I do not understand exactly how the below formula has been found. The formula allows to calcul the output capacitor (Cout) of a PFC boost converter working in CCM :

enter image description here

Here is the definition of the hold up time if needed, enter image description here

The formula for the capacitor is approximately equal to :

$$C = \frac{Ic*\delta t}{\delta V}$$ (I did not suceed in doing the greec letter "delta")

So the hold up time is 20 ms : $$\delta t = 20 ms$$

Then during the hold up time the output voltage must decrease at the maximum from Vo to Vo(min). So $$\delta V = Vout-Vo(min)$$

So we have at the moment :

$$C = \frac{Ic*HoldupTime}{(Vout - Vout(min))}$$

So Ic would be equal to $$Ic = \frac{2*Pout*(Vout-Vout(min))}{(Vout^2-Vout(min)^2)} = \frac{2*Pout}{Vout+Vout(min)} = \frac{Pout}{\frac{Vout+Vout(min)}{2}} = (approximately) Iout ???$$

But why ? Ic would be equal to Iout ? During the hold up time which is a portion of time during the recitfied line voltage drops, the current will flow into the capacitor and out to the capacitor according to the duty cycle and the switching frequency. Vout is roughly constant even for a PFC, so the average current through the capacitor must be close to 0 more than Iout.

The formula is coming from here https://www.infineon.com/dgdl/an-1166.pdf?fileId=5546d462533600a40153559aabdf1128 (page 9)

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    \$\begingroup\$ There isn't any energy flowing into the capacitor during the hold-up time. The rectified line input is zero. The capacitor is providing all the current and its voltage ramps down accordingly. The capacitor is sized to provide the required current for the required period of time without discharging too deeply to cause the downstream DC/DC converter to lose regulation / trip out due to excessive input current. \$\endgroup\$ Aug 4 '20 at 17:14
  • \$\begingroup\$ You re right. I was wrong about the definition of the hold up time. \$\endgroup\$
    – Jess
    Aug 4 '20 at 17:47

the average current through the capacitor must be close to 0 more than Iout

That is entirely true and why should it be a surprise?

We know this: -

$$I = C\dfrac{dV}{dt}$$

And we know that "v" rises (during charge) and falls (during discharge) but, on average remains at a constant value therefore, the average current into the hold-up capacitor is zero.

Ic would be equal to Iout ?

Not true except when C is providing the current during the hold-up period but, after that period ends, the capacitor is recharged and thus, the average current becomes zero again. Current flows out of the capacitor and current flows back in.

  • \$\begingroup\$ Thank you for your comment :) How do you explain the formula given if Ic is not equal to Iout ? My "detailed" calcul goes into the wrong direction ? \$\endgroup\$
    – Jess
    May 15 '20 at 14:10
  • \$\begingroup\$ @Jess I've just amended that part. \$\endgroup\$
    – Andy aka
    May 15 '20 at 14:12
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    \$\begingroup\$ Thank you I am sorry ! \$\endgroup\$
    – Jess
    May 15 '20 at 14:13
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    \$\begingroup\$ Haha don't apologize @Jess \$\endgroup\$
    – Andy aka
    May 15 '20 at 14:23
  • \$\begingroup\$ Are we done with this @jess \$\endgroup\$
    – Andy aka
    May 16 '20 at 19:52

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