Circuit-break locator for the holidays


Posted on Feb 7, 2014

The circuit illustrated here comes to the rescue by sniffing out defective bulbs by capacitively sensing the electric fields they produce ( see the figure ). To understand how it works, consider a string of bulbs with an opencircuited filament, labeled in the schematic as String Under Test.  The intact filaments on either side of the burned-out bulb complete continuous circuits from the ac supply


Circuit-break locator for the holidays
Click here to download the full size of the above Circuit.

all the way around to the opposite ends of the broken filament. The ac voltage differential between wires leading to any intact bulb is zero, but the differential between the connections to the defective bulb will be the full 110 V. The presence of 110 V ac on a wire produces large electrostatic fields that are easily detected through plastic insulation. The high-impedance CMOS inputs of XOR  U1 can do just that. Resistors R1 and R2, in addition to delivering operating power to the circuit, effectively suspend U1 midway between the ac supply rails. Thus, equal-amplitude (160 V p-p) but opposite-phase voltage differentials will exist between U1 and both ac rails. R4 and C1 reference U1 pin 2 to ac neutral,  while R2 and the Test  electrode are used to capacitively probe the voltages present in the wires of the String Under Test. U1 then performs as a phase-sensitive detector of those ac voltages. Suppose the Test electrode is held close to point A. Because A is continuous with the hot  rail, the signals at U1`s inputs will have opposing phases. One input will be high (logic one) whenever the other is low (logic zero). This causes XOR U1 to hold pin 3 high, so D2 glows. If the probe is moved to points B or C, the same phase relationship will persist and D2 will still glow. But if the probe is brought near D, the sensed phase will reverse because the circuit break lies between D and the hot rail. Consequently, neutral...




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