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This question already has an answer here:

I would like to drive 144 leds with a battery power supply. I want to drive all the leds all the time. 12 leds in series on each string. The Led specs are:

PD 100 mW Max. 
VR 5 V
IAF 25 mA
IPF 85 mA
VF at 20mA 1.8 Typ, 2.1 Max
IR at 5V 100 uA

Where do I start for something with a low cost?

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marked as duplicate by Phil Frost, Dave Tweed, Nick Alexeev, Matt Young, placeholder Jul 15 '13 at 18:50

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Where do I start for something with a low cost?

You start by understanding what you need. If the forward voltage of each LED is 1.8V and there are 12 in series in each string, you need a supply voltage that is at least 12 x 1.8V = 21.6V.

But you will need a little bit of overhead to control the current through the string of 12 so I'd go for a 24V DC supply and a resistor in series with the LEDs. The resistor will develop 2.4V across it (24 - 21.6) and have 20mA flowing through it. The value of resistance that meets those demands is 120 ohms.

The power taken by one string is 24V x 0.02A = 0.48W.

The power for all twelve strings is therefore 5.76W.

The current is 12 x 0.02A = 240mA so you should possibly look for a wall-wart that produces 24V and has a power rating in excess of 6W.

In case your LED voltage might be higher (such as 2.1V as per the details in your question) then the max voltage will be 25.2 and you'd realistically be looking at a 30V dc supply.

If you think you need a higher current than the 20mA then use the method shown in my answer for recalculating \$ R_{series} \$

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  • \$\begingroup\$ I like Andy's approach. A few notes: \$\endgroup\$ – Eric Gunnerson Jul 13 '13 at 19:28
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    \$\begingroup\$ If you want batteries, two 12V batteries in series will work for you. Or 16AAs in series. Note that the values you list are typical for the LEDs; LEDs have a fairly wide manufacturing tolerance and you may see some differences in voltage and therefore current. If you could split the strings in 24 half-strings, you could drive them from a single 12V source, which would be easier. \$\endgroup\$ – Eric Gunnerson Jul 13 '13 at 19:31
  • \$\begingroup\$ Never use the typical Forward Voltage, use the max, 2.1v in this case. \$\endgroup\$ – Misunderstood Mar 4 '17 at 17:07
  • \$\begingroup\$ @Misunderstood I don't think that's entirely correct especially when there is a string of LEDs but, please leave an answer and justify why the maximum voltage should always be used. \$\endgroup\$ – Andy aka Mar 4 '17 at 17:12
  • \$\begingroup\$ See the TI App Note Design Challenges of Switching LED Drivers, Section Number 2, Output Voltage Changes when LED Current Changes ti.com/litv/pdf/snva253a \$\endgroup\$ – Misunderstood Mar 4 '17 at 17:20
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The question does not specify size and weight constraints (batteries can be heavy) nor expected run-time per battery set. Hence, making some "reasonable assumptions" for power source:

  • at least 1 hour run-time per battery set
  • weight under 250 grams for portability

As has been computed by @AndyAka in another answer, each string of 12 LEDs would need, at maximum rated forward voltage, 12 x 2.1 = 25.2 Volts plus some headroom, say 30 Volts. Current per string is 20 mA, 240 mA total.
Taking into account dissipation across the current limiting resistors per string, the actual power consumption would be 30 V x 20 mA = 600 mW per string, hence 7.2 Watts at 30 Volts total, not just ~6 Watts.

To obtain the proposed 30 Volts without using a large number of batteries, or very heavy sealed lead-acid (SLA) 6 or 12 Volt batteries, use an inexpensive DC-DC boost converter and normal flashlight cells. Such boost converters provide around 80% efficiency:

  • Power demand for providing 7.2 Watts @ 80% efficiency = 9 Watts.
  • For 1 hour of operation, battery capacity needed = 9000 mAh

Boost converters such as this 4-35V, 2.5 Ampere device cost under $4 including international shipping, from sites like ebay:

Boost converter


Power can be provided by AA or D cells, either rechargeable or primary chemistry.

Some Battery Options:

Energizer E91 Alkaline AA, 8 cells:

  • 8 x 1.5 = 12 Volts nominal
  • Weight = 23 x 8 = 184 grams
  • Capacity available at ~750 mA load = 6,000 mAh, not quite enough for 1 hour operation

Energizer L91 Lithium AA, 8 cells:

  • 8 x 1.5 = 12 Volts nominal
  • Weight = 14.5 x 8 = 116 grams
  • Capacity available at ~750 mA load = 22,000 mAh, more than ample!
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