0
\$\begingroup\$

I have a question when I am working on MP2639A 2-cell LIPO charger (+8V4) from USB +5V. The recommend schematic uses P-MOSFET to protect reverse current from Load back to a USB plug.

enter image description here taken from MP2639A datasheet Page34

The signal /ACOK is an open-drain output signal. It will pull-low when the USB-power is OK.

From my understanding to turn-on P-MOSFET Vgs > Vgs-threshold. However, when we use Vg and Vs have the same voltage ( +5V ) How could the USB power turn on MOSFET? even the Vgs-threshold is a negative value. If the P-MOSFET can turn on, the current that can flow from Drain to the Source is the minimum current ( near 0 )

I am not sure that I understand correctly about the P-MOSFET using in this situation

\$\endgroup\$
0
\$\begingroup\$

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.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.