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BQ76930充电激活电路疑问

阅读TI官方开发板的图纸,关于充电激活电路有三个疑问:

1. 图中的Q56是起什么作用的,是否可以去掉

2. Q14作为源极跟随,是否可以去掉,靠后端加强保护

3. 为何图中要用D29,D30,D34,D37四个二极管累加电平,是否可以用电阻分压+稳压管保护的方式来输出电平

Star Xu:

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Star Xu:

回复 Star Xu:

1. Q56 provides a switch with Q16 of the current to the diode string.  It looks like the voltage limit is 33V with the zener and gate voltage limit of Q56.  Q56 was a later addition from its number, it must have had some significance to the circuit.  With a quiescent system it would seem a single FET should handle the Vgs for the VC5X derived voltage at Q14 source.  You might simulate to determine if it is needed in your application.

2. Q14 provides a limited voltage to the circuit with current from the full cell stack.  Removing it with connection to BAT+ would overvoltage the circuit.  Connecting to VC5X_B would pull unexpected current from the node and shift the operating voltage of the part.

3. Four diodes will provide a more definite voltage than a low voltage zener as well as block reverse voltage (which would not seem to be an issue here).

您可以用 http://www.ti.com/tool/TINA-TI 模拟一下您的设计电路。

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