask. Manufacturers like Kemet, Murata, and Samsung have comprehensive parametric search tools on their site that let you look up performance parameters for specific bias, temperature, etc. on different parts. \$\endgroup\$
    – Polynomial
    Commented Feb 22, 2023 at 18:16
  • \$\begingroup\$ "The effective capacitance of a capacitor is reduced as you increase the DC bias voltage" but this only applies to certain type of ceramic capacitors right? Tantalum, Electrolytic and Class I ceramic capacitors do not suffer from this issue of decreasing capacitance with increasing bias voltage. \$\endgroup\$
    – gyuunyuu
    Commented Jun 29, 2023 at 8:26
  • \$\begingroup\$ @gyuunyuu Correct. Tantalum & aluminium are immune from that. Class I MLCCs use calcium zirconate dielectrics which are paraelectric, whereas Class II MLCCs use barium titanate which is ferroelectric. Class I MLCCs (C0G/NP0) have a much lower dielectric constant, so can only offer small capacitances (typically pF) but they don't suffer DC bias derating. Class II MLCCs (Y5R, X5R, X7R, etc.) have a much higher dielectric constant, so they can offer high capacitance density, but the spontaneous polarisation reversal and magnetic domain wall heating lead to DC bias derating and increased AC noise. \$\endgroup\$
    – Polynomial
    Commented Jun 29, 2023 at 14:44
  • \$\begingroup\$ @gyuunyuu I talked about this subject in depth here: electronics.stackexchange.com/questions/655619/… \$\endgroup\$
    – Polynomial
    Commented Jun 29, 2023 at 14:48

Like I explained in your previous question, X7R caps need to be derated by how much capacitance it will have when some given voltage is applied to it.

If by that metric you ask which one is better, then by experience, in general, the larger package will have more effective capacitance than the smaller package, given that the voltage is approaching the rated voltage.

But that is not supposed to be the only metric or being better. It depends on what you need. For instance, the effective capacitance may be irrelevant as long as it is enough for the job it needs to do. And different caps have different ESR values too. Certainly different ESL too, so they may have wildly different self-resonance frequency, if that is important to know which is better.

Sometimes putting two cheaper caps in parallel is still better or cheaper than having a single good enough capacitor.

So no, neither cap is better unless you know what you need it for.


Such generalizations don't work. So, no. You need to read the actual specifications looking for the aspects that actually matter to your use case. Other than the size, these two capacitors would seem identical, and they might well be in practice. But if your notion of "better" cares about something that is not captured by "50V 100nF X7R ±10%", then you'll definitely will have to study datasheets.

  • \$\begingroup\$ Capacitor data sheets often give generalizations which help to guide you. After you study the data sheets, you also need to do measurements in your lab on the parts you intend to use. \$\endgroup\$
    – qrk
    Commented Feb 22, 2023 at 16:59
  • \$\begingroup\$ @qrk I'd generally agree, but honestly: it's a capacitor. If you're using some property of the transistor you can't read from its specification table and derating curves, chances are you're doing either something very high-speed or something strange; the former is often an instance of the latter ;) \$\endgroup\$ Commented Feb 22, 2023 at 17:01
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    \$\begingroup\$ Yeah, but you are not including the fact that if you select a "50V 100nF X7R ±10%" capacitor, you just need to be aware that it approximately has only 10nF effective capacitance at 50V even without looking at the data sheet. So you need to know if you must have 100nF of capacitance, or if you simply can put a 100nF capacitor if you know 10nF capacitance is enough. Like you have to take temperature ot ESR into account with electrolytics, or surge current ratings of tantalums, etc. That's not strange, that's stuff you need to know in everyday life (as an electrical engineer, anyway). \$\endgroup\$
    – Justme
    Commented Feb 22, 2023 at 17:40
  • \$\begingroup\$ @Justme 100% agree, but I'd say "not suitable if you need 90 nF or more at 48 V" is included in "50V 100nF X7R ±10%". \$\endgroup\$ Commented Feb 22, 2023 at 17:43
  • \$\begingroup\$ Most cap datasheets don't include impedance curves or capacitance loss due to DC bias. So there are significant differences that aren't shown on the datasheet. \$\endgroup\$
    – Drew
    Commented Feb 22, 2023 at 17:54

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