Help, my fuse doesn't work

Discussion in 'The Projects Forum' started by Darkstar, Jun 30, 2011.

  1. Darkstar

    Thread Starter Member

    Sep 3, 2010
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    Ok, it's an electronic fuse, see pic.

    I've seen this posted in 3 different places, by 3 different people. Assuming they are not all wrong, then I must be making the mistake somewhere.

    These are the instructions for what's supposed to happen:

    When connected, nothing happens until the start/reset button is pushed. Current continues to flow unless the current drops below a certain level.

    The current flowing through SCR T1 sinks below the holding level when the current is rerouted through T2 (2N3055). T2 and RS are built into the electronic fuse circuit for this purpose. If the voltage drop at RS exceeds the base-emitter-diode trigger voltage of T2 (~0.7V), the transistor conducts thereby bypassing the SCR T1. The resistance value of RS must be at least 0.2 Ohms. The resistance of RS must be such that RS multiplied by the fuse current shut-off value equals 0.7 volts. (E=IR)

    Once T2 bypasses T1, the current flowing through the SCR sinks below the holding level and T1 shuts off. This in turn causes the voltage drop at resistor RS to sink below the base-emitter trigger voltage of T2 and the transistor shuts off. The end result is the shutting off of the whole circuit. The DC fuse can be reactivated by pressing the start/reset button.

    The value of resistor R1 is dependent on the supply voltage. Multiply the supply voltage by 1K Ohms to get the value of R1. Connect the DC electronic fuse circuit to the PLUS line of the load. The voltage drop of the circuit is less than 1 volt.

    This is what actually happens:

    It does turn on when the button is pushed.

    It will not turn off when more current flows than is supposed to or when more voltage is supplied. I've gone up to 3x the shut off current.

    It will turn off if the circuit is broken, until reset is pushed again...

    I've tried simulating the circuit at www.falstad.com/ but this does not work either. When the voltage drop across RS goes above 0.7V, the fuse does not shut off.

    I'm not familiar enough with transistor circuits to spot the error. I can't find a fault in my wiring. My pinouts are correct as are the input and output polarities.
    Any explanations why this isn't working as it should?
    Thanks
     
  2. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    The problem is this:
    As the load current increases to the point that the transistor is conducting, it does indeed tend to limit the current through Rs. The problem is, it does not reduce the current through Rs. As the load current increases further, the current through Rs (and the SCR) still increases, because Vbe will increase slightly as Ie increases.

    Where did you get this circuit?
     
    Last edited: Jun 30, 2011
  3. Darkstar

    Thread Starter Member

    Sep 3, 2010
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    The current through RS is not supposed to be reduced. RS is not the load. The larger current (i.e. a short in the load) causes a larger voltage drop across RS which starts the shut down process. See explanation below.

    I got this from http://electroschematics.com/2923/dc-electronic-fuse/
    I've also seen it at
    FreeCircuitDiagram.com
    Simple-Circuit.blogspot.com

    These are the instructions given at the last site:

    "The first operation is beginning with initially the load current flows through SCR and resistor R1.The value of R1 (RS) is so selected that, the maximum load current multiplied by the resistance of R1(RS) is equal to 0.7 volts. When the load current exceeds the maximum value the voltage drop across R1 becomes more than 0.7V and switches transistor (2N3055) Q1 ON. Now the transistor completely bypasses the load current (meaning the current bypasses the SCR- my words) and the current through triac falls below the holding current. This makes the triac turn OFF. When SCR is OFF there will not be any current flow through R1 and so the voltage across it falls to 0.This makes the transistor turn OFF, completely isolating the load circuit. The fuse can be reset by pressing S1.When S1 is pressed the SCR is again triggered and remains latched to conduct the load current."
     
  4. CDRIVE

    Senior Member

    Jul 1, 2008
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    The TIC106 holding current (IH) = 8mA. This means that the 3055 has to shunt the SCR to the point that there is less than 8mA flowing through the SCR before it will shut off. Doubtful that this is going to happen.
     
  5. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    My original explanation is correct.
    The transistor does shunt current from the SCR, but it does not reduce the SCR current. Therefore the SCR will not turn off.
     
  6. Darkstar

    Thread Starter Member

    Sep 3, 2010
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    I was hoping to use this for low voltage and very low current work, like 100 uA to 100 mA and less than 20 volts. Will this be possible?

    Ron-
    I don't understand why the transistor, in parallel with the SCR, does not reduce the SCR current? The SCR has the resistor RS in series with it, the transistor does not.
     
  7. CDRIVE

    Senior Member

    Jul 1, 2008
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    If you're building a bench supply for circuit experimentation you would be far better off building a 'Current Limited' supply. You can make one with adjustable current limit.
     
  8. BillB3857

    Senior Member

    Feb 28, 2009
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    You might try installing a diode or two in series with the SCR anode. That would allow the ON voltage of the transistor to be lower than the required voltage to maintain current through the SCR. Of course, that would also cause the output voltage to be reduced by the diode drops.
     
  9. Darkstar

    Thread Starter Member

    Sep 3, 2010
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    Cdrive---
    The power supply is already built (as well as some things I want to use with the fuse.) This fuse itself was an afterthought. I thought it would be nicer to have something tell me when a limit was exceeded rather than just continue working with without knowing what limits were reached.

    BillB---
    I'll try the diodes. That seems to be a simple solution. I can always adjust my power supply to compensate for their voltage drop.

    Thanks
     
  10. SgtWookie

    Expert

    Jul 17, 2007
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    1,728
    The SCRs' cathode is the only current source for T2's base.

    The only way to turn off the SCR, once on, is to decrease the SCR current below the 8mA holding current. But, with the SCR being the only current source for T2's base, there is just no way for the current to fall that low unless the actual load current falls below 8mA - which, by the way, precludes the circuit from working with your 100uA-to-8mA circuits, unless you want to hold the RESET button down all the time.
     
  11. CDRIVE

    Senior Member

    Jul 1, 2008
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    Ron, no argument here. Personally, I think it's an awful design and Darkstar is going to be chasing his tail.

    For the record, I tried to spice this to the point of extreme irritation because I knew I was attempting a bad concept from the get-go. I did get it to work by replacing the BJT with a FET, plus other mods, but it's still awful!
     
  12. Ron H

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    Apr 14, 2005
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    Here is a simulation.
     
  13. Darkstar

    Thread Starter Member

    Sep 3, 2010
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    Wookie...
    Oh. So much for that idea. Thanks.

    Cdrive & Ron H...
    Thanks for trying to Splice the circuit and relaying what you encountered.
    I guess I'll just chalk this up to experience and move on.

    Thanks for letting me know that it's nearly hopeless to try to get this circuit to work. You guys saved me a lot of aggravation.
     
  14. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    I designed and simulated the attached circuit. I used a MOSFET which is specified down to a supply voltage as low as 4.5V. You could safely go as high as about 35V. If you decide to try this, you might want to use a different MOSFET.
    This would be much more efficient if a comparator were used to set the trip voltage across Rs. Rs could be much lower in value, giving less voltage drop and less power dissipation. The safety limiter (Q3) would have to be rethought.
     
  15. Darkstar

    Thread Starter Member

    Sep 3, 2010
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    Wow, thanks a bunch Ron. I didn't expect anyone to design a new circuit for me. I'll defintely study this circuit.
    Thanks again.
     
  16. CDRIVE

    Senior Member

    Jul 1, 2008
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    Ron, that's sweet!;)
     
  17. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Thanks. I think it will need some filtering to minimize sensitivity to current transients. A cap across the switch would probably do it.
     
  18. pilko

    Active Member

    Dec 8, 2008
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    20
    Hi Ron,
    I am trying to understand your schematic and formula.
    If the V in your formula is supply voltage,then V/Rs
    would be 111 Amps.
    If the V in your formula is the volt drop across Rs
    then Iout can never be less than V/Rs.
    Regards
    pilko
     
  19. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    I see why you are confused. The "V" in the equations means volts. Itrip=0.57/Rs, e.g., if Rs=0.27Ω, then Itrip=2.1 Amps.
    Q1 and Q2 form an SCR-like structure, only with the gate being on the PNP instead of the NPN, as it would be in an SCR. The 0.57V is the "gate" trigger voltage, being the Vbe of Q1 required to trigger the "SCR".
     
  20. pilko

    Active Member

    Dec 8, 2008
    213
    20
    Thanks Ron,
    I had interpreted your formula as :- 0.57 x V / Rs.

    Regards

    pilko
     
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