Battery power is often used on small products, so it is necessary to charge the battery.
Here record the charging and power supply scheme of 1S battery 3.7V.
Paste the schematic first:
Let's briefly analyze:
1. The final terminal for supplying power to the system is VCC
2. Single battery power supply: When the USB is not plugged in, the G pole of the P-channel MOS tube is kept at a low level due to the pull-down resistor. At this time, the MOS tube is turned on, and the VBAT of the battery passes through the MOS tube to the switch and then to VCC.
3. Single USB power supply: No battery is inserted at this time. After inserting the USB, the G of the MOS tube is extremely high, the MOS tube is turned off, and the VUSB reaches the switch through the diode and then to VCC.
4. The battery and USB are connected at the same time: at this time, VUSB is high, so the battery will not supply power to the system through the MOS tube. work, USB charges the battery at the same time.
One thing to note here is that the diodes do a lot, assuming the diodes are shorted or there are no diodes:
1. When the USB and the battery are powered at the same time, the D of the MOS tube will be extremely VUSB, so the voltage of the battery VBAT is lower than the VUSB, and the current flow of the MOS tube is from S to D, so what will happen? condition.
2. When the battery is powered by itself, the battery assumes that the power supply is normal, then the VS of the MOS tube is close to VD, and the absence of a diode will cause VD to flow back to VG, and the MOS tube will be turned off again, which is also unable to supply power. So this diode is a must.
3. In the case of USB power supply, since the current to the system must pass through the diode, the current size of the diode needs to be considered. Here, different types of diodes need to be selected according to the size of the current. Only the unidirectional conduction property of the diode is used here. The voltage difference of the diode also needs to be considered. For a voltage difference of 0.7V, 5V through the diode is also about 4.3V. The voltage drop of AMS1117 is about 1.2V, so it needs an input of more than 4.5V to stabilize the voltage to 3.3V. The 5819 used here is also determined according to the situation, and the measured pressure difference is 0.3V.
By Urien February 28, 2018 15:58:35