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I have some christmas lights powered by 3 x 1.5v AA batteries, but the label shows they need 3v 0.06w. Taking apart the very small circuit board I can see they are stepped down with a resistor. Other questions here say quite simply - don't do this as you cannot guarantee the load etc. which makes sense.

My problem has been that the lights will be at differing brightness when they all have fresh batteries, plus they only last a few days before I need to replace the batteries when I have another set of battery powered LED lights that will last for a month (at 6 hours per day). I'm partly guessing the quality of the resistor is highly variable - but electronics is really not my area of expertise!

So the questions: does the use of a resistor have a detrimental effect on battery life? Would replacing the pack with a simple switched 2 x AA battery pack improve things (so no need to try and drop voltage)? Or how can I power more than one set of these lights from a transformer?

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  • \$\begingroup\$ Presumably these are not LED lights? \$\endgroup\$
    – awjlogan
    Commented Nov 11, 2017 at 11:51
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    \$\begingroup\$ Yes they are LED lights. The label says "30 bulbs x 3V 0.06W" \$\endgroup\$ Commented Nov 11, 2017 at 11:59
  • \$\begingroup\$ The label is a little ambiguous, but I've left an answer below. A 60 mW LED is reasonable, but if you have 30 of them your AA battery shouldn't last as long as it does..! \$\endgroup\$
    – awjlogan
    Commented Nov 11, 2017 at 12:19
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    \$\begingroup\$ @DavidLester: Do you have access to a multimeter? If so, can you take a current reading and measure the resistor value and include the results in your question? \$\endgroup\$
    – Transistor
    Commented Nov 11, 2017 at 12:21
  • \$\begingroup\$ As the voltage drops the current draw drops, and the life of the batteries increases. Days is not unusual. \$\endgroup\$
    – Passerby
    Commented Nov 11, 2017 at 17:20

3 Answers 3

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does the use of a resistor have a detrimental effect on battery life?

Partly. There is a voltage drop across the resistor which means wasted power. But this is required for proper operation in your setup.

Would replacing the pack with a simple switched 2 x AA battery pack improve things (so no need to try and drop voltage)?

Yes and no. Remove the resistor and if the voltage is less than the VF at 20mA, then the leds will be fine. You may have a shorter battery life as you essentially removed 1/3 the power, but it may be longer. You have to measure to see.

Or how can I power more than one set of these lights from a transformer?

Get a 4.5V or 3.3V power supply, and wire it in place of the batteries. That's it. A 4.5V supply will look brighter. Or use a usb power supply. The leds will be even brighter, as the current increased, so the led life may drop, or you can add a resistor or 1n400x diode to make it go down by 0.7V or 2 for 1.4V. Perfect for your string.

I've done this, a usb port to a diode to one of the led strings with a timer. Running for a year straight no issues.

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  • \$\begingroup\$ I think a transformer is the way to go - the question for me is - how? Can I run all from a single transformer? I have one that is adjustable showing 2500mA at 3/4.5/5v. Can I just plug that into some kind of wiring loom and have running to 4 sets? I'm guessing I need it wired in parallel? \$\endgroup\$ Commented Nov 12, 2017 at 16:22
  • \$\begingroup\$ Yes, wire them in parallel. Since the board expects 4.5V you dont have to do anything else either. \$\endgroup\$
    – Passerby
    Commented Nov 12, 2017 at 22:22
  • \$\begingroup\$ I'm pretty sure that the lights are no where near 2.5 amp so that adapter is perfect. Ideally you would want to measure them to be sure, but meh \$\endgroup\$
    – Passerby
    Commented Nov 12, 2017 at 22:32
  • \$\begingroup\$ So I think I will want use the lead that holds the CR2032 cable. It's got a handy plug on the end that goes into the main battery unit but bypasses the resistor so I guess I'd run that as 3v? Just need a selection of wires, a box to hold it, and a socket on the box that will take the adapter I guess! \$\endgroup\$ Commented Nov 13, 2017 at 17:07
  • \$\begingroup\$ I just make a notch in the battery cover and use that. No need to bypass the resistor. \$\endgroup\$
    – Passerby
    Commented Nov 13, 2017 at 19:11
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The resistor isn't there to drop voltage, it actually limits the current. LEDs (to a first approximation) effectively become short circuits once the voltage across them reaches a certain value. The current through the LEDs will drop as the battery voltage drops with just a resistor, so the intensity of the LEDs will gradually go down.

The resistor does waste power, but for this usage case it would be difficult or impossible to use a more efficient solution without significantly increasing the cost. Resistor quality is not really a consideration, unless the resistor is truly terrible and then it would probably just pop.

Two AA batteries probably won't have high enough voltage to make the LEDs conduct at all so you can't do that. A 5 V adaptor would probably be fine, although you'd have to check the power rating of the resistor.

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  • \$\begingroup\$ Thanks - although that leaves me more confused! Why does it state they are "3v" and yet I have 4.5v from the batteries? Also - there is a "tester" battery (from when they were in the packaging - hold the button and see them light up) that connects via a seperate lead. This is powered by a single CR2032 3v battery and i can see that this will bypass the resistor. \$\endgroup\$ Commented Nov 11, 2017 at 12:22
  • \$\begingroup\$ Ok, so when the voltage across the LED is 3V, it will light up but looks like a short circuit and so a large current would flow and it would go pop (sometimes literally). You have 4.5 V, and the resistor. You can calculate the current (and hence light level) using those figures. \$ I = \frac{V_{battery} - V_{LED}}{R} \$. Does that make sense? \$\endgroup\$
    – awjlogan
    Commented Nov 11, 2017 at 12:25
  • \$\begingroup\$ CR2032 can power maybe 1 or 2 LED at low current and has internal resistance and wont last long. ( very Dim maybe >24hr after a couple hr half-bright) If they use 4.5V then they must have resistor in series added. (hidden) \$\endgroup\$ Commented Nov 11, 2017 at 15:31
  • \$\begingroup\$ Your are mistaken to claim that "The resistor does not impact the efficiency in this simple usage case" - in fact it very much does. The amount of power wasted may be small, but the efficiency is most definitely reduced by this, likely by quite a lot. \$\endgroup\$ Commented Nov 11, 2017 at 22:49
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    \$\begingroup\$ @awjlogan Actually the resistor does both, drops V and limits current and Ohms Law can be used as I have shown. \$\endgroup\$ Commented Nov 13, 2017 at 6:28
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  1. does the use of a resistor have a detrimental effect on battery life?
    1. Would replacing the pack with a simple switched 2 x AA battery pack improve things (so no need to try and drop voltage)?
    2. Or how can I power more than one set of these lights from a transformer?

  1. only if it is the wrong value?
    1. It would save wasted power from series drop of 1 to 1.5V to 3V or 30% to 50% loss but then may not be full brightness
    2. Much wiser solution. Buy a brick matched to voltage and current of the load, which has yet to be defined except ~3V with roughly 30 LEDs in some disarray

Engineering wisdom

A string of LEDs will not last very long on Alkaline batteries. If you expect 30 5mm LEDs to be bright for 6hrs a day for a month, it wont happen with a few Alkaline AA batteries, rather 8 day and dimming slowly during that day.

Consider AC-DC supply only or Lithium CR123A (Primary 3.0V) or LiPo (secondary 3.6V with small R depending on number of LEDs in parallel.

White LEDs

  • 3.1V full brightness equivalent to 2.8V + If[20ma] * 16 Ohms (per LED)
  • 2.9V half bright
  • 2.8V dim
  • 2.7V nearly off

One 5mm White LED is rated at 3~3.1V @ 20mA = 0.06 W. How many do you have sharing this power?

If 0.06W per LED x30 implies total power of 1.8W at full brightness. AA Alkaline cells are about 3Wh using 100mA rates but voltage declines to <1V so how many hours did you expect and at what power level? Thus you might expect a few hrs at this power range then dim quickly and be sustained for a much longer period. With 30 LEDs I expect them to be arranged as 1S30P so the series R for 30 parallel strings of say 10mA each to drop 1V is 1V/0.3A= 3.3 ohms

300 mA ^2 * 3.3 ohms = Pd = 0.3W so use a 0.5W R

When batteries are fresh the Vdrop=1.4 to 3.1V LEDs with 3.3R in series is then 1.4V/0.3ohm= 420 mA /30 = 14 mA or <50mW per LED so this works for while.

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  • \$\begingroup\$ So the label says 30 LED's but hard to count them as they are arranged hap-hazardly on a star shape. \$\endgroup\$ Commented Nov 12, 2017 at 16:09
  • \$\begingroup\$ I wonder who and why some disagreed with my information. It is obvious one needs to regulate current and a resistor can be a simple but less than perfect method. The choice of batteries is poor. (does not meet spec for time) \$\endgroup\$ Commented Nov 13, 2017 at 6:22

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