An elementary electronics question.

Gremlins, not fairies!

I thought the general consensus was use a resistor and suitable values were offered.

regards Martin

If you have multiple lights required you can put a few in series to do the voltage dropping, you only need one ballast resistor per series string. For 5 LEDs at 2.2v each it would be 2.2v * 5 = 11v, so then you only need to drop a volt or so using the ballast resistor meaning less heat dissipated.

Apologies if I have the wrong end of the stick.

The OP stated the led string was a unit.

The point many people don’t appreciate is that you need a current source to drive LEDs. The resistor is there to limit the current not to drop the voltage. The 2.2V is defined by the LEDs.
regards Martin

True, but I find it easier to calculate the value of resistor from the known voltages: we know the input is 12V and that the LEDs hold the line at 2.2V, so the resistor has to drop 9.8V at 0.27A. Rearranging Ohms Law V=IR, R=V/I so R=36.3ohms.

I think that's close enough for government work.

If 9 LEDs draw 30mA each, and one of them fails, then the current needs to be regulated at 0.24A. The resistance needed to do this is 40.8ohms, not 36, so by pulling the line down to 2.2V the remaining 8 diodes each get 34mA, a 13% overload. If a second LED pops, 7 LEDs take 39mA each, which is a 30% overload. Six LEDs take 45mA each and are 50% overloaded.

The point is that powering a line of parallel LEDs from a single resistor puts a progressively worse overload on the remaining LEDs if one happens to die. Chances are if one LED fails, then all the others will pop in turn. This could happen if the battery is 13.8V rather than 12V and the nearest preferred value of 33ohms is used rather than the calculated 36 ohms. It would be safer to go up the preferred range to 39ohms, or perhaps parallel four 150ohm resistors for 37.5ohms. Best of all, use electronics smarter than a single resistor!

Dave

Don't forget that the fairy lights already include a 2.7 ohm resistor, so an additional 33 ohm in series gives the calculated 36 ohm required (close enough, etc...).

Not terrified yet Joe, or running away. I'm clueless about practical electronics I know, and was hoping to learn something (which I have) by starting this thread.

I have now tried with a 5W 33ohm resistor in series with the fairy light chain and it works fine ( albeit running from a bench PSU set at 12.0V rather than the battery). However the resistor gets pretty hot and I'm not happy with that.

Martin - thanks for your link to the AL5809, I've been looking at the Farnell website but have so far failed to find that device listed. I'll keep looking and trying to understand!

Dave (SoD) - I'm not sure your analysis is quite right. As I understand it forward biased diodes don't conduct at all (well, maybe a little bit!) until the 'diode drop' voltage, after which they have (almost!) zero resistance. So driving with a 13.8V volt battery through a 33 ohm resistor the drop across the resistor is 13.8 - 2.2 = 11.6V, not 8.2V.

V=IR then gives the current through the resistor (and therefore the circuit) as 11.6/33 = 0.352 A. I haven't actually counted the LEDs, but there must be at least 20, so taking that as a lower bound gives <=17.5mA per LED. Which seems to be OK - a bit of research suggests that ~ 20mA is is the upper limit for red LEDs. I have cranked up the current on the PSU to 400mA (voltage is steady at 2.2V) and no failures.

Martin again, I've been reading/replying to posts in sequence, from what you say my understanding is correct.

I should say that in the context of the making this particular prop all this is probably insane - there are easier ways. It's just that the it gives me the stimulus and opportunity to learn some stuff. Back in the day, when I got my first lathe, I made a prop ship's wheel complete with spring-loaded detent to allow the wheel to be speedily demounted from the support. Nuts, but educational for me!

Robin.

I am still struggling to understand why the DC to DC converter that I mentioned is not the optimum solution.

It’s even available from a U.K. seller at £2.35 with free postage

… probably not worth putting a link here though.

.

If the answer is “because I want to learn stuff”

My response is “read the data-sheet”

MichaelG.

You could indeed use the 3V3 regulator, and more generally if you had a current sense resistor in the feedback of the adjustable version you can create a very good current regulator for driving LEDs. There are a number of buck and boost converters specifically designed for driving LEDs especially at higher currents.

regards Martin

Given that the Fairy Lights are already ‘designed’ to work adequately with a pair of AA cells, I would be happy to just use that converter as-is

MichaelG.

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Edited By Michael Gilligan on 05/11/2022 09:25:20

That’s what I said.

.

True … You said that and more

So that’s maybe two of us agreed

MichaelG.

The series resistor creates a crude current source. The larger the voltage drop across the resistor compared to the voltage across the LED the closer the arrangement is to an ideal current source.

Andrew

Exactly Andrew, I have been labouring the point to get people to think in terms of current sources not voltage sources for driving LEDs.

regards Martin

Whilst nearly all electrical things are current driven I think it's usual when designing circuits to ask 'what voltage is needed to produce the required current?'

I'm probably missing something (again!) but how exactly does thinking in terms of current sources help calculate Robin's resistor?

Dave

The straight answer is power supply voltage less the led forward voltage divided by the required led current = resistor value.

The more involved answer.
When designing led drivers especially for higher currents, current regulators usually switching to reduce dissipation with current feedback. That way you get to drive the thing with a large range of input voltage so is more convenient in terms of power supply selection as well as simply programmable current and pulse width modulated dimming.

Beginners often think you just need to connect a voltage source and get in a muddle so designing for current even with a simple resistor as a voltage to current converter is a helpful way to think right from the start.

regards Martin

I was one of the proponents of the "stick a resistor in and drive from 12V" approach, although this is only the "simple" solution if you already have a suitable resistor (or resistors if you need to parallel them for heat dissipation) to hand. If you are ordering something, then the converter is not such a bad move.

However, there is one good reason why the converter is a better long-term solution, perhaps. What if a LED fails? Simple resistor solution is going to increase current through the remaining LEDs (assuming open-circuit failure) where the dc-dc converter will be OK, a constant-voltage source feeding via the existing per-LED resistors.

Where does it say that each LED has it's own series resistor?

Andrew

Thanks for further discussion. Tonight I had another bash at at analysing the Zener circuit in my opening post and hypothesised that I could work out the value R1 by adding say 20mA (to keep the Zener happy) to the current drawn by the LEDs. Then I re-read the thread and saw that was (essentially) what Clive said in the first reply. Sorry for not taking that on board earlier Clive. Doh!

Working it out that way I don't gain much in terms of battery drain / heat dissipation in the limiting resistors by using this configuration. Wrong thinking on my part.

Looks like a buck converter might be the way to go. I had never heard of such things until now.

Or I could just notice that the switch I'm planning to use is actually DPST, so I could just go with the original AA powered fairy lights and the 'other gubbins' on 12V being operated by one  switch, which was my starting point. But where's the fun in that?

Anyhow, thanks to all for input. For me this is really more about the journey than the end, and I like to take the scenic route.

Robin

Edited By Robin Graham on 06/11/2022 01:15:36

Shown in the very first post - I assumed that it was integrated into the fairy light string as a current-regulator when running off battery.

My apologies to SOD, though - I found his post while going back to check my assertion and see that he gave a much fuller analysis of my failure scenario in a post which I had missed (I find this forum's lack of a "next unread" button a bit frustrating!)