Electronic – How does this non-isolated circuit create 12VDC from mains supply

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?

Either can work correctly if designed properly. If you have a dumb rectifier supply feeding a 7805, then all the rectifier part needs to do is guarantee the minimum input voltage to the 7805 is met.

The problem is that such a power supply only charges up the input cap at the line cycle peaks, then the 7805 will drain it between the peaks. This means the cap needs to be big enough to still supply the minimum 7805 input voltage at the worst case current drain for the maximum time between the peaks.

The advantage of a full wave rectifier is that both the positive and negative peaks are used. This means the cap is charged up twice as often. Since the maximum time since the last peak is less, the cap can be less to support the same maximum current draw. The downside of a full wave rectifier is that it takes 4 diodes instead of 1, and one more diode drop of voltage is lost. Diodes are cheap and small, so most of the time a full wave rectifier makes more sense. Another way to make a full wave rectifier is with a center tapped transformer secondary. The center is connected to ground and there is one diode from each end to the raw positive supply. This full wave rectifies with only one diode drop in the path, but requires a heavier and more expensive transformer.

A advantage of a half wave rectifier is that one side of the AC input can be directly connected to the same ground as the DC output. That doesn't matter when the AC input is a transformer secondary, but it can be a issue if the AC is already ground-referenced.