Indicator light for 110V outlet

[Curl]

I know some 110V extension cords have a little orange light that shows you if it is plugged in.
How do these work? I am trying to build an indicator light like that but typical LEDs draw about 20mA, and off 110V that's still about 1 watt of power (assuming I'm using only half of the AC).
So it seems like a big resistor + big diode + LED in series is not a good option. Is there any clever and inexpensive/simple way of achieving this? I need to make about 10-15 of these that's why I want a simple/inexpensive way of doing it.

Thanks guys!

 

[Carl Pugh]

Google neon bulb.
A neon bulb and resistor works great.

 

[eq1]

Is 1W a problem? Or does a 1W rating on the R make it too big? If not I think LED + resistor is a pretty simple and effective solution.

Not all LEDs require 20mA. For example this one only requires 2mA.
http://www.digikey.com/product-detail/en/HLMP-7000/516-1400-ND/637661

 

[Curl]

but if I just use LED+Resistor won't it blow through the diode when the voltage switches? I not knowledgeable in this but I doubt a small LED can hold back 110v, or can it?

 

[Bobbywhy]

Curl, If you apply 110 Volts AC to an LED it will surely POP/BURN/SELF-DESTRUCT. LEDs operate on Direct Current (DC) and that current must be limited, or otherwise...more smoke! Please see Carl Pugh's suggestion in Post # 2. That is the most common, the simplest, and most reliable method to visualize AC power in on or off.

Bobbywhy

 

[eq1]

Assuming US outlets: 110Vac is ~170Vp. 170V/0.002A=85K will limit the peak forward current to 2mA. So use a 91K,5% resistor in series with the LED. Pretty much all the voltage will be on the resistor so if we just figure 110V*0.002A=0.22W so use a 1/4W R. Should work ok depending on how many lumins you need. The LED will blink at 60Hz but that's a pretty decent frame rate. Note: 2mA is just the test current so they'll guarantee there is some light at that current. If it is not enough you'll need to make the resistor smaller.

Although I think the LED and resistor will be safe, the extension cord might not be. You should really take care to make sure any exposed metal is well insulated from any possible outside touch. Personally I would recommend just buying them from the store. I doubt you'll save much money in the end, assuming you don't already have extension cords, and you'll definitely save a lot of time.

 

[eq1]

but if I just use LED+Resistor won't it blow through the diode when the voltage switches? I not knowledgeable in this but I doubt a small LED can hold back 110v, or can it?

LED is not holding anything back, the resistor is. The resistor will limit the current as predicted by ohms law: V=IR.

 

[willem2]

LED is not holding anything back, the resistor is. The resistor will limit the current as predicted by ohms law: V=IR.

Reverse breakdown voltages of LED's are low. While the resistor will still limit the reverse current, it's better to use a normal diode in series that can handle 200V or so.

 

[eq1]

Reverse breakdown voltages of LED's are low. While the resistor will still limit the reverse current, it's better to use a normal diode in series that can handle 200V or so.

Just curious with what's wrong with running a diode in reverse when the current is limited? According to the datasheet, the manufacturer does it to every LED at least once in its life.

 

[Curl]

when you run it in reverse there is very low current meaning the resistor has very low voltage drop - this means that the LED takes almost all of the 110V which means it will blow. That's what I was worried about in the above post.

 

[eq1]

According to the LED datasheet Vr = 12V(typ) so there will not be a significant difference in the LED current in either direction. Just to bound it Prms of the LED will be less than 12V*2mA = 24mW.

 

[skeptic2]

I believe you are worried about the 1 W dissipation and resulting heat of the resistor and rightly so. Your resistor should be rated at, at least 2 W. However if you used a 0.47uF capacitor instead and used an opposite polarity diode in parallel with your LED, your power dissipation would be only about 30 mW.