I am looking to build a light stick.

I am quite new to electronics and have just been experimenting so far. My light needs to be portable so I am using a 3.7v LiPo battery (I am into RC so have some lying around). I initially thought that wiring my LEDs in parallel would be best as it requires no voltage boost but I seems like it will use more battery that way.

I looked at linking them up in series but I have about 16 and need a voltage of around 51-52V to run them in series. From what I have read this uses fewer amps for better efficiency but I don't know if this is correct.

The voltage regulators seem to get quite bulky when you up to a higher voltage. What I was thinking of was creating segments of LEDs in series and put the segments in parallel using multiple step up voltage converters in one project.

Would it be ok to use more than one and would you recommend using a pre-made one or incorporate it into my design and have it on the board.


It's generally ill-advised to place two diodes (including LEDs) in parallel due to the negative temperature coefficient of a diode - a thermal runaway can result.

Two diodes in parallel will have the same voltage applied across them (i.e. forward voltages are the same), but due to differences in the diodes (manufacturing variation, thermal path differences, etc), one will draw more current than the other. The negative temperature coefficient will cause the diode drawing more current to draw even more current, and so forth, eventually causing it to overheat if left unchecked.

Boosting from 3.7V to ~50V could be tough, so you may have luck with using a few smaller strings of LEDs. A tri-channel boost LED driver such as the LT3797 would keep the boost ratio (V_out / V_in) reasonable.

  • \$\begingroup\$ Didn't know that now looking to create some strings of leds using MC34063 as a step up regulator and do some more testing. \$\endgroup\$
    – bobthemac
    Mar 15 '16 at 14:52

I initially thought that wiring my LEDs in parallel would be best as it requires no voltage boost but I seems like it will use more battery that way.

Not really.

Assuming you have 16 leds with a 3.2V Forward Voltage @ 20mA, in series you have:

16 * 3.2V = 51.2V
51.2V * 0.02A = 1.02 Watts

In parallel, you have

16 * 0.02A = 0.32A or 320mA

320mA * 3.2V = 1.02 Watts

Same power usage.

But now you have to factor in efficiency. A boost circuit will not be 100% efficient. You are normally looking at 80~90% efficiency with a switching boost circuit. Plus the current limiting resistor used (figure 0.8V dropped.

0.8V * 0.02A = 0.016 Watts
(1.02 Watts + 0.016 Watts) / 0.85 = 1.22 Watts

And for the parallel leds, we do the same for the resistors used (figure 0.5V dropped).

0.5V * 0.02A * 16 = 0.16 Watts
0.16W + 1.02W = 1.18 Watts

In the end, we have the higher voltage, series led string as wasting slightly more power from the battery as the lower voltage, parallel led strings.


As you mentioned I step up led driver would be the best option. You could use a IS31BL3508A since it could support 26V led strings, so you could build 2 strings of leds but you need to be sure that the current of the led does not exceed maximum output current (in this boost converter is 20mA). In the datasheet of the converter there are several example circuit. Regards.


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