LED Christmas Lights Conundrum!

[JohnW2]

Hi there. I have a number of sets of ‘battery-operated’ (although I don’t use them with batteries) LED ‘Christmas lights’, each consisting of 20 lights. Although they have various different coloured envelopes, I suspect/imagine that the LED elements are actually white-ish.

The ‘lamps’ are wired together in a string with just two conductors between them (twisted pair). The ‘control box’ which comes with each set contains just three AA batteries (hence 4.5V), a switch and a 16Ω resistor (in series with the 2-conductor output from the box). When one of these sets is running, the total current is about 88mA, with almost exactly 3.0V across the outgoing supply to the LEDs (which is credible for the Vf of white-ish LED elements), and hence around 1.5V across the resistor. I therefore can but presume that the situation is as per the diagram below, with all the LED lamps (which could contain resistors, but I doubt it) wired in parallel, each drawing about 4.4mA.

So far, so good. However, on ‘checking’ them in anticipation of the upcoming festive season, I find that, for one set (which was working fine last Christmas) the first 4 LED are not working. Try as I may, I can think of no way in which they could have been wired such that the death of one LED would affect any others, yet for the first 4 (and no-others) to die during a period of storage would seem to be almost beyond belief as a ‘pure co-incidence’.

Am I missing some possible explanation (other than an almost unbelievable ‘co-incidence’)?

Kind Regards, John

 

[AndyPRK]

sometimes these cheap LED things are wired in series, having 3 LEDs in series and then another 3.
Helps with voltage drop,
maybe 4 in series in your case

I think you doing to have to measure across the faulty leds on diode mode.

1) see if all 4 conduct.
2) you should see them light up dimly

 

[JohnW2]

sometimes these cheap LED things are wired in series, having 3 LEDs in series and then another 3.
Helps with voltage drop, ... maybe 4 in series in your case

Under other circumstances (higher supply voltage), that would be the obvious explanation.

However, the Vf of each element is likely to be at least 2.5V, so even two in series would not light with only 3V across them. Indeed, even without the resistor, they would only be getting 4.5V (the battery voltage) - so, again, even two in series would probably not light, and more than two certainly wouldn't.

Kind Regards, John

 

[Nozzle]

Vf of a diode is ~0.7V..

Or do you mean "...of each element" to mean 4 LEDs as a single element?

Nozzle

 

[bernardgreen]

Vf of a diode is ~0.7V..

that is true only for normal rectifying diodes.

Lighr Emitting diodes have Vf as high as 6 volts

 

[JohnW2]

Vf of a diode is ~0.7V..

That's an ordinary silicone diode - for a white-ish LED element, it will be well over 2V, possibly over 3V.

Edit: Ah, I see that bernard has posted some 'chapter and verse' - thanks!

Kind Regards,
John
Edit: typo corrected

 

[DetlefSchmitz]

I'm sure you will be embarrassed when you notice the typo, John!

 

[JohnW2]

I'm sure you will be embarrassed when you notice the typo, John!

OK, having just read the whole thread again twice, I give up (and the more times I read it, the less likely that I will notice my own mistake'!)! Where is this embarrassing typo?

Kind Regards, John

 

[DetlefSchmitz]

Oh sorry, it stuck out for me - 'silicone diode'. Minor point unless you have OCD!

 

[JohnW2]

Oh sorry, it stuck out for me - 'silicone diode'. Minor point unless you have OCD!

Oh, I see. It proves what I said - I read it at least three times (probably far more, since it was the most recent post before your comment), without being able to 'see' my own typo! Thanks. Now corrected!

Kind Regards, John

 

[John D v2.0]

I am guessing you've looked carefully at the circuit, but a possible explanantion is that there is one dropper resistor for every 4 leds. If one LED goes open circuit the other 3 will get extra current which would burn them out quickier. Or if one goes short circuit it would drop the voltage.

Back nearly 20 years ago I used to make bike lights with loads of the small super bright leds packed into an old bike light on some strip board and run them off a dynamo with a lead acid battery floating across it. I think they were some series parallel combination with one dropper resistor for the lot. I noticed that after a while a few LEDs would become dim, and before long most of them would start to lose brightness.
I'm assuming the underlying issue was the slight variations in forward voltage would take out the higher current ones, which would redistribute the current elsewhere. The solution was probably to have more resistors, but on christmas lights you're not going to return them the year after.

 

[JohnW2]

I am guessing you've looked carefully at the circuit, but a possible explanantion is that there is one dropper resistor for every 4 leds. If one LED goes open circuit the other 3 will get extra current which would burn them out quickier. Or if one goes short circuit it would drop the voltage.

As I said, there's just a single 'visible' resistor, and that is in the feed from the battery 'to everything'.

As far as I can see, with only two conductors between each (every) pair of LEDs, none of the 'obvious' answers based on LEDs being somehow in groups of 4 (whether in series and/or parallel) are really possible.

Much as I'm always suspicious of what appear to be 'amazing co-incidences', I fear that I'm not going to end up with any better explanation for what I'm seeing! ... with only two conductors between lights and only 3V (after VD in the resistor) going to the whole set of LEDs, I really don't think there scope for any other explanation - unless I'm being dim!

Kind Regards, John

 

[John D v2.0]

maybe the conductors have a noticable resistance, and those ones were slowly cooking unnoticed last year as the voltage drop across the wires was making the closer ones take a lot more current.

 

[JohnW2]

maybe the conductors have a noticable resistance, and those ones were slowly cooking unnoticed last year as the voltage drop across the wires was making the closer ones take a lot more current.

... but the conductors between those first four LEDs are the same ones that are still satisfactorily powering the other 16 ('further downstream') LEDs. It could, of course, be the connections between those wires and the LED elements (which I can't inspect, since everything is 'moulded') - but it would again be an incredible co-incidence if that had happened to just the first four LEDs.

... I I don't think that too much 'cooking' is likely to have happened, at any point in time, with only 88mA flowing (even though the CSA is probably very small).

Kind Regards, John

 

[John D v2.0]

I didn't mean the conductors were cooked, I meant the voltage drop along the length of the cable would be greater as you get further away.
So the LEDs which all have the same forward voltage, would get a greater current where the voltage is higher, i.e. close to the power source.