Share a classic switching circuit.
Using a MOS tube as a switch for battery power, the original 5V power supply can be either USB or 5V for an external adapter.
It works like this, when the VUSB is powered on, the PMOS is cut off, although the body diode current flows here, but the VUSB voltage will be higher than the Vbat, the Vgs of the PMOS is >0, the PMOS body diode is cut off, the load is powered by VUSB, when the VUSB is out of power, the PMOS is turned on, and the load is powered by VBAT.
It should be noted here that the direction of the PMOS tube is not reversed, because the battery voltage passes through the body diode to the S pole, which makes Vgs <0 and the MOS tube turns on to power Vout when there is no VUSB.
So why not choose the S pole of the PMOS tube against VBat, and the D pole against Vout?
Or because of the body diode, if there is a VUSB, through the body diode, the VUSB voltage will directly reach the battery, which is not okay.
What if you accidentally turn PMOS around?
Let's make a modification to the circuit above.
The working principle is the same as before, but here a D2 is added, which is to prevent the VUSB from potentially charging the battery. In other words, there is an additional D2 drop here.
Both circuits are suitable for situations where the Vbat must be smaller than the VUSB.
So what if two voltages are equal?
Let's use the MOS tube instead of D2 to try, the voltage difference between VUSB and VBat here is relatively not very strict, when VUSB is powered on, Q2, Q1 cut-off, VUSB power supply, when VUSB is out of power, Q2, Q1 conduction, VBat power supply.
Here the switching can be done at the same voltage.