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AFE3010: Solenoid Operation with GFCI

Velani Mayank
Prodigy 20 points
Part Number: AFE3010

Tool/software:

  • Solenoid Operation with GFCI:

    • Can you clarify whether a solenoid connected to a GFCI outlet remains continuously energized when the GFCI is plugged in? Does the GFCI affect the solenoid’s operation in any way, or is it only responsible for disconnecting power in the event of a ground fault?

  • SCR Triggering for Solenoid:

    • In a circuit diagram I have, the solenoid is shown as normally open and is only connected to an SCR. How does the SCR maintain current flow to keep the solenoid energized? Specifically, once the SCR is triggered, what mechanism ensures that it continues to conduct current through the solenoid to keep it in the energized state?
  • 0
    TI__Mastermind 32395 points

    Hello Appreciated Engineer,

    I am looking over this question and will respond promptly.

    Best,

    Peter

  • 0
    TI__Mastermind 32395 points

    Hello,

    1. During normal operation the solenoid would not be energized because the SCR would remain off. Solenoid is simply there to provide a mechanism to open mechanical load switch.

    2. The use of the SCR and solenoid is meant for a many electromechanical GFCI receptacles. The solenoid is not a relay and the SCR is not transistor. SCRs are best suited for quickly conducting temporary, large current transients, which would be necessary to actuate a solenoid, and thus mechanically latch open the load switch. 

    If using a relay to open load switch, you would probably not need the solenoid and then add extra circuitry to create an electrical latch signal enabled from SCR pin of AFE3010. This latched signal would then maintain the constant energization needed for a relay’s input coil.

    The only problem is finding a way to supply power to the relay as they can require significant current. Using the 20V VDD pin is tempting, but it is not recommended because this rail is not usually robust enough to supply a large DC current and tolerate worst case conditions in UL 943 where one of the current limiting resistors is opened up. While you add capacitance to VDD, you also need VDD to ramp quickly.

    Also at the same time you would need this relay  power supply to ramp up quickly in order to satisfy fault trip times when device is powered up.

    I understand there is plenty to consider. Please post back for any questions or thoughts.

    Sincerely,

    Peter

  • 0 in reply to Peter Iliya
    Prodigy 20 points

    thank you for your response much appreciated

    1. Could you please clarify if by "normal condition," you are referring to a scenario where no faults occur and the supply to the load is continuous?

    2. Does this mean that the solenoid is in a normally closed position?

    3. Could you provide a flow chart illustrating the working process?

    4. When using a latching relay in a system, what precautions or arrangements need to be taken?

  • 0 in reply to Velani Mayank
    Prodigy 20 points

    Hello Peter,

    we're awaiting your response.

  • 0 in reply to Velani Mayank
    TI__Mastermind 32395 points

    Hello Velani,

    My apologies for the delay.

    1. Yes. Normal condition mean the GFCI system is powered up and device is passing self tests. Load switch may be open if closed, but if closed GFCI could be supplying current to the load normally (continuously). No ground faults or self test failures are present.

    2. During normal conditions no current flows through solenoid and load switch is closed/open. The solenoid is not necessarily a relay and thus does not have a normally open or closed. The position of solenoid must also be paired with a mechanical load switch to complete any opening or closing of load switch.

    3. Device in normal operation and load switch is closed -> ground fault occurs -> AFE3010 determines there is a ground fault -> AFE3010 drives rated current out of SCR pin for at least one half line cycle.

    After this point what happens if dependent upon mechanical/electrical architecture you have chosen, but if we assume the typical schematic shown, then it goes:

    -> SCR IC turns on and pulls high current through solenoid -> solenoid creates magnetic field -> solenoid magnetic core is actuated -> solenoid core hits a mechanical component (e.g. spring) which releases a mechanical latch (load switch) -> load is disconnected from line and fault current stops.

    4. Well what i described above essentially accomplishes what a latching relay would do: one single pulse from SCR latches open load switch (relay).

    The main caution you need to take with relays are where you get the current to drive coils inputs. You don’t want to pull significant  DC current directly from AFE3010 VDD pins unless you have a robust 20-V source driving them. If they are in series with load they need to be rated up to this, usually 20A. 
     
    Lastly, you need to consider how the load switch will be closed. Safest thing to do is to perform a self test via the PTT pin before closing the load switch. Additionally if the self test fails, how prevent load switch from closing.

    See Figures 11 and 12 of datasheet on use of PTT pin. 

    Please post back with any questions you have.

    Sincerely,

    Peter