Understanding PFC chips

rtarbell

Member
2005-09-16 2:39 am
Greetings all!

A question regarding power factor correction chips for switchmode power supplies (specifically, boost converters):

Many power factor ICs (example: L4981 from ST) include a pin for input voltage feedforward. The designer is supposed to apply a scaled version of the rectified AC input voltage to the feedforward pin.

I understand that the purpose of this feedfoward pin is to help keep the DC output voltage constant if any sudden change in the AC input waveform happens (I.E. a power surge, or brownout).

Question: Why does this feedforward voltage get squared?

For PFC, the duty cycle equation is something like:

Duty Cycle to comparator =
K * Iin(ac)*vout(dc) / (Vff(ac) ^ 2)
where K is a simple constant
 
as far as I know, that pin is to sense where in time AC voltage is on input. You need that, so that you can scale current drawn from mains in almost the same way as if load was only resistance, current same shape and phase as voltage.

That pin is not to keep DC voltage flat, for that you have feedback from output

Why it is squared? can't say, but probably is because mathematical relation between those 4 values
 
The term “power factor correction” is really a misnomer. What we really want to correct is the harmonic content of the current waveform. However, the classic power factor equation (watts / (V x I)) works fairly well in this case so the name stuck. The regulatory standards address harmonic content however and not power factor.

In order to eliminate unwanted harmonics in the current waveform we really want the load to behave like a resistor. This would be possible if the modulator only used a representation of the AC wave and a DC error signal but the gain would be very non-linear with respect to the input voltage. This is where the equation for power (P= V^2 / R) comes into play. If you think of the power factor stage as a resistor placed across the AC line, the value of that resistor needs to increase with respect to the square of the voltage in order to keep the power constant. Multiplying the representation of the AC waveform by the DC error voltage gives us a “resistor” and dividing the resistor by the square of the AC voltage gives us a constant power resistor.

Cheers,
Kevin
 
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