I’m designing a PCB that is powered by a 3-cell LiPo battery. The battery’s +12V output is stepped down to +5V using an LM2596S-5 switching regulator, and then further stepped down to +3.3V with an LP38693SD-3.3 LDO.
When programming the MCU, I want to power the PCB via a USB Type-C connection instead of using the battery. This would allow me to program the firmware without needing to connect or recharge the battery each time.
At the input of the LDO, I want to seamlessly switch between USB power, battery power, or both being connected simultaneously, without causing any conflicts. To achieve this, I’ve used two RB058L-30DDTE25 Schottky diodes. These diodes have:
Forward voltage: 680mV (I assume at 3A),
Forward current rating: 3A,
Repetitive reverse voltage: 30V,
Reverse current: 2.5µA.
Since the LP38693SD-3.3 LDO has a dropout voltage of approximately 250mV, I need a minimum input voltage of +3.55V to maintain regulation. With the maximum 680mV forward voltage drop across the Schottky diodes, the design ensures I still have sufficient input voltage, so this configuration should work well.
However, I have two concerns about the design:
VBUS Capacitance Compliance:
If a Schottky diode is placed between the buck regulator and its external circuitry, and the USB port (VBUS), does C14 (330µF) still count as capacitance on the VBUS line? My understanding is that the capacitance on the VBUS line must not exceed 10µF, and if C14 is counted, it would far exceed this limit.
Reverse Leakage Current from Diodes:
The diodes on the +5V and VBUS lines have a reverse current of 2.5µA. Could this small reverse current cause damage to any part of the circuit? Additionally, if both the battery and USB are connected simultaneously, could this reverse current damage the USB host (e.g., the computer)?
The power circuit:
