Mains indicator LEDs

[JohnD]

Bottom line is, buy a ready made 240v AC indicator, there may be some small enough for your application.

Yes, that's what I'm trying to do.

 

[briwire]

There seems to be plenty available, have you seen those from Farnell, ebay, amazon etc? Will they fit?

 

[JohnD]

There seems to be plenty available, have you seen those from Farnell, ebay, amazon etc? Will they fit?

I have looked at quite a number that are the wrong voltage or the wrong size.

Please tell me if you know of one suitable.

 

[RandomGrinch]

I love LEDs, but I have to say, neon indicators are so perfectly fit for purpose - they can connect to the mains with a single low wattage resistor; have a running voltage of 50V - the resistor isn't taking the full voltage drop; they have a quoted lifetime of around 20'000 hours (and as in BigClives vid, they can be under-run for a longer life); and they have a running current of 0.6mA.
Apart from the limited brightness and colour choice, neons can still be a decent choice.

 

[Deryck Tintagel]

What you need is something like this

Capacitive power supply - Wikipedia

en.wikipedia.org

or this

https://www.nomad.ee/micros/transformerless/index.shtml

 

[JohnD]

What you need

But not what I want.

 

[JohnW2]

I have several Crabtree switches with indicator neons that have gone black. I can still get replacements, but I'm hoping there is a source of LEDs that will last longer and perhaps be brighter.

One certainly would expect an LED to last longer, and maybe be brighter but, unusually, in most other respects a neon is probably preferable.

Having said that, if you decided to go with an LED, then I imagine that LED+diode+capacitor would be preferable to LED+diode+resistor, since that would negate most of the downsides of an LED in this situation

 

[DetlefSchmitz]

since that would negate most of the downsides of an LED in this situation

...except if the physical size of a suitable X2 rated capacitor causes an issue (they're quite big). Without doing the experiment, I believe that an led will need much less current than people who first met them in the 80s seem to assume. As a matter of interest, the mains led indicators I bought for my wiring centre light up the whole back hall at night, whereas I could hardly see the neons they replaced. Over the years the leds we used on the equipment we sold seemed to get brighter every time we bought a new batch of the same part number, and we were forever changing the series resistor to compensate.
Anyway I believe that a suitable 400-500 volt resistor dissipating a power of about an eighth to a quarter of a Watt would easily be bright enough and meet the safety requirements, but without trying I can't be sure.

 

[JohnW2]

...except if the physical size of a suitable X2 rated capacitor causes an issue (they're quite big).

Yes, that could be an issue. However, we would probably be talking about a tiny capacitance (probably in the ballpark of 0.01 pF), so a pretty small one might be available.

Indeed, we're talking about such a small capacitance that, although I've yet to attempt any calculations, a couple of bits of insulated wire twisted together might be sufficient.

 

[DetlefSchmitz]

we would probably be talking about a tiny capacitance (probably in the ballpark of 0.01 pF),

Mmm... I made it about 10nF. I could have made an error. And I don't think they make X2 caps in such unusual sizes.

 

[plugwash]

probably in the ballpark of 0.01 pF

Nowhere near it. 0.01uF might be more in the ballpark but probablly still on the small side.

And on top of your capacitor you typically want a couple of resistors, one in series for spike protection and one in paralell to drain the capacitor when unplugged.

And you need some kind of reverse polarity protection, typically a diode (or a second LED if you want more light output) in anti-paralell.

 

[JohnW2]

Mmm... I made it about 10nF. I could have made an error.

That does sound more credible. My mental arithmetic has been known to be pretty faulty, and it sounds as if I may have lost/found several zeros Let me check ...

Yep, 10nF at 50 Hz is about 318k - so about 0.75 mA at 240V (probably half that, because of the rectification) - i.e. roughly the right ballpark. Apologies on behalf of my brain

And I don't think they make X2 caps in such unusual sizes.

There are plenty of 10 nF (and also 22nF, 33nF etc, if more current were needed) 250V+ X2s out there. Some are certainly fairly large, but I'd have to look around to see if there were any small ones!

 

[JohnW2]

Nowhere near it. 0.01uF might be more in the ballpark but probablly still on the small side.

Indeed- see my subsequent post. Never trust my mental arithmetic It is often said, perhaps with some truth, that the more mathematical education one has had, the worse becomes ones ability to do simple arithmetic!

And on top of your capacitor you typically want a couple of resistors, one in series for spike protection and one in paralell to drain the capacitor when unplugged.

I use capacitors fairly extensively for 'switched levels' with LEDs (dimming LEDs being such a nightmare), and I never bother about either

And you need some kind of reverse polarity protection, typically a diode (or a second LED if you want more light output) in anti-paralell.

Sure. I said diode, but a second parallel LED would be an alternative approach.

 

[plugwash]

I think reducing the brightness of an existing LED bulb is a different case from protecting a bare LED.

With no resistive component in the path, the initial current at switch-on will be limited only by the internal resistance of the capacitance and LED. I can't see that being good for the LED if switched on at the peak of the cycle.

 

[JohnW2]

I think reducing the brightness of an existing LED bulb is a different case from protecting a bare LED.

Yes, on reflection, you are probably right - since, as you go on to say ...

With no resistive component in the path, the initial current at switch-on will be limited only by the internal resistance of the capacitance and LED. I can't see that being good for the LED if switched on at the peak of the cycle.

As above, I can't really argue with that.