I'm not sure if this is the right group but here goes, I want to use 2 led's as indicators but they will be on 240v ac and I don't know how to work out the resistance needed or what size or type of diode I would need. I have looked at different site that sell them but they are too large for my requirements which rules out using transformers to drop and rectify the voltage, and I can't find any other information on how to wire them. Trevor Smith
Surely neon lamps like
david
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Thing is I need one blue and one red. Trevor Smith
Firstly, consider a neon indicator -- it's a better fit to main voltage.
If you stick with LED, then 5mA is enough current for an indicator, or make that 10mA for half wave rectified...
1N4004 LED---/\/\/\----|>|-------|>|--------- 24k | | 1.5W --/\/\/\-- 10k 0.25W
The 24k resistor drops the mains at 10mA, but gets hot. The 1N4006 diode rectifies the mains. The 10k resistor ensures reverse mains voltage is all across the 1N4006, and not the LED which couldn't handle it.
We could reduce power consumption by dropping most of the voltage across a capacitor rather than the resistor. Can't completely get rid of the resistor as the inrush current to charge the capacitor if swithed on mid-cycle would destroy the LED.
LED
---||-----/\/\/\--------|>|--------- 0.068µF 6.8k | | 400V 0.25W ---|
In article , Trevor Smith writes
Ok, diagram below but please think about safety before implementing this circuit. Neon cases are designed to be safe when used at mains voltages but LEDs are not. An LED mounted in a clip and poked through a panel or even in a panel bezel is not safe when used with this circuit. Unless you are prepared to totally enclose the LED in something like a salvaged neon housing then please don't implement this circuit.
Anyway, lecture over:
C is 400V ac rated 22nF for 1.6mA, 100nF for 8mA, 220nF for 16mA
View with a fixed width font:
| | | --- --- C (value) | / \ / 470R \ | ----- | | | | --- --- 1N4007 / \ \ / LED --- --- | | | | ----- | | |
Resistor values (ohms) = (Vs-Vf)/If
where Vs is supply voltage Vf is the forward voltage drop across the LED If is the forward current through the LED
Vf and If are shown in the specs for each LED in the Maplin etc catalogue
For AC supply connect a diode eg 1N4148 in inverse parallel with the lED and halve the value of the resistor
(says the Maplin Catalogue from long enough ago when it contained Useful Information)
The whole lot will have to be insulated for 240V.
Owain
Oh, and put a 1M resistor across the capacitor too, so it can't retain a charge after use and zap you.
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Search for 'neon blue' on rswww.com (Radio Spares)
Ready made mains LED indicators are available and might be the easy answer as regards safety etc. I've got some which are identical in diameter to a normal 5mm LED plus bezel with the electronics mounted inline behind and the whole lot sealed with flying leads for connections. Overall length about 1.5 inch.
On 14 Nov 2006 15:13:25 -0800 someone who may be "Trevor Smith" wrote this:-
Why do you "need" a blue one?
Maplin sell snap-in neons in red, amber, green *and* blue
Neons simply don't look as good as LEDs, though. They're also a source of RF interference which might matter on sensitive equipment.
Probably the best thing to do is to use a capacitor in series with them
- resistors to get to say 20mA will be large and bulky and get hot (5W). Then you want a diode across them so they don't get reverse voltage across them.
Working on 20mA for half the time - so 40mA in all, you need around
6Kohms impedance at 50Hz..I make that about a 0.5uF 400v AC capacitor you need, and any old diode like a IN4148..The other otion is an inductor - 50mH..but that is a fairly large chunk of iron and a lot of fine wire.
Ah,. Pretty much what I came up with..except you have forgotten that the current through the LED is half wave..I made it .5uf for 20mA average..and I see no need for a high voltage power diode like a IN4007..it doesn't have to take more than 5v reverse across it after all..the LED takes care of that and clamps it. The 470R is good to limit transients.
I had another thought on this..it might be possible to use a triac as a fairly crude 'chopper' - like a basic dimmer type circuit - and then a resistor and capacitor to smooth it out a bit. More expensive, but more efficient.
That's probably what they use in these
Probably available from RS components and the like
Thanks for the advice and I have decided to go for ready made indicators. Trevor Smith
Doesn't work for AC.
Not even where it says "halve the value of the resistor for AC"?
Owain
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