# trying to adapt warm white LED to candles

B

#### bubbas

Jan 1, 1970
0
I recently purchased the light string shown here:

http://www.amazon.com/Transparent-White-Clear-Christmas-Lights/dp/B0069VI8ZS/ref=pd_sim_sbs_hg_5

The hope was to replace existing 120 V lamps in window candles with an LED
light from the string, but I am having some difficulty. As I suspected when
I cut off one of the lights from the string and, still fused, plugged it
into the 120 v outlet, the fuses blew. So, it looks like a voltage drop is
needed. The total voltage/ current of the entire 25 bulb string, according
to the link above, is 0.027 amps (3.24 watts) @ 120 V. A tag on the string
says that each bulb is 3.1V @ 0.062 amps. So, which approach is best:

1) Use a dropping resistor for each bulb I want to use? What value and
wattage if so? Would the resistor stay cool enough to hide it in the
candle?

2) Use an AC plug in the wall type transformer (rated for 3.1 V AC), which
I assume would be quite hard to find because most of them are DC (I tried
placing 3 V DC across the LED lamp and it didn't work)?

Appreciate any help here. The candles I placed in the windows recently all
use what I thought were cool, smaller standard 120 V incandescent bulbs but
there was enough heating to discolor the window blinds they are up against.
The wife liked the "warm white" of the LED string, so I wanted to replace
the incandescents with those and I also want to use a much thinner power
cord to the candles if possible, something I couldn't really get away with
using the higher powered incandescent.

Thanks,
Bud

N

#### N_Cook

Jan 1, 1970
0
bubbas said:
I recently purchased the light string shown here:

http://www.amazon.com/Transparent-White-Clear-Christmas-Lights/dp/B0069VI8ZS
/ref=pd_sim_sbs_hg_5

The hope was to replace existing 120 V lamps in window candles with an LED
light from the string, but I am having some difficulty. As I suspected when
I cut off one of the lights from the string and, still fused, plugged it
into the 120 v outlet, the fuses blew. So, it looks like a voltage drop is
needed. The total voltage/ current of the entire 25 bulb string, according
to the link above, is 0.027 amps (3.24 watts) @ 120 V. A tag on the string
says that each bulb is 3.1V @ 0.062 amps. So, which approach is best:

1) Use a dropping resistor for each bulb I want to use? What value and
wattage if so? Would the resistor stay cool enough to hide it in the
candle?

2) Use an AC plug in the wall type transformer (rated for 3.1 V AC), which
I assume would be quite hard to find because most of them are DC (I tried
placing 3 V DC across the LED lamp and it didn't work)?

Appreciate any help here. The candles I placed in the windows recently all
use what I thought were cool, smaller standard 120 V incandescent bulbs but
there was enough heating to discolor the window blinds they are up against.
The wife liked the "warm white" of the LED string, so I wanted to replace
the incandescents with those and I also want to use a much thinner power
cord to the candles if possible, something I couldn't really get away with
using the higher powered incandescent.

Thanks,
Bud

If you like the flicker effect of candles try mixing some orange neons ,
perhaps overdriven, fed from DC rather than AC.

over plate glass. Missing the point entirely , that the visual effect of
leaded glass windows is not the lead itself but the multi-facit reflections
off the individual small panes.

B

#### bubbas

Jan 1, 1970
0
N_Cook said:
If you like the flicker effect of candles try mixing some orange neons ,
perhaps overdriven, fed from DC rather than AC.

These original candles don't flicker (they have a standard incandescent) and
the hopeful LED replacements won't work from DC, which is why I am asking
the best way to power them.

B

#### Bob Engelhardt

Jan 1, 1970
0
First, & very importantly, LEDs are INHERENTLY DC devices. You have a
string that uses 120v AC because there is a rectifier built into the
string, or each bulb has 2 LED's, in opposite orientation.

So, you need to determine which it is. Here's how: get a DC supply of
more than 3.1v. You can use 3 AA or AAA batteries, end to end, or a 6v
DC wall wart. Connect a 100 ohm resistor in series with the supply.
This limits the current to 30 ma, so as to not burn out the LED.

Then connect the supply to the LED, both ways (both polarities). If the
LED lights with both connections, there are 2 LEDs in your bulbs and
you'll need an AC supply. If it only lights on one polarity, there is
just 1 LED and you'll need a DC supply.

Report back with your findings and we'll go from there.

Bob

B

#### bubbas

Jan 1, 1970
0
Bob Engelhardt said:
First, & very importantly, LEDs are INHERENTLY DC devices. You have a
string that uses 120v AC because there is a rectifier built into the
string, or each bulb has 2 LED's, in opposite orientation.

So, you need to determine which it is. Here's how: get a DC supply of
more than 3.1v. You can use 3 AA or AAA batteries, end to end, or a 6v DC
wall wart. Connect a 100 ohm resistor in series with the supply. This
limits the current to 30 ma, so as to not burn out the LED.

Then connect the supply to the LED, both ways (both polarities). If the
LED lights with both connections, there are 2 LEDs in your bulbs and
you'll need an AC supply. If it only lights on one polarity, there is
just 1 LED and you'll need a DC supply.

Report back with your findings and we'll go from there.

Ah, the walwart I initially tried wasn't plugged in completely so that's why
the LED's wouldn't light. My findings are that each bulb has 1 LED as it
only lights the one way.

Looks like I should just be able to use a cheap walwart to power these if
I'm not mistaken. Should I include the 100 ohm resistor to each LED?

Thanks,
Bud

H

#### hr(bob) [email protected]

Jan 1, 1970
0
Ah, the walwart I initially tried wasn't plugged in completely so that's why
the LED's wouldn't light.  My findings are that each bulb has 1 LED as it
only lights the one way.

Looks like I should just be able to use a cheap walwart to power these if
I'm not mistaken.  Should I include the 100 ohm resistor to each LED?

Thanks,
Bud

- Show quoted text -- Hide quoted text -

- Show quoted text -

Do you want just one bulb in each candle? Are these individual
candles, or are they the "8 candles in a row" of 4 ascending and 4
descending height?

B

#### bubbas

Jan 1, 1970
0
"Do you want just one bulb in each candle? Are these individual
candles, or are they the "8 candles in a row" of 4 ascending and 4
descending height?"

Thanks, one bulb per candle. I'll have two candles per room and I hope to
use a separate power source for each candle pair as there are 3 rooms.

B

#### Bob Engelhardt

Jan 1, 1970
0
bubbas said:
"Do you want just one bulb in each candle? Are these individual
candles, or are they the "8 candles in a row" of 4 ascending and 4
descending height?"

Thanks, one bulb per candle. I'll have two candles per room and I
hope to use a separate power source for each candle pair as there are
3 rooms.
OK, here's the next thing about LEDs: they're current devices. I.e.,
you design your circuit for the current through them, rather than the
voltage across them. And LED brightness is a matter of how much current
you put through them. For your lights, I would use the .027A (27 ma).
The .067 is bogus: how can a bulb draw more than is going through the
whole string? Although the .067 might be a maximum current.

So, assuming a 6v DC supply and 3.1v dropped across the LED, there has
to be 2.9v dropped across the resistor. With .027A needed, a 2.9/.027 =
107 ohms. 100 ohms is a common value - use that. .027A & 2.9v gives
..08 watts dissipated by the resistor - it will not heat up the candle
stick. And you can use the smallest resistor (1/8W).

Each LED will need a resistor.

BTW - it you want to use a separate wallwart for each, you can get them
for $2.59 each (shipped) here: http://www.ebay.com/itm/170898516293 If you want to carry this further, you can experiment with the brightness by changing the 100 ohm resistor. Smaller resistor equals brighter LED. But shorter life. Too small means zero life. If the ..067A LED rating is a maximum current, you couldn't use a resistor smaller than 2.9/.067 = 43 ohms. And at that, the life would probably be very short. IME Have fun, Bob M #### mike Jan 1, 1970 0 "Do you want just one bulb in each candle? Are these individual candles, or are they the "8 candles in a row" of 4 ascending and 4 descending height?" Thanks, one bulb per candle. I'll have two candles per room and I hope to use a separate power source for each candle pair as there are 3 rooms. There are a bunch of issues here. Have you verified that the optical properties are acceptable? Light comes out of a LED in one direction. It's non-trivial to adapt an incandescent fixture to LED. You might need to do some grinding on the LED plastic or use more than one led to get the look you want. Those lamps look like the right stuff, but can't be sure from the picture. Try it. This project appears to be all about aesthetics. Make sure you like the look before you invest in a bunch of power supplies. Even number of LEDS means you can use AC. Your power choices depend more on what power supplies you can get. Without going into all the gory details... Go to a couple of garage sales and pick up some 25-cent cellphone chargers. Use a 5V charger and a resistor at least 68 ohms 1/2W. Larger resistor for lower brightness is no problem. For the nitpickers reading this, you're probably gonna put the resistor in heat-shrink without heat sinking inside some plastic thingie, so a higher wattage resistor is warranted. At 100 ohms, 1/4W should be big enough. When you're done with the prototype, run it a while and make sure nothing gets hot. READ THE LABEL. Not all cellphone chargers are 5V. Some are 5.4V and need larger resistors. Check the current rating on the label. One charger should easily supply two LED-Resistor combinations...or more...check the label. You mentioned smaller wire. The size of the wire is relatively unimportant. BUT You want something that won't be damaged and shorted by flexing, slamming the window on it, cat chewing on it. If it shorts and your wall-wart is plugged in behind the curtain... have your fire insurance paid up. Chances of a short AND just the right combination of PS factors to cause overheating are very small, but the consequences are catastrophic. Save your good luck for something else. B #### bubbas Jan 1, 1970 0 OK, here's the next thing about LEDs: they're current devices. I.e., you design your circuit for the current through them, rather than the voltage across them. And LED brightness is a matter of how much current you put through them. For your lights, I would use the .027A (27 ma). The .067 is bogus: how can a bulb draw more than is going through the whole string? Although the .067 might be a maximum current. So, assuming a 6v DC supply and 3.1v dropped across the LED, there has to be 2.9v dropped across the resistor. With .027A needed, a 2.9/.027 = 107 ohms. 100 ohms is a common value - use that. .027A & 2.9v gives .08 watts dissipated by the resistor - it will not heat up the candle stick. And you can use the smallest resistor (1/8W). Each LED will need a resistor. BTW - it you want to use a separate wallwart for each, you can get them for$2.59 each (shipped) here:
http://www.ebay.com/itm/170898516293

If you want to carry this further, you can experiment with the brightness
by changing the 100 ohm resistor. Smaller resistor equals brighter LED.
But shorter life. Too small means zero life. If the .067A LED rating is
a maximum current, you couldn't use a resistor smaller than 2.9/.067 = 43
ohms. And at that, the life would probably be very short. IME

Have fun,
Bob
Sounds like a good plan, thanks, except I'm a little concerned about one
thing: actual wallwart voltage. Very often, they're marked as 6V, but
output will be a volt or two higher. In that case, I'm assuming I'd just
need to increase the resistance slightly to each bulb, correct? If 7V, then
3.9v dropped across the resistor so 3.9/.027= 144 ohms (I would use the
closest standard resistor, probably 150 ohms). So power would be a little
more, but I could still use a 1/4 watt resistor. Please correct me if I'm
wrong. Been years since I've done much with electronics calculations, but
they're coming back to memory somewhat.

Bud

B

#### Bob Engelhardt

Jan 1, 1970
0
bubbas said:
Sounds like a good plan, thanks, except I'm a little concerned about one
thing: actual wallwart voltage. Very often, they're marked as 6V, but
output will be a volt or two higher. In that case, I'm assuming I'd
just need to increase the resistance slightly to each bulb, correct? If
7V, then 3.9v dropped across the resistor so 3.9/.027= 144 ohms (I
would use the closest standard resistor, probably 150 ohms). So power
would be a little more, but I could still use a 1/4 watt resistor.
Please correct me if I'm wrong. Been years since I've done much with
electronics calculations, but they're coming back to memory somewhat.

That's all correct.

Bob

B

#### bubbas

Jan 1, 1970
0
Bob Engelhardt said:
That's all correct.

Bob

Working out good so far. I've had whatever walwarts I've had in the junk
box and have had to readjust the resistance accordingly. You'd be surprised
how much more some of the actual voltages are than are stamped on the units.
One I have said "12 volts", but in actuality, it was 19.5V! Makes me wonder
how much electronics in general burn out because of mislabeled adapters.

Bud

B

#### Bob Engelhardt

Jan 1, 1970
0
bubbas said:
Working out good so far. I've had whatever walwarts I've had in the
junk box and have had to readjust the resistance accordingly. You'd be
surprised how much more some of the actual voltages are than are stamped
on the units. One I have said "12 volts", but in actuality, it was
19.5V! Makes me wonder how much electronics in general burn out because

Some over-voltage is to be expected: it's the compensation for the
supply bogging down under load. But 19.5 vs 12 is a bit much. If
you're curious, you could load it to its rated current & see what
happens to the voltage.

If you are going to be using 12v supplies, you might need higher wattage
resistors. E.g., (12 - 3.1) * .026 is just a 1/4 watt, so using 1/2
might be a good idea. Of course, if that "12v" supply is really 19.5,
you'd need an even larger one.

Bob

B

#### bubbas

Jan 1, 1970
0
Bob Engelhardt said:
Some over-voltage is to be expected: it's the compensation for the supply
bogging down under load. But 19.5 vs 12 is a bit much. If you're
curious, you could load it to its rated current & see what happens to the
voltage.

If you are going to be using 12v supplies, you might need higher wattage
resistors. E.g., (12 - 3.1) * .026 is just a 1/4 watt, so using 1/2 might
be a good idea. Of course, if that "12v" supply is really 19.5, you'd
need an even larger one.

Yes, I used resistors up to 1 watt today. I aimed to use double the wattage
required as a safety margin. All the candles are now in the windows and are
lit nicely. My only concern now is LED life. I understand it varies
greatly according to manufacturer. I had really hoped I wouldn't have to
touch these for years. Perhaps so, but I'm not sure.

Bud

G

#### gregz

Jan 1, 1970
0
bubbas said:
Working out good so far. I've had whatever walwarts I've had in the junk
surprised how much more some of the actual voltages are than are stamped
on the units. One I have said "12 volts", but in actuality, it was 19.5V!
Makes me wonder how much electronics in general burn out because of mislabeled adapters.

Bud

It's 12 volts at specified amps.

Greg

B

#### bubbas

Jan 1, 1970
0
gregz said:
It's 12 volts at specified amps.

Greg

Oh, ok, well that makes sense now. Didn't think about it that way. Thanks.
In other words, the label is correct when the unit is, say, plugged into
whatever device it was made for, and the device drawing the same current as
on the label. Bud

B

#### Bob Engelhardt

Jan 1, 1970
0
bubbas said:
... All the candles are now in the
windows and are lit nicely. My only concern now is LED life. I
understand it varies greatly according to manufacturer. I had really
hoped I wouldn't have to touch these for years. Perhaps so, but I'm not
sure.

I'm sure that life does vary with mfg'r. It also varies greatly with
the current used to drive them. But that affects the brightness, too.
You could try this: use larger & larger resistors until the bulbs are
too dim. A pot would be the easiest. It might surprise you as to how
low you can go & still get the desired affect.

Bob

B

#### bubbas

Jan 1, 1970
0
Bob Engelhardt said:
I'm sure that life does vary with mfg'r. It also varies greatly with the
current used to drive them. But that affects the brightness, too. You
could try this: use larger & larger resistors until the bulbs are too dim.
A pot would be the easiest. It might surprise you as to how low you can
go & still get the desired affect.

Today, I went back and measured the current for all the lights. All were
from about 20-27 mA. I then increased resistance for each one until current
was from about 13-16 mA. There was slight dimming, but I think still
acceptable. I'll find out for sure after dark.

It has been fun modifying the original incandescent candles into these.
I've also had a chance to revisit electric circuit study, something I
haven't done for nearly 15 years.

Bud

B

#### Bob Engelhardt

Jan 1, 1970
0
bubbas said:
Today, I went back and measured the current for all the lights. All
were from about 20-27 mA. I then increased resistance for each one
until current was from about 13-16 mA. There was slight dimming, but I
think still acceptable. I'll find out for sure after dark.
...

That's really good. If the dimming was only slight and it's acceptable,
your bulb life will be greatly extended. How much, exactly, I couldn't say.

Bob

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