I have a hobby project of putting together a light curtain of 100-150 LEDs (3V, 20mA). It is meant to be used in an event outdoors without mains power. It could be operated with bunch of ordinary batteries and it should operate around two hours.

It would have a main rail without lights and then strips of 10 LEDs hanging down from it. Any ideas for construction?

Edit: LEDs are simply all on, no controls needed.


1 Answer 1



Use 25 + Watt hours of battery.
Boost to 35V - apply to main rail. Boost converter - buy made up or use eg MC34063. Run 10 LEDs in series with 150 ohm 1/4 Watt resistor per string.
Hang strings off main rail with plugs or clips or sliding contacts or ...



  • Are LEDs all "just on" or modulated in some way and/or able to be turned off in groups etc.

  • White presumably.

3V x 20 mA x 150 LEDs x 2 hours = 18 Watt hours.
Allow 25 - 30 Watt hours of battery capacity.

1 x AA 2500 mAh ~= 3 Wh or less = 10 of

1 x SLA 6V 7Ah = 42 Wh = ample.

1 x 18650 LiIon = 3.6V mean x 2Ah = 7+ Wh = 4 of.

10 LEDs in strip = 30V+ (typically 33 - 35V depending on LEDs) = more than usually used but doable.

5 LEDs = 15V at 3V each or < 20V in practice.


Use of a boost converter and constant current drivers is a good idea.
Could run off several batteries directly with buck converters or linear regulators.
eg 2 x 12V SLA = 24 V = 8 LEDS badly, = 7 LEDs maybe, = 6 LEDs OK.
If all are on then arranging in strings of 10 physical but 6 logical is annoying.

Easiest organisationally if all on is to boost whatever battery to say 35V and run strings of 10 with constant current drive or even (probably) just a resistor per group.

Say LEDs worked out at 3.2V mean = 32V for 10.
Headroom at 235 V = 35-32 = 3V or about 10% of LED string voltage ~~= OK. R = V/I = 3/0.020 = 150 R.
Power_R = V x I = 3 x 0.02 = 60 mW = almost anything.

The reason to use a boost converter is to ease driving LEDs from low volatge with lots of current control electronics. A 6V battery will drive one LED properly with linear or buck regulator. A 12V battery will drive 3 LEDs directly. A 24 volt battery will drive 7 LEDs.
LEDs are said to be 3V and some white LEDs are, but at 20 mA many are in 3.0 - 3.6V range. At even 3.2V you get few LEDs in series from 12 or even 24 V.

As you want strings of 10 the simolest hing to do is to have a common bus that drives 10 LEDs. Voltage needed = 10 x VLED + a few volts headroom. Say 3.2 x 10 + 3 = 35 V. Adjust to suit.

The 62 cent/1 workhorse MC34063 will drive all 150 LEDs directly!.
MC34063 datasheet

I've included two versions of essentially the same circuit below as one has example PCB layout and other shows IC internals. I'd recommend using a 12V 7 AH "brick" SLA battery as it has far more capacity than you need, you will not "deep cycle " it in normal use, you can use it several times in an emergency and can charge it with a small automobile car charger.
Examples show 28V out. Change R2 to adjust voltage out. Use >= 10 x VLED + 3V BUT Voutmax should be somewhat less than 40V.

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This product from ebay found by @baaggage_lump looks useful.
He said:

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For your LED use it looks good but is close to maximum rating for 10 LEDs in series. Better probably to use 5 LEDs in series electrically and 10 in series physically. Run rail at say ~= 5 x 3.2V + 3 =~ 19 V with a series resistor per 5. Actual bus voltage depends on worst case voltage of 5 LEDs in series at 20 mA. At 19V rail and 3.2V avg per LED R = (19 - 16)/0.02 = 150 ohms.

It uses the NatSemi LM2577 boost converter adjustable version datasheet here worth $6.63/1 AMD NatSemi LM2596 buck converter IC worth about the same datasheet here . That board is extremely good value for money.

The boost section should suit your need. Typical cct and probably very close to what they are using:

enter image description here

  • \$\begingroup\$ Good amount of information! I feel pretty confident I'll get something done :-) \$\endgroup\$
    – Petteri H
    Commented Mar 1, 2012 at 14:12

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