My understanding (replace the drain d with the source s): B_VCC is USB power supply and is 5V. When the DC interface is not inserted, PSELF is grounded, then Ugs=-5V, PMOS is turned on, VCC=B_VCC; when the DC interface is inserted, PSELF is floating, S_VCC=5V, then Ugs=0, PMOS is turned off, VCC=S_VCC. In this way, the power supply can be switched.
I don’t know if it’s my understanding or the circuit?
Giancarlo Posted on September 29, 2020
This is no problem, but the body diode of the PMOS tube is borrowed. In this case, when there is no external power, the B_VCC will be powered by the body diode of the MOS tube. After the external power is connected, the PMOS tube will be closed under the influence of VGS, and the external voltage is greater than B_VCC-VF If the voltage is higher, external electricity will be used.
Edison Posted on September 29, 2020
Simulation test.
Annette Posted on September 29, 2020
What does the two VCC mean? A little dizzy.
Stuart Posted on September 29, 2020
DC is the main power and bat is the auxiliary power.
