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I dont know how to find an answer to this Quetion. how much power in watts does a small L.E.D. inicator use? one like this PANELTRONICS LED 120 VAC 1/4 INCH GREEN 048-016, **broken link removed** how much would it cost per L.E.D. per month? any help is Appreciated. Glenn
The first thing to do is look for a real data sheet for the product from it's manufacturer.
**broken link removed**
Since they offer nothing substantial, you'd have to contact them to receive a good lie or measure a sample device.
From what I know of these, it would have a single LED with a reverse protection diode and a limiting resistor and would pass ~15mA on every half cycle so 120 * 0.015 / 2 = 900mW. You might want to be conservative and call that 1W for energy budget calculations.
The cost of operation on a monthly basis would depend on electric rates where installed but for example, if your rate were $0.25/kWh, it would cost $0.25 to run for 1000 hours. An average month has 730 hours. I'll leave the rest of the math to you.
If you used a transformer or switcher you could reduce that even more.
The led itself assuming its a standard green will use 15ma at 2v approx which is 30mW, a transformer supply assuming you used a 3v trans would use without losses about 45mw, you could assume alltold with losses about 100mw.
In the uk a kw hour is 15pence and that would run this led for 10,000 hours.
This is just a rough guide.
Thanks Guys: the reason I asked is I’m on a solar system with only 19Kw of battery backup and I want to put these on my 007 circulators. For the boiler. And every Watt. Is essential. But It looks like It isn’t enough to worry about. Thanks again Genngg
A green or red LED as shown has a forward voltage of between 1.8vdc and 2.5 vdc. LED emit light on DC current. The 120vac LED likely has a capacitor ballast to provide the voltage drop from 120 VAC to 2 volts to run the LED.
This may throw you a little but there is appearent power and true power, the ratio being the power factor of the device. The true power is in the neighborhood of 2 volts x 15 mA or 30 milliwatts. With a capacitive ballast the series cap value would need to provide 118v drop for 15 mA of average current. This is about 7870 ohms of capacitive reactance or about a 0.33 uF cap for a 60 Hz AC grid. Appearent power or V*A magnitude would be 0.015mA X 120 vac or 1.80 VA. Notice I called it VA not watts.
Power factor would be 0.03 watts/ 1.8 VA or 0.016, which for all practical purposes is just a 0.33 uF cap put across AC line.
Most all residential grid power meters charge for true watts with no surcharge for poor power factor. You would pay for the 30 milliwatts X time used not the 1.8 VA X time used. Some European countries are regulating power factor by putting requirements on manufacturers to produce a good power factor on some larger power draw appliances, where good is >0.9.
A dropping resistor could be used for a ballast but this is very unlikely as the resistor would dissipate near 1.8 watts which would make such a small package get too hot.
Judging by the small size of the case behind the LED, it's possible that they're using a resistor and running a high-output LED at e.g. 2mA - then only a standard 0.25W resistor is needed. The power consumption in this case is 0.25W.
Another option for are Neon indicators; these use less power (e.g. 100uA or 0.012W at 120V) and are more suitable for use on mains.
Wow: I asked for answers and I got them. Now you guys are making my head spin. L.O.L …I don’t understand a lot of what you guys say I’m not an electrician just a crane operator.
I do appreciate all the help. And I do have some neon indicator bulbs from Radio Shack. https://www.radioshack.com/product/index.jsp?productId=2102793 if you think these would be better, I will be installing them in parallel strait on the circulators. Thanks again Glenngg
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