Connect with us

Circuit to detect blown fuse

Discussion in 'Electronic Design' started by [email protected], Dec 24, 2006.

Scroll to continue with content
  1. Guest

    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.
  2. Baron

    Baron Guest

    A number of blown fuse indicators just use a neon bulb and 120K
    resistor in series wired across the fuse. If its lit its blown !
  3. Eeyore

    Eeyore Guest

    Talking of overkill, that reminds me of the first product I designed with a
    'fault' LED indicator. It monitored certain internal functions and illuminated
    when relevant.

    The first pre-production unit to be built by the client company somehow got
    miswired slightly.

    On application of power, there was a loud bang, a bright flash and a large cloud
    of smoke.

    After the smoke had cleared, the fault indicator was glowing helpfully.

    Cracks me up everytime that one. I have more too !

  4. Baron

    Baron Guest

    Nice one ! ROFL
  5. Put a neon test lamp across the fuse.

    Assuming there is some reasonable load connected, the neon lamp lights
    with a blown fuse.

    Standard procedure for checking a 35Z5.

    Many thanks,

    Don Lancaster voice phone: (928)428-4073
    Synergetics 3860 West First Street Box 809 Thatcher, AZ 85552
    rss: email:

    Please visit my GURU's LAIR web site at
  6. Jamie

    Jamie Guest

    you could use a Neon lamp, stick it in a little tube with
    a photo detector circuit to fire up a low voltage circuit
    where you can display LED's...
    Also makes for a good lightning detector. :)
  7. me

    me Guest

    You already have one, the lights don't work...
  8. Guest

    I thought of this and I could use it but I was hoping not to have to go
    this route. I guess it gets tricky when you have to bring AC into a DC
    circuit but I know there has to be a way.
  9. Brian

    Brian Guest

    Check out the circuit at
    Brian Ellis
  10. 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
    Between half periods C3 will charge again via the 330 Ohm resistor with about
    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 :)

    Circuit is not tested, so use at your own risk.
  11. Jamie

    Jamie Guest

    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.
  12. Baron

    Baron Guest

    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 !
  13. Small correction, added a resistor R2 of about 10k els red LED always on
    via base emittor of T1.
    You already noticed that of course :)
  14. neon


    Oct 21, 2006
    use LEDS using two LED two resistors green LED connects after the fuse and a red LED connects across the can get more elaborate but that will work.
  15. Guest

    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
    breaker, and run your indicator off the unused section. Or, look into
    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
    to use a contactor (relay) for overcurrent protection, and adding
    light contacts to a relay design is very easy.
  16. Ken Moffett

    Ken Moffett Guest

    wrote in
    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.

    Just and observation.

  17. 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.
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day