I bought this 5V battery powered LED strip on amazon for a project of mine. It wants 3 AA batteries. Due to size constraints, I would have difficulty fitting in even 3 AAA batteries. However, 2 CR123A batteries would fit fine. I have heard that LED's are more sensitive to current than voltage. So should I have any problems powering it with CR123 batteries? And also would 1 (3v) or 2 (6v) be preferable?
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\$\begingroup\$ It depends on how the batteries are wired, parallel or series... And how long you want to run the strip. \$\endgroup\$– Ron BeyerCommented Dec 28, 2020 at 23:53
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\$\begingroup\$ That little 3-button controller thingy that comes with it contains electronics we don't know anything about, so no-one can say for sure. 3.7V might be too low for it to operate, 7.4V might damage it. \$\endgroup\$– UnimportantCommented Dec 28, 2020 at 23:56
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\$\begingroup\$ @RonBeyer, I can wire the batteries in parallel but for simplicity would prefer series. The runtime of the strip isn't really important. \$\endgroup\$– RafaelCommented Dec 29, 2020 at 0:36
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\$\begingroup\$ LEDs are more sensitive to voltage their response is a dramatic change in current. \$\endgroup\$– Jasen Слава УкраїніCommented Dec 29, 2020 at 3:53
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2 Answers
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I have heard that LED's are more sensitive to current than voltage.
These are probably WS2812B LED strips. Their voltage range is typically about 4.0V to 5.5V. The control electronics probably has a similar voltage range.
Exceeding 5.5V is probably not a good idea. They will work with slightly less than 4V, but the constant current drivers may not function properly and the brightness of the LEDs may be noticeably different.
Two 3V batteries in series with a diode in series to drop the voltage slightly would probably work. These strips can draw a lot of current, a 3A diode is probably needed.
answered Dec 29, 2020 at 0:18
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\$\begingroup\$ Thanks for the answer! I just reallized (and tested) that the voltage of CR123A batteries is actually 3V, not 3.7. Does this make a difference? 2 in series is not that far off from 6V. \$\endgroup\$– RafaelCommented Dec 29, 2020 at 0:39
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\$\begingroup\$ I wouldn't try 6V. But, a diode in series might work, see revised answer. \$\endgroup\$ Commented Dec 29, 2020 at 1:14
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\$\begingroup\$ Acc. to this datasheet that I found, they can handle up to 5.7V. I don't want to put any power dissipating elements in due to thermal issues. But if should there be any problems if I pre-drain the batteries to lower the voltage a bit? \$\endgroup\$– RafaelCommented Dec 29, 2020 at 3:07
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\$\begingroup\$ Pre-draining might work, but the discharge curve is not linear, there may not be much life left at 5.7V. \$\endgroup\$ Commented Dec 29, 2020 at 9:07
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\$\begingroup\$ Acc. to these discharge curves, they very quickly drop to a bit above 2.5 volts. Thanks! \$\endgroup\$– RafaelCommented Dec 29, 2020 at 12:33
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As long as you select a proper current-limiting resistor to put in series with the LEDs, the voltage you choose doesn't matter... as long as it's greater than the LED threshold voltage, so I would be wary of using 3V.
You can't easily tweak the resistance that I assume is built into the light strip controller, but if you select 6V and figure that the operating current of this strip is specified to be 0.6A (from the product listing on Amazon), then in order to burn off the excess power and present 5V to your lights, you could use a R = V/I = (6V - 5V)/0.6A = 1.67Ω resistor in series with the batteries.
Note that P= I2R = (0.6A)2 ✕ 1.67Ω = 0.6W, which may be more than your average kit resistor can handle without exploding.
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\$\begingroup\$ Thanks for answering! Turns out CR123A batteries are actually 3V. How much of a difference would this make? \$\endgroup\$– RafaelCommented Dec 29, 2020 at 0:40
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\$\begingroup\$ Same theory applies, only now you will need a (6V - 5V)/0.6A = 1.67Ω resistor. \$\endgroup\$ Commented Dec 29, 2020 at 0:45