图上，这个电压过压保护的原理 有大神能够分析一下吗！感谢

正常电压是5.6V+0.7V等于6.3V，当输入电压低于6.3V，Q3截止，Q2有输出，当输入电压大于6.3V，Q3开始导通，使Q3的E、C压降降低，当输入电压超过一定值时，Q3E、C压降小于0.7V（这个和Q3的放大倍数有关），使Q2截止，停止输出，起到过压保护的作用。

就是这个截图，，，，，此图是TI的一个方案里面看到的，

      中午好，如果右边是输入电压的话，Q2这个PNP三极管是不能导通吧，电源方向不对。
      以下是自己的分析，觉得是左边在给右边充电，当充电电压过高或者电流过大，则稳压管D3会被反向击穿，把Q3的基极电压钳位在5.6V,而Q3集电极钳位在5.6+0.7V的样子（要看Q3的发射极和基极间的开通电压，这里以0.7V举例），从而进行过压保护。

当输入电压低于5.6V时，Q3截止，Q2导通，VCC约等于Vin
当输入电压高于5.6V是，Q3导通，Q2截止，VCC=0

The main pass element in the protection circuit shown in Figure 17 is the PNP transistor Q2. Take care
when selecting this part because any drops in the power supply voltage will be determined by the
characteristics of this transistor. The FMMT720T device has been used for this duty. The FMMT720T is
one of a family of devices that exhibit very low VCE saturation voltage values. This minimizes the voltage
drop induced by the presence of the protection circuit. The transistor Q3 acts as the control element for
Q2 and will turn on (turning Q2 off) when the voltage at the power supply input is equal to the sum of the
Zener voltage due to diode D3 and Q3's own VBE voltage at a collector current of about 650 μA. Q3 and
D3 together produce a typical trip voltage of 5.85 V at 25°C. Approximately 0.53 V of this is due to the
VBE voltage of Q3. The remaining 5.32 V is produced across D3. Note that the Zener diode D3, although
a nominal 5.6-V device, is being operated at a very low reverse current, about 200 μA, as defined by the
VBE of Q3 together with the 2.7-kΩ resistor. At this current, the Zener voltage is below the characteristic
"knee" and is therefore less than the rated value. The 6.8-kΩ resistor connected to the base of Q3
provides the current necessary to keep Q2 turned on under normal circumstances.

vcc提供的是多少v呢？

我认为看作是稳压电路更准确。D3把Q3的基极电源稳定在5.6v，当外电路电压升高时Q3的基极电压不变，发射极电压升高导致集电极电压也升高，这个电压又是Q2的基极电压升高，导致Q2的导通减小，输出电压降低。输入电压减小时过程相反。