Share a classic overvoltage protection circuit on the net.
We have three cases, the input voltage is less than 5.1V, the input voltage is between 5.1V and 5.7V, and the input voltage is greater than 5.7V.
The input voltage is less than 5.1V
When the input voltage is 5V, the D1 regulator is 5.1V, because the input voltage is less than 5.1V, D1 has not yet entered the stable state, there is no conduction at this time, and the transistor is not on. (The b and e poles of the transistor are 5V, Vbe=0).
The input voltage is between 5.1V and 5.7V
When the input voltage is 5.4V, which is greater than 5.1V of D1, D1 turns on and regulates at 5.1V.
At this time, the Vbe of the triode = 5.1V-5.4V = -0.3V, the on-voltage of the PN junction requires Vbe to reach -0.6V, and the triode Q2 is still not conductive.
The g pole of the MOS transistor Q1 is pulled to 0V by the resistor R3. The Vgs of MOS transistor Q1 = -5.4V, so the MOS transistor Q1 is on, and the Vout voltage is equal to 5.4V, which supplies power to the post-amp circuit normally.
The input voltage > 5.7V
There are three current paths
- R1两端电压为0.7V（5.8V-5.1V），流过0.7mA的电流。
- The Vbe of the triode Q2 is -0.6V (5.2V - 5.8V) and the triode Q2 is turned on
- Transistor Q2 is turned on and the current flowing through R3 is 1.23mA (5.8/4.7)
At this time, the Vgs of MOS transistor Q1 = 0, and MOS transistor Q1 is not on, that is, the input voltage is now too high.
At this time, the output voltage Vout is turned off, which protects the load circuit behind.
If you want to change the overvoltage protection circuit to a different voltage, you can directly replace the regulator tube with different voltage regulation values.