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A Circuit Breaker is an electronic circuit that trips in the event of a short circuit (or overcurrent). This circuit helps prevent the power supply and load from sustaining damage by cutting of the supply to the load when an overcurrent is detected. Usually, a relay is used to break the circuit, but in this design, a MOSFET is used to break the circuit in case of a fault. The advantage of using a MOSFET over a relay is the absence of moving parts and low power consumption. A relay's is mechanical, due to which it could fail quickly due to ware and tare. Moreover, it's coil consumes power. With a MOSFET this additional power consumption is eliminated.
Starting from the right, the "upside down" op amp is used as a differential amplifier (or subtractor) to amplify the voltage drop across the current sense resistor (100mohm). The output of this op amp is fed to a comparator (bottom op amp). This compares the amplified drop across the 100mohm resistor against a 2.5V reference/threshold voltage. If the amplified drop exceeds 2.5V, the comparator outputs 15V. This is fed to a latch (the last of the 3 op amps, pointing up). Initially, the latch's output voltage is 0V. The PMOS stays on. When an overcurrent triggers a pulse, the latch's non inverting input sees 15V. That turns it's output on, which is fed back to the non-inverting input, thus keeping it on even after the pulse goes away. Since the output of the latch is 15V, it turns the MOSFET off. The diode near the latch plays an important role here. After the MOSFET turns off, the drop across the 100mohm resistor is zero. So the comparator outputs 0V. The diode prevents the comparator from pulling the input of the latch to ground and turning the MOSFET on.
The latch is reset by pushing the push button at the bottom, which pulls the latch's input down to ground. That turns the MOSFET on again.
Thanks to kiani for the idea.
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