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I thought the only thing that mattered was absolute voltage across a capacitor puncturing the dielectric, but there are some that have an AC rating, too. What does this mean? Why does the value change at different frequencies?

I see Vrms vs frequency graphs, this equation

  • \$V_{rating} \geq {1 \over k} \cdot V_{PP} + |V_{min}|\$

etc. but I'm not sure I understand it.

How do you rate a capacitor if it will have both DC and AC superimposed? If, for instance, I'm putting 200 VDC average across a 33 nF capacitor, with a superimposed 70 kHz AC waveform that occasionally reaches 1200 Vpp (so 424 Vrms, 200 + 600 = 800 Vmax, and 200 - 600 = -400 Vmin), then what does it need for DC and AC ratings?

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    \$\begingroup\$ I'd consider the DC as the MAX acceptable voltage across the terminals to play it safe. That means if (AC + DC) > DCMAX, you're exceeding the voltage spec. \$\endgroup\$ – Joel B Nov 7 '11 at 20:56
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Summary

Based on the guidelines in Selection of Capacitors for Pulse Applications
the required voltage rating is liable to be surprising and annoying.

  • Capacitor voltage rating = DC volts + AC component / Kfactor.

  • Kfactor is dependant on frequency and <= 1. Value as per this chart (from above reference).
    At 70 Khz K ~= 0.35 so AC voltage component is multiplied by a factor of 1/0.35 = 2.9 !

  • For polypropylene K ~~= 1.16 - 0.16 x log(f)
    (Numerical values were correct. Formula has been corrected ). (log base 10) - for 10HZ < f < 1 MHz.
    (empirically based on graph below)

eg
at 1 MHz multiply any AC component x ~= 5
at 100 KHz multiply any AC component x ~= 3
at 10 KHz multiply any AC component x ~= 2

For this specific example

enter image description here

  • Kf at 70 kHz ~= 0.35
  • Veffective = Vdc + (Vpeak-Vdc)/kf
  • = 200 + (800-200)/0.35 =~ 2000 Volt capacitor required !!! enter image description here

This is more applicable for pulse applications or very high frequency AC (which your example is a case of), although it's worth noting that at 100 HZ the scaling factor is already down to 80% of the DC capacitance value.

The example graphs you have given are for Polypropylene film dielectric.
Thye numerical values will vary with dielectric type.

The reason given is that the dielectric strength of the film decreases with increasing frequency.

  • The explanation behind the reason, which doesn't need to be known to apply the formulae, starts to get into deep magic and arcane physical properties but appears to relate to the increase of dissipation factor with frequency and the increasing likelihood of internal corona discharge with increasing thickness of material (or "effective thickness" with increasing frequency).

    This interesting (or boring depending on ones interests) document
    Mylar film - Product informationm from Dupont Teijin offers some insights for polyester / Mylar which can be expected to be generally applicable to other plastics. Figure 8 shows increasing dissipation factor with frequency (hence lowering of resistance to applied voltage and corona discharge)

Application of the formula is easier than understanding the reason.

(a) Solution for:
+ve DC voltage with
+ve going pulse
or added AC such that Vmin >= 0V.

This applies to a capacitor with a (say +ve) DC offset and an added +ve going pulse OR DFC with an added AC waveform such that V is always > 0.
For AC offset by a DC component such that the waveform still crosses 0 Volts see (b) below.

  • Calculate a k multiplier value based on frequency.
    From table K <= 1.
    This is a derating factor for the AC part of the waveform.

  • Calculate minimum voltage = Vmin

  • Calculate Vpp = Vmax - Vmin.

  • Calculate effective voltage of the AC component

    Vac effective = Vpp/k.
    (Wghich will always be >= Vpp)

  • Add DC and AC values

    Veffective = Vdc_applies + Vac = Vdc_applied + Vpp/k.

    QED.

(b) Solution for Vdc + Vac such that combined waveform still crosses 0v twice per cycle

  • Vmin = 0

  • Vpp = Vpeak [[= VAC_peak_to_peak/2 + Vdc]]

  • Obtain k from table as above.

  • Veffective = Vpp / k.

In your example case (a) applies.

Vdc = 200V
You report that Vmax = 800V so Vpp = (Vmax - 200) = (800-200) = 600v.
K calculation from referenced WIMA document.

enter image description here

K for 70 kHz =~= 0.35

Veffective = 200 + 600/0.35 = 1914v

2 kV capacitor required !!!!!!!!!!!!!!!!!!!!!!!!!

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    \$\begingroup\$ Did you just say 200 V = 600 V? :) \$\endgroup\$ – endolith Nov 7 '11 at 21:54
  • \$\begingroup\$ I did :-). But should have said.| Vpp = (Vmax - 200) = (800-200) = 600v. \$\endgroup\$ – Russell McMahon Nov 7 '11 at 22:10
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Capacitors have a maximum voltage they can hold as you say, but also have a maximum current they can handle. This is usually referred to as the ripple current spec. Since it's the current that matters, it can also be expressed as a maximum AC voltage at particular frequencies.

In your case, you will have a 1200V p-p 70 kHz sine wave on your capacitor. The fastest rate of change of that signal is at the zero crossing, which is 600V * 2 * Pi * 70kHz = 264 V/µs. The current thru a capacitor is dV/dt * C. Let's use 1 µF for example. 264 V/µs * 1µF = 264 Amps peak, 187 Amps RMS, which is the ripple current the cap must be able to sustain.

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