I've come across some graphs comparing the impedance of a capacitor over frequency and it understandably declines as frequency increases -- up until a certain point. Afterwhich, the impedance begins to increase, like an inductor.


What exactly is going on here? Why do larger capacitors have a more gradual shift from decreasing to increasing impedance while smaller caps have a sharper change?

I'm sure this is something basic but I'm having a hard time finding anything about it.

  • 2
    \$\begingroup\$ Does anyone else feel that "dielectric loss" and "electrode loss" should be swapped? \$\endgroup\$ – Henry Crun Apr 23 '18 at 3:43
  • \$\begingroup\$ Any conductor, including the silvered plates inside a SMT cap, has inductance. I use the rule-of-thumb that 1mm length is 1nanoHency inductance. \$\endgroup\$ – analogsystemsrf Apr 24 '18 at 5:43
  • \$\begingroup\$ No the dielectric loss is high at low frequency and decreases with increasing frequency.While the ohmic part is fixed So at low frequencies the dielectric part dominates while the at high frequencies the ohmic part dominates \$\endgroup\$ – Mohammed Hisham Apr 30 '18 at 19:12

The behavior of a realistic multi-layer ceramic capacitor is determined by its construction. It is made of ceramic pieces with conducting surfaces, which are connected together at collector electrodes.

Unfortunately, every conductor posess some self-inductance, which begins to play dominant role at higher frequencies. One good article about equivalent models of MLCC is presented by Taiyo Yuden corporation, as illustrated below:

enter image description here

For larger-size (more layers) the equivalent circuit grows up (see the article), so the effective characteristics change accordingly.

  • \$\begingroup\$ Incredibly comprehensive, thank you. My only remaining question is: at what point does the inductor-like behaviour overtake the capacitor-like behaviour? Why do capacitors with different capacitances seem to have different points? \$\endgroup\$ – Capn Jack Apr 23 '18 at 0:33
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    \$\begingroup\$ The point where the inductive behavior overtakes the capacitive behavior is the definition, or one of them, of the self-resonant frequency. As for the second question, in short, it's because of the different geometry of different capacitors. Capacitors with longer and thinner metal sections will have a higher inductive component to them, because longer and thinner wires have higher inductance. \$\endgroup\$ – Hearth Apr 23 '18 at 1:01
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    \$\begingroup\$ @CapnJack, a simpler model of the capacitor is an ideal capacitor with a series resistor and a series inductor. The self-resonant frequency occurs at the resonant frequency of the ideal cap and series inductor (which form a tank circuit with near zero impedance at resonance). Once you go above resonance frequency, the series inductor dominates the impedance of the component, and the capacitor impedance is so low as to be negligible. This is a simplification of the more elaborate model provided by Taiyo Yuden. But the basic behavior is the same. \$\endgroup\$ – mkeith Apr 23 '18 at 1:19
  • \$\begingroup\$ @mkeith but then how is this self-resonance frequency point determined? Is there an equation to find the frequency a given capcitance will begin increasing in impedance? Are there more paremeters involved here than just capacitance in determining this point? \$\endgroup\$ – Capn Jack Apr 23 '18 at 1:40
  • \$\begingroup\$ If you can find the detailed datasheet for the specific cap, it will list the SRF. \$\endgroup\$ – mkeith Apr 23 '18 at 3:10

Why do larger capacitors (more capacitance, same package) have a more gradual shift from decreasing to increasing impedance while smaller caps have a sharper change?

The sharpness is the Q of the resonance. (sharper = more Q)

L is a function of the size, so L is roughly constant for a given capacitor body/lead size.

So L is constant, More C = lower resonant frequency

Lower frequency means the reactances XL and XC at resonance are lower.

Q is XL / Rloss. So if the loss R were constant, and we know that XL,XC is lower for bigger C, then Q is lower for bigger C, and the resonance is less sharp.

Quite likely loss R increases with bigger C values (different dielectrics, thinner metalisation, thinner layers = higher field = higher loss) so that makes it worse.

enter image description here

X = reactance (the red and blue dotted lines in the graph).

Z= impedance (complex number) = complex sum of XC+XL+R

Resonance will happen when XC = XL (ie. red and blue lines intersect)

At the left (low frequency) the impedance (Z) is made up of XC (red).

At the right it is all XL (blue)

As inductive and capacitive reactances have opposite signs, they cancel each other out at resonance XC=-XL so XC+XL=0, and you are just left with the Z being the loss Resistance (the black dotted line is the loss resistance part)

So as you can see, at resonance the Z curve = loss R curve

  • \$\begingroup\$ Sorry, what's XL and XC here? \$\endgroup\$ – Capn Jack Apr 23 '18 at 1:41
  • \$\begingroup\$ reactance. see above \$\endgroup\$ – Henry Crun Apr 23 '18 at 2:02

Inductance is a function of length to width (L/W) ratio and wire as thin as circuit traces is about 1 nH/mm .

For low L SMD caps with higher SRF (MHz) they have an L/W ratio <1 rather than the typical 2:1 ratio

When caps increase in value by a longer path length by cascading more and more thin layers of dielectric between conductive layers connected by alternating edges , the result is a lower SRF because \$\omega =\dfrac{1}{\sqrt{(LC)}}\$ so frequency drops with both C and L rising.

The same is true for rolled metallized turns of electrolytic capacitance. Although lower dielectric density parts such as metal film caps tend to be more ideal with lower C and lower ESR due to wide metal film and thus even if L was the same for a similar L/W conductor e-cap the lower C value raises the SRF significantly but at the trade off of a much bigger part.


This used to be called the "dual" but the modern term is "Parasitic Inductance" The stray inductance comes from the capacitor's construction (its plates and leads) the real world equation of the capcitor formula is impedance parasitic

enter image description here

where Z is the impedance of a capacitor exhibiting parasitic inductance (but not exhibiting parasitic resistance). the static esr is not included here, but added to this equation in real world analysis because its resistance is constant in both ac and dc. But for this thread, the formula will be focusing on just the ac components that effect the impedance at different frequencies.

You'll notice (as the op noticed by observation,) that the impedance of the capacitor drops as it reaches its fq (resonate frequency), it reaches the lowest impedance at that frequency. Now as the frequency increases past this resonate point, the capacitor's parasitic inductance start to react to frequency because the dielectric is at a lower impedance state than its stray inductance. The stray inductance causes the impedance to increase due to its inductive reactance increasing.

To give you an example of this, I will post an example from this article: http://www.capacitorguide.com/parasitic-inductance/

Let’s assume an angular frequency of 1Mhz (approx. 6.2·106 rad/s), a capacitance of 0.1 µF and a typical parasitic inductance for ceramic capacitors, approximately 1nH. In the absence of any parasitic effects, the impedance of such a capacitor would be approximately -j·1.591 Ω. If parasitic effects are considered, the impedance is now -j·1.585 Ω. Not a big deal, since the effective impedance is only 0.37% less when a parasitic inductance is present.

However, at larger frequencies, parasitic inductance becomes a bigger problem. Let us now increase the frequency to 10MHz and repeat the calculation. The angular frequency is now approximately 6.2·107 rad/s. In the absence of parasitic effects, the impedance of a 0.1 µF capacitor would be approximately -j·0.1591 Ω. If we introduce parasitic impedance, the impedance is now -j· 0.0963 Ω. The effective impedance is now reduced by 40%! At higher frequencies, this becomes an increasing problem and at some point the impedance becomes positive and the capacitor in facts starts acting like as an inductor

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    \$\begingroup\$ Link-only answers are not very useful, since the link can go away or the content change at any time. \$\endgroup\$ – pipe Apr 23 '18 at 7:41
  • \$\begingroup\$ ok, @Form Review, I'll add some dialog from them with reference. \$\endgroup\$ – drtechno Apr 23 '18 at 13:33
  • \$\begingroup\$ There must be some sort of jealousy. Seriously, we should be above this: Bickering about the way the correct awnser is displayed. If you have the points up, help out and mod the awnser, otherwise the bickering should be silent. \$\endgroup\$ – drtechno Apr 23 '18 at 14:11

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