On 28 Sep 2020 at 11:49:53 BST, ""Dave Liquorice""
wrote:

On 28 Sep 2020 10:04:56 GMT, Roger Hayter wrote:

I think the secret is low current consumption. To drop (most of) 240V
at, say, 5mA needs only to dissipate 240 x 5 x 10^-3, that is about 1.2W
and the large resistor can probably do that even air cooled.

Agreed but I think "capacitive droppers" are more effcient.

Maybe so, but I took the OP's description of the module to exclude the
presence of a mains-rated capacitor.



Only few low power components are needed to voltage limit, smooth and
regulate the low voltage supply.

Good regulation probably isn't needed as the load will be known and
thus the voltage determined with component values.

There should be an electrolytic capacitor though, unless anyone can
give a clue how to avoid it.

A capacitive dropper uses a non-polarised capacitor IIRC but normally
a wire ended device rather than surface mount.


--
Roger Hayter