I am working on a 5kW grid-connected inverter where I need to find the DC bus capacitance value. I understand that the capacitor serves 2 purposes: 1) to provide a low impedance path for the high frequency currents from the device, and 2) To reduce voltage ripple in the DC bus.

Some specifications:

DC bus voltage = 800 V

Phase voltage = 240 V

So I first thought of finding out the ripple current that flows into the capacitor, so that it can be used to find the required capacitance for a given voltage ripple specification. I found the equation for the DC bus capacitor current in SPWM inverter from a paper. I have 2 questions, one mathematical and one relating to the design.

Question 1:

Here, \$ \hat{A}_{0n} \$ and \$ \hat{B}_{0n} \$ are zero (proved by the paper for SPWM inverters). The resulting equation has a dc component and a bunch of other components. Now, if I were to find the frequency spectrum of this series, will the magnitude of \$ i_C(t) \$ at frequency \$ m \omega_c + n\omega_o \$ equal to \$ \sqrt{\hat{A}_{mn}^2 + \hat{B}_{mn}^2 }\$? Given that I wanted to find the rms ripple current, I wanted to find the rms value of all the components except the DC term.

Question 2:

After finding the rms ripple current through the capacitor, I used: $$ \frac{i_{ripple, rms}}{C} = \frac{dV}{dt} $$ If the voltage ripple specification (\$ dV \$) is given, can this be rearranged as: $$ C = \frac{dV}{i_{ripple, rms} \cdot f_{sw}} $$ where \$ f_{sw} \$ is the switching frequency?

  • \$\begingroup\$ If you are working on a design then sure, it may be of some interest to some folk to derive a simplistic version of the math but, why aren't you using a simulator to model the circuit and find all the nuances that maths isn't going to show up. \$\endgroup\$
    – Andy aka
    Commented Jul 1, 2023 at 15:08
  • \$\begingroup\$ @Andyaka I am trying to do it on Simulink and am working on that model as well, but thought whether this theoretical design would make sense. \$\endgroup\$ Commented Jul 1, 2023 at 15:31
  • \$\begingroup\$ What are A and B? Just Fourier coefficients, no waveform yet defined? \$\endgroup\$ Commented Jul 1, 2023 at 20:14
  • \$\begingroup\$ @TimWilliams They are Fourier coeficients that depend on a bunch of factors related to the inverter; I am able to evaluate them (there were equations for the coefficients in the paper I mentioned). \$\endgroup\$ Commented Jul 2, 2023 at 1:03

1 Answer 1


If the voltage ripple specification (\$ dV \$) is given, can this be rearranged as: $$ C = \frac{dV}{i_{ripple, rms} \cdot f_{sw}} $$ where \$ f_{sw} \$ is the switching frequency?

No -- you lose the harmonic information when taking RMS (the step for which was not specified, but it will be Parseval's theorem). Therefore the frequency-dependent impedance can't be factored in. Or, this works only for the fundamental (for which Parseval's is trivial). (I also assume you meant a factor of 2π in there. Or maybe not, if the waveform is square.)

Swap the steps around: use the impedance per harmonic first, and then sum over harmonics.

Which, since the harmonics are AC sinusoidal steady-state, you should use that form: \$\frac{V}{I} = \frac{1}{2 \pi F C}\$.

Since SMPS waveforms are generally square to triangle in nature, the harmonics drop off fairly quickly (1/N or 1/N2), so this is a reasonable approximation for design purposes; the actual value may be ∼10% higher.

  • \$\begingroup\$ Do you mean evaluate \$ C_h = \frac{I_h}{2 \cdot \pi \cdot f_h \cdot V_{ripple}} \$ and then sum the capacitor value over the range of harmonics? \$\endgroup\$ Commented Jul 2, 2023 at 1:12
  • \$\begingroup\$ @curious_direwolf I'm just correcting the mathematical method. Presumably you want to derive a relation between Vrms and Irms in terms of C, and then solve for C. (C per harmonic wouldn't be very meaningful.) \$\endgroup\$ Commented Jul 2, 2023 at 2:31

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