An elementary electronics question.

[Robin GrahamParticipant @robingraham42208]

It's Panto time again – this time I have to make a prop which involves some electrical parts running from 12V DC and also some LED fairy lights which run from a 3V (2xAA) battery pack. It would be good to have a single (12V lead acid) battery and a single switch to run  the whole thing.

This is the fairy light circuit:

OK, the numbers don't work out exactly, just measurements with what I have, but ballpark.

My first thought was to use a Zener diode to replicate the 3V from the AA batteries:

but (a) I'm unsure as to how to calculate the value of R1 (b) I'm not sure if I need R2 – it's just there to limit the current through the LEDs, but presumably that could be done by a correct choice of R1? And (c), maybe this is the wrong approach altogether – I'm an ignoramus about this sort of thing. Maybe a simple potential divider would be easier?

Any advice would be welcome.

Robin

 

Edited By Robin Graham on 01/11/2022 23:08:38

Edited By Robin Graham on 01/11/2022 23:09:57

[Robin GrahamParticipant @robingraham42208]

[Clive FosterParticipant @clivefoster55965]

R4 simply has to be sized to carry the LED string current, plus a bit more to set the Zener safely in its operating range, with a nominal 9 volts across it.

But a fiver or so will get you a DC-DC 12 volt to 3.3 volt step down converter off E-Bay which will be rather more efficient than a resistor and zener set up.

Or you could use a linear regulator IC such as the venerable LM317 et al.

Back in the day I did a similar job, albeit 12 to 6 v, using a voltage controlled PWM oscillator, power transistor and decent size capacitors. Stiil have the circuit somewhere. Quite simple but I'll bet its not cost effective against an E-Bay special DC-DC device.

Clive

[HuubParticipant @huub]

You want the led running on 12V. The current = 0.27A. Add a second resistor (serie) that has a voltage drop of 9V at 0.27A. That is a resistor of (9 / 0.27) 33R0.

So you don't need a Zener.

You could replace the 2R7 resistor by a 39R0

[Joseph Noci 1Participant @josephnoci1]

Robin,

Is your sketch representative of the inside of the device? – ie, is there only one LED fed from the 2.7ohm R? If so, 270mA is very high for basic single LEDs, even hi-bright types. Max is typically 80mA , and often down to 20mA.

As LEDs start conducting ( and glowing) , the actual knee voltage of the LED is well defined. A few 10's of mV above that knee and the current through the LED increases dramatically. That knee is typically 2V for red leds, and 3v for green and white leds. I suspects your leds are white? The 2.7ohm resistor may almost just as well not be there – @270mA it drops 0.1v, hardly limiting the led current, and that is why the current is so high. A small variation of the two cells voltage will cause big variations of LED current in this case.

The better options is as has been suggested, but if it is a single led after the 2.7 ohm resistor, then rather limit the current to maybe 20-30mA – so @ 12V supply, 3v across the led, needs a 270 to 330 ohm resistor.

If there are leds in parallel, then you need to work is so that there is say 30mA per led and use total current to get the resistor value.

Led's in parallel are not recommended, some will be brighter that others, but maybe the fairies won't care…

[Michael GilliganParticipant @michaelgilligan61133]

I have had very satisfactory results from modules that use the LM2596 Adjustable Buck Converter

… available, with or without LED voltage displays, from numerous ebay sellers

Just search for LM2596 and browse through the offerings

MichaelG.

.

Ref. Datasheet [for the chip] is available here

https://www.ti.com/product/LM2596
 

Edited By Michael Gilligan on 02/11/2022 08:15:01

[NealebParticipant @nealeb]

I would say that it's a case for an engineering approach, not high-tech! The need is for a short-term solution for an environment where there is plenty of power for the duration needed – the lead-acid battery is already there, can be topped up between performances if needed, and I'm guessing that it will only run for a few hours a day anyway. So, stick an appropriate series resistor in (as already shown and calculated) and that will do the job fine for minimal time and cost.

In other circumstances a DC-DC converter would be the ideal solution, but not here, perhaps.

But check the wattage rating of that series resistor…

Edited By Nealeb on 02/11/2022 08:25:56

[Andy_GParticipant @andy_g]

I'm with Huub – make R1 (in your sketch) 33 ohms and delete the zener diode.

Note that it will be dissipating about 2.5W so will need to be suitably rated (say 5W minimum) and is likely to get warm even so.

[Michael GilliganParticipant @michaelgilligan61133]

Posted by Michael Gilligan on 02/11/2022 08:10:12:

I have had very satisfactory results from modules that use the LM2596 Adjustable Buck Converter

… available, with or without LED voltage displays, from numerous ebay sellers

Just search for LM2596 and browse

.

Not arguing with any of the recent posts, but I would just mention that the cheapest I found, just now, was £1.21 with free postage.

MichaelG.

[Clive FosterParticipant @clivefoster55965]

Michaels last post in yet another example of how hard it is for older penguins like me to keep up with the price performance ratio of modern electronics from China.

These days a proper engineered solution can often be gotten for less than the simpler DIY, minimal discrete components, solution that used to be all that ordinary folk could afford.

One minor issue is that there are reports of the "LM2596" chips in cheaper examples being re-badged "LM2576" versions. Fundamentally the same but running at a lower switching frequency, 50 kHz rather than 150 kHz, which generally matters not at all but efficiency may well be lower. Re-badging has apparently been worth it because the 2596 chips can be sold at higher prices than the 2576.

Hardly seems worth the effort to me but the economics of mass production, especially Chinese mass production, make my head hurt.

Clive

[Martin KyteParticipant @martinkyte99762]

For this particular application a simple series resistor to limit the current to 270mA.

However just for general information there are a number of 2 pin constant current diodes available in at various values of current such as the following offering, that make for simple LED driving ccts.

**LINK**

Thy can be run in parallel to increase the current drive or at a pinch in series to share power dissipation and increase the voltage range.

regards Martin

[Robin GrahamParticipant @robingraham42208]

Thanks for replies. I think that this, for me, is an example of not seeing the wood for the trees. I became focused on the idea of making a 3V source from a 12V battery without thinking about the actual application.

It's been educational – I didn't know that DC-DC voltage converters were a thing!

I now see that a 33 Ohm resistor in series with the fairy lights will do the job. Boringly!

Robin.

[SillyOldDufferModerator @sillyoldduffer]

There is a problem with the simple circuit, which is a risk of catastrophic failure.

The circuit is required to feed 2.2V at 0.27A to a fairy light string of parallel LEDs. Fed by a 12V battery, the resistor is calculated to drop 12-2.2 = 9.8V at 0.27A. From Ohms Law R=V/I so R=9.8/0.27 = 36.3ohms. Also, because Watts = VI, the resistor has to be big enough to dissipate at least 9.8V*0.27A = 2.65W

The dissipation isn't critical, anything bigger will do : I'd use a 5W type.

36 ohms isn't a standard value, so it's tempting to use the nearest resistor which is 33ohms.

What could possibly go wrong? What follows is a simplification, see Joe's comments above.

1. V=IR, so 0.27A * 33 = 8.2V, putting 3.8V on the LEDs rather than 2.2V, causing them to draw more current and get hotter. Exactly how much hotter depends on the characteristics of the LEDs, but they are all being over-driven.
2. How accurate is the 12V battery? A nominally 12V Lead Acid Battery is actually 13.8V when fully charged. If so, the 33ohm resistor only drops 8.2V, putting 5.6V on the LEDs, further over-driving the LEDs and making them even hotter
3. Sooner or later one of the over driven LEDs is likely to fail. This reduces the current being drawn through the resistor, causing the voltage on the LEDs to rise again. The extra stress is likely to pop the next weakest, putting yet more volts on the survivors. A chain reaction is likely.

The risk would be reduced by using a 39 ohm resistor, but a potential divider circuit is really too simple if any level of reliability is needed. Should be OK for a theatrical production provided the plot doesn't depend on them! Definitely not acceptable in an emergency lighting system.

A better answer is a voltage or current regulating circuit, like those already listed. These ensure that a chain of LEDs aren't over-driven by input voltage changes or as a result of LED failures. The switching type are also more efficient, not wasting a couple of watts in the dropper resistor.

Most educational this forum. I'd no idea the 'diode' devices identified by Martin existed! They're purpose made for this problem – some rather complicated electronics packaged inside a 2-pin device.

In practice, for convenience in a non-critical situation, I'd be inclined use a 39 ohm resistor and make sure the battery is 12V, not 13.8!

Dave

 

 

 

 

 

Edited By SillyOldDuffer on 04/11/2022 10:03:59

[Joseph Noci 1Participant @josephnoci1]

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

[Michael GilliganParticipant @michaelgilligan61133]

Posted by Martin Kyte on 02/11/2022 10:07:14:

[…]

However just for general information there are a number of 2 pin constant current diodes available in at various values of current such as the following offering, that make for simple LED driving ccts.

**LINK**

Thy can be run in parallel to increase the current drive or at a pinch in series to share power dissipation and increase the voltage range.

.

That’s a very useful looking device, Martin … thanks for the link

MichaelG.

.

Edit: __ I have previously tried the infinieon-bcr402r but yours looks simpler to implement.

Edited By Michael Gilligan on 04/11/2022 10:48:43

[Nicholas FarrParticipant @nicholasfarr14254]

Hi Dave, you can always use resistors in series to obtain the correct value, e.g. two 18-ohm resisters would give 36 ohms, and you could also include a 0.33 one to make it closer to the calculated value, or a 0.47 one if the two 18 – ohm ones are on the lower side of their tolerance. If the battery is at 13.8 volts, just use a 39 and a 4.7- ohm resisters.

Regards Nick.

[Martin KyteParticipant @martinkyte99762]

Dave’s cct is safer than he thinks as the LEDs are reverse biased as drawn so will not take any current to speak of.

Apart from the somewhat bad practice of running LEDs in parallel which generally means some are brighter than others due to mismatched forward voltages the first parameter of interest is the forward voltage across the string. The OP states this as 2.2V so the current limiting resistor will see 12 – 2.2 = 9.8V

For 270mA we need 36.3 ohms dissipating just under 2.7 watts. A series of 6 6R2 0.6W resistors would fit the bill. In the event of a led failing the forward voltage of the string will remain the same so the available current will be the same albeit shared between one less led. Not likely to cause great harm assuming the LEDs are not being driven to the max current.

However if a lead acid battery is to be used then the above calculations need to be redone with a max terminal voltage of 13.8V The current limiting resistance needs to increase to about 43 ohms

adding an extra 6R2 will do it.

regards Martin

[Martin KyteParticipant @martinkyte99762]

It’s amazing how many convenient devices have been produced for the automotive industry which doesn’t like complexity. Add one led and a series current limiting diode and its job done. There are also negative temperature coefficient heater devices in a range of temperatures for jobs like De-misting mirrors. Select your 10 degree flat disc device, clamp to rear of mirror add 12V and off you go. It just runs up to temperature and sits there.

regards Martin

[Martin KyteParticipant @martinkyte99762]

Incidentally Dave’s opening statement is misleading. The requirement is to supply 270mA to a string of diodes whose forward voltage is 2.2V

regards Martin

[Neil WyattModerator @neilwyatt]

I'd use a current limiting resistor for each individual LED, and power them all using a cheap variable voltage module.

Or to make life easy, use strings of four LEDs at ~ 8.8V with a current limiting resistor oif the 12V supply – but calculate the resistor value according to the actual voltage from the supply. If it's fairly accurate or a little high, 15 ohms will probably be near enough for jazz – it’s a prop not a NPL project!

Neil

[Michael GilliganParticipant @michaelgilligan61133]

Posted by Neil Wyatt on 04/11/2022 13:41:42:

I'd use a current limiting resistor for each individual LED, and power them all using a cheap variable voltage module.

Or to make life easy, use strings of four LEDs at ~ 8.8V with a current limiting resistor oif the 12V supply –

[…]

.

Surely not [?]

Have you seen how strings of LED fairy lights are constructed ?

The end result would not justify the work involved.

MichaelG.

[SillyOldDufferModerator @sillyoldduffer]

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..

…

Gremlins, not fairies!

[Martin KyteParticipant @martinkyte99762]

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

regards Martin

[Jon LawesParticipant @jonlawes51698]

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.

[Martin KyteParticipant @martinkyte99762]

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

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