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可能是我问题没描述明白。BQ76952的CHG输出高电平是Vbat+11V,低电平是Vbat。而BQ77216触发后控制红框中的MOS,把BQ76952的CHG往GND拉,怎么做到保证BQ76952的CHG被拉到Vbat而不是GND。而且BQ77216触发时,BQ76952的CHG可能还在输出高电平Vbat+11V。
您好,this configuration pulls down the CHG gate to GND. This is why the image you shared also shows DOUT and COUT being connected to DFETOFF and CFETOFF respectively. This ensures that the BQ779216 turns off the FET drivers of the BQ76952 and avoid discharging the charge-pump.
The pull-down resistance must be chosen to not draw too much current from the battery.
If they wanted to pull down the gate without using the DFETOFF/CFETOFF pins, then they would likely need additional circuitry to ensure the BQ76952 charge pump does not become discharged.
十分感谢,这个控制逻辑我明白了。还有个小问题,图上不能直接这样连接吧?BQ76952的CFETOFF脚最大输入电压是5.5V,而BQ7721602的COUT输出最低6V电压,MCU的IO口一般也就5V耐压,图上三个信号是直接接在一起的,即使FETOFF能靠稳压管钳位,DCOT引脚也和MCU的IO口接在一起了,这样不会损坏MCU的IO吗?
您好,You are correct that the recommended max for CFETOFF is 5.5-V. I've seen customers use 5-V or 5.5-V Zeners in order to limit the voltage into the pins. The figure is an example, it does not necessarily need to connect to the MCU.
They could simply connect the same clamped voltage into the BQ76952 and the MCU through the same line.