Electronic – What type of power supply is this and how does it work

The opamp acts to make the junction of R6 and R7 the same voltage as Vc. And the R6/R7 junction is a fixed proportion of Vl. So

• Vl = Vc x (R6 + R7) / R7.

Because - the opamp works to set it's input terminals equal when negaive feedback is applied. Non inverting input is set to target voltage by PWM. If R6/R7 feeback point (call this Vf) is too low then opamp output will increase positively to increase Vadj on LM317 which will increase Vl and thus Vf. Opposite applies when Vf is too high.

All the rest is "engineering" (or not :-) )

The opamp is effectively forming a dynamic value of the "datasheet's R2" in series with R4 on the circuit.

R5 drops 1.25V by design of the IC so R4 drops whatever else is required to allow the opamp to balance things as above.

This allows the opamp output to operate at a lower voltage than Vl. This is not especially needed here as the opamp operates from V2 which is at least 3+ Volts above the mzimum level Vl can get to (due to LM317 design) so R4 actually limits the lowest voltage that can be achieved by Vl (due to the division of R5 and R4.)

For best flexibility here R4 = 0 ohms!. Some opamps will not go to full Vdd but here there is ample headroom for almost any opamp as Vs-Vl = LM317 Vdropout_min + 1.25V or about 3.5V. An eg LM324 or LM358 would work OK there.

R3 is something of a mystery - it is not needed to make the circuit work. It adds a positive offset to the PWM voltage which should be unneeded. The circuit designer may have had something special in mind when he added it. Can you provide a link to the original article?

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.