Electrical – Li-Ion battery for RGB strip

dc/dc converterled striplithium ionpower supplyrgb

I am looking to power up my Teensy LC and my RGB strip from the same power supply, which is a 3.7V Li-Ion battery 3000mAh. The teensy can work perfectly fine with 3.7V. enter image description here

The problem is when it comes to the RGB strip. Since I have about 40 LEDs, and it says that every LED takes 0.2W at 12V, I guess I will have around 0.7A current draw.

I looked up the internet for DC/DC converters, but they all have low output current. I need at least 0.7A and max 1A. Do you think I will have to boost to 3.7 to 5V and then 5V to 12 V ? Or is there another way ?

Thank you

Best Answer

You haven't looked hard enough if you didn't find a 3.7V->12V step-up for >=1A.

I've built stronger things based on SMPS controller IC reference design circuits years ago. Since this is a design, and not a shopping recommendation site:

Try going to www.ti.com, use the (mobile) webbench power supply designer (it's on the home page), and design a 3.7 V -> 12 V converter. It works.

Also note that it's quite possible that you'd not want to use just any LED strip in this case – after all, they often contain series resistors themselves, which just wastes the voltage between the actual forward voltage of the LEDs and 12V. Meh.

With a bit of thought (generally: take a step-up controller, and instead of using a voltage divider to give the feedback voltage to achieve the a fixed output voltage, use a shunt resistor to achieve a fixed output current), you can design an actual constant-current LED driver instead of a constant-voltage voltage source, and get rid of the series resistors.

Also note that, depending on the currents, bijunction transistors as used in your block diagram (it's really not a schematic) might be suboptimal and you might want to go for MOSFETs, or just switch the step-up on and off (which requires controlling of some ENABLE pin or some secondary comparator, instead of actually low-side switching, and thus might be more effective and easier).