Current Transformer thanks

"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

Congratulations.

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.

Yes thank you.

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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.

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. :-)

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Yes, I concur.

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

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 ----- From: "Bordon"

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

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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.