The impending death of incandescent bulbs

[Joseph]

They are DC devices, but worse in terms of power factor is that they do not conduct until the potential of their silicon junction is exceeded. They will draw power from the peaks of the AC power cycle like rectified power supplies with filter capacitors even if they don't employ filter capacitors in their power supplies.

All that applies so long as their power supply is simple rectified AC, even from a transformer.

[MrAl]

Hi Joseph,

That's a very interesting observation (about the power consumption
being mostly at the peaks of the sine wave). Something i had not
thought about until now.

There is the potential for a problem if everyone was running
their LEDs from transformers and the whole country switched
to LED lighting, but we know that wont be the case because
harmonics will show up and force power supplies working with LEDs
to be more power line friendly. In other words, yes LEDs will
draw current at restricted angles for a while, but then standards
will be put into place that begin to govern what types of LED
circuits can legally be connected to power lines and these would be
power supplies that are corrected for harmonic content.

Bad conduction angle isnt always the case either, for example, an LED
driven through a resistor going to a rather high voltage acts more like a
constant current than a load that draws from the peaks alone (rectified
power supply). Unfortunately i dont think this will become the
de facto standard either because of efficiency.

On the other hand, linking several LEDs in a string to power directly
from the power line with a bridge rectifier is going to draw current
mainly at the peaks, which is not good for the power company.

I think we will have to wait to see how the industry handles this.

[dyarker]

Hmmmm, for bug lighting the patio I've also been thinking about putting two strings of 50 yellow LEDs in a clear vinyl tube, with just a full-wave rectifer and current limiter in plastic box at one end of tube, other end sealed. Fully enclosed, so transformer not needed.

Another "also" is that I also had not thought of power factor.

So maybe four strings of 25 LEDs run by a 48V switching supply. The trouble is a 48V, 1A switcher (like Meanwell brand) is over $60.

1A is way overkill, regulation needn't be that good, and mains isolation isn't needed if everything sealed. Does anyone have a schematic for a power factor friendly 100mA supply?

C U L -

P.S. (added): I DID think about the heat of all those LEDs inside a tube. One fix would be vent holes at far end, and a fan in the box.

[Joseph]

You could get a ferrite core that you can wind a lot of turns onto by hand. It would be something like a small e-core or pot core. The easiest core to obtain to use for the task is one of those clamp-on chokes. Remove the halves from their plastic housing. Then insulate half with tape, wind enough turns onto it so that it will be impedance limited, and drive it with a MOSFET. Then rectify the output and feed your string of LEDs.

[Joseph]

D,
Alternately, use a low power SMPS and limit the duty cycle to give the constant or maximum power output you need. For powering LEDs, it would be a small duty cycle.

But I prefer going more toward impedance limiting the transformer as this tends to lower EMI more because the rise time of the transformer drive will tend to be much longer.

[MrAl]

Hi again,

Dale:
A 1k resistor in series with the AC side of the bridge rectifier seems to
drop the harmonics quite a bit as well as add some high-line/low-line
regulation. If you notice any blinking you can
add a 50uf cap across the LED string. The number of LEDs in the
string should be 34 with 1k resistance.
The 1k needs to be 1 watt, so perhaps put two 1/2 watt 470 ohm
resistors in series as these are easy to get.

Let us know how it works out.

[Joseph]

Maybe I am mistaken on an observation. I opened up one of these http://www.harborfreight.com/cpi/ctaf/D ... D=13567690 LED flashlights and I didn't see any dropping resistor between the batteries and the LEDs. I have never liked a flashlight more than this one.

I am purchasing these http://www.goldmine-elec-products.com/p ... ber=G16402 which are rated for between 3.2 and 3.4 v. My thought is to run them off of 3.2v with no resistor. It is a 6.25% jump up to 3.4v, which would be the safety margin. Do we really need the resistor which will burn some power?

[smariotti]

Maybe I am mistaken on an observation. I opened up one of these http://www.harborfreight.com/cpi/ctaf/D ... D=13567690 LED flashlights and I didn't see any dropping resistor between the batteries and the LEDs. I have never liked a flashlight more than this one.

I am purchasing these http://www.goldmine-elec-products.com/p ... ber=G16402 which are rated for between 3.2 and 3.4 v. My thought is to run them off of 3.2v with no resistor. It is a 6.25% jump up to 3.4v, which would be the safety margin. Do we really need the resistor which will burn some power?

I bought those same LED's. They're great!

And you CAN burn them out with too much current. I've done it.

Forgive me if this sounds pedantic, as I'm just now learning these things via trial and error and reading various things so when I actually know something, I'm prone to run on about it:

The flashlight specs are the clue to why there's no resistor needed. The flashlight runs off of 3 AA cells. If they're alkalines, that's 1.5v each, presumably in series, which creates 4.5v for the cells. That's enough voltage to power an LED. Of course, if you hooked up an LED to that it'd burn out instantly. If you connect two led's in series to that, they wouldn't light up (4.5v divided by 2 leds is 2.25v each.) If the 5 LEDs were connected in series to the batteries that'd be (4.5v divided by 5 or) 0.8v per LED and they wouldn't light either (they need at least 3.2v-3.4v as you said.)

But by putting them in parallel, you can provide that 4.5v to all 5 of them, and that chops the current down (to a fifth.) Most LED's can operate at between 20mA and 40mA. If you give them more than that, they flame out (quite brightly with those 12000mcd white LED's too!)

So whatever the total current those 3 AA cells generate, divided by 5 for the 5 LED's, it must be below 40mA. I'm guessing it's around 150mA?

The reason the resistor is in most LED circuits is to chop the current down below 40mA or so to prevent flame-out.

Forgive me if this is all already known and I'm misunderstanding you. Like I said... I get excited and pedantic when I can apply my knowledge.

[jollyrgr]

MrAl,

I know your posting was done some time ago but I thought I'd add this link for you:
http://www.cappels.org/dproj/ledpage/leddrv.htm

It allows you to replace a #222 lightbulb with a white LED and associated driver.

Then you can also buy LED versions of the #222 bulb here:
http://www.tek-tite.com/src/product_info.php?id=2731
(WOW! $30 for an LED!)


Now let's do some math here. The LED #222 bulb cost $30. The #222 comes in a two pack at Radio Shack for $1.59. For grins lets make it even and say you can buy 18 packs or 36 bulbs for the same cost as ONE LED lamp. As much as I like LED flashlights (especially the human powered ones) I cannot justify buying a $30 bulb.

CFC lights and even full blow Fluorescent fixtures are not that great. Try using one of the CFCs in an outdoor light in the colder climates. They don't even work in a cold garage let alone outside in winter. Even when the weather turns cold and it gets 60 degrees F inside my kitchen lights (3 60W "equivalent" CFCs) are dim and flicker until the heat is turned on. There are two eyeball "floods" on my fireplace mantle I'd like to replace with CFCs. Since this is in my TV room I like to dim the lights. But a SINGLE dimmable flood CFC costs as much as ten incandescent floods.

I love LED flashlights and have many. For close up work an LED version works great. My coworkers are constantly borrowing my 9 LED light (works on three AAA batteries) or my crank flashlight. But these don't hold a candle (pun intended) to my 3D and 5D Maglite incandescent flashlights as far as brightness is concerned. Though I can say I've dropped my Maglites and lost a bulb. This never happnes with the LEDs.

As far as LED landscape lights; I got fed up replacing lamps in my Malibu 12V system. They have been slowly (and as of last weekend) almost comletely replaced by solar white LED versions. This past summer (after being in service since 1999) I had to replace my first batch of NiCads in the originals. Not in ALL of the orignal solar lights installed in 1999 needed new batteries; only about four of the dozen or so. There are two low voltage lamps left in place; 50 watt (each) flood lamps that light up my US flag (on a 20 foot pole). These bulbs last about six months to a year. If I could find LED replacements cheap, I do so.

[Joseph]

I bought those same LED's. They're great! And you CAN burn them out with too much current. I've done it.

Great, thanks! I was wondering.

Forgive me if this sounds pedantic,

No prob, we're here to learn and exchange.

The flashlight specs are the clue to why there's no resistor needed. The flashlight runs off of 3 AA cells. If they're alkalines, that's 1.5v each, presumably in series, which creates 4.5v for the cells. That's enough voltage to power an LED. Of course, if you hooked up an LED to that it'd burn out instantly. If you connect two led's in series to that, they wouldn't light up (4.5v divided by 2 leds is 2.25v each.) If the 5 LEDs were connected in series to the batteries that'd be (4.5v divided by 5 or) 0.8v per LED and they wouldn't light either (they need at least 3.2v-3.4v as you said.)

But by putting them in parallel, you can provide that 4.5v to all 5 of them, and that chops the current down (to a fifth.) Most LED's can operate at between 20mA and 40mA. If you give them more than that, they flame out (quite brightly with those 12000mcd white LED's too!)

So whatever the total current those 3 AA cells generate, divided by 5 for the 5 LED's, it must be below 40mA. I'm guessing it's around 150mA?

The reason the resistor is in most LED circuits is to chop the current down below 40mA or so to prevent flame-out.

Forgive me if this is all already known and I'm misunderstanding you. Like I said... I get excited and pedantic when I can apply my knowledge.

I considered this thought that the internal resistance of the batteries was limiting current. Yet I saw two counter indicators to this idea.

AA alkalines can pump out at least a couple amps of current. So it would take about 40 to 100 LEDs in parallel to just begin to load them down to where the LEDs began to protect themselves.

The other thing is that the batteries in this flashlight last decades of hours. If they were being loaded down, that wouldn't be possible. I could be missing something.

I think the answer lies in the internal resistance of the LEDs. Maybe extra is added on purpose.

[rshayes]

The current-voltage plots that I have seen for the "white" LEDs seem to be that of a diode with a definite amount of series resistance. The most common way of fabrication Gallium Nitride and Indium Gallium Nitride devices seems to be epitaxial growth on a sapphire substrate. The active layers may be just thick enough to bring the series resistance down to a tolerable level rather than an insignificant level. This series resistance will increase the heat dissipation of the diode and reduce its efficiency.

I can well imagine a manufacturer buying high resistance diodes from the LED manufacturer at a cheaper price, testing them, and then selecting strings of diodes to have the series resistance needed for current limiting.