Power-on LED supplied by 240V AC

Ok yet another qucik electronics question. For my current DIY project I need an LED to indicate that the power is on. Fairly rudimentary stuff I'm sure, but I need to power the LED from the main 240V AC supply. Google didn't return much of use so any suggestions on how this can be achieved?

 

farnell part num 461581

they have others aswell

 

Cheers for the suggestion. The only drawback I can see is the length of the unit, which would make it difficult to place in my layout. But I'll hunt through their range and see if there's anything suitable.

 

RS also have some 240V AC LEDs:

http://rswww.com/cgi-bin/bv/browse/...eID=ukie&3261383135=3261383135&stockNo=194802

Michael.

 

After some more digging around on Maplin's site I've found these VW69A and BK55K which look like they'd fit my requirements.

 

why not use a normal led with at least 2 resistors in series, the total resistor value = supply voltage - required led voltage( normally around 2.2 v)/current required for led (normally 20ma) ,
the wattage of the resitor is (supply voltage - led voltage) x led current

 

themadhippy, so am I right in thinking that if were to use this LED from Maplin (N46AT) the value of the resistor would be 240-2.5/0.03 = 8k approx. and the wattage would be 240-2.5*0.03 = 7.1 W, so I'd need two of these H4K7 (10W 4.7k resistors from Maplin)?
This was my original plan, but the guy in Maplin got me confused by recommending that I have a half value resistor and diode in parallel with the full value resistor and LED, with the diode arranged with opposing polarity to the LED.
Are there any drawbacks to using just resistors to run the LED, i.e excessive heat generation?

 

why use a diode? after all a led is a light emitting diode.as for the resitors ,yep 2 x 4.7k ohms will do nicely,the resistors will get hot as theve got to disapate the wasted voltage.