OT: resistor in LED light string
So we try to get on the green bandwagon and get LED Christmas lights to save precious electricity. But early failure of what is supposed to be a long life product irks me. Heck, I'm persistent enough to go through a string of mini-lamps and find the bad bulbs. I'm my wife's hero! So I cannot toss out a string of LED's without investigation.
So we notice this stink coming from the mini Christmas tree and the LEDs are off. I find this lump on the cord and cut it open and see what appears to be a stinky resistor burned up. It appears to be about the 1/4 watt power variety. But I have no idea what resistance it should have. Maybe I could replace it?
Why would something like this 'burn up'? The application seems pretty straight forward......but it seems like someone, somewhere (well, China actually), under-engineered something. This was only one string of LEDs, without any further strings daisy-chained on.
Sounds like a basic lowest budget dollar store product......
Each LED 'drops" a certain amount of "forward" voltage as it works. How much depends on what kind it is. Old-time LEDS are about 1.25 to 1.5 V per LED. Newer high brightness types can be 3 or 4 volts.
Evidently, the remainder of the 170V peak/120Vrms voltage was "dropped" by the "resistor", assuming it was one. Other things are that size also, such as small rectifiers. No telling what the chinese designer thought about, assuming thought was actually used to "design" that.
At this point, I'd not bother further. I suspect so many things are wrong, starting with the basic design, that you are better off to bag it. But you can, if you want, count up LEDS, and see how much voltage each had to handle.
I say "had", since I very much suspect the "resistor", diode, or possibly fuse, opened when enough LEDs failed.........
Yet another reason not to leave the house or go to sleep while a cheap chinese electrical device is still plugged-in.
Actually, the resistor is there to limit the current flowing through the LED. At least with DC power, that is. If the light string was using AC power stepped down, and the manufacturer tried to use a simple reisitor to limit current, it's no wonder the string fried the resistor. But I can't see even the Chinese putting an AC current through LEDs. I might see them using a half-wave rectifier, but that would make the LEDs flicker as the clipped part of the waveform passed through them.
The simplest explanation is that the Chinese simply used to small a resistor in terms of wattage dissapation and it cooked. Replace it if you want, using Ohm's Law to set the value. I don't know how much current the new ultrabright LEDs want to see, but regular LEDs used to want to see only about 5 mils.
Buy another set and cut that one out to identify it........?
That is the same thing........ "drop voltage ~= limit current in this case......
Originally Posted by vraptor
1) that is all that was in the "lump"...
2) it ever worked for more than 3 minutes
Then it wasn't "undersized"..... so it failed for some reason, and the reason is probably the failure of some other part(s)
Even if there was more stuff in the "lump", burned parts don't "just happen".
My advice is to take that dollar store incendiary device to an appropriate disposal receptacle. I consider burned resistors to be a far better option than burned houses.
For super-cheap, the chinese, and even some US folks might well consider almost ANY crazy scheme. I've seen quite a number of them. Bad design, worse design, whatever. It just depends on how experienced (and so suspicious vs optimistic) the "designer" was.
But I can't see even the Chinese putting an AC current through LEDs. I might see them using a half-wave rectifier, but that would make the LEDs flicker as the clipped part of the waveform passed through them.
35 LEDs is what I count in the string.
Take a 0-100K ohm pot and put it in the circuit set to 0. Turn it on slowly. When the LED's come back on, remove one leg of the pot and measure it.....
I use LED's on my model railroad but only one at a time. I figure a voltage drop of 1.4 VDC for each one and then just use Ohm's law to calculate the needed resistance. The resistors I use are only 1/4 watt. In your case, with 35 LED''s, obviously the Chinese designer miscalculated the power factor. You probably need at least one watt.
I try to keep each LED below 30 ma. I've yet to burn one out and some of them have been running for 25 years.
Last edited by Newman109; 12-11-2010 at 12:05 PM.
This was my quickest solution to installing LED indicators as well. You can even decide how bright you want the light. The light will decide the mean time before failure.
Originally Posted by Newman109
Just to clarify. Set the pot to the maximum resistance (vs. zero) and then decrease the resistance slowly. Otherwise, you'll fry the circuit as soon as you turn the power on.
Originally Posted by Newman109
In general, however, I agree with JST. Resistors, usually, don't get burned for no reason. There's something else going on.
Originally Posted by MichaelP
Good catch.That's what I meant to say. Most pots have two sides. Start with the high resistance side to prevent problems as you say.
Could be that something happened to the bridge that they have in the circuit...maybe just a zener diode/1/2 wave rectifier that got burnt out and took the resistor with it.
Originally Posted by HuFlungDung
The voltage drop across LED's varies, from ~1.7V for red, 2V for green, 3V for blue, 4 for white. Roughly, anyway. Better data here Light-emitting diode - Wikipedia, the free encyclopedia. So the LED string itself uses up from 59 to 140 volts.
The Chinese may have put a series diode in that failed. About 5V reverse voltage will toast an LED.
35 red LEDs would drop ~60V at say 10 milliamps (20 is probably safe). So 60V needs to be dumped; if a resistor was used on a nominal (rms) 120V line, then you'd need 60V/.020 amps = 3KΩ. (20 mA since the ac will only flow for a half-cycle; then the average will be 10 mA).
That average 10 mA x 60V is .6W; a fairly hefty resistor.
35 white LEDs would have about no compliance in the budget; you'd need to drop a very small voltage, hard to say what resistance but really small.
Is there an DC adapter block (wall wort)? There ought to be, in which case reading off the DC output would be a helpful start.
Nope, these run straight off the AC.
What (if anything) is done to protect the LED from reverse breakdown during the "off" half of the AC cycle? LEDs generally have a very low PIV rating, and wouldn't like the high inverse voltages present when reverse biased across a 120VAC power line.
The chinese designer may have added up the PIVs of all 35 lights and concluded that he was safe...........
That's what I don't know about the the component that burned up. I've seen diodes that have about the same shape as resistors.
Originally Posted by N2IXK
I assume that the LEDs are in series. If one fails, I guess it shorts, instead of becoming open? But then the forward voltage becomes higher for all the remaining ones, does it not, or is there still a residual resistance in the shorted one?
I suppose there is no way to build in protection of the string with a fuse? At least I don't see a fuse, unless there is something built into the molded plug.
The forward voltage drop of an LED is more or less independent of current (up to a point). If LEDs in a series string were to start failing shorted, the current in the remaining LEDs would go up, but the voltage dropped across each one would remain pretty much the same. The difference in voltage would get dropped across the series resistor, possibly overheating it.
Many holiday light strings have a tiny glass fuse installed in the plug.