Reduce voltage after AC-DC rectification before step-down

The LM2596 is probably a bit better than what you think. If you look at figure 8 in the data sheet: -

It tells you that for a 1A load (at 25ºC) the drop-out is about 0.9V. Also note that the output is now just slightly out of regulation (Vout = Vreg - 50mV).

If you are running less than 1A this figure will improve.

However, take note of figure 11: -

It is telling you that at 25ºC the minimum operating supply voltage is about 3.6V (and this id for producing a miserly 1.23V on the output).

There are better devices than this I suspect and you need to consider what your maximum load current is when selecting one.

The LP2950 is a linear regulator and if you look at figure 10 it tells you that for a 50 ohm load and 4V inputted you will get about 3.6V coming out. Also look at figure 12 because this gives a different viewpoint in terms of output current.

In the LM317 datasheet you can find and interesting scheme showing how to use two LM317s as a linear regulator with preregulation. The first LM317 maintains the dropout voltage of the second at about 5V.

If you can find a step-down IC with a sense input which works the same as the LM317, you could do the same trick:

The important thing to note is that the LM317 works by maintaining a fixed reference voltage (~1.25V) between OUT and ADJ terminals. Find a step-down chip that keeps a fixed reference voltage between its sense input and its output and you are done (the sense input must draw negligible current).

In that schematics the calculations are done in this way: the current in R1 is set by the reference voltage across ADJ and OUT:

$$ I_{R1} = \dfrac {V_{ref}} {R_1} $$

neglecting the ~40μA absorbed by ADJ \$I_{R1}\$ flows through R2, causing a voltage drop:

$$ V_{R2} = R_2 \cdot I_{R1} = V_{ref} \dfrac {R_2} {R_1} $$

Add the reference voltage across R1, i.e. Vref, and you get the dropout voltage of the second regulator:

$$ V_{DO} = V_{ref} + V_{R2} = V_{ref} \left(1 + \dfrac {R_2} {R_1} \right) = 1.25V \left(1 + \dfrac {720 \Omega} {240 \Omega} \right) = 5V $$

EDIT (to report more findings)

You can get some ideas from the following sources.

This sites sports a circuit that does what you want: LM317 Adjustable 20V, 1.5A Supply with Simple SMPS Tracking Preregulator (it warns the circuit was only simulated and not built).

Some application notes on the subject that can be relevant or interesting:

• High Efficiency Linear Regulators (Jim Williams - Linear Technology).
• Switching Regulator Circuit Collection (John Seago - Linear Technology). Note in particular the pages where switcher preregulators for linear power supplies are shown. For example, this:

The EEVblog video #260 talks about simulating a switching preregulator in the context of Dave Jones μSupply project (there are other videos in that series).