Blown Fuse Indicator

Discussion in 'The Projects Forum' started by doug3460, Dec 14, 2008.

  1. doug3460

    Thread Starter Active Member

    Oct 19, 2008
    87
    0
    Hi all. Please check the attached & let me know if I'm on the correct path.

    I want to add a fuse status monitor to the ATX power supply panel I'm putting together (the BrakeFlashPulser Circuit is on hold while I get my test equipment built ;)).

    I've spent many days reading about transistor switches, researching the web, etc. I know there are very simple ones out there, but I have a bad feeling about connecting across the fuse.

    My thought is that the Green Led is lit as long as current is flowing thru the fuse & the transistor is cut-off by the same supply: when the fuse blows, the Green LED goes off & the PNP goes saturated by the 5V supply, turning on the Red LED.

    Am I even close?
     
    Last edited: Dec 14, 2008
  2. eblc1388

    Senior Member

    Nov 28, 2008
    1,542
    102
    Your design will work. However, you may encounter difficulty for 3.3V monitoring.

    I have moved the series resistor for bi-color LED to individual LED pin so you can calculate both values independently. One usually requires a lower value series resistor for Green LED than Red. I also added labels to your drawing.

    Be careful when you calculate R1 = 5 / 0.796mA = 6.28KΩ not ohm, and this is an easy mistake to made.
     
    Last edited: Dec 15, 2008
  3. Alberto

    Active Member

    Nov 7, 2008
    169
    36
    the one attached is simpler (avoid to use the +5V)
    It will work only with load present. (as yours anyway).
    As L. Chang already said be careful with the base resistor calculation, 10 ohms on the base will burn your transistor.

    Alberto
     
    Last edited: Dec 15, 2008
  4. doug3460

    Thread Starter Active Member

    Oct 19, 2008
    87
    0
    I edited the drawing to make the needed corrections & have attached it.

    (I can't believe I forgot to move the decimal point - what a goof :rolleyes:)

    Thanks to both eblc1388 & Alberto for your review & comments.

    eblc1388 wrote:
    I know it's close, but the LED is pulling 2.5 on the green side, so that's .8V remaining. I figured a 40Ω, 1/8ω resistor, so it should be okay.

    Alberto wrote:
    I understand your suggestion, but as I posted originally, I do not want to go across the fuse & the ATX has plenty of 5V source lines.

    BIG QUESTION NOW IS:

    What do I need to change if the source rail is negative? I have fused all my outputs from the ATX & given each it's own ground per Sgt Wookie's recommendations elsewhere in the forum. So I have -12V & -5V too that need indicators.

    Thanks again.
     
  5. eblc1388

    Senior Member

    Nov 28, 2008
    1,542
    102
    Yes. But don't over do it. :)

    At 5V, a 5KΩ resistor will give 1mA, a slightly larger value one will give 0.8mA.

    Post your design for the 3.3V monitoring and we'll check it.

    Negative voltage monitoring usually requires one additional transistor to invert the logic state on existing circuit topology. Why don't you think on it a bit and see if you can come up with something workable?
     
  6. doug3460

    Thread Starter Active Member

    Oct 19, 2008
    87
    0
    After further research & much pondering, it occurred to me that I was making this overcomplicated for what I really wanted. Attached is a simpler solution that will tell me if the binder posts are "hot."

    In regard to the negative supply question for my original design, I think I would skip adding a transistor & simply reverse mirror the components. If my understanding of the circuitry is correct, that should work with a change to the LED to common anode.

    But I think there is another problem in that circuit. If both the Supply Voltage & the PNP collector voltage are the same, they'd cancel each other out, so I'd need to switch to 12V on the collector for the 5V lines.

    I think. lol.

    Anyway, thanks for the reviews & assist.
     
  7. kathapurushan

    Member

    Dec 17, 2008
    14
    0
    i think this is one simple circuit without transistors. i have added additional diodes D1 & D2 for additional protection.

    Principle :-
    Case1 - Fuse OK
    The GREEN and RED portion are energized. Bicolor LED will be showing "SOMEWHAT" GREEN color since the RED portion is getting lower voltage compared to GREEN portion.

    Case2 - Fuse Blown
    Only the RED portion is energized with +5 volts via R2.The bicolor LED will be glowing in RED colour.

    Note :- Select the values of R1 & R2 after calculations and trial and error methor using prototypes before final. Variable resistors will help you.The only de-merit of this circuit what i think is - you will have to say a strange color as "GREEN" (when the fuse is OK)
    kathapurushan
     
  8. Wendy

    Moderator

    Mar 24, 2008
    20,765
    2,535
    Another time honored technique...

    [​IMG]
     
  9. Pennybags

    New Member

    Sep 21, 2009
    3
    0
    So I am looking at doing something like this but with a 12V 400 Amp ANL fuse. What kind of circuit do I need with this high amp type application?
     
  10. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    See Bill Marsden's post right above yours.

    The voltage is what matters in this case, not the load current.

    If your supply is 12V DC, then it's practically in the bag.

    Looking at Bill Marsden's schematic, the +12v source would be connected on the left, load connected to the right.

    Once you've selected the LED you want to use, look at it's typical Vf @ current rating; it might be something like 2.5v @ 25mA. (Vf=forward voltage) You use that to determine the value of the LED current limiting resistor (we'll call that Rlimit), shown above the fuse symbol on Bill's schematic.

    The basic formula is:
    Rlimit >= (Vsupply - Vf(LED)) / DesiredLEDCurrent
    So, just plug in the numbers:
    Rlimit >= (12v - 2.5v) / 25mA
    Rlimit >= 9.5/.025
    Rlimit >= 380 Ohms
    380 Ohms is not a standard value of resistance. The closest standard value >= 380 is 390 Ohms.
    A table of standard resistance values is here: http://www.logwell.com/tech/components/resistor_values.html
    Use the yellow and/or green columns. You can usually get those values locally, but you'd have to order E48 and higher values.

    The table is the same for values from fractions of Ohms up to 10 million Ohms. Just shift the decimal point right or left as your needs dictate.
     
  11. Pennybags

    New Member

    Sep 21, 2009
    3
    0
    Thanks SgtWookie, I figured that but I guess what I am stuck on is if a 400 amp fuse blew what is preventing the motor to get the power through the resistor/LED side. Isn't there going to be a lot of heat on that resistor and cause it to pop? Also, isn't the LED only going to light when the motor is called upon? (Such as a spike that caused the fuse to blow. No short anywhere)
     
  12. Pennybags

    New Member

    Sep 21, 2009
    3
    0
    Anyone have a response???
     
  13. BillB3857

    Senior Member

    Feb 28, 2009
    2,400
    348
    The current going to the motor with a blown fuse will be limited by the LED current limit resistor. True, if the motor is being fed through a contactor (BIG relay), the LED would only be on while the motor is being told to run. A fix for that problem would be to put a resistor from the load side of the fuse to the negative side of the power bus. Something in the order of 100 ohm, 2 watt should do it. With a good fuse, that dummy load resistor would burn an extra 1.44 watts.
     
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