An elementary electronics question.

[SillyOldDufferModerator @sillyoldduffer]

Posted by Martin Kyte on 04/11/2022 17:22:13:

…

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

[Andy_GParticipant @andy_g]

Posted by SillyOldDuffer on 04/11/2022 18:16:14:

R=36.3ohms.

…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!

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…).

[Robin GrahamParticipant @robingraham42208]

Posted by Joseph Noci 1 on 04/11/2022 10:27:40:

Amazing how many an unsuspecting bloke asks a 'simple' question on this forum, and about his only option is to run away, terrified..

That's what happens when you believe in Fairies..

Edited By Joseph Noci 1 on 04/11/2022 10:28:16

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.

[Michael GilliganParticipant @michaelgilligan61133]

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.

[Martin KyteParticipant @martinkyte99762]

Posted by Michael Gilligan on 05/11/2022 07:12:23:

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

[Michael GilliganParticipant @michaelgilligan61133]

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.

.

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

[Martin KyteParticipant @martinkyte99762]

That’s what I said.

[Michael GilliganParticipant @michaelgilligan61133]

Posted by Martin Kyte on 05/11/2022 09:57:15:

That’s what I said.

.

True … You said that and more

So that’s maybe two of us agreed

MichaelG.

[Anonymous]

Posted by Martin Kyte on 04/11/2022 17:22:13:

…The resistor is there to limit the current….

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

[Martin KyteParticipant @martinkyte99762]

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

[SillyOldDufferModerator @sillyoldduffer]

Posted by Martin Kyte on 05/11/2022 13:00:09:

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

[Martin KyteParticipant @martinkyte99762]

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

[NealebParticipant @nealeb]

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.

[Anonymous]

Posted by Nealeb on 05/11/2022 20:56:20:

…via the existing per-LED resistors.

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

Andrew

[Robin GrahamParticipant @robingraham42208]

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

[NealebParticipant @nealeb]

Posted by Andrew Johnston on 05/11/2022 21:30:17:

Posted by Nealeb on 05/11/2022 20:56:20:

…via the existing per-LED resistors.

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

Andrew

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!)

[Robin GrahamParticipant @robingraham42208]

Posted by Nealeb on 06/11/2022 07:52:22:

Posted by Andrew Johnston on 05/11/2022 21:30:17:

Posted by Nealeb on 05/11/2022 20:56:20:

…via the existing per-LED resistors.

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

Andrew

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. […]

Fair enough – my initial diagram wasn't clear enough, my bad.

To draw a line under this (for me, at the moment) all replies have been useful, I can now see at least 4 (and counting) ways of doing what I wanted to do. That was never entirely the point though – I wanted to use the project to learn some basic electronics. I actually invested in a 3A 18V bench PSU which is one of the best toys ( per &pound I've ever bought.

I've spent some time drawing out simple circuits with non-linear components and checking them in practice with the PSU and a multimeter. I now realise why Horowitz and Hill called electronics an art!

Robin.

[Jeff DaymanParticipant @jeffdayman43397]

For basic electronics and circuits, there are a series of little handbooks available by Forrest Mims III from the USA. Radio Shack stores in USA and Canada sold these books from the late 1960's to the early 1990's. If you google Mr Mims' name you can find many of the books as PDF downloads. Worth a read, and lots of very basic circuits are included for learning. Just food for thought.

[Anonymous]

Posted by Nealeb on 06/11/2022 07:52:22:

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

Shown in the very first post….

I don't think the original schematic is representative. Fairy lights implies a number of LEDs. It seems unlikely that each LED is drawing 270mA. Conversely if each LED is taking 10mA or so, then each 2R7 series resistor will only drop a few tens of millivolts.

Presumably we are talking a consumer product here, so it seems unlikely the manufacturer would waste time and money adding a resistor to each LED when one will do.

Andrew

[SillyOldDufferModerator @sillyoldduffer]

Posted by Robin Graham on 08/11/2022 00:10:26:

Posted by Nealeb on 06/11/2022 07:52:22:

Posted by Andrew Johnston on 05/11/2022 21:30:17:

Posted by Nealeb on 05/11/2022 20:56:20:

…via the existing per-LED resistors.

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

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. […]

Fair enough – my initial diagram wasn't clear enough, my bad.

…

At least Robin's circuit didn't have all the LEDs upside down like mine did!

[Anonymous]

Posted by SillyOldDuffer on 08/11/2022 11:00:06:

At least Robin's circuit didn't have all the LEDs upside down…

If it makes you feel better I've been there, done that, too.

Andrew

[Michael GilliganParticipant @michaelgilligan61133]

Posted by Andrew Johnston on 08/11/2022 10:55:11:

[…]

Presumably we are talking a consumer product here, so it seems unlikely the manufacturer would waste time and money adding a resistor to each LED when one will do.

.

… which is, incidentally, why driving the whole string, including its resistor, from an appropriate voltage source should be no riskier than using the designer’s Qty. 2 AA cells

MichaelG.

[NealebParticipant @nealeb]

I think the "2R7" value might have been down to an inaccurate measurement – based on measuring the whole string? Seemed reasonable to me that there were resistor/led pairs along the string for current-balancing based on a variable battery voltage, and I'm pretty sure that that is what I saw on some led tapes I installed for workshop lighting – surface-mount devices embedded in the tape next to the leds themselves, although that might have been one resistor for 2-3 leds as it runs off 12V. Given that the Chinese produce the buck converter modules already discussed for less than the price (including postage) we would pay for the components, the incremental cost of adding SM resistors can't be that much, even at a very price-sensitive end of the market!

A resistor in series is a pretty crude current source but better than a constant voltage source in this case?

[Michael GilliganParticipant @michaelgilligan61133]

Looks like we need to see the string of lights that Robin is using

A picture is worth a thousand assumptions !

MichaelG.

[Martin KyteParticipant @martinkyte99762]

LED tapes are a slightly different animal. They are often designed to slice off the length you want every 3 or so LEDs so the devices are bunched together with their own current limiting resistor and the supply running down the whole length with the clucters in parallel. Fixed strings are designed to be run as a unit so can have all sorts of internal arrangements dependent on what is convenient and or cheapest.

regrds Martin

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