[求助] 求解 BQ76930 20串电路 驱动电路的问题

图片是TIbq76930-20串电池管理系统的demo板驱动电路（两个BQ76930串联使用）

bottom device是 第一串至第十串电池，upper device 是第十一串至第二十串电池

电路要达到的目的是：不管DSG_U和DSG_B,两个信号的任一信号为高电平约为12V，都会驱动底下的Q38,Q40,Q42关闭，以实现保护功能。

官方介绍是这样的：Given that the voltage at BAT+ in Figure 3 is 80 V, and both DSG outputs of upper and bottom devices assert upon initializing with no fault detected. With the DSG output of upper device labeled as DSG_Uasserts, the voltage of DSG_U to the ground reference GND_U for upper device would be ≈ 12 V;because the ground reference of the upper device is in the middle of battery stack, the voltage between GND_U and GND_B is ≈ 40 V, as the GND_B is the ground reference of the bottom device. Then the voltage of DSG_U to the GND_B is ≈ 52 V, with the typical VGS(TH) of Q27 being about 1.7 V; the voltage at the source of Q27 is about 52 V – 1.7 V = 50.3 V. With this voltage at the source of FET Q28, this

FET is on, and the voltage at Q32 drain is also 50.3 V. With DSG_B asserts at ≈ 12 V to GND_B, the voltage at source of Q32 is also clamped at≈ 10.3 V, subtracting the forward voltage of D41 and VGS of Q33 summed at 2.6 V, the voltage at the gates of Q38, Q40, and Q42 is about 7.7 V,thus the DSG FETs are all turned on.

大致意思就是GND_B为参考地，假设  BAT+=80V，那么GND_U=52V，那么Q27的G极DSG_U=52V，Q27的S极=50.3V，这个50.3V加在Q28的S极，Q28导通，

我的问题来了：电路的驱动过程是怎么样的？或者说 是如何实现驱动效果的？   

我的理解是：当Q27的G极DSG_U为高电平时， DSG_B为低电平时，此时Q28并未导通，Q32也未导通，Q27的S极是悬空状态。相反，当Q32的G极DSG_B输出为高电平时，DSG_U为低时，Q27和Q28都未导通。驱动电源都无法到达底部功率MOS管Q38,Q40,Q42的G极。

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你理解的应该没问题吧。

DSG是放电控制端

注意查看查看BQ76930说明

楼主的这个图是官方demo板电路？

英文是原版demo电路的说明， 没看明白