You have 2 constant current arrangements in series - all the volt-drop will be imposed on whichever regulates at the lower current.

The authors of "Step-down Rectifier..." state that their circuit functions as an approximate DC current source when the input RMS voltage is very much greater than the (DC) output voltage. In the case of my circuit, the output and input voltages are about 43V and 120V. That means, I think, that U1 is needed to keep current thru the string of LEDs properly regulated.

 

ronv- The step-down rectifier will be permanently installed in a plastic case, so I'm treating a bleeder resistor as unnecessary. If it were necessary, then that would also mean some additional loss of efficiency.

ian field- The equation that I used (from the paper by Sokal and others) in calculating capacitance of C1 determines the output voltage of the step-down rectifier taken at the DC voltage output of bridge rectifier D1. This is referring to the schematic in post #37. So I would say that first there is voltage step-down/ rectification followed by current regulation by U1 (LM317T). This is also the way the step-down rectifier is presented in the paper, as a means for obtaining a certain reduced DC output voltage.

If you are saying that the step-down rectifier portion of the circuit is actually a current regulator, can you explain?

Regards,
Pete

The wattles dropper between the isolation transformer and the bridge is a current regulator - so is the LM317 with the reg pin tied to the other end of the I sense resistor.

Most of the volt drop in that circuit will be across whichever sets the lowest current - if that turns out to be the LM317, it won't be current limiting for long!