Fuses are essential parts of power-distribution systems because they prevent fire or damage to electronic equipment. Fuses have the disadvantage of requiring replacement after every burnout, but they have the advantages of being inexpensive and widely available. It is difficult to determine the failure time of fuses with ceramic or sand-filled bodies to prevent arcing. This Design Idea presents a simple circuit that solves this problem (Figure 1). It visually and audibly indicates ac-mains-fuse failure; in most cases, audible indication is sufficient. The circuit works with a range of loads, and you can change its components to adapt to particular ac mains and load specifications.
Figure 1. This circuit visually and audibly indicates ac-mains-fuse failure.
When a fuse is in good order, the indication circuit is off because the fuse shunts it. When a fuse burns out, the indication circuit starts working. Capacitor C1 reduces the ac-mains voltage, and bridge diode D1 rectifies the ac voltage. Resistor R1 limits inrush current when capacitor C1 is discharged. Zener diode D2 and capacitor C2 form a dc voltage to operate a buzzer- and blinking-LED network. The blinking LED flashes, and buzzer B1, which has a built-in generator, sounds.
Like most other simple circuits, this circuit also has a disadvantage: It is incompatible with some load-power and ac-mains-voltage values. When a fuse burns out, the load stays connected to the ac mains, and the ac voltage divides between the circuit and the load. When the load is highly resistive or the ac-mains voltage is 110V rather than 220V, the circuit’s operating voltage may be too low to drive the circuit. In that case, decrease the value of capacitor C1 to 47 or 68 nF, after which the circuit’s resistance rises. With the component values in Figure 1, the tested circuit operated with resistive loads of 20 to 200W. With higher-power loads, the circuit operates well because, with higher load-power values, the circuit’s load resistance is lower.