This circuit is used to light an LED if the fuse breaks.
The fuse is set to breakdown at 1A. The lamp below is designed to run @12V at 13W (current >1A).
When the fuse is not broken, the voltage at the right of D2 is slightly under 12V, but the voltage difference across the LED is not higfh enough to light it up.
Once the fuse breaks, the diode D2 is not in parallel with the LED anymore, and therefore the circuit is simply the LED and its current-limiting resistor. It is recommended to use a low current LED, as the current flowing through the LED is only 2mA, as the current limiting resistor is high in order to minimize the used current.
There are currently no comments