电路断路器提供过流和精确过压保护

In the event of an overload when ILOAD exceeds its maximum permissible value, the increase in voltage across R6 results in sufficient base-emitter voltage to turn on Q2. The voltage on R4 and, hence, the reference voltage now pull up toward VS, causing D2’s cathode voltage to fall to approximately 2V. D2’s output transistor now sinks current through R7 and R8, thus biasing Q3 on. Q4’s gate voltage now effectively clamps to the supply voltage through Q3, and the MOSFET turns off. At the same instant, Q3 sources current into R4 through D1, thereby pulling the voltage on R4 to a diode drop below the supply voltage. Consequently, no load current flows through R6 because Q2, whose base-emitter voltage is now 0V, has turned off. As a result, no load current flows through R6, D2’s output transistor latches on, and the circuit remains in its tripped state in which the load current is 0A. When choosing a value for R6, ensure that Q2’s base-emitter voltage is less than approximately 0.5V at the maximum permissible value of the load current.

As well as responding to overcurrent conditions, the circuit breaker also reacts to an abnormally large value of the supply voltage. When the load current lies within its normal range and Q2 is off, the magnitude of the supply voltage and the values of R3 and R4, which form a potential divider across the supply rails, determine the voltage at the reference input. In the event of an overvoltage at the supply voltage, the voltage on R4 exceeds the 2.5V reference level, and D2’s output transistor turns on. Once again, Q3 turns on, MOSFET Q4 switches off, and the load becomes effectively isolated from the dangerous transient.

The circuit now remains in its tripped state until reset. Under these conditions, Q3 clamps Q4’s gate-source voltage to roughly 0V, thereby protecting the MOSFET itself from excessive gate-source voltages. Ignoring the negligibly small voltage across R5, you can see that the reference voltage is VS×R4/(R3+R4) in volts. Because D2’s output turns on when the reference voltage exceeds 2.5V, you can rearrange the equation as R3=[(VST/2.5)–1]×R4 in ohms, where VST is the required supply-voltage trip level. For example, if R4 has a value of 10 kΩ, a trip voltage of 18V would require R3 to have a value of 62 kΩ. When choosing values for R3 and R4 to set the desired trip voltage, ensure that they are large enough that the potential divider will not excessively load the supply. Similarly, avoid values that could result in errors due to the reference-input current.

When you first apply power to the circuit, you’ll find that capacitive, bulb-filament, motor, and similar loads having large inrush current can trip the circuit breaker, even though their normal, steady-state operating current is below the trip level that R6 sets. One way to eliminate this problem is to add capacitor C2, which slows the rate of change of the voltage at the reference input. However, although simple, this approach has a serious disadvantage in that it slows the circuit’s response time to a genuine overcurrent-fault condition. Components C1, R1, R2, and Q1 provide an alternative solution. On power- up, C1 initially discharges, causing Q1 to turn on, thereby clamping the reference input to 0V and preventing the inrush current from tripping the circuit. C1 then charges through R1 and R2 until Q1 eventually tur