One off cct board creation

I think you are missing the point that it is the instantaneous switch on surge as the C charges up, not the steady state.

Still 31 mA is an awful lot for an LED these days, it was an awful lot 20 30 years ago, when the "rule of thumb" design guideline was 20 mA...

Modern LEDs seem to be somewhat better at the conversion of amps into lumens. Tried to modify a network disc drive case that had a blue power LED on the front that would flood light the room at night. The thing would noticeably glow with the leakage through my, dry, fingers. I can't remember the value of the R I ended up using on the

It's not actually difficult to knock up your own prog for drawing out a Vero layout. With a free CAD prog like DraftSite. Once you've drawn each component it can be saved in a library for later.

Something like this:-

But once you've geared up for producing simple PCBs it's just as easy to do them. If you enjoy that sort of thing.

A company which attends several of the lighting exhibitions I go to is Spirit Circuits, and one of their claims is free prototype boards, and they keep emailing me. The free boards are copper clad and drilled only, no solder mask, printing, gold plating, etc.

I've never used them, but I made a mental note.

Another possibility would be a local maker lab which you could join.

Other cheapo PCB places:

In PCB terms you pay per square cm, so smaller is better.

Theo

not for long :)

most indicator LEDs are 20mA max, so your rms rectified current should be max 14mA.

First pick your resistor to limit inrush at 330v. 330v 33mA = 10k.

Then pick your capacitor to determine rms run current, not forgetting that R and C act at right angles mathematically. If you wanted 12mA on 240v that's total reactance of 20k. Xc squared + Xr squared = 20k squared, so Xc = 17.3k

Xc = 1/ (2 pi f c) so C = 1.9 e-7 = 0.19uF. NPV are 0.1, 0.22uF.

Now check resistor dissipation: 13mA 10k = 130v. P= 130x0.013 = 1.7w. Oops.

Try 0.1uF: Xc = 31.8k Xtotal = root(1024+100) = 33.5k i= v/r = 240/33.5k = 7.16 mA rms Vr = 7.16m x 10k = 71.6v Pr = v i = 71.7 x 7.16m = 0.51 w.

You'll likely find you need less current though, you can calculate for 47nF.

NT

Bare neons with resistor would be better. Poke the tip into a 2mm hole.

NT

I have the 0.47uF and the Diode rectifiers ... what would the resistor need to be ? Interesting that the cct provided features on several project web si6tes

- must be blowing a lot of LED's

I have the 5mm LED's now ... so want to use them

I think cost wise I'm back to veroboard ... it's only a one-off

As my calcs showed you wont get an ok result with 0.47uF.

With 1.5A peak... probably. LEDs often tolerate huge overloads, but not always. I cant see any sensible reason to run them at 30x rated current.

NT

I know Andrew was going to look at this .... hope he has a suggestion. I followed this:

So that is where the cct came from and component values.

Having bought all, except resistors .. keen not to waste what was purchased.