Why aren't many / most LED light bulbs dimmable?

In , Jeff Thies wrote in part:

White LEDs fade mainly, almost entirely, from the phosphor degrading.

BTW, the usual white LEDs have phosphor over blue LED chips, not UV ones.

I am also noticing many LED traffic lights with some LEDs out. That appears to me to be, as you say, broken connections. I don't know where they're breaking. One thing I notice is that affected LEDs are disproportionally at the edges of the traffic lights around where I live. This makes me suspect stress concentrated at the edge of a PCB that the LEDs are on, so I wonder how well the PCBs are fitted to what they are mounted to.

Reply to
Don Klipstein
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This is the LED flashlight guts. You can see there are no resistors.

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I would always use a resistor in anything I made, even if it was fairly low resistance.

I broke one of those LED lights that were all the rage on TV for a while. The ones that respond to a magnet. There are 2 LEDs, 2 transistors, a reed switch and 4 SMT resistors. The 2 transistors are in a darlington and 2 of the resistors are in series with the 2 LEDs. This seems to be a fairly well thought out design.

Reply to
gfretwell

It's not heat. The I-V curve is an exponential, just like any other (semiconductor) diode. Heat will flatten out the ideal I-V curve further but that generally isn't seen until the device is operating outside of normal. The effects of heat can be seen if you change current (and measure the voltage) quickly.

Experiment: Take your diode and a resistor, connected in series with a variable voltage source, as such.

+---------------+----------o + | | | | | .-. | | I | | | R Vr + | | | | | V '-' | /+\ | | V ( ) +----------o - \-/ | + | V -> | | - Vled | | | === +----------o - GND | === GND

Vary the voltage source and plot Vr and Vled in a spreadsheet. Calculate (using the spreadsheet) I (=Vr/R) and then, again using the spreadsheet, graph I (Y-axis) against Vled (X-axis) using the "scatter" option. This graph shows the characteristic I-V curve of this diode and can be used to set the bias resistors for any supply voltage (if the current is within the range of the graph). Note that each diode will be somewhat different and diodes from each batch (or "lot") can be significantly different. You might notice that LEDs of the same part number and manufacturer may be "grouped" by brightness (and even color). Two LEDs with the same specs may look entirely different if put next to each other.

The internal resistance of the battery is the ballast. Measure the unloaded voltage (open-circuit or Voc, below) of the battery and the voltage and current (this will be difficult) of the battery/LED combination. The internal resistance of the battery can be calculated from (Voc-Vled)/I.

Reply to
krw

The battery *is* the resistor.

Reply to
krw

HF buys their stuff from where? ;-)

The problem that I have with it is counting on the internal resistance of the battery. As long as the same *type* of battery is used there isn't a real problem. If another type is used things can go very wrong.

Reply to
krw

Oh, certainly. Good catch.

Reply to
krw

That's a current-sharing nightmare waiting to happen. I suppose the LEDs are all from the same batch, so at least in theory, it works. ...well enough for cheap Chinese junk, anyway.

Thinking about it a little, the temperature coefficients are in the right direction so at least the thing isn't going to run away.

Reply to
krw

I still don't want to lose "night sight" when I stumble from one room into the next, at night. The light doesn't alter my Braille ability, though. ;-)

Reply to
krw

I thought you wanted task lighting under kitchen cabinets. That has to be quite bright, even in a well lit room. A couple of watts isn't going to cut it.

Series strings of LEDs with a ballast.

I don't like them because some day someone is going to put the wrong battery in the things.

Agreed, but too simple is not. ...primarily because people are. ;-)

The *waste* heat may go *up* as dimming starts. The waste heat is V^2/R.

They don't. Voltage regulation would serve little purpose.

Right.

You should easily be able to see a switcher's signature. If you have a scope, you can also see the effect of heating by driving the LEDs from an AC source.

Yep, a few million of them will really screw with the power factor. ;-) You'll see some oddities with LED lamps, too.

Reply to
krw

Just gotta control the PWM better. ;-) A PWM current source is typically used.

Reply to
krw

This is a Fluke 8060. I doubt that is a big deal and it was constant between the LED with and without the pot. I suppose there is still some drop tho. I have a hand full of Hi Intensity LEDs from those TV wonder lights that I am going to play with tonight. I made a light bar out of some wood scrap that matches my cabinet and I am going to see what happens. I do believe, if you are dimming a LED array, low voltage may be your friend, both from an application standpoint and a safety/UL listing standpoint. Things are pretty lenient on the load side of a listed class 2 power supply. I figure I will start out with a ballast resistor selected to apply about 15ma "all in" and dim from there. From my experience with the flashlight I think a 25 ohm pot will do a pretty good job and I have one.

Reply to
gfretwell

If the voltage drop across the LEDs is somewhat fixed at a hundred and sumpin and your ballast is picking up the slack, that slack can vary quite a bit with the normal tolerance in power. A LED string that works in the browned out north (108v or so) is going to be running pretty hot here with my nominal 124v. That ballast needs to give you a fairly reasonable current over a wide swing. I almost have to believe they have real solid state current or voltage regulation, not just a resistor. If so, that is the problem with dimming them.

Reply to
gfretwell

Looking at the spec (for the 8062, looks to be the same as the 8060A), the burden voltage on the 2A range (next down is 200mA and you said 211mA) is .9V (.3V on the 200mA range). That's at FS, but I'd want to measure it to see how much of that is resistive and how much is fixed. That'll certainly throw a monkey wrench into your readings.

DC is your friend. PWM it. ;-)

Reply to
krw

But there is your mistake. It is *NOT*.

Why? All they need to do is cover the maximum, so it doesn't burst into flame.

~:-/

Reply to
krw

" snipped-for-privacy@att.bizzzzzzzzzzzz" wrote in news: snipped-for-privacy@4ax.com:

Communist China. :-)

they come with "heavy duty" AAA carbon-zincs,but I also tried alkalines with them,and they were brighter with the carbon-zincs. they may be UNDERdriven. I still have two more coupons for free ones,and the local HF store is only a short walk away! Plus,I see the ads are still coming out with the coupons.

they're nice little flashlights for the "price"! B-)

Reply to
Jim Yanik

" snipped-for-privacy@att.bizzzzzzzzzzzz" wrote in news: snipped-for-privacy@4ax.com:

Well,you KNOW this guy is not going to use a PWM circuit with his wallwart,just a pot(rheostat).

Reply to
Jim Yanik

" snipped-for-privacy@att.bizzzzzzzzzzzz" wrote in news: snipped-for-privacy@4ax.com:

the usual "bright white" LED found in these flashlights is a ~0.1W light. ~3.5 @20ma.

26ma seems a bit high.

don't forget that 120VAC rectifies out to ~170 VDC,not 120 VDC. for that string and a 350 ohm resistor,your LEDs are not going to last more than a millisecond. BTW,the "bright white" T1-3/4 LEDs work out to ~3.5v forward voltage.

not gonna happen;only 3 AAA cells will fit,there's a plastic battery holder inside the flashlight barrel.

Reply to
Jim Yanik

" snipped-for-privacy@att.bizzzzzzzzzzzz" wrote in news: snipped-for-privacy@4ax.com:

120VAC is RMS voltage,not PEAK voltage. multiply by 1.414 if you want to find the DC voltage after rectification.
Reply to
Jim Yanik

Only with a capacitor filter. It'll still have a 170V peak but the RMS will still be ~120V. The current waveform will be messed, though.

In *your* freebie HF flashlight, perhaps. In the case of the HF flashlights, the zinc cells may be the worst case. I don't see evidence that this is the case in every flashlight out there.

Reply to
krw

Well, he's not going to PWM the wall wart (voltage source), either.

Reply to
krw

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