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Circuit breaker, clearing a fault current

  1. Mar 18, 2013 #1
    hi could someone please explain the attached circuit to me in detail.

    From what i understand:

    the open circuit caused by the opened switch in series with the resistance causes a fault current in the system.
    This higher than normal current across the 200 ohm resistor causes a high voltage to flow through the mosfet and into the inverting input.

    the input is then compared to a reference( using a schmitt trigger config Threshold values set to 1 an 5V. )

    when input increases past the high threshold value set the output is low, when input is below high threshold output is high.

    The low output will turn the mosfet off and the high output will turn it on.

    When the mosfet turns on, a voltage is sent to the switch closing the switch.

    This switch should not close until the fault current is cleared and the switch doesnt re-open unless reset.

    based on what i explain is my design correct.

    Think of what i am doing as a way to clear a fault current using a schmitt trigger configuration+ a mosfet+ a mechnical switch

    please help.
     

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  3. Mar 19, 2013 #2

    NascentOxygen

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    An open switch doesn't cause a fault current. The closed switch causes higher than normal current through the MOSFET.
    It is not apparent how this last action happens. In any case, after clearing the fault you want the switch to OPEN and the MOSFET to turn ON and that puts you back to normal operation.
     
  4. Mar 19, 2013 #3
    Thank you, so when the switch is closed is what causes the more than normal current.
    When the mosfet turns on how does it make the switch to turn on?

    thank you though you explain things really well.
     
  5. Mar 19, 2013 #4

    NascentOxygen

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    When the switch is closed, it places a 5Ω resistor in parallel with the normal 200Ω, so current can be approximately x40 larger.
    We aren't told. Presumably, someone or something will independently turn the switch back on or off.

    The MOSFET itself will be automatically turned back ON immediately after being turned OFF, the speed of the Schmitt trigger causing this. The MOSFET switching will repeat over and over until the switch is opened to remove the fault current. The circuit doesn't seem to accomplish much.
     
  6. Mar 19, 2013 #5
    hey how do i get a 0.2V as my reference if i know that R1 is 10K

    i would like to get a 0.2V voltage set but i cnt figure out how to do it please help.

    based on the posted schematic.
    how do i get my mosfet to switch on and off?

    it doesn't seem to switch, because the voltage at the source of my mosfet is just a straight line.

    Would my voltage varry when the switch goes from an open position to a closed position? if so why isn't my mosfet reflecting this.
     
    Last edited: Mar 19, 2013
  7. Mar 19, 2013 #6

    NascentOxygen

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    The Schmitt Trigger comparator has two switching levels. Do you want to make the lower one 0.2V, or the higher one?
    With R5 being 0.1Ω, and the comparator requiring 5V to switch, you need 50A through R5 to trigger the MOSFET OFF.
    What value is your voltage at R5 when the switch is (a) OPEN, and (b) CLOSED?

    Are you constructing this using hardware, or are you using a computer simulation?
     
  8. Mar 22, 2013 #7
    hi, in the the schematic below i would like my upper trigger point to be at 3 and my lower trigger point at 1. How do i choose my resistors appropriately?
    i can't seem to get the hysteresis right, so my output keeps switching rapidly.

    i am going to construct this design on a veroboard after i run a few simulations and i am satisfied with it.

    My R5 (currently R6) has been changed to a 1W power resistor now. Wouldn't want to burn anything.

    i am currently supplying my comparator with +15V
     

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  9. Mar 22, 2013 #8

    NascentOxygen

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    If you construct this on breadboard, you may find the tracks or contacts are damaged by 5A.

    If you supply the comparator with a single-ended 15V supply, not ±15V, with [strike]R9=120k, R8=22k, and R10=68k, the upper switching level will be 5V.[/strike]

    EDIT:
    With R9=560k, R8=22k, and R10=100k, the upper switching level will be 3.1V
     
    Last edited: Mar 22, 2013
  10. Mar 22, 2013 #9
    Thanks, built it today. Burnt a few things, even though i was sure only 4A was supposed to run through my system.

    thank you let me try that now.

    actually those values don't work even if i only use a single ended 15V supply
    My trigger output switches too quickly from its low state to high state and then takes progressively longer to switch from high back to low state.

    what formula did you use?
     
  11. Mar 27, 2013 #10

    NascentOxygen

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    With R13 being 5Ω, if you were to have 4A through it that amounts to 16*5=80watts
    so R13 would need to be a big power resistor, or else something like 15 resistors, each 10W and in parallel.
    Maybe it is working, but just not how you'd imagined it would?
    I expect the MOSFET will switch on and off a few hundred times per second. When the comparator switches, it turns off the MOSFET causing current to fall towards zero. As the level of current falls below what will trigger the comparator, the comparator output returns high and turns on the MOSFET. Current returns, and triggers the comparator, etc..

    ▻ When the comparator output is +15V, the R8-R9-R10 resistor network forms a simple potential divider where two of the resistors return to +15V.

    ▻ When the comparator output is 0V, the R8-R9-R10 resistor network forms a simple potential divider where two of the resistors return to ground.

    The voltage trigger levels are 3V and 0.5V with R8-R9-R10 as 22k, 560k, and 100k, resp.
     
  12. Mar 28, 2013 #11
    my comparator has 2 output states though
    a high of +15V and a low of -15V. i am using dual supply
    i can't find any examples of a dual supplied schmitt trigger.

    i need to use one so that my on and off periods are equal, with a single supply i noticed that the off period is always quite abit longer than the off period.

    also slowing the period and pulse width improved my output response.
     
  13. Mar 28, 2013 #12

    NascentOxygen

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    I doubt that you do need to change the Schmitt trigger. Are you still using the basic circuit in your post #7? Because, in that circuit there is no designed "on period" or "off period". There is no monostable multivibrator. There is no periodic signal whose duration you have considered or calculated.

    Maybe you are referring to the switching rate you see the MOSFET is being subjected to? But that is not something you have designed for; it's due to the slew rate of the op-amp and the battery/load characteristics, and is something I expect you were not anticipating. If you want to control the rate of automatic switching, you will have to incorporate a monostable multivibrator, as was used in one of your original schemes. (I'm curious, how many times per second does your circuit, as it stands, switch the load?) What delay would you like before the load is switched by the MOSFET?
    What do you mean by "output response"?
     
    Last edited: Mar 28, 2013
  14. Apr 3, 2013 #13

    hey sorry, to be honest i was confused for a while and i was doing the wrong thing

    here is what my complete circuit design looks like.

    The mosfet starts on at t=0
    At t=1ms the switch closes causing fault current which gets cleared.
    At t=3ms, the breaker is reset.

    How do i design a drive circuit supplying ±20V to my mosfet in order to turn it on or off.

    The input coming into the push pull circuit attached below is ±15,V i need to boost the 15V input to give a 20V output.
     

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    Last edited: Apr 3, 2013
  15. Apr 3, 2013 #14

    NascentOxygen

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    Who or what performs the 3ms timing for the breaker reset? (No circuit is shown.) That is a very short time within which an overload can be expected to be removed.

    Isn't the op-amp's output driving your MOSFET switch? I'd have thought 15V and 0V should suffice.

    Are you sure you need this?
    Anyway, if you correctly copy a working amplifier circuit from somewhere, it should work.
     
  16. Apr 3, 2013 #15
    I need something to boost the ± 15V coming into the Mosfet
    because when i am building it the Mosfet i am using requires ±20V
    thats why.

    the reset button is the switch timed at t=3ms connected to 0V.
     

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  17. Apr 3, 2013 #16

    NascentOxygen

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    I doubt it. I think you should begin building parts of this circuit, starting with just the MOSFET switch and a drain load of the LED and associated resistor. Don't bother about the Schmitt trigger, leave that out at this stage, and instead manually switch the gate of the FET to either 0V or 15V and see whether that switches it. (Keep a voltmeter constantly connected between drain and source to monitor how well the FET is switching.) If this works well, then close U2 and keep your finger on the FET to make sure it doesn't get hot. It shouldn't get hot whether it's ON or OFF.

    Don't leave the gate connected nowhere; it should go to either 0V or 15V. If you leave it floating it may half turn on and cause the FET to get hot and burn out. The safest approach would be to put a resistor of a few 100kΩ from the gate to ground, by default that connects the gate to 0V. Leave the resistor there, and whenever you need 15V just connect 15V to the gate.

    Report back what you find.

    BTW, what is your reason for adding R6 in series with R13?
     
  18. Apr 4, 2013 #17
    put R13 in series with R6 because i wanted the R13+Diode connection to be my fault indication
    From what i now, know R13 and the diode should be in parallel with R6 is this correct.

    I will report back, however your suggestion will only work with a Mosfet that requires less than 20V to turn on or -20V to turn off. the IRF9530N requires 20V.
     
  19. Apr 4, 2013 #18

    NascentOxygen

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    That's an impressive power MOSFET you've got hold of. Go cautiously, don't want to damage it. For the IRF9530N, its drain should be negative, so reverse your 20V supply.

    Vgate of ±20V is the maximum rating. From Fig. 1 (see datasheet), a gate voltage of –15V should be ample to turn it fully on. It will be fully off with about –4V, but try it with 0V since that level may be more convenient. (Looks like the Schmitt trigger will need to be modified.)

    data sheet: http://alltransistors.com/pdfview.php?doc=irf9530n.pdf&dire=_international_rectifier
     
  20. Apr 4, 2013 #19
    so it would seem. what is weird is that i cannot find any good examples of dual supplied schmitt triggers. To achieve the 0V to turn the Mosfet off, i would have to change my current schmitt trigger to a single supply.

    One thing i noticed with my design is that after my fault current has been cleared it, the current doesnt return to its original value

    on another note: what do you know about zero current/voltage crossing??
    I am trying to design my circuit breaker to have this feature.

    To me zero crossing is when the voltage or current crosses below zero a switching device will switch. Zero switching wont happen before that. is this correct??
     
    Last edited: Apr 4, 2013
  21. Apr 5, 2013 #20

    NascentOxygen

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    With your load powered by DC there will be no zero crossings.

    Only when the load is purely resistive does the zero crossing of the AC voltage co-incide with the zero crossing of the current. With a partly inductive load if you can switch at a zero crossing of the load voltage, this is not a zero crossing of the current. I think you should defer that enhancement until you have perfected the basic switch.

    While testing the current design, keep the load resistive. If you add inductance you may damage the device. For inductive loads steps need to be taken to deal with voltage spikes; I think you had this in the very first (of your many) schematics.
     
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