B
Bordon
- Jan 1, 1970
- 0
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.
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.