Idea for system that shows which fuse is blown

• 7th January 2012, 07:55
itskannali
Idea for system that shows which fuse is blown
Question…
The railways are using hrc fuses in series. They have more than 30 fuses in series connections. If any one of the fuse is blown, they want to know which fuse is blown bcoz manually checking each fuse and detecting the faulty one is time consuming.

My idea…
The given circuit. Before each fuse, we have to make a parallel connection containing a high resistance, relay, led which is grounded. If any of the fuse is blown, the high current passes through the respective parallel circuit & the relay coil is energised. When the relay coil is energised, a closed circuit is formed between relay and LED, and so LED is ON. Thus detection works.

http://images.elektroda.net/14_1325919213_thumb.png

is the circuit a blunder...? suggestions..?
• 7th January 2012, 08:59
umery2k75
Re: fuse alarm system...
Select Resistor that can limit DC current in LED to around 10mA depending upon on high the DC voltage is, attach is the diagram that will work.

http://images.elektroda.net/38_1325923167.jpg
• 8th January 2012, 05:40
itskannali
Re: fuse alarm system...
bt in this diagram even if the fuse is blown the current will pass through the parallel led circuit which damages the load... when the fuse is blown, the circuit must get open and we must get an indication....
• 8th January 2012, 08:43
Gauthier
Re: fuse alarm system...
around 10 mA will pass in your load if one of the fuse blow, is it enough to damage the load?
• 9th January 2012, 12:00
umery2k75
Re: Idea for system that shows which fuse is blown
@itskannali
You can attach a resistive load in parallel with your load. In this case you can attach resistor with proper wattage in parallel with your load. When the fuse blows, particularly LED is bypassed and very small mA would flow through the resistor attached in parallel with your load. I don't know if your load is very sensitive that it cannot even sink few mA to operate an LED, that's why I ask you to put resistor in parallel with your load. It completes the loop.

---------- Post added at 12:00 ---------- Previous post was at 11:59 ----------

@Gauthier
I don't even think few mA would damage the load.