Solid state DC circuit breaker, used to protect a load

  1. My choice of solid state Dc circuit breaker is the Mosfet, easily available( i got loads!!!)
    i want the mosfet to interrupt the circuit to protect the load from overvoltages
    This is what ive done so far, please help its not working the way i want it to.

    the resistor connected to the inductor is the load i want to protect
     

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    Last edited: Feb 27, 2013
  2. jcsd
  3. NascentOxygen

    Staff: Mentor

    You have breadboarded this circuit, or are you saying the simulation doesn't work?

    You have managed to get the DC supply for the op-amp mixed in with the 24V for the load. Separate these. You intend operating the 324 on a single supply?

    Is that really a 10 Ohm resistor in series with the FET's gate?!

    I suggest that you remove the inductor and its associated zeners, etc, and the 24V supply, and concentrate on getting just the op-amp and the MOSFET working. Perhaps use a LED and a 2.2k resistor to represent the load, powering it off the 10V, so you have a visual indication of the MOSFET switching correctly. Only when that is working and the voltages all tally with your designed values, should you bring in the load and its associated power supply.
     
  4. I obviously suck at this!!! thank you.
    i was trying to design a mosfet switch to protect a load from an overvoltage by interrupting the circuit
    I will do as you have recommend
    but if u were to design such a switch how would you go about doin it
     
  5. NascentOxygen

    Staff: Mentor

    I wouldn't. I would do a google search for a similar arrangement, and base my design on that. Better brains than mine have succeeded at this task.:smile:

    I noticed that your circuit does nothing to sense the overvoltage, but I presumed you were concentrating on getting the basic switch working first.
     
  6. yes i was
    however i haven't found anything on google that might help
    what do you think about this
    http://www.linear.com/product/LT4356-3
    this is the best design i have found

    however my pspice package doesnt have LT43256 or any surge protector so i guess ill just start building
     
  7. hi please could you give me a quick sketch of the type of circuit you just described in this post
    10v to to power the mosfet and the opamp and the LED being there to give me an indication that my mosfet is switching.

    thank you.

    this is what ive got so far, are my connections correct.
    i want the gain of my differential amplifier to be 2.
    so i can decide R2 ,R1, R3 and R4 pretty easily, my question is though, is it necessary to have those resistors in place?
     

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    Last edited: Mar 6, 2013
  8. NascentOxygen

    Staff: Mentor

    I think you will need those 4 resistors around the op-amp. Don't forget that the op-amp needs to be supplied with power, too.

    You may need to experiment when you build it, to get the switching working how you want it to.
     
  9. I want V1 to be where i introduce my voltage pulse
    so does that make V2 where i introduce for 12V power supply?
     
  10. NascentOxygen

    Staff: Mentor

    The power supply pins are often not shown on basic schematics. V2 is not for the power supply (although you may need to connect V2 to the power supply as well, to allow the op-amp to operate off a single supply.)

    Your original attachment showed power supply connections. Download that op-amp's data sheet and check that you have it correct.
     
  11. Following your advice i made my load the LED and a 2.2kohm resistor and connected it, the LED turned on and off indicating i connected them properly to my 10V supply.

    Having moved on from that could you please tell me if the design attached is correct
    i.e does it function as a circuit breaker protecting the LED and the 2.2kohm resistor

    I am trying to design it in such a way that when the LED remains off there is no fault condition, there is no difference between the overvoltage and the reference voltage (feedback from the mosfet source resistance back to the negative input of the comparator) no signal is sent to the 555 timer, which leaves the mosfet off.

    When there is a voltage difference at the comparator a signal is sent to the 555 timer, a square wave is generated and the 555 timer determines when the mosfet turns on or off.
    the LED, would flicker indicating that there is a fault condition being resolved

    When the LED remains ON till it burns out this means, that the fault condition could not be resolved. (Unlikely scenario)

    please send me your email NascentOxygen if interested the file size is too big to attach
    ive tried everything.

    edit: ps: why is the LED connected a 2.2kohm resistor? how would this affect my mosfet is the resistor was 2Kohm?
     
    Last edited: Mar 6, 2013
  12. in this circuit how do i prevent the voltage at the source of my Mosfet from being distorted

    i am trying to feed it back to the inverting input of my opamp but i can't do that for now because it is awfully distorted
     

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  13. NascentOxygen

    Staff: Mentor

    I would have said the source is grounded. Maybe you mean the MOSFET's "top" terminal? But that's not what you want to feed back either. When the FET is ON its drain voltage is low and it likely stays low even when the 15V shows some voltage spikes. The drain voltage will never trigger overvoltage protection.

    Are you really powering pin 8 of the 555 from your OP-AMP output? Is that intentional?

    What is the purpose of the pair of 2N2222's?

    Don't you want the MOSFET to be ON and conducting during normal operation? Then if the voltage supply (shown as 15V here) goes too high the comparator switches the MOSFET OFF and no current flows from the 15V through the LED to ground. That's how I picture it to operate.
     
  14. sorry should have been more careful before i sent a circuit design
    the pair of 2N2222's is actually supposed to be 1 NPN and 1 PNP transistor to give a push-pull output, this serves to source some current to the mosfet to kick start it.

    the LM324 is supposed to be a LM393 but i dont have that comparator in my package.(should have stated that!! sorry)

    yes the output of the comparator is to be connected to pin 8 and 4, if its a comparator the output signal should be strong enough to connect it directly

    this is how i see it working.

    The surge at the inverting input( is going to be replaced with the feedback from the source of the mosfet) when Vin(overvoltage) > Vref the output at the comparator is low
    vice-versa the output is high
    when the output is high a signal is sent to the 555 timer which sends it to the mosfet

    in short, this means that when the input voltage is above the Vref no signal is sent from the comparator( this config limits the current in the load=2.2K+LED )

    seeing that an overvoltage would cause a fault current in the load, i think the design might be able to respond to an overvoltage and protect the load from a fault current
     
  15. NascentOxygen

    Staff: Mentor

    I can't picture the MOSFET requiring a "kick start", whatever that is. Its gate draws no current, so why do you think you need a push-pull stage to drive its gate? Where did you get the idea of needing a push-pull stage here?
    If the 555 sits there inoperational during normal use of the circuit, what will be the signal level at its pin 3 for driving the next stage?
    I can only repeat what I said earlier. The voltage at the source of the MOSFET switch is not going to show any significant change when there is a spike on the 15V supply. Shouldn't you be sampling the 15V supply instead?
     
  16. you are right, the 15V should be sampled instead. that is my reference voltage. i was confused.
    the push pull should be removed as well, really dont remember why i put that in there. more confusion am guessing. This is what my current circuit looks like, its almost all good except for one thing, the feedback from the mosfet source (input voltage) is much lower than the reference voltage. How do i boost the input voltage??

    also i have introduced an Ac current in the load, causing the fault current condition.
    the 555 is not inoperational during normal use, my mistake if i implied such. However during normal use, the 555 would be at a low, i.e sending a low signal to the mosfet thus turning it off.
     

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  17. NascentOxygen

    Staff: Mentor

    Sampling the voltage across that 100 Ohm resistor is actually sampling the load current. I doubt that is what you would ever want to do. Load current is [in general] no indicator of load voltage.

    You have again got the 15V supply for the control circuit intertwined with the load supply voltage. I assume you want these to be kept separate: the source powering the load should connect only to the load, nowhere else.
    Have you got this wrong, or am I misunderstanding what you are trying to achieve? Surely the FET is ON during normal operation, and it turns OFF to interrupt current to the load upon detection of abnormally high voltage?

    You want to interrupt the load current for a short preset period, and then reconnect the load, is that the principle of operation? Are you sure that is going to be acceptable? What is the nature of the load are you planning to power?
     
  18. I am confused though, if the mosfet turns off to interrupt the current to the load, that leaves an open circuit( high resistance) wont the current just be exerted through out the load.
    Shouldn't the mosfet turn on to interrupt the current, i.e allowing it to flow through it, and into ground to be drained?

    My load is simply the LED connected to the resistor ( i want the LED)
    to be a signaling light when there is a fault

    this is what i thought about when i designed the circuit: The input voltage is the feedback voltage from the Mosfet source, when the feedback voltage is above the reference, the output at the comparator would be at a low. This low output from the comparator leads to a low output in the 555 timer, which leads to a low voltage turning the Mosfet off. (input voltage into the mosfet must be > than threshold, the low voltage would be below this value, which prompts the mosfet to turn off.)
    The Mosfet would only be turned off when the current in the load is no longer at a fault. The mosfet turning off causes an open circuit, current in the load will now only be the current being supplied from the power source. The LED should be in its off state.
     
    Last edited: Mar 14, 2013
  19. NascentOxygen

    Staff: Mentor

    At this stage, don't worry about including a LED to indicate a fault condition. That's merely a decoration which can be added in later. You are confusing me and yourself by discussing two LEDs. Well, I think you are.

    In your original sketch you showed an inductor and resistor representing the load. I suggested that in the preliminary stages you should substitute a LED and its current-limiting resistor for the L-R load as a visual indicator of load current status. The LED will glow when load current is flowing because in this simplified arrangement the LED current IS the load current; the LED is the load.

    The MOSFET is in series with the load. When the MOSFET conducts (i.e., is ON), current flows through the load and through the MOSFET to ground, so the load gets powered. If an over-voltage is detected, I believe you want the MOSFET to become non-conducting (i.e., turn OFF), so no current flows through the load or the MOSFET.

    Why are you reluctant to indicate details of the load you are wanting to protect?
     
  20. lol, sorry i dont know the load yet. Havent decided. not reluctant at all, you've been so helpful. Any suggestions?

    There is only 1 LED, represented by the diode in series with the 2.2kohm resistor

    My issue is my if the mosfet conducts, i.e is on how does that interrupt the current flowing through the load? wont the mosfet being on, simply let the current flow through the mosfet?? there is no break in the circuit if the mosfet turns on to protect the load( at least i dont think there is)
     
  21. NascentOxygen

    Staff: Mentor

    When you want to interrupt the current, you turn the MOSFET OFF. MOSFET current stops, so does load current because the load is in series with the MOSFET.

    You will probably end up connecting multiple MOSFETs in parallel to produce a switch with sufficiently high current rating. I think there should be no issues with doing so.
     
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