Last time, we published an issue of "Power Supply Automatic Switching Circuit Design" and received many questions from friends. For example, in the second circuit , why do we need MOS tubes when diodes are used in both circuits? In this issue, we will answer some of these questions and further explore these three automatic switching circuits.

First of all, let’s talk about the question mentioned at the beginning. Why do we need a MOS tube when there are two diodes? These two diodes are actually easy to understand . One of them is what we mentioned before, which is to protect VUSB to prevent it from charging the battery, and the other one is to protect the MOS tube .

The second question is whether this circuit (the first circuit) can achieve seamless automatic switching?

 That is to say, when VUSB and battery are both providing power, can the load maintain normal operation if one of VUSB or battery is removed ?

There are many determining factors, and seamless automatic switching can be achieved by adjusting components according to specific circumstances. For example, the parameters of the MOS tube, the resistance of the R2 resistor at the GND of the MOS tube. The smaller the resistance, the faster the MOS tube will turn on. However, it should be noted that R2 consumes power continuously . When the resistance is too small, the additional power consumption will be lost.

In addition, there are filter capacitors at the Vout end, capacitors at the Vin end, and load power consumption.

Third, why can't the two circuits be used when the voltages are the same?

These two circuits are only applicable to the case where Vbat must be smaller than VUSB. The circuit using two MOS tubes does not have such strict requirements on the voltage difference between VUSB and VBat, but of course not all voltages are suitable.

When the main and auxiliary input voltages are required to be equal, and the output voltage is also the same, and there cannot be too much voltage drop, in order to solve this problem, Wei Bi found another circuit on the Internet and shared it with you:

Vin1 is the main power supply and Vin2 is the backup power supply.

When both Vin1 and Vin2 have electricity, Vin1 will be used. As long as there is Vin1, Q1 is turned on to connect the G pole of Q2 to the ground. Q2 is also turned on, and the G pole of Q3 is connected to Vin1, and the S pole is also Vin1. Then Q3 is turned off, and Vout comes from Vin1.

That is to say, when Vin1 = 3.3V, regardless of whether Vin2 has voltage or not, Vin1 outputs voltage through Q3, and when Vin1 is disconnected, Vin outputs voltage through Q2. Since the Rds of the MOS tube is very small, the voltage drop is also very small, so Vout is basically equal to Vin.

Well, that’s all for this issue. Thank you for reading!

Some of the above pictures and information are from the Internet