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。 最佳答案

就是这个截图，，，，，此图是TI的一个方案里面看到的，

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？