Hello,
I'm working on a BLDC motor controller using a STM32F446 for the MCU and a DRV8301 (
http://www.ti.com/lit/ds/symlink/drv8301.pdf) for the gate driver. The DRV8301 has a built-in buck converter that I'm using to bring the motor source (24V) down to 5V or 3v3 depending on the best solution. I'm will be communicating with the MCU via an embedded FTDI chip. I want to be able to flash the MCU and test some peripherals without hooking up the 24V source needed for the buck converter, so ideally I'd just use the 5V from USB with a 3v3 LDO. But when the 24V source is plugged in, I want the buck converter output to take preference. Given that, I think these are my two options, but I don't fully understand how to implement either of them.
1. Set the output of the buck converter to 5V, and run the USB 5V through a Schottky diode connecting to the buck converter 5V. That combined 5V rail would go to a 3v3 LDO. I think this would work because the forward voltage of the diode would drop the USB 5V to 4.5-4.7, putting it below the buck converter output which would then supply all the power. What I don't know is if it is bad to apply the USB 5V to the output of the buck converter when the 24V isn't present.
2. Set the output of the buck converter to 3v3. From the perspective of powering the system independent of USB this works because the only 5V peripheral is the Hall effect sensor which can run off the 24V source. The issue I see with this method is that I don't see how to make the 3v3 preferential to the USB 5V. Is there some sort of "switch" that can automatically activated when the 24V source is present?
Hopefully I explained my dilemma well. Please offer any suggestions or corrections to my assumptions! I'm here to learn
![ThumbsUp :-+](https://www.eevblog.com/forum/Smileys/default/icon_smile_thumbsup.gif)