# How does one compute the ripple currents seen by an active PFC output capacitor?

Suppose one is designing a single-phase power-factor-corrected power supply, with a known average current load:

The ripple current seen by the output capacitor is critical. If that current is too high, the capacitor will heat up, and its lifespan will be reduced. But how does one compute the ripple seen by this capacitor?

The ripple current seen by the capacitor has two components: the fundamental, and the high-frequency switching. First, the fundamental:

Assuming the PFC to be perfect, the current output through the diode is a rectified sine wave. Over any given 180-degree period:

$$1: I_d = I_{pk}sin \Theta$$

The average load current is known and fixed. The instantaneous current seen by the capacitor is the difference between these currents:

$$2: I_{cap} = I_d - I_{load}$$

We can find the load current in terms of the peak current, by taking the average of a sine wave over a half-period:

$$3: I_{load} = \frac{2I_{pk}}{\pi}$$

Substituting (1) and (3) into (2), then factoring:

$$4: I_{cap} = I_{pk}sin \Theta - \frac{2I_{pk}}{\pi}\\ 5: I_{cap} = I_{pk}(sin \Theta - \frac{2}{\pi})$$

The RMS current seen by the capacitor:

$$6: I_{RMS}=\sqrt{\frac{\int_{0}^{\pi}{I_{cap}^2}d\Theta}{\pi}}\\ 7: I_{RMS}=\sqrt{\frac{\int_{0}^{\pi}{I_{pk}^2(sin \Theta - \frac{2}{\pi})^2}d\Theta}{\pi}}\\ 7: I_{RMS}=I_{pk}\sqrt{\frac{\int_{0}^{\pi}{(sin \Theta - \frac{2}{\pi})^2}d\Theta}{\pi}}\\ 8: I_{RMS}=I_{pk}\sqrt{\frac{\int_{0}^{\pi}{sin^2 \Theta}d\Theta - \int_{0}^{\pi}{\frac{4sin \Theta}{\pi}}d\Theta + \int_{0}^{\pi}{\frac{4}{\pi^2}}d\Theta}{\pi}}\\ 9: I_{RMS}=I_{pk}\sqrt{\frac{\frac{\pi}{2} - \frac{sin{2\pi}}{4} - \frac{0}{2} + \frac{sin{0}}{4}+ \frac{4cos \pi}{\pi} - \frac{4cos 0}{\pi}+ \frac{4}{\pi}}{\pi}}\\ 10: I_{RMS}=I_{pk}\sqrt{\frac{\frac{\pi}{2} - \frac{8}{\pi}+ \frac{4}{\pi}}{\pi}}\\ 11: I_{RMS}=I_{pk}\sqrt{\frac{1}{2} - \frac{4}{\pi^2}}\\$$

Solving (3) for peak current and substituting into (11): $$12: I_{RMS}=I_{load}\frac{\pi}{2}\sqrt{\frac{1}{2} - \frac{4}{\pi^2}}=I_{load}\sqrt{\frac{\pi^2}{8}-1}\\ 13: I_{RMS}\approx.4834 I_{load}$$

The high-frequency switching component is more complex. Starting with this question, we can see that the RMS current seen by the capacitor will vary as the duty and input voltage change across the sine period. Unfortunately, this function becomes exceptionally complex, making an exact integral impractical. Numerical approximations give a ripple of .96 the load current, with voltage, frequency, and inductance having little effect at all practical values.

• It all looks good but is Trms the ripple current? – Andy aka Jul 11 '13 at 17:43
• That covers the line-frequency ripple, but what about the switching-frequency ripple, which is significantly larger? – Dave Tweed Jul 11 '13 at 18:02
• Quite right, I will edit the question. – Stephen Collings Jul 11 '13 at 19:04
• @Andy aka Yes, I_RMS is ripple current seen by the cap. – Stephen Collings Jul 11 '13 at 19:05