pcbpcb-designpower supply
Here's what I think the circuit is doing:
My question is, how does the BCM856BS (PNP/PNP matched double transistors) with the AO3415A MOSFET switch (I'm assuming it's for switching) between the two power supplies?
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.
This circuit should do what you want.
The two diodes steer either the output of the boost regulator or the external supply to the load. The typical V\$_{F}\$ of the diodes is 100 mV at a 100 mA load, which should not have much of an effect on your circuitry.
The two diodes are available in surface mount.
Best Answer
That whole circuitry with the PNPs and the PMOS work as a diode.
The PNPs in your circuit work like this input stage to a Raspberry Pi:
Reference
The circuit inside the box is what you have for PNPs and the PMOS. That is called an "ideal diode" circuit. Essentially it will allow current flow from the input to the VIN node in your circuit, but will protect from current going in the other direction. The advantage of this circuit is that the voltage drop across the "ideal diode" much lower than if you were to use a silicon diode and even an Schottky.
You can read the link I referenced for more in-depth explanation of how the circuit works. But in a nutshell, when the voltage on the left side of the circuit is greater than what's on the right, the PMOS will initially conduct through its body diode, but soon after it'll fully turn on because the PNP on the output node side will be cut off, thus the PMOS gate gets pulled down to ground through the resistor, while the source is at ~VIN minus the diode drop. Once it turns, the current flows through the RDS(on) of the PMOS.
If the voltage on the right is greater than the one of the left of the circuit, the body diode is reverse biased, so no current (except for leakage) can flow through that. Also, the PNP on the right side of the circuit turns on, keeping the PMOS gate and source relatively at the same potential, and therefore off.