Want more power from this circuit? No problem...

Clearly, the third calculation - (full wave rectified high voltage rail from RMS value) - shown in the previous post tells us that rectifying both sides of the sine wave is beneficial.

Lets see how much current we can squeeze out using full wave rectification.

XC = 1/ 2Pie(FC)
= 1 / 6.283185307 x (50Hz x 0.00000047F)
= 1 / 6.283185301 x 0.0000235
= 6772Ω

Now, there's a 470K bleed resistor in parallel with the cap.

XC = 1 / (1/ 6772) + (1 / 470K)
= 1 / ( 0.000147666 + 0.000002127)
= 6675Ω

High voltage DC rail - (full wave rectification)...

= 240VAC x 1.414
= 339.36VDC

Absolute max current (short circuit)

= 339.36VDC / 6675Ω
= 50.84mA

Absolute max current @ 5V

= (339.36VDC - 7.5VDC) / 6675Ω
= 331.86VDC / 6675Ω
= 49.71mA

Absolute max power @ 5V

= 49.71mA x 5
= 248mW

% More power

(248mW / 156mW) x 100
= 59%

That's quite a significant increase. You could also push things even further by
decreasing XC - use a larger cap or parallel a few together. Practical limitation of such a circuit is about 1W. Takes up too much real estate and defeats the cost-effectiveness of it otherwise.

All told, it's better to use a transformer...

Trent Jackson