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This question already has an answer here:

I'm talking about devices for tracing hidden AC mains wires in walls like this one: https://www.ebay.com/itm/3-in-1-LCD-Wall-Detector-Stud-Center-Finder-AC-Live-Wire-Metal-Detector-Scanner-/322510160930

Is it correct that it uses capacitive coupling and can be approximated by the following circuit?

enter image description here

SPICE tools require a common ground between the mains side and the device side (resistance in this case) but it reality the device is isolated from the mains, so I'm not sure if this is still a valid approximation because of this.

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marked as duplicate by RoyC, Sparky256, Dave Tweed Jan 16 '18 at 5:51

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • \$\begingroup\$ I saw the other question and the answers do mention capacitive coupling but it mostly go into answering the specific question about why the tester fires on a USB charger cable. \$\endgroup\$ – axk Jan 13 '18 at 16:20
  • \$\begingroup\$ Look at How internal capacitor stud finders work? for an explanation of how the stud finding works, then consider what effect a live wire would have. \$\endgroup\$ – τεκ Jan 13 '18 at 17:15
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Your ground can be anywhere, it just a reference point. So you could draw your circuit like the first one. Remember air can be modeled as resistance and the metal in the wall and in your detector will form capacitance.

One word of warning is the spice simulator may not like having a 1GΩ resistor in between it and the circuit, so the second circuit might be kinder on the solver. Its easier for the solver matrix to have more row zeroed out and makes it easier to determine a solution.

Secondly it might be a better model to put the cap in parallel with the 1GΩ resistance that simulates air, off the top of my head I'm not sure which would more closely model this situation, I'd have to run some numbers.

Any two conductors with an electric field between them can be modeled as a capacitor.

Any thing with current running through a material (even air) can be modleled as a resistor.

Anyway hope that helps.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ In my understanding there would be no current flowing through the wall if its dry. Wouldn't it be purely capacitive coupling? (as in the current through the wall is so small that it is undetectable) \$\endgroup\$ – axk Jan 13 '18 at 20:49
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    \$\begingroup\$ Lets see, Drywall is made from gypsum and has a conductivity of max 1000Ω (although it could be more but I couldn't find any info on that) with PVC insulation it will be around 1MΩ. Humid air (60%+) at 120V can have leakage into the nA range. At 90% almost into the uA range, so no it's not negligible. Yes the capacitance will probably be a larger value. \$\endgroup\$ – Voltage Spike Jan 14 '18 at 3:29
  • \$\begingroup\$ (Source for resittivity of air: google.com/…) \$\endgroup\$ – Voltage Spike Jan 14 '18 at 3:29

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