Description
This circuit will detect AC line currents of about 250 mA or more
without making any electrical connections to the line. Current is
detected by passing one of the AC lines through an inductive pickup (L1)
made with a 1 inch diameter U-bolt wound with 800 turns of #30 - #35
magnet wire. The pickup could be made from other iron type rings or
transformer cores that allows enough space to pass one of the AC lines
through the center. Only one of the current carrying lines, either the
line or the neutral should be put through the center of the pickup to
avoid the fields cancelling. I tested the circuit using a 2 wire
extension cord which I had separated the twin wires a small distance
with an exacto knife to allow the U-bolt to encircle only one wire.
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 on the AC line. The
output 1458 op-amp will only swing within a couple volts of ground so a
voltage divider (1K/470) is used to reduce the no-signal voltage to
about 0.7 volts. An additional diode is added in series with the
transistor base to ensure it turns off when the op-amp voltage is 2
volts. You may get a little bit of relay chatter if the AC load is close
to the switching point so a larger load of 50 watts or more is
recommended. The sensitivity could be increased by adding more turns to
the pickup.
Circuit Schematic Diagram
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