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Current Transformer thanks

Discussion in 'Electronic Basics' started by Bordon, Sep 12, 2005.

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  1. Bordon

    Bordon Guest

    After some mis starts my Current transformer works fine. I went the AC
    adaptor route.

    I want to thank John P., Jason B, and Gordon W for their comments.

    I have a question about a circuit found at Bill Bowden's Site.

    http://ourworld.compuserve.com/homepages/Bill_Bowden/page8.htm#aclatch.gif

    The circuit "AC Line Current Detector" has a confusing description, which
    may be wrong.

    I feel like fool for suggesting this like an ant telling Einstein that he
    theory of relativity is wrong. But I think there is a mistake in the
    description of the circuit.

    Can you read it and tell me if I'm wrong?
    .....
    The magnetic pickup (U-bolt) produces about 4 millivolts peak for a AC line
    current of 250 mA, or AC load of around 30 watts. The signal from the pickup
    is raised about 200 times at the output of the op-amp pin 7 which is then
    peak detected by the capacitor and diode connected to pin 7. The second
    op-amp is used as a comparator which detects a voltage rise greater than the
    diode drop. The minimum signal needed to cause the comparator stage output
    to switch positive is around 800 mV peak which corresponds to about a 30
    watt load o
    .....

    Does Mr Bowden means to say pin 1 instead of pin 7 in the above statement.
    Since pin 7 does not connect to a capacitor.
    Or is my ignorance showing.

    I'm also confused does this circuit produces 12 Volts DC ??

    Is the theory of relay to turn on say a light or something when it detects
    voltage? I'm probably not going to make the circuit but just trying to
    understand it a bit.

    Thanks.
     
  2. Danni

    Danni Guest

    "Does Mr Bowden means to say pin 1 instead of pin 7 in the above
    statement. Since pin 7 does not connect to a capacitor.
    Or is my ignorance showing.
    I'm also confused does this circuit produces 12 Volts DC ??
    Is the theory of relay to turn on say a light or something when it
    detects voltage? I'm probably not going to make the circuit but just
    trying to understand it a bit."
    ____________________________________
    Re;
    I think you're right. It is OP amp A that does the "detecting" or
    amplifying of the current signal. The signal is then integrated and
    filtered at it's output; namely pin 1; by the 10uF cap paralleled by the
    100k bleeder resistor. OP amp B then acts as a "switch" or comparator,
    to activate the relay through the 2N3053 buffer transistor.
    The relay is activate when current of sufficient magnitude passes
    through a wire "monitored" by L1.
    I did note that the article said that the relay may chatter when
    monitored current is near the setpoint of OP amp B. This could be
    remedied by inserting some hysteresis into OP amp B (the comparator);
    something the circuit woefully lacks. I hope that helps...

    -Dan Akers
     
  3. You caught him. The chip us a dual opamp, one half used as an
    amplifier and the other as a comparator. Evidently he wrote the
    description with the second half as the amplifier, then drew the
    diagram with the first half as the amplifier. They are interchangeable.
    It doesn't. there is a separate, external, 12 volt supply connected
    to this circuit from an unspecified source. This circuit just
    switches that 12 volts off or on to a load.
    At a particular level of line current, the relay is energized. What
    you do with the relay contacts is your choice.
     
  4. Bordon

    Bordon Guest

    Yes thank you.
     
  5. Bordon

    Bordon Guest

    Thanks again. I get it now. Its just that the relay was not drawn with as
    much detail as I like . With one side showing the 12volts coming in and the
    other showing the relay. When your a novice you like to see it spelled out.

    Thank you.
     
  6. Guest

    But fleshing out obscure descriptions and simplified drawings, and
    figuring out other people's mistakes is part of how you get past being
    a novice. :)
     
  7. Bordon

    Bordon Guest

    Yes, I concur.

    Knowledge, Knowledge Knowledge, Its also a good way to prevent oneself from
    killing themselves.
    Regards.
     
  8. Richard

    Richard Guest

    I just realized I wrote a response to your question, then forgot to post it
    to the group. Senility is no longer creeping up on me, its is running at
    full speed.

    Anyway, mabe better late than never (though it says the same thing as
    everyone said)..

    By the way, I sent a note to Bill Bowden. He concurred that there is a typo
    and stated it would be fixed.

    Richard

    ----- Original Message -----
    It appears you are correct; pin 7 was probably just a typo.
    No, 12 VDC is the source voltage. When the transistor turns on, the current
    path from 12 VDC, through the relay coil, and to ground is completed.
    Instead of "detecting" the voltage, it is more that the relay is used to
    perform a function when there is current, or lack of current, through the
    coil. Could use a normally-open or normally-closed relay, depending on the
    application desired. The relay would be only the front end. Opening, or
    closing, the relay could be used to drive additional circuitry to sound an
    alarm, change a computer input, turn off the equipment, etc.

    I hope this helps.

    Best of luck in your pursuit of additional knowledge.

    Richard
     
  9. Bordon

    Bordon Guest

    Thanks Richard. All the information is helpful.
    Good of Mr. Bowden to make the adjustments. Good information to start off
    with prevents misunderstanding down the road.

    Thanks.
     
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