Skip to main content
Electrical Engineering

Return to Answer

added 151 characters in body
Source Link
AnalogKid
AnalogKid
  • 23.2k
  • 1
  • 14
  • 40

When ACOK- is high and the external FET is off, the body diode within the FET still is in the circuit and can conduct current from the USB port to the chip. The chip then evaluates the voltage level from the USB port to determine if it is good enough to run the system. If yes, the chip turns on the FET, which effectively "shorts out" its own body diode to decrease the voltage drop to essentially 0 V, increasing the USB voltage at the chip to the full 5 V.

When the USB cable is unplugged, the voltage at VL decreases, the chip wakes up and begins using battery power through the boost converter circuit.

When ACOK- is high and the external FET is off, the body diode within the FET still is in the circuit and can conduct current from the USB port to the chip. The chip then evaluates the voltage level from the USB port to determine if it is good enough to run the system. If yes, the chip turns on the FET, which effectively "shorts out" its own body diode to decrease the voltage drop to essentially 0 V, increasing the USB voltage at the chip to the full 5 V.

When ACOK- is high and the external FET is off, the body diode within the FET still is in the circuit and can conduct current from the USB port to the chip. The chip then evaluates the voltage level from the USB port to determine if it is good enough to run the system. If yes, the chip turns on the FET, which effectively "shorts out" its own body diode to decrease the voltage drop to essentially 0 V, increasing the USB voltage at the chip to the full 5 V.

When the USB cable is unplugged, the voltage at VL decreases, the chip wakes up and begins using battery power through the boost converter circuit.

Source Link
AnalogKid
AnalogKid
  • 23.2k
  • 1
  • 14
  • 40

When ACOK- is high and the external FET is off, the body diode within the FET still is in the circuit and can conduct current from the USB port to the chip. The chip then evaluates the voltage level from the USB port to determine if it is good enough to run the system. If yes, the chip turns on the FET, which effectively "shorts out" its own body diode to decrease the voltage drop to essentially 0 V, increasing the USB voltage at the chip to the full 5 V.