Electrical – In = 3.7V / Out = 5.2V @ 2A Switching Regulator

dc/dc converter

I'm setting up a Step-Up Boost circuit for my main board using the MT3608 CI. It will be powered by a 18650 3.7v battery and need to output 5.2v @ 2A.

Reading the datasheet ( http://prom-electric.ru/media/MT3608.pdf ) of MT3608 I saw that it regulates its output through R1, R2 and the Schottky diode.

I wonder if R1 and R2 are correct and what is the value of my diode to get 5.2v output?

My Schematic:
enter image description here

Best Answer

The formula for setting the output voltage is given on page 6 of the data sheet, the values of R1 and R2 you've chosen will result in a very high voltage. Values of 47k and 6.2k will be closer to getting 5.2v output with a 0.6v reference.

The schottky diode is not part of the voltage regulation, it's required for the basic operation of the converter. Diode selection is also covered on page 6.

You are unlikely to get 2A out of this converter, notwithstanding its headline figure is 2A and the switch current 4A. Read the data sheet in detail, and once you've read the efficiency curves and power dissipation limits, and applied them to your specific operating point, I'm sure you'll agree. Don't forget that an 18650 is not a 3.7v battery, you will probably want to use it down to 3v, 2.7v if you're hardcore.