How to realize automatic switching in dual power supply circuit

In real life, dual power supply switching is widely used. Let’s briefly take a look at how to use relays and contactors to achieve automatic switching!

working principle:

1. When there is BAT1 3.8V and no VIN_5V:

The voltage passes from the D pole of U1 to the S pole through the body diode of the MOS tube, S≈3.6V (assuming the body diode voltage drop is 0.2V), because VIN_5V does not exist, so G=0V, Ugs=-3.6V, the MOS tube will conduct After the MOS tube is turned on, because the conduction voltage drop is very small, the body diode will be cut off, and the S voltage is equal to 3.8V;

2. When there is VIN_5V and no BAT1 3.8V:

U1 is cut-off, VOUT=VIN_5V-Vdf (Vdf is the conduction voltage drop of diode D1). It should be noted that R1 should choose a larger resistance value as much as possible. VIN_5V will consume current to the ground through R1;

3. Sometimes both BAT1 3.8V and VIN_5V are available:

Assume that the voltage drop of the body diode of U1 is 0.2V, the conduction voltage drop of diode D1 is 0.3V, 3.8V passes through the body diode, S is 3.6V, VIN_5V passes through D1, so that the S pole becomes 4.7V, Ugs>0, MOS The tube is turned off, so in the end VOUT=4.7V, that is, VOUT uses the VIN_5V power supply;

That is to say, when the battery has both BAT and VIN_5V, VIN_5V power will be used first. Then let's think about it again. When VIN_5V is suddenly powered off, the G pole of PMOS will become 0V, Ugs=0-3.6V=-3.6V, the MOS tube is turned on, and VOUT=3.8V, so the advantage of this circuit is that VIN_5V is off. When powered on, the power of BAT1 will continue to ensure that VOUT does not lose power.

Several points to note about this circuit:

1. Select the appropriate turn-on voltage for the MOS tube;

2. The forward voltage drop of D1 needs to be considered;

3. Confirm the working voltage of the VOUT back-end load to avoid causing the voltage to be too high or too low.