I'm looking for an alternative power source to illuminate several LEDs. I'm using a 9V battery to power between 12 and 20 3mm/20mA white LEDs. They are wired in series parallel, everything works fine but a 9V battery will only last approx 2 hours. I'm looking for a rechargeable option that would last 3-4 or 6 hours. Any suggestions?

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    \$\begingroup\$ You should probably privide more restrictions as there are a lot of options. You have rechargable 9 V batteries or AAA, AA, C, and D batteries (6 in series). In any of these cases, if you dont have enough capacity wire more in parallel. Or just use a wired DC power supply. \$\endgroup\$ – Matt Aug 13 '18 at 20:21
  • \$\begingroup\$ @Matt it's generally better to use larger capacity cells, not parallel cells. \$\endgroup\$ – Chris Stratton Aug 13 '18 at 21:22
  • \$\begingroup\$ There are some cell sizes that merit series/parallel arrangements due to price. Lithium ion is probably your best bet and there are a wide array of options for size and shape of battery. They offer better charge density and much higher rated current than a 9V, and if you need a voltage higher than 3.7V, you can buy series arranged lithium ion packs, or use a boost mode led driver. How are the LEDs arranged? What is the series/parallel arrangement and are there inline resistors? \$\endgroup\$ – K H Aug 14 '18 at 5:45
  • \$\begingroup\$ The LEDs you're describing are basically the most primitive of the family of "power LEDs", and if you are choosing an array, be aware that power LEDs have an efficiency tradeoff with heat as well as with drive current, so if you have 12 LEDs, and you adjust them to 20mA, and then you have 20 of the same LED on another board and you adjust those to the same total brightness, they will be running at a lower current and use less power at the same brightness. Looking at it another way, at 3.5V, 20mA, those LEDS are 1.4W at full load. You can get a single 3/5W LED if you want and run it at 1.4W. \$\endgroup\$ – K H Aug 14 '18 at 6:18

9v batteries (commonly known as 'transistor radio' batteries) are inherently inefficient (and expensive) because you are essentially using a "stack" of six 1.5v cells in series to get 9v. The internal resistance of this stack is eating up a lot of your power. 9v batteries are (should only be) used when something in the device needs high voltage at low current.

Meanwhile, 20 x 20ma = 400ma or .4 amperes. This is too much big a continuous load on a 9v battery. I would guess that if you measured the temperature of your battery you'd see a substantial temperature rise, which is causing the battery to burn up much of it's potential energy as heat inside the battery itself.

Ideally you need an op-amp(s) to drive your LEDs. Remember that you need to drive LEDs with CURRENT, not VOLTAGE. If you can't use op-amps, then use a single 1.5v "D" cell with an appropriate size of current limiting resistor, and wire all of your LEDs in parallel. You'll still be burning up some of your power as heat in the resistor(s), but this heat will be dissipated outside the battery instead of inside the battery.

Lastly, if you really want to continue using expensive 9v batteries in your present series/parallel circuit, use several of them in parallel to keep the continuous current from each battery below about 50ma per battery (my best guess). Again, measure the temperature rise of the batteries. The batteries need to dissipate internal heat faster than they create heat.

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    \$\begingroup\$ There are plenty of simple constant current LED drivers that should be more efficient than an opamp. Also 20mA is a very high current for modern efficient LEDs. \$\endgroup\$ – David Aug 13 '18 at 21:19
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    \$\begingroup\$ I don't see how op-amps would help here, and a single "D" cell won't provide sufficient voltage to drive a visible LED without additional circuitry. \$\endgroup\$ – Peter Bennett Aug 13 '18 at 21:19
  • \$\begingroup\$ There are a few things I didn't mention, one being size, it needs to fit in a space that's only 1-1/2" and second has to be completely portable and independent of any other source \$\endgroup\$ – dre157 Aug 13 '18 at 21:35

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