I am looking for a circuit that will detect whether or not a 120v fuse has blown. I would like a green LED to show the fuse is not blown and a red LED to show when the fuse has blown. I am new to circuit designs but I have assembled many circuits in the past.
On a sunny day (24 Dec 2006 10:45:49 -0800) it happened firstname.lastname@example.org wrote in :
LEDs need something like 10mA to glow. So a LED circuit across the fuse to show 'open' will have 10mA flowing still, and 10mA is deadly 9assuming you are playing with main fuses this can be a nasty surprise. So in that case you need it powered from before the fuse.
This design should do, but of course 2 Neons would work too :-) I have worked in a place where some units had a neon that came on when the fuse was blown, and some would be on to indicate power.. and there were several hundered of these modules in racks....
Use something with high beta for the transistors, for example BC109C. Ib peak will be 120 x sqrt(2) / 470 000 = 361 uA, Ic = x say beta 500 = 180mA, but only 10mA can flow, this should saturate T2, light RED LED, and discharge C3. Between half periods C3 will charge again via the 330 Ohm resistor with about
10mA. To keep the charge voltage below .7 V so T1 does not conduct, in 1/120 sec, C3 = i . t / U = .01 x .0083 / .7 = 120uF.
120V AC ------------------------------------------------------ fuse
--------------- | | | | | |a | | | e| NPN --- e| NPN diode D2 | |a | \\| === \\| |k | 5V | C2 |----| C3 |----------===------------- zener --- /| | /| T2 470k |k === c| T1 |___c| | D1 |+ | | |- a diode k -| [ ] 330 [ ] 330 | | | | | | k k === C1 | GREEN LED RED LED | | a a | |______|___________| | C1: | 12mA @ 120V -> Z = 10kOhm, | @ 60Hz 1/jwC = 10000 = 1 / (6.28 x 60 x C) ->
| 4080000 C = 1 -> C = 1/ 4080000 = .000 000 242F -> 330nF | (rest of the current into zener). [ ] C1 should be specified at about 400 VAC. |R1 D2 is Vbe protection. |
------------------------------------------------------------------------------------ Now calculate C2, voltage rating about 10 V: for 1 V discharge in 1/120 second (half period) at 10mA load (LED), C x U = i . t -> C = i . t / U = .01 x .0083 / 1 = .000 083F = 83uF. So use 100uF / 10V.
R1 is a fusible resistor to limit current in case C1 shorts, maybe 10 Ohm.
For DC somebody else :-)
Make sure you use gold plated fuses in case of audio :-)
ok, if you want to get simply, use a LED with a R in series and a Diode across the LED to suppress the reverse voltage. That will take care of the Open fuse. To show good voltage, connect the same thing across the AC lines after the fuse.
I actually have working LED's from a 120 volt source with out the use of the reverse diode, has been working for almost a year now, others will argue this fact, I must be lucky... They are just simple 20 ma leds i had in my junk drawer.
"I\'m never wrong, once i thought i was, but was mistaken"
Real Programmers Do things like this.
Why lucky ! If you don't exceed the current rating or reverse voltage they will be fine ! In my original suggestion, you could just as easily used a LED. The resistor value would have been different but it would have to dissipate more power, thats all !
While the wire-a-neon-across-it technique is valid, some fuses are part of safety-critical circuitry and you aren't encouraged to wire ANYTHING around them (so, use a fusible resistor in the neon lamp circuit). Be aware that the lamp circuit can cause electric shock even with the fuse out...
While maintaining safety, it would be permissible to use a two-gang circuit breaker, and run your indicator off the unused section. Or, look into the many 'indicating' fuse designs (one nice one uses a spring-loaded flag, and when the fuse wire opens, it releases the spring). It's also possible to use a contactor (relay) for overcurrent protection, and adding indicator light contacts to a relay design is very easy.
email@example.com wrote in news: firstname.lastname@example.org:
Along with pin-indicating fuseholders:
Bussmann also has the neon-indicating type:
If they were not considered safe I doubt that they would be making and selling them. The RMS current available between the circuit side of the neon fuse holder (with the fuse open) to neutral or ground is approximately equal to the line voltage minus the voltage drop across the neon lamp divided by the series limiting-resistor in the fuse holder. With a dead short to ground the available current is:
(Vline-Vneon)/Rlimit, so (120v-90v)/47000 = 0.638mA
If I recall my medical electronics electrical safety numbers correctly UL's limit on consumer equipment is a maximum allowable leakage (to any "externally" accessable parts of a device) of 5.0 mA. Since the available current "internally" is only about 1/10th of the "safe" external limit (fuse blown and neon lighted), I doubt that it is too great a risk.
That said, poking around inside of any device, with it still plugged-in, even with the fuse blown and/or the power switch off, presents a risk. And, if the OP is modifing an existing commercial device, then (s)he is assuming the risk (as seen by a lawyer) for others' safety.
simply buy a fuse with an indicator - if the fuse is the resettable kind, they can be had with a set of free contacts that open when the fuse trips. The simple fuses are available with a pin that pops out and activate a microswitch in the fuse socket.