Hi,
a typical L7805 with no load with input at 12 volts ... will use 2 MA +/-
and a typical buck converter at input 12V with no load will use 5 MA +/- plus a ripple
Yes that's a good summary (except that strictly MA is
Mega Amps and I think you mean milli).
It happens that a few years ago I needed a 3 volt supply at a few amps and the exact buck converter you linked in #8 looked to be a nice simple solution. However, in practice it ran
very hot and the chip/board is not really designed for easy heat-sinking. So I thought it was just a typical Chinese "fake" or at least "optimistic" specification and went for a different solution.
However, looking at the manufacturer's specification of the chip they only
expect about 70% efficiency at 3 volts output. Even if the output diode is a Schottky type it will drop around 400 mV, so with a 3v output that's about 15% loss straight away. The output stage is actually an emitter follower (for good reason) which will drop around 1 volt, so more than 10% loss even from a reasonable input voltage like 9v. So actually that module is not too bad, although they have replaced the chip manufacturer's recommended Tantalum electrolytics with normal Aluminium electrolytics and replaced the 5A diode with a 3A (SS34), so it might be rather less efficient than the chip data sheet.
So I recently took the module out of my "sin bin" and actually tested it: At 9 volts, the Quiescent current is 6 mA, rising slightly at lower input voltages and dropping at higher voltages (but not to below 5 mA), so some of the losses are after (or in) the output stage. Probably worth using from a 12 volts input supply if your load is above about 10 mA continuous.
Then I tested for the "break even" point where the input current is the same as the output current (so an efficient non-switching regulator could be just as good). With an accurate 9 volts input, the break-even current measured as 20 mA, but that rises to 40 mA at 7.5 volts input (which might be expected from an Alkaline battery at ~50% discharge). Bear in mind that this break-even point is the
average current drain over the whole time that the system is operating, not just the "peak" when lots of LEDS are alight or a transmitter (RF or IR) is operating. The regulator must of course be capable of these latter peak currents as well.
So, IMHO there won't be many PICaxe applications that would justify the use of a buck converter from a 9 volt battery, particularly a normal PP3 with its limited current capability. However, a
boost converter from a single Lithium cell (3 or 3.7 volts) might be a good solution, provided that the Quiescent is sufficiently low. For example see
this charge/discharge buck/boost battery controller which claims only 50uA in boost (5v output) mode. BUT note that it automatically shuts down after 30 seconds; follow their link to the "continue to output current" product and that appears to require 500 uA (still maybe a good solution though).
Cheers, Alan.