This electronic scheme is a very simple circuit that acts as an electronic fuse. This electronic fuse is designed with three transistors, which cuts off the battery of a load when power consumption exceeds a certain limit. Fuse can be reset by pressing a button.
Transistor T3 is connected to the power line, which will block when the circuit is energized. When you press S1, T2 will receive the current through the voltage divider R5-R6. Therefore, it opens T2 T3 through R3, so the LED D1 will light, indicating the presence of output voltage. The third transistor of the circuit, T1 base-emitter junction is connected to the resistance R1, which is inserted the negative supply line.
Once the voltage on R1 is about 0.6 V, and bypasses T1 begins to conduct current in the base of T2 (leading to blocking of T2 and T3 and the disappearance of the output voltage).
If R1 = 10ohmi, the circuit is operated at a current of about 60mA output. Other levels can be obtained with the formula: R1 = 0.6 / I (Ohms), where I will be expressed in amperes (A).
Series transistor used in the scheme is a 2N2907A and lead currents exceeding 200 mA (if higher values of current estimates, the transistor will be replaced with another type).