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I have a led string like this enter image description here

I want to power it from batteries and not from the outlet.

The string consists of 3 parallel strings of 28 leds, which I can power separately.

Originally, the output of the controller is 220 v and on the first 5 leds of each string there is a resistor of 4.2k, which gives us ~21k.

From a test with batteries I performed on a single led from the string, it has a voltage drop of 2.01V. It suggests that the operating current was (220-33*2.01)/21k=7.3mA, which seems reasonable - with this current I get a reasonable light from the led and no burning smell. With about 16 mA I get about the same light and a burning smell.

The straightforward solution I think would be to connect 2 leds + 200 Ohm resistor in a series to a 6v battery pack to get ~10mA operating current, and to connect many such units to the batteries. Probably there is a better solution?

How can I drive more then 2 leds in chain with a 6v battery pack?

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  • \$\begingroup\$ blocking oscillator comes to mind, but something less crude on the lines of a boost converter might serve as well. \$\endgroup\$
    – Ecnerwal
    Sep 6, 2017 at 2:20
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    \$\begingroup\$ 99 LEDs drawing 30mW nominal at half rated current is 3 watts or 0.5 amp at 6V even neglect a boost converter losses, which will not last long with a small 6V low Ah battery. What's the point? \$\endgroup\$
    – Tony Stewart EE75
    Sep 6, 2017 at 2:27
  • \$\begingroup\$ I can use separate batteries for each one of the 3 strings, which leaves us with 33 leds. Actually, I disposed of the 5 leds with resistors on them, which brings us down to 28 leds per string. The current they draw will be likely to the tune of 10mA, which brings us to 20 mW * 28 leds = 560mW. On 6V, the current will be 93mA. If I use 4 D cells, it will be well within the recommended current, and the capacity of 12000mAh, they will last ~130h, which suits my needs. The question is how I feed 28 led string with 6V battery array. \$\endgroup\$
    – alex440
    Sep 6, 2017 at 18:12

1 Answer 1

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Assuming 28 series leds at 2Vf @ 10mA, then you are looking at 56 V * 0.01 A = .56W. At 6V, that's 0.56 / 6V = 93 mA. This you already know.

So you have a few options. Multiple batteries in series, or a boost converter Is needed. There are premade modules you can buy for this. Or you can make a primitive one by hand. These are called joule thieves. Often just a ferrite bead, wire and a transistor.

Or buy an led string made for parallel drive.

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