Indicator circuit for power supply 220V blown fuse

This is the indicator circuit that indicates a blown fuse, it will show the condition of the fuse via an LED. This circuit is suitable for fused 220V power supplies. This compact circuit is very useful and reliable. It is a low-cost circuit using few components. Often, when an electronic device indicates no power, the underlying cause could simply be a blown fuse.

Under normal conditions (the fuse is OK), the voltage drop on the first arm is 2V+(2x0.7V)=3.4V, while the second arm is only 2V. So current flows through the second arm, i.e. through the green LED, causing it to glow; while the red LED remains off.

When the fuse blows, the power to the green LED is blocked, and since there is only one LED in the circuit, the red LED glows. In the event of a power failure, both LEDs remain "off".

LED indicator light:

Red LED: fuse blown

Green light: normal, fuse is normal

Both LEDs are off: power failure, mains shutdown, or circuit damage.

This circuit can be easily modified by adding an alarm/buzzer circuit to generate an alarm in the event of a blown fuse (see Figure 2). Optocouplers are used to trigger the alarm. When the fuse blows, the red LED lights up. At the same time it "turns on" the siren.

Instead of a bicolor LED, two LEDs, red and green, can be used. Similarly, only one diode can be used instead of D1 and D2. Two diodes are used to increase the voltage drop since two LEDs may produce different voltage drops.