The power factor is managed ("corrected" is really the wrong term, although its the common one) by making the current follow the voltage. In your schematic, the bus voltage will be a bit higher than the peaks for the AC waveform. The inductor, FETs, diode, and capacitor form a boost converter. This converter takes the rectified AC input voltage and makes the bus voltage.
If the control system only regulated the output voltage, there would be no PFC happening. What it does instead is regulate the average current thru the diode to be proportional to the instantaneous rectified AC input voltage. Remember that the ideal load from a power factor point of view has the current in phase with the voltage. Another way of looking at it is that load on the AC line needs to look resistive. Just like a real resistor, you want to keep the current proportional to the voltage.
Of course that is at odds with regulating the bus voltage. This is handled by having a fast response to the AC input voltage but a much slower response to regulating the bus voltage. In other words, the AC line still sees a resistance, but the resistance value is slowly changed as needed to keep the bus voltage near its target value.
You can check out my Digital PFC Control writeup for more background on PFC and a way I came up with to keep the current proportional to the voltage without having to measure the current. I've got a patent on that, which also includes using digital computation to control the bus voltage more accurately. With a little computational power, you can know what ripple is caused on the bus due to following the AC line voltage, then use that to determine what changed due to varying demand from the load. This allows adjusting to load changes more quickly than the conventional approach but without defeating the PFC function.
PF := RealP / S -- by definition of power factor PF
S := Vrms * Irms -- by definition of apparent power S
If you have those two waveforms (the instantaneous voltage and instantaneous current waveform), you can
- find Irms, the RMS flow of electrons through the load over a full cycle.
Electric power supply companies prefer to make
Irms as small as possible, so their power lines don't melt from the I2R heating caused by this current.
(They can successfully reduce this number while keeping the real power the same by using very high voltages, mandating power factors closer to 1, etc.)
- find Vrms, the RMS voltage across the load over a full cycle. (In your case, this is pretty close to 120 VAC).
- find the apparent power
S := Vrms * Irms
- find a third waveform: the instantaneous real power waveform. At any one instant, the product -- of the instantaneous voltage and the instantaneous current -- gives the power going into the load at that instant. With some loads -- such as inductive motor loads -- the power waveform alternates between positive (power going into the load) and negative (power coming out of the load).
- find the average real power RealP over a full cycle (the plain average, not the RMS).
- find the power factor PF := RealP / S
As far as I know, that's the only way to calculate the power factor for non-sinusoidal loads.
(All other formulas I've seen for power factor end up doing the same or more calculations, perhaps in some other order with other names for the intermediate terms, or else are wrong).
unnecessary rant
Some textbooks imply that calculating the power factor involves first taking the cosine of something, perhaps something related to zero-crossings. That's doing it wrong. It may happen to give the currect result for perfectly linear loads with ideal sine-wave power, but not for nonlinear loads, or linear loads with our less-than-ideal not-exactly-sine-wave power. Instead, good power monitoring devices use the above equation, which gives the correct power factor even with nonlinear loads or non-sine-wave power or both.
Some textbooks imply that one can calculate the real power by multiplying the power factor times something. That is an unnecessarily round-about method of getting the real power, since the device must have already calculated the real power in order to get the power factor.
None of the formulas for calculating power factor in terms of phase shift will work for you.
Those "phase angle" formulas only apply when both the voltage and the current are sinusoidal.
(I ranted about this before, in places like
Power monitoring 240VAC devices
).
Best Answer
If you have a wattmeter you can take the ratio of real power to apparent power (VA). You might be able to use a current probe, voltage probe and multiply function on an oscilloscope to get true power.
A much cheaper option (accuracy might be open to question) would be to use one of the ubiquitous "Kill a Watt" consumer products (as, I think, @ tgun926 suggested), which actually have a PF indication. The specifications are somewhere between sketchy and non-existent. "0.2% Accuracy" (of what?), but the price is right at less than $20 street price.