2.3 Current Sensor - Short Circuit Cutoff

High-Side Current Sense Cut-Off Circuit The circuit detects over current and short circuits. It uses a 12 mOhm Shunt Resistor. There are two OP Amplifier use in the circuit. The first one acts as a Differential Amplifier (DA) to detect voltage across the Shunt Resistor then it sends out amplified Voltage to the second OP Amplifier that acts as Comparator (COM). The COM sets with reference voltage of 4 Volts to its negative input terminal. When COM received voltage on its positive input terminal equal or greater than its reference voltage (i.e. 4 Volts), it sends out 10v from its output terminal to power the Transistor and consequently powers the Relay to cut the connection of Battery supplying power to Lamp. The DA is set to sense max 12.9 Amps of current across the Shunt. The setting is made through 430 Ohms Resistor connected to its positive input terminal. The rest of the resistor is set to 1 kOhms. There is a momentary switch place before the relay terminal. This will restore connection when a disconnect occurred provided the load wattage or current is reduce. To simulate, increase the lamp wattage settings to about above 95 watts, this will draw current more than 12.9A and disconnect will occurs. Or the switch on the lamp can be close to Short the line.

published 1 year ago