Electronic – arduino – How does this design work

arduinopower supplyvoltage-regulator

I'm looking at the Adafruit Feather 32u4 schematic.

power schematic

It auto-switches between battery and USB power. From looking at the schematic, when the USB power is not connected, the battery power VBAT goes through a MBR120 Shottky diode, then a SPX3819 regulator to power the Atmega32u4.

From looking at the datasheets, the MBR120 has a 0.45V drop, and the SPX3819 has a 0.25V dropout at 50mA. This would be a total voltage drop of 0.7V. Since the regulator outputs 3.3V, the minimum input voltage would have to be 4V.

Since a LiPo battery runs down to about 3.5V when getting low, that would mean that the circuit could not be powered except when the battery was very full.

Am I missing something, or is this how it works? It would seem strange that you would be unable to use most of the battery power.

Best Answer

You have to read the MBR120 datasheet carefully, that Vforward of 0.4 V is at 1 A. This circuit is not going to draw 1 A.

My guess is that it draws less than 100 mA, now let's look at the datasheet again.

At 25 Celcius Vforward is less than 0.2 V. Only under very cold conditions will it increase (but in the cold the battery will give up also so this is a moot point).

So this means that the minimum required Battery voltage will be around 3.3 + 0.25 + 0.2 = 3.75 V, a lot better than your 4.0 V.

Even if the battery is slightly below 3.75 V, it does not mean the circuit will stop working immediately. The regulated 3.3 V might drop a bit and that is no problem either as the Bluetooth module can work down to 1.8 V and the Atmega32 can work down to 2.7 V.

My guess is that the module will only stop working if the battery voltage drops below 3.3 V or so. And then the battery is empty.

Also, have a look at a discharge curve of a LiPo battery:

enter image description here

We discharge the battery slowly so we can use the black 0.2C curve. Note how you would have used 90% of the capacity if you would discharge to 3.6 V (which is a safe value).

So it all looks reasonable to me !