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LED Lighting Project - Help Please !

Discussion in 'LEDs and Optoelectronics' started by harvey66, Dec 29, 2010.

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  1. harvey66

    harvey66

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    Dec 29, 2010
    I have bought several remote control "LED lights" for down the stables. They have 10 superbright LEDS in the light which run off 3 x 1.2v AA NiMh batts. I cannot find a data sheet so don't know what current they draw

    http://www.amazon.co.uk/Solar-Mate-Point-5-SM005/dp/B001Q3YIXK

    I have a main 100w solar panel & regulator which charges a 12v leisure battery for other things such as my electric fence

    I want to remove the batteries/solar panels from the lamps and wire them into the 12v circuit direct from my main solar/batt regulator - but this is 12v and the LED lamps are running off 3.6v as far as I can see

    Can anyone advise me what I would need to do to make this possible. I have been advised to use a 7805 regulator (and 1 capacitor on input and output) in each lamp to drop the circuit voltage to about 5v but would it really be that simple.....?

    Thanks for any help
     
  2. cj_elec_tech

    cj_elec_tech

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    Oct 7, 2009
    In principle, it could be that simple, but if you were to use a 7805, you would still need a resistor to drop the voltage to 3.6v
    The resistor would generate a little heat, which is waste power (not so good if battery life is an issue) but it would ensure some *current* regulation happens - this is important as LEDs are current dependent and you would be using a voltage regulator.

    Actually, now that I've thought about it, you could use a 7805 in it's constant current configuration (still need a resistor though), but you would need to ensure the input voltage doesn't go below about 8V for it to work.

    In both cases you can only calculate the resistor value once you know how much current the LEDs draw at 3.6V - you will need to measure it to get it absolutely right.

    I noticed recently that people sell pre-assembled voltage regulator units on ebay heaps cheap, so that might be a quick and easy solution for you.

    For instance, a linear regulator based on the LM317 (not 7805) is only $10 delivered!
    http://cgi.ebay.com.au/Voltage-Regu...900?pt=LH_DefaultDomain_0&hash=item3ca5c8012c

    You can adjust it for whatever voltage you want - a little over 3.6V would be good
    and again use a small series resistor to provide some current regulation.
    It will take DC or AC input voltages and supply up to 1.5A output.

    Being a linear regulator it wastes some power though, so if this is an issue, you can get a unit using switch-mode regulation (with almost the same specs) for $11
    http://cgi.ebay.com.au/NEW-AX3022-B...=AU_Video_Cam_Accessories&hash=item5196bdca5d

    and if you don't need 1.5A (which you probably won't), here's another one that will supply up to 0.5A for only $4!!!!
    http://cgi.ebay.com.au/MC34063-Base...lectrical_Equipment_Tools&hash=item519516b86a

    so it's hardly worth building it yourself :)

    ok hope this lot helps, please let us know how you go.

    CJ
     
    Last edited: Dec 30, 2010
  3. cj_elec_tech

    cj_elec_tech

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    Oct 7, 2009
    for the constant current circuit, have a look at figure 8 at the top of page 4 of:
    http://www.national.com/an/AN/AN-181.pdf
    (FYI: the application shows a LM117 which is the military version of the LM317 - it works the same though)
    The circuit to do the same thing with a 7805 is the same, but the resistor will be different - I'm searching for a circuit now - keep checking back here!

    CJ
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
  5. harvey66

    harvey66

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    Dec 29, 2010
    Thanks very much for your very detailed replies...I have had a look at some of the pre-assembled units on ebay and might order one to try it out. The problem I have is I'm not sure how to measure the current which the light unit will draw as it's been along time since I did anything with electronics (Lone before LEDS were used as lights!!). Everything I've read about the 7805 indicate that 1A is maximum, Do you think 10 superbright LEDS would draw more than 1A..?
     
  6. harvey66

    harvey66

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    Dec 29, 2010
    Can you twll me what switch mode regulation does...? Thanks
     
  7. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Knowing the LED current is the same as knowing a lamp's voltage.

    You wouldn't ask us what voltage you should use with a lamp because that's pretty much the basic information you need to know about it.

    Likewise, you can't ask us about the current requirement for a LED. You need to look at the specs and determine what a safe current is. Without any other information, I'd say that 20mA would be safe, but max currents range from significantly lower than this to significantly higher.

    The rest of your questions are answered here (and I'm not going to point you there again).

    Switchmode regulators are simply a very smart type of regulator that can lower the voltage without wasting the excess power as heat. They can also output a voltage higher than their input.

    Linear regulators, such as the 7805 (ignoring inefficiencies) have the same input current as output current, and thus the power dissipated can be no less than the difference between the input and output voltage multiplied by the output current.

    Switchmode regulators (ignoring inefficiencies) have an input power (voltage times current) equal to the output power (voltage times current). Practically, switchmode power supplies can be 60 to 95% efficient, so between 5% and 40% of the input power is wasted as heat (compare that with a 7805 operating from 12V -- even with 100% efficiency in the regulator, it dissipates 58% of the input power as heat).

    Switchmode power supplies are typically used where efficiency is very important, or where the input and output voltages are significantly different *and* currents are large. Switchmode regulators operating from the mains do not have bulky transformers. A computer power supply, if linear, would weigh probably 5 to 10 times as much, and generate enough heat to be used as a room heater (well, almost).

    edit: OK, you say you can't get the specs on the LEDs. I would go with 20mA initially. If you can measure te current that they draw in the application you have I would reduce that by 30% and go with that (on the basis that most such lamps overdrive the LEDs)
     
    Last edited: Dec 31, 2010
  8. cj_elec_tech

    cj_elec_tech

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    Oct 7, 2009
    About the only thing I'd add to steve's comments is that it is quite likely, and common with these type of multiple LED lights, to have some (or all) LEDs in parallel with each other - and this would affect the current draw (i.e. 20mA x number of leds).

    However in some cases where high power LEDs (say, 1 watt ones) are used - EACH LED draws about 300mA at 3.3 Volts.

    So to prevent any uncertainty, you *really* need to measure the current draw to be sure.
    Do you have access to a multimeter with current measurement?
    If so, whack it in series with the LEDs and check the current being drawn.
     
  9. harvey66

    harvey66

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    Dec 29, 2010
    Thanks for all your help

    I have opened the units up, the batts are wired in series and are 3.83v fully charged. I've tested the current on a multimter and the light unit is drawing 0.168 amps

    I have looked into the AX3022 switching regulator to step down 12v to 5v (max 10 watts 1.5amps) but I have a question.... If I used one of these with 10 lamps at 0.168 amps at 3.83v this will give me 6.43 watts - is that correct or would I need to work it out on 12v which would be much higher and too many watts for one unit..?

    http://cgi.ebay.co.uk/ws/eBayISAPI.dll?Vie...e=STRK:MEWNX:IT

    Many thanks everyone
     
  10. cj_elec_tech

    cj_elec_tech

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    Oct 7, 2009
    Frequently some (most?) manufacturers of LED lights will achieve extra brightness by running the LEDs at a higher current level and whilst this works ok, the *long term* life of the LEDs will be shortened quite a bit. So if I were doing this job, I'd be sticking to a conservative LED current, primarily so the light doesn't started burning out LEDs after 2 months (ok, that's probably an exaggeration, but hopefully you get my drift).

    The manufacturers also tend to rely on the inherent 'battery resistance' to provide the current limiting for the LEDs - you NEED to remember that LEDs are current devices NOT voltage devices and therefore you *should* current-limit your lights now that you are not using tiny NiMH batteries.

    The link that Steve posted (see above) talks about *why* current limiting is important and it also highlights a problem - it sounds like you will have the problem but probably don't know about it.?

    It would appear that the 10 LEDs in your light are connected in parallel with each other - just like it says NOT to do in second method of Section 2 of Steve's link (see the pic with the big red X on it? that's probably what you have).
    Now the BEST way to fix this is to rewire the 10 Leds into a better (safer) configuration as per Steve's link - but I guess this will be too problematic for you?

    So assuming you want the SIMPLEST solution, I would suggest you get one of the Switching Regulators I posted links for (see above) and just use a series resistor to limit the current to your LEDs - I emphasise that this is NOT the ideal solution.

    Now I'm not sure what you meant by the 6.43 watts you mention???

    Your lights use 0.168 amps at 3.83 volts = 0.168 x 3.83 = 0.643 Watts (for all 10). As you can see, this is way way less than the ax3022 series could supply and still less than the MC34063 series could supply too (and as these are only $4 each, I know which one *I* would be buying :) ).

    As we've said, the important value is the current - and your 0.168 Amps is in the right ball-park - however I suspect that the combination of the battery resistance and the Multimeter resistance in the circuit when you measured the current probably LOWERED the current slightly - as Steve said, 20mA per LED is norm.
    In an ideal world, you would have used 2 multimeters at the same time - one to measure the current and the other to measure the voltage *directly* across the LEDs (i.e. after the multimeter measuring the current).

    So.... IF we assume 20mA per LED and 10 LEDs, then 0.02 x 10 = 0.2 Amps and that is the value you should probably use to calculate the series resistor.

    (see my next post for the rest of my diatribe) :)
     
    Last edited: Jan 4, 2011
  11. harvey66

    harvey66

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    Dec 29, 2010
    Thanks for your reply.

    Sorry I think I've confused you. Yes there are 10 LEDS per lamp but I will be running up to 10 lamps in total (so 100 leds), and was hoping to run them all off one circuit with the AX3022 - so each unit is 0.643 watts x 10 lamps = 6.43 watts. The AX3022 is only GBP 6.29 delivered to UK from China so if this will drive all 10 lamp units then it's great value. I would of course use a resistor in each lamp to drop the 5v to 3.83v which (using your figure of 0.2 amps) I think each resistor should be about 6 ohms (10 ohms nearest?), but I'm only learning so if you could confirm this then that would be great.

    Sorry if I appear a bit 'thick' - but asking questions is the best way to learn I always find!

    Thanks
    Chris
     
  12. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    6.8 ohms would be the closest value. You'll need a 1/2 Watt resistor.
     
  13. cj_elec_tech

    cj_elec_tech

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    Oct 7, 2009
    Ahh ok I didn't understand you wanted to use 10 light units. Sorry about that.

    Your wattage calculation is correct, however the ax3022 can only supply a maximum current of 1.5 Amps and your 10 units will (ignoring losses) consume
    10 lights x (10 Leds x 0.02 Amps) = 2 Amps i.e. more than 1 ax3022 can supply.
    EDIT: ah ok, more like 10 x 0.168 Amps = 1.68 Amps which is still more than 1.5 Amps
    So maybe you need to set this lot up with 2 ax3022 units, each running 5 Lamps.

    As Steve says, you will need a 1/2 watt resistor for each lamp unit. However, the value of the resistor is not too critical - 4.7, 6.8, 8.2 or 10 ohms is the ball-park you want.
    I say this because the important thing is to adjust the current to achieve the 0.168 Amps you measured before - the voltage from the ax3022 module (that I posted the link for) is adjustable, so this is easy to achieve, and by using the same multimeter when setting this (the current will rise a little when the multimeter's resistance is removed from the circuit), you *should* then have the same brightness level you had with fully charged NiMH batteries.

    A higher resistance value will give you better current regulation, but will 'waste' more power doing it (and therefore getting slightly hotter), 10 ohms > 0.4 watts wasted per resistor (and you will have 10 of them), so 4 watts (in total) will be wasted as heat.
    So maybe the lower value that Steve suggested (6.8 ohms) would be a more efficient solution (2.7 Watts for all 10 units) - my gut feeling about using 4.7 ohms (1.9 watts in total) is that it may not provide much current regulation, which could result in dead LEDs.

    Hope this all helps, please let us know how you go. :)
     
    Last edited: Jan 7, 2011
  14. cj_elec_tech

    cj_elec_tech

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    Oct 7, 2009
    Oh, I realised I had neglected to say a couple of things in my last post:
    - as a precaution against accidentally killing ALL of your light units, I would suggest you adjust the current with only ONE light set connected, and only connect the other 4 (per ax3022 unit, WITH their individual resistors) when you have the current set right.
    but before even connecting ANY light units
    - I would suggest you adjust the output voltage of the ax3022 units to the minimum value (about 2.5 volts from memory?) - that way you will be adjusting the current UP to the value you want, thereby minimizing the risk of accidentally starting at too high a current (too high a voltage) and potentially destroying what lights you *have* connected to the regulator unit.
     
    Last edited: Jan 28, 2011
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