Robin Graham | 01/11/2022 23:08:02 |
1089 forum posts 345 photos | 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 |
Clive Foster | 01/11/2022 23:44:15 |
3630 forum posts 128 photos | 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 |
Huub | 01/11/2022 23:52:51 |
220 forum posts 20 photos | 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
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Joseph Noci 1 | 02/11/2022 07:36:14 |
1323 forum posts 1431 photos | 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...
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Michael Gilligan | 02/11/2022 08:10:12 |
![]() 23121 forum posts 1360 photos | 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 |
Nealeb | 02/11/2022 08:24:56 |
231 forum posts | 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_G | 02/11/2022 08:25:17 |
![]() 260 forum posts | 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 Gilligan | 02/11/2022 09:08:00 |
![]() 23121 forum posts 1360 photos | 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 Foster | 02/11/2022 09:41:20 |
3630 forum posts 128 photos | 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 Kyte | 02/11/2022 10:07:14 |
![]() 3445 forum posts 62 photos | 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. 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 Graham | 03/11/2022 01:13:30 |
1089 forum posts 345 photos | 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. |
SillyOldDuffer | 04/11/2022 10:01:58 |
10668 forum posts 2415 photos | 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.
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 1 | 04/11/2022 10:27:40 |
1323 forum posts 1431 photos | 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 Gilligan | 04/11/2022 10:32:54 |
![]() 23121 forum posts 1360 photos | 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. 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 Farr | 04/11/2022 10:43:11 |
![]() 3988 forum posts 1799 photos | 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 Kyte | 04/11/2022 10:46:39 |
![]() 3445 forum posts 62 photos | 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 Kyte | 04/11/2022 10:55:36 |
![]() 3445 forum posts 62 photos | 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 Kyte | 04/11/2022 10:59:46 |
![]() 3445 forum posts 62 photos | 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 Wyatt | 04/11/2022 13:41:42 |
![]() 19226 forum posts 749 photos 86 articles | 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 Gilligan | 04/11/2022 14:06:15 |
![]() 23121 forum posts 1360 photos | 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. |
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