AC DETECTOR

Description This circuit will detect AC line currents of about 250mA or more without making any electrical connections to the line. Current is detected by passing on of the AC lines through an inductive pickup (L1) made with a 1 inch diameter U-bolt wound with 800 turns of #35 magnet wire. The pickup can 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. This is most important is very difficult to achieve. The best method is to make a short extension cord with the three conductors separated from each other. If you make a 3-turn loop with say the active line, and pass a straight rod such as a metal bolt, containing 400 or more turns through the centre of the 3-turns, you will produce a very sensitive pick-up. The magnetic pickup produces about 4 millivolts for AC line current of 250mA, or AC load of around 30 watts. The signal from the pickup is increased 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 800mV which corresponds to about a 30 watt load on the AC line. The output of the 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 can be increased by adding more turns to the pickup.

Circuit Diagram