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Thread: Leviton GFCI nuisance tripping and circuit analysis

  1. #161
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    Quote Originally Posted by gar View Post
    190119-0850 EST

    tersh:

    The PNP is only rated 65 V. Can't put PNP, 4.7k, and LED in series across an AC 240 V line. Things will blow up. Not one of the components could tolerate this. Both the forward and reverse directions are a disaster.

    .
    Are all PNP rated 65V? Someone at stack exchange identify the tiny transistor as PNP. Maybe another model of PNP. Note in the actual pcb picture, you can see the emitter of it (or whatever it is) connected to the AC black line.

    Also in the Rline of 43K in series to the diode you asked me to remove. In a brand new PCB. I only measure 12volts in the Rline leading to the violet dot.



    The Reset is connected to the top of the receptacle. It's on the right. The two red wire is connected to the button. In the left Test button, one is connected to load line, while there is a resistor going to outside line (as is how Test button) connected.

    I can easily connect the sense coil of the 4th gfci pcb and it will be working but can't be put back in service. And I don't have the skill to look at it furthermore and learn the real source of the transient or current imbalance and how it affects the sense coil and pin 1,2,3. So maybe I can just send it to you along with the water shaded pole motor? I will never use it anymore. I avoid anything above 3mG. And the motor is about 200mG from my brain.

  2. #162
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    https://www.ebay.com/itm/50pcs-SMD-T...-/291878530763

    I just found the exact transistors at google. It's 400V capacity and NPN, not PNP. Will redraw and relabel the diagram. I'm doing all this just to have confidence in the product that will be with me for many decades to come.





  3. #163
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    190119-0916 EST

    tersh:

    No matter what the switching device is NPN, PNP, SCR, FET, or mechanical switch, the LED in series with 4.7 K can not tolerate a source voltage of 340 V peak in either polarity, nor the resistor.

    47 V / 4.7 k is 10 mA which is very bright for todays LEDs used as a pilot light. More likely voltage is down in the 10 V range for a 4.7 k series resistor.

    Use an ohmmeter to verify where some of those leads go.

    .

  4. #164
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    Quote Originally Posted by gar View Post
    190119-0916 EST

    tersh:

    No matter what the switching device is NPN, PNP, SCR, FET, or mechanical switch, the LED in series with 4.7 K can not tolerate a source voltage of 340 V peak in either polarity, nor the resistor.

    47 V / 4.7 k is 10 mA which is very bright for todays LEDs used as a pilot light. More likely voltage is down in the 10 V range for a 4.7 k series resistor.

    Use an ohmmeter to verify where some of those leads go.

    .

    Actually I was thinking about that too. I could light up the red led by simply putting my multimeter terminals to them. But in the actual circuit. The negative or pin 1 of the red led is really connected to the AC terminal at bottom. While pin 2 is connected just as I indicated in the circuit.




    This is back side again:



    This is the datasheet of the 400v NPN

    http://www.secosgmbh.com/datasheet/p...23/MMBTA44.pdf

    This is the NPN symbol with the arrow reversed: I'll quadle check everything tomorrow. Many thanks.


  5. #165
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    190119-1059 EST

    tersh:

    I like the NPN as a more likely device. The 10k base to emitter is OK. Emitter going to AC power black seems OK, that is the common for the IC chip.

    Something wrong with LED location, orientation, and it can not go to the AC power red with with just 4.7k current limiting.

    In proper orientation and with 4.7k and 10 V would be OK.

    .

  6. #166
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    100119-1124 EST

    Something wrong. My last post does not show up.

    By submitting this post my previous one now shows.

    Addition to previous post --- there must be some current limiting going to the transistor base. The rede line to the base requires some adequate resistance.

    .
    Last edited by gar; Today at 12:32 PM.

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