# Current Transformer thanks

B

#### Bordon

Jan 1, 1970
0
After some mis starts my Current transformer works fine. I went the AC

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
.....

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.

D

#### Danni

Jan 1, 1970
0
"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

J

#### John Popelish

Jan 1, 1970
0
Bordon said:
After some mis starts my Current transformer works fine. I went the AC

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

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
....

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.

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

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.
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.

At a particular level of line current, the relay is energized. What
you do with the relay contacts is your choice.

B

#### Bordon

Jan 1, 1970
0
Danni said:
"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

Yes thank you.

B

#### Bordon

Jan 1, 1970
0
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.

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.

J

#### [email protected]

Jan 1, 1970
0
Bordon said:
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.

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.

B

#### Bordon

Jan 1, 1970
0
out.

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.

Yes, I concur.

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

R

#### Richard

Jan 1, 1970
0
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 said:
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.

It appears you are correct; pin 7 was probably just a typo.
I'm also confused does this circuit produces 12 Volts DC ??

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.
Is the theory of relay to turn on say a light or something when it detects
voltage?

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.

Richard

B

#### Bordon

Jan 1, 1970
0
Richard said:
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.

Richard

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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