Running LED lights from 110 cord

[MikeBliss]

It will be obvious that I am not a certified Electrical Engineer. I am trying to create a small display and using LED lights to "accent" the display. I was wanting to use 6 to 9 LEDs in a circular fashion. What I am not certain of is: Can I run 6 - 9 LED's just by using a power cord and resistor(s) rather than buying a transformer? I am trying to keep the cost down because these ( the displays ) might be mass produced. Any help is GREATLY appreciated.

 

[NobodySpecial]

LEDs typically drop 1-2V so unless you are using 50-100 of them then runnign them from 110V is going to involve some extra components.

If you are intending to mass produce them it's almost certainly going to be cheaper to include a little 1:10 transformer than meet the electrical safety and insulation standards of having the whole string of LEDs 'live'

 

[skeptic2]

What do you mean "...in a circular fashion."

 

[jambaugh]

LED's are rated at a certain current not voltage. What with voltage spikes and such you're more likely than not to burn them out this way.

If you're very careful so as not to exceed voltages (and can filter out voltage spikes caused by say your refrigerator's compressor kicking on) then you still will get heating of your series resistor. Consider the relative voltage drop across the resistor and the LED's Ball park assume 10 resistors with a 1V drop is 10 volts, the other 110 volts is across the resistor and you'll have the same (albeit small) current. This means the resistor must dissipate 110/10 = 11 times the power of the LED's as heat. It is both inefficient and a potential a fire hazard.

It's all together a very bad idea. You're better off powering the display with some rechargeable AA batteries. Even then you still may need to buffer the current with a resistor so you don't overload the LED's.

If you use a transformer you should also use a voltage regulator as line power can be quite variable and "spiky". But you shouldn't need to buy one. Don't you have some old wall charger for some old piece of electronics which you no longer use or no longer works? If you don't then I'm sure you can find a friend who does.

EDIT: I didn't see the part about "mass produced" I'd then strongly suggest using batteries or finding a supplier of cheap low power wall adapters. You can easily find them in bulk for under $5 a piece. A quick google found http://www.robotshop.com/sfe-power-supply-9vdc-650ma-1.html"

 

[pantaz]

... I am trying to keep the cost down because these ( the displays ) might be mass produced. Any help is GREATLY appreciated.

You can find low cost transformers, and even pre-wired LED strings, at many surplus electronics dealers.

A few that I use are:
http://www.allelectronics.com/
http://www.goldmine-elec.com/
http://www.herbach.com/
http://www.jameco.com/ (Not a surplus dealer, but worth a look)

 

[yungman]

I wonder what is inside those LED lightbulbs that are selling in the stores, those that you replace the normal light bulbs at home. Do they use a transformer inside or they have some other ways. Maybe you should get on in the store, take it apart and learn. The manufacturers must have gave this a lot of thoughts and doing it the cheapest way.

They must have some ways to regulate the current without using resistors because any resistors will hurt efficiency by burning power. Maybe they not use a small transformer, instead they have a constant current online switching regulator to regulate current to get constant intensity.

 

[vk6kro]

We have some ceiling fans that use switched series capacitors to allow a choice of fan speed.

You could do this with LEDs. You have to put them in a full wave configuration because they have a poor reverse voltage rating.

Like this:
[PLAIN]http://dl.dropbox.com/u/4222062/Mains%20LEDs%20with%20C.PNG
The diodes shown are LEDs

However as always I have to advise anyone against playing with mains wiring unless they have enough experience to do it safely and legally. All mains wiring should be enclosed in a sealed box before power is applied.

I haven't built this, so it is only a suggestion. It simulates OK.

You could consider putting equally sized resistors across each LED so that the reverse voltage is shared equally between the LEDs.

 

[Phrak]

Like this:
[PLAIN]http://dl.dropbox.com/u/4222062/Mains%20LEDs%20with%20C.PNG
The diodes shown are LEDs

However as always I have to advise anyone against playing with mains wiring unless they have enough experience to do it safely and legally. All mains wiring should be enclosed in a sealed box before power is applied.

I haven't built this, so it is only a suggestion. It simulates OK.

It's similar to one of my designs. The only one I ever modeled in SPICE.

There's an embarassing issue. When the circuit it turned on at peak, the full line voltage is applied across the LEDS. They eventually pop after enough turn-ons. A negative coefficient thermistor in series with the source solved the problem---after they hit the field.

 

[mdjensen22]

if you can guarantee that safety isolation won't be an issue, this can be done with a resistive or capacitive power supply.

http://ww1.microchip.com/downloads/en/AppNotes/00954A.pdf
http://www.daycounter.com/Circuits/...Supplies/Transformerless-Power-Supplies.phtml

I would recommend the capacitive supply with the full bridge. Don't forget the resistor to limit the inrush of the supply.

 

[vk6kro]

There's an embarassing issue. When the circuit it turned on at peak, the full line voltage is applied across the LEDS. They eventually pop after enough turn-ons. A negative coefficient thermistor in series with the source solved the problem---after they hit the field.

Good point. That is something you don't forget in a hurry.

I have modified my diagram (above) to show a varistor that would hopefully clip off any startup transients.

Normally, the capacitor will drop the voltage to about 20 volts for each LED string. The LEDs would probably survive a pulse going up to 30 volts because they have series resistors, so a 25 volt varistor is chosen to clip off the very brief startup pulse at 25 volts.

You can get low voltage varistors here:
http://www.hokuriku-europe.com/pdf/eng/e031101.pdf
or here:
http://cgi.ebay.com.au/QTY-42-39ZR-10D-SERIES-LOW-VOLTGE-RADIAL-VARISTOR-H637-/200560731148?pt=LH_DefaultDomain_2&hash=item2eb259e40c

 

[Phrak]

I have modified my diagram (above) to show a varistor that would hopefully clip off any startup transients.

Normally, the capacitor will drop the voltage to about 20 volts for each LED string. The LEDs would probably survive a pulse going up to 30 volts because they have series resistors, so a 25 volt varistor is chosen to clip off the very brief startup pulse at 25 volts.

You can get low voltage varistors here:
http://www.hokuriku-europe.com/pdf/eng/e031101.pdf
or here:
http://cgi.ebay.com.au/QTY-42-39ZR-10D-SERIES-LOW-VOLTGE-RADIAL-VARISTOR-H637-/200560731148?pt=LH_DefaultDomain_2&hash=item2eb259e40c

Your circuit looks better than a simple thermistor in series (In my own defense, adding the thermistor as a redesign solution wasn't my idea. I'm innocent! And besides the whole fiasco was the fault my custumer engineer to whom I gave the basic topology on gradis then didn't do the research to make it a reality solution.), though I've never had the opportunity to use a varistor in a circuit.

What does your spice model tell you about the peak current for a 0.47 uF in series with a 25 volt varistor for a half cycle of 60 Hz?. Your first link says you can dissipate 1.7 joules in 2ms with the 25VAC device. This means slightly better than 1.7 Joules over a single half cycle. For initial conditions the voltage across the capacitor isn't necessarily zero volts.