Connect with us

Fuse and LED design question

Discussion in 'Electronic Basics' started by Junkmail789, Oct 8, 2003.

Scroll to continue with content
  1. Junkmail789

    Junkmail789 Guest

    I have a fuse that cannot be checked by visual, it must be ohm'd out. If I
    put an LED and resistor in parallel with it and run this second leg to
    ground, the LED should light when the fuse blows and NOT supply power to the
    other side of the fuse.

    Am I correct?

    Thanks,

    Matt
     
  2. It will supply the LED current to the rest of the circuit.

    You will have to know the circuit voltage and I assume this is a DC circuit
    fused.

    --
    *
    | __O Thomas C. Sefranek
    |_-\<,_ Amateur Radio Operator: WA1RHP
    (*)/ (*) Bicycle mobile on 145.41, 448.625 MHz

    http://hamradio.cmcorp.com/inventory/Inventory.html
    http://www.harvardrepeater.org
     
  3. John G

    John G Guest

    I do not understand what you mean by "second leg to ground".
    The Led and resistor should just be across the fuse.
    This was a common ploy back in the High voltage days of tubes and a neon was
    used instead of a LED.
    You will have to know the voltage and the effect of the LED current on the
    circuit after it has blown the fuse.
    Tell us a few facts and we might offer a bit of advice.
     
  4. John Fields

    John Fields Guest

    ---
    Probably not.


    Why not do this:

    DC>---[FUSE]--+------+
    | |
    [R] |
    | [LOAD]
    [LED] |
    | |
    GND>----------+------+

    The LED will go out when the fuse blows.
     
  5. You are looking at a circuit like this:

    |
    |
    |
    +----+------+
    | |
    .-. |
    ( X )LED |
    '-' |
    | |
    | |
    .-. ,-.
    | | |||
    | |R ||| Fuse
    '-' '-'
    | |
    +---+-------+
    |
    |
    ===
    GND
    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

    If so, then yes, the LED will not light unless the fuse blows. However,
    there WILL be current through the LED/resistor pair after the fuse blows.
    You should limit it by making the resistor as large as it can be while still
    glowing sufficiently under failure conditions.

    If you use an NPN transistor where the base connects to the + side of the
    fuse, you can separate the two circuits more cleanly as so:

    VCC
    +
    |
    +---------------+
    | |
    .-. |
    | |
    | | Your Load Goes Here
    '-'
    | |
    | |
    \| ___ |
    |----|___|- --+
    <| |
    | ,-.
    | |||
    .-. |||
    ( X ) '-'
    '-' |
    | |
    +---------------+
    |
    ===
    GND
    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

    That way, when the fuse blows, the transistor will turn on, but your load
    won't be powered through the LED/resistor pair. The base resistor should be
    chosen low enough so that when the fuse blows, the voltage drop across the
    base generates enough current to saturate the transistor, but not low enough
    to cause the PN junction to pass too much current and fry either the LED or
    transistor.

    Regards,
    Bob Monsen
     
  6. John Fields

    John Fields Guest

    ---
    Usually, when the fuse blows, it's desirable to remove the source of
    power from the load instead of just removing the load from ground.

    That can be done like this, and have the LED light when the fuse blows
    as well:

    +DC>--+--[FUSE]--+------+
    | | |
    E | |
    PNP B--------+ |
    C | |
    | | [LOAD]
    [LED] | |
    | [R] |
    [R] | |
    | | |
    GND>--+----------+------+
     
  7. JeffM

    JeffM Guest

    A much better idea is to take it from the fused side (load side, not supply side)
    thru the resistor to return (ground).

    This is called a pilot light.
    The indication will be the inverse of what Matt wanted,
    but the results will be more predictable.
     
  8. The Al Bundy

    The Al Bundy Guest

    ---
    Maybe it's wise to put a diode between B (A) and E (K) of the transistor, to
    protect it if the suddenly drop of current causes high(er) voltage spike at
    the load.

    Al
     
  9. John Fields

    John Fields Guest

    ---
    Good idea, but since the only way the load could generate a spike is if
    it looked inductive when the fuse blew, this might work better:

    +DC>--+--[FUSE]--+------+---------+
    | | | |
    E | | |
    PNP B--------+ | |
    C | | |K
    | | [LOAD] [DIODE]
    [LED] | | |
    | [R] | |
    [R] | | |
    | | | |
    GND>--+----------+------+---------+
     
  10. Junkmail789

    Junkmail789 Guest

    Let me clarify a few points, now that I see the other ideas presented.

    1. I don't want any load on the circuit at all. That's why I was thinking of
    doing it in parallel, but I was unsure of the size of the resistor in order
    to get the least amount of current, but then if you get the resistor high
    enough so that the LED doesn't light does this leg actually conduct?

    Didn't want to go across the fuse itself because then the circuit would
    still get power through the R and LED.

    Didn't want to go in series because of #1.

    So let's try this question--

    Is there any way to check the status of a fuse without putting a load into
    the circuit?

    To give you guys an idea of my problem. We use gas analyzers that contain
    thermal fuses. I don't want to interfere with any of the internal circuitry
    but would like to be able to have a LED or alarm like a piezo to indicate
    that the fuse has blown open. As it stands we have to remove the cover,
    unplug the fuse and ohm it out.

    Thanks in advance you guys are a great help.


     
  11. John Fields

    John Fields Guest

    ---
    I'm confused with what you mean by "parallel" and "series" and about
    what you mean when you say that that you don't want to interfere with
    the circuitry. If you want to modify the instrument to include this
    blown fuse detector you're going to have to mess with _something_, no?

    If you can add circuitry around the fuse you should be able to use the
    circuit shown below which will put about an extra 120µA load on the
    supply when the fuse is intact and about a 2mA load on the supply when
    the fuse blows if you use a high-efficiency LED like an HLMP-4700.

    Plus, when the fuse blows and the LED comes on there will be absolutely
    _no_ current flowing into the load.

    Some other info would be handy, like how big is the fuse (amps) and is
    it really on the DC side of the supply?



    +DC>--+--[FUSE]--+------+
    | | |
    S | |
    PCH G--------+ |
    D | |
    | | [LOAD]
    [LED] | |
    | [100K] |
    [R] | |
    | | |
    GND>--+----------+------+
     
  12. ChadMan

    ChadMan Guest


    Will this type of circuit do this work? Does it have too much overhead
    in parts? I am a newbie and would like to know. This looks like it
    should light up the LED on event failure (blown fuse). Could another
    device like a FET or Transistor do the same thing (allow no current
    to flow until the base looses power) as the relay?




    o----------------------------o---------o
    | |
    | LED/LAMP |
    | ___ .---. |
    GND>--o---|___|--| X |o-----o |
    R1 '---' | |
    | | | Relay
    o /o _|_
    / -|_\_|
    / |
    o |
    | |
    o------------------o |
    | |
    | ____ |
    +DC>------o--|_--_|-o---------------o---------o

    Fuse

    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de


    ChadMan
     
  13. The Al Bundy

    The Al Bundy Guest

    Well it would work, only the relay should be activated at all time. Only
    when the fuse blows it goes off and switches on the LED. In normal operation
    (fuse ok) the relay draws some current and generates heat. I guess you do
    not want that.
    So at any time this circuit will draw a current.


    The idea with a FET that John Fields drawed is a good one (last posting of
    him). Only some diodes must be placed to prevent that the FET goes up in
    smoke when the fuse blows, beteen G-S and between the Gate/resistor
    connection and the load.
    When the load has a huge capacitance (buffer) it is possible that the load
    voltage drops slower when the fuse just went. This means a high voltage
    across the G-S of the FET appears and can damage it. The G-S diode protects
    it from high voltage spikes caused by interrupting a high current.

    In this schematic the resistor between the gate/load and GND defines the
    current that will be drawn through the fuse. This is much less then a coil
    of a relay.
    When a Mosfet is turned off the resistance between Drain and Source is very
    high (>10meg), so no current goes through. (Be aware of the body diode of
    the mosfet, between Drain-Source)
    So in total only the current through the resistor OR through the LED is
    drawn.

    Hope this helps.

    Al
     
  14. The Al Bundy

    The Al Bundy Guest

    In this way it protects that no reverse voltage can appear across the load.
    However when there is some inductance in the load (wires, etc) and the
    current is suddenly interrupted, the inductance generates a voltage(spike)
    on top of the voltage that is still across the load. On the right side of
    the fuse a higher voltage (spike) appears then the input DC voltage, and so
    a too high reverse B-E voltage at the transistor. The diode across the load
    wont stop this.

    Al
     
  15. John Fields

    John Fields Guest

    ---
    Yes, it will, since the high voltage spike will be negative and will
    largely dissipate in the diode. During the time the diode is conducting
    there will be about -1V applied to the base because of the drop across
    the diode, but all this will do is turn the PNP on harder while any
    inductance in the load is discharging.
     
  16. John Fields

    John Fields Guest

    ---
    No. If the load looks partly like a huge capacitance in parallel with
    the rest of the load there is no way a voltage more positive than +DC
    can appear on the gate when the fuse blows.

    +DC>--+--[FUSE]--+------+--------+
    | | | |
    S | | |
    PCH G--------+ | |
    D | | |+
    | | [LOAD] [BFC]
    [LED] | | |
    | [100K] | |
    [R] | | |
    | | | |
    GND>--+----------+------+--------+

    Notice that while the fuse is intact the BFC will be charged up to +DC.
    The instant the fuse blows the BFC will still be charged up to +DC, and
    cannot become more positive than +DC. When the fuse blows, only the BFC
    will be supplying operating current for the load and for the 100K ohm
    gate resistor. As current is drawn from the BFC the voltage across it
    will decay until the point is reached where its voltage will no longer
    be sufficient to keep the P-Channel FET turned OFF, then as the gate
    voltage falls closer and closer to GND the FET's channel resistance will
    decrease until eventually, when the gate-to-source capacitance is fully
    charged, the drain-to-source resistance will fall to as low a value as
    it can.
     
  17. N. Thornton

    N. Thornton Guest

    LEd/resistor across the fuse is almost certainly good.

    If thats no go for some odd reason, try an opamp monitoring the V at
    each end of the fuse.

    Regards, NT
     
  18. The Al Bundy

    The Al Bundy Guest

    But with this circuit is it possible (long wires, high current) that
    _spikes_ appear when the fuse blows:

    +DC>--+--[FUSE]--+-////-+--------+
    | | | |
    S | | |
    PCH G--------+ | |
    D | | |+
    | | [LOAD] [BFC]
    [LED] | | |
    | [100K] | |
    [R] | | |
    | | | |
    GND>--+----------+-////-+--------+

    //// = wire inductance, long wire

    These will be higher the the applied DC voltage (which respect to ground).


    Al
     
  19. John Fields

    John Fields Guest

    ---
    Yes, of course it's possible, but the spikes won't be _higher_ than +DC,
    they'll be more _negative_ , and the way to cure that problem is to do
    this:

    +DC>--+--[FUSE]--+-----+---////-+--------+
    | | | | |
    S | | | |
    PCH G--------+ | | |
    D | |K | |+
    | | [DIODE] [LOAD] [BFC]
    [LED] | | | |
    | [100K] | | |
    [R] | | | |
    | | | | |
    GND>--+----------+-----+---////-+--------+



    I just tried it using a couple of transformers for inductors, like this:

    HV
    ____ |
    +DC------O O---+--+ +----
    S1 )||(
    )||(
    )||(
    +--+ +----
    |
    |
    [8R]
    |
    |
    +--+ +----
    )||(
    )||(
    )||(
    GND>---------------+ +----

    I adjusted +DC until I got 1A flowing through the circuit with S1
    closed, then when I opened S1, HV went to about -300V!

    Then I connected a diode from HV to GND with the anode to GND and closed
    the switch. This time when I opened the switch HV went to about -1V.
     
  20. John Fields

    John Fields Guest


    Actually, more like this:

    HV
    ____ |
    +DC------O O---+--------+ +----
    S1 )||(
    )||(
    )||(
    +--------+ +----
    | |
    | |+
    [8R] [10000µF]
    | |
    | |
    +--------+ +----
    )||(
    )||(
    )||(
    GND>---------------------+ +----
     
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

-