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I have a 2v-2.5v DC LED and I want to use it to indicate if the line fuse is working.

ON fuse is ok

OFF fuse blown up

What is the proper way to connect it since obviously a direct connection would fry it instantly (and using a dedicated AC-DC adapter just for a led would be crazy)?

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Well if you want to keep it really simple, you could achieve an inverted logic(led off when fuse is okay and on when fuse is blown), you could simply connect the led circuit in parallel to your fuse. Your led circuit would be simple, your LED in series with a higher reverse voltage diode(something like 1n40007 could work) and a resistor. Your led, the diode and led would all be in series and would typically stay off if the potential difference across the fuse is 0v(fuse is intact). The potential difference would rise to line voltage when the fuse is blown and hence your led would light up. You may find a bit of flicker in led due to line voltage frequency but It shouldnt be a major challenge.

If you certainly want the led to be on while your fuse is intact, you would have to wire up the led circuit between live and neutral. The led, diode and resistor would be connected between live(live after fuse) and neutral. From an execution perspective, solution 1 would be simpler to wire up and execute.

Hope this helps.

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  • \$\begingroup\$ From the reality perspective, however, there is no guarantee that the load will still pass the current through after the fuse is blown. Which makes solution 1 totally unreliable. Not only that, one wouldn't want mains voltage to be present on faulty load, even if it passes through diode and resistor. \$\endgroup\$ – Maple Jan 25 '20 at 19:04
  • \$\begingroup\$ @Maple I find no reason why solution 1 will be unreliable. When the fuse is blown, The load would certainly pass current to keep the LED on. If the load wouldn't pass current, the voltage on the output of the led circuit would rise to line voltage due to infinite impedance and hence the load would be forced to conduct a few milliamps to keep LED even if its flickering. \$\endgroup\$ – Nouman Qaiser Jan 27 '20 at 18:53
  • \$\begingroup\$ You assuming that the load survived the surge. Note, that the fuse is usually there not to protect the faulty load, but to protect others from it. You missing the safety point. It's like "the fuse is blown, there must be something wrong with the load... let's put some mains voltage to it and use it to light the LED" \$\endgroup\$ – Maple Jan 27 '20 at 19:10
  • \$\begingroup\$ @Nouman Quaiser Could you detail better your solution? Probably I'm missing something, but If I try to put the 1n40007 and a resistor in series with the led and whole led circuit in parallel with the fuse, according simulator, in case fuse blows up, led doesn't turn on but blows up as well due to overvoltage. \$\endgroup\$ – AndreaF Jan 28 '20 at 0:18
  • \$\begingroup\$ @Maple, Yes that is correct. We apply the mains voltage to test if the load can still pass current. The load experiences a line voltage only until it passes a very small current, as early as it does, the resistor in series with the LED drops most of the voltage and the load actually gets a very small share of voltage(it forms a potential divder with the resistor in series to the LED.) \$\endgroup\$ – Nouman Qaiser Jan 28 '20 at 5:53

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