Blown fuse monitor -/+24vdc supply

Discussion in 'The Projects Forum' started by mcgyvr, Oct 15, 2009.

  1. mcgyvr

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    Looking for a circuit to monitor the state of up to 2 fuses and energize a relay if either one of the fuses blows but not energize the relay if only one fuse is installed in the fuse holder and the other position is not filled (unused).
    Would also like a failure led for each fuse position.
    Trickier part is the supply voltage is - or + 24vdc. I need the circuit to function with either polarity and not have 1 for plus and 1 for negative.
    Thanks in advance for any help you provide.
     
  2. KMoffett

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    What type of fuse holder are you intending to use...PCB clips?...panel mount?...???? Not sure, but without some unique fuse-holder design, I can't think of a circuit that distinguish a blown fuse from an empty fuse holder?

    ken
     
  3. mcgyvr

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    panel mount holder.
    I would think you would monitor for a load attached to the fuse output.

    Load present and voltage present at the fuse holder output lead means fuse present and not blown = do not energize led or relay.

    Load present and no voltage at the fuse holder output means blown fuse = energize led and relay.

    No load/no voltage present at output means no fuse installed and do not energize relay or led.
     
  4. KMoffett

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    Are you fuses at the outputs of the +/-24V supplies, or at the input. A schematic of your system might help. The addition of whether a load is present or not was not in you original "spec' ".

    Ken
     
  5. SgtWookie

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    This is pretty basic, nothing fancy. The LED won't light unless there is a load on the output.

    [​IMG]

    Current limit resistor gives about 20mA current thru the LED if the output is a dead short, correspondingly less otherwise.
     
  6. mcgyvr

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    SqtWookie, I need the LED to turn ON when the fuse is blown. Not ON when the fuse is good.
     
  7. KMoffett

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    Sgt, That was my first idea too. But that doesn't distinguish between a blown fuse with load present, and an empty fuse holder with a load present. I was hoping for more details of the OP's system.

    Mcgyvr, The fuse shorts out the added circuit. The curcuit works when the fuse blows...or when there is no fuse.

    Ken
     
  8. mcgyvr

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    simplified schematic
    battery-----fuse---load---return (battery could either be - or + polarity)
     
  9. mcgyvr

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    attached is the basic system schematic
     
  10. SgtWookie

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    The LED will illuminate when the fuse is either blown or not present, and there is a load of a few hundred Ohms or less. The input voltage polarity does not matter.

    If the fuse is present and not blown, the LED will not illuminate.
    If there is no load, the LED will not illuminate, regardless of the fuse condition.
     
  11. SgtWookie

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    Yes, Ken - but that's probably good enough. If an LED is lit next to a fuseholder, it really doesn't matter if there is a fuse in the holder or not - someone still needs to put a good fuse in the holder.

    [eta]
    If our OP really wants the LED to come on whether or not there is a load present, they could always add a 2.4k 1/2 Watt resistor across Rload. That'll only give about 6.3mA through the LED, but it should still be visible.
     
    Last edited: Oct 15, 2009
  12. mcgyvr

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  13. SgtWookie

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    Well, what I threw together is basically the same idea as Bill Marsden's schematic in reply #8, but with a bridge rectifier thrown in to take care of the polarity issue.

    If there is a voltage across the fuse, the fuse is missing or blown.

    [eta]
    If you really want a more complex solution, I suppose that could be arranged. ;)
     
    Last edited: Oct 15, 2009
  14. mcgyvr

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    last questions.. Are there any safety hazards associated with this? What about if it was used for 75 or 100 vdc? I notice I still measure supply voltage at the output of the fuse even though no fuse is installed?
    I just don't want to pull the fuse to hookup another load and get a shock.

    Even the power indicator LED on my load was still on and the cooling fan on the load was spinning "slowly" with the fuse removed.

    Should I just reduce the current to the LED and sacrifice brightness?
     
  15. SgtWookie

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    Not with a 24VDC supply.

    A different approach should be used for higher voltages.

    At 24v, you shouldn't get shocked. If you go above 50VDC, a different approach would be needed.

    That's OK. There will be a small amount of current through the "Blown Fuse" LED indicator, but as I mentioned before - even if the load was a dead short, it'll be limited to 20mA by the 1.1k resistor. Unless, of course, you wired something wrong.

    You could do that if you'd like. Increase the resistor to 2.2k; that'll cut LED current to a max of 10mA.
     
  16. KMoffett

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    This is not possible with the schematic you posted....or you're measuring the input of the fuses...or the load has another phantom source of power. Are you sure that the schematic shows all the parts of your system?

    Ken
     
  17. mcgyvr

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    Is this a "better" solution to go up to and past 50vdc..
     
  18. SgtWookie

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    Nope, that won't work.

    Do you still have a requirement to work whether the input is connected up backwards, or not?

    Does your load require the polarity to be correct?
     
  19. mcgyvr

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    yes, requirements are the same. Dual polarity operation. Load needs same polarity as supply.

    Why do you say that doesn't work?
     
  20. SgtWookie

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    The schematic has many problems. For one, the base-emitter breakdown voltages would be exceeded.

    Here is a more or less generic schematic that would work at 24v DC no matter the input polarity. A bridge is required for the supply voltage to the comparator and reference voltage. Also, a bridge is required for each fuse to be monitored.

    [​IMG]

    R1 needs to be adjusted for voltages other than 24v; basically R1=(Vsupply-10v) / 20mA; Vsupply must be >11v.

    R2 and R3 can be replaced by fixed resistors connected as voltage dividers. The junction of the voltage dividers should measure a volt or two higher than the junction of R4/R5 with Vsupply applied. The voltage divider resistors should be sized so that current flow through them is between 0.1mA and 1mA, basically 1k to 10k Ohms per volt.

    Ignore the symbol V2 over to the right. That was simply for testing the circuit in simulation; it causes F1 to blow after a few moments by increasing the current through Rload1 over 1A.
     
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