Here is a simple scheme for adding a blown fuse indicator to your existing power supply circuit. This is done by just adding a resistor and LED to the existing circuit. The LED and resistor are connected in series and this combination is connected in parallel to the existing fuse. When the fuse is intact, it offers a low resistance and so the voltage drop across it will be not sufficient enough to glow the LED. When the fuse is blown off, it is equivalent to infinite resistance and so the entire power supply will drop across the resistor LED series combination and makes the LED glow. The resistance R1 is used to limit the current through LED.
The component values of other components are not given in the diagram because it depends on your specific power supply requirements. Any way for a conventional 12V power supply the component values are as follows: 1N4007 for D1,D2,D3 and D4; 230V primary,12V secondary ,2A step-down transformer for T1, 2A fuse and a 1000uF/25V capacitor for C1.
Circuit diagram with Parts list.
- Assemble the circuit on a general purpose PCB.
- The circuit will not work if the load is not connected.
- A highly resistive load may also impart the working of the circuit.
- The components for this circuit can easily obtained from your electronic junk box.
Will this circuit / components be ok for a 20>50 vdc load ?
I want layout of this circuit.
Can you pls mail the layout of this circuit.