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I made a hidden wire detector from this tutorial. It works, but I'm not quite understand why. Here is the circuit: enter image description here I used 2SC1815 transistors and the value of R was 1K0. My antenna was made of 300 mm length copper wire with diameter 0.3 mm. I've measured its inductance - it is about 2 uH.

First of all, I've measured the base-emitter junction voltage of the first transistor - its amplitude was about 80 mV. I believed that BJT transistors require about 0.7 V to open, but the circuit was able to detected the wire. How is that possible? The base-emitter junction voltage was 10 times lower, so I thought the transistor should stay closed.

The second question might be silly, but I was not able to find the answer. I realize that electromagnetic field develops the voltage across the antenna. But the antenna is connected to a circuit with only one side. So, which voltage do we use as input? My suggestion is that voltage may only occur when the resistance is high and the field does not need to develop a high current to maintain such voltage. So, maybe the equivalent circuit actually includes the Ra resistance, which is as high as the resistance of open circuit? See the circuit below:

enter image description here

If I'm right, the input resistance of the circuit must be very high too. So maybe this is the key for my first question and the transistor does not open at all? But in such case how on Earth this circuit can work?

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  • \$\begingroup\$ Well, the link isn't in English so if you want an explanation, please translate. \$\endgroup\$
    – Andy aka
    Commented May 26, 2020 at 8:06
  • \$\begingroup\$ @Andy aka the link only explains how to build the device, and it is clear for me. The question is about how the above circuit work \$\endgroup\$
    – msmirnov91
    Commented May 26, 2020 at 10:25

2 Answers 2

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A better way to look at this is that your mains voltage is hundreds of volts wrt ground, and only a small capacitive coupling will drive enough current through the base to illuminate the LED. There is plenty of voltage compared to the few hundred mV required to forward bias the base sufficiently.

schematic

simulate this circuit – Schematic created using CircuitLab

enter image description here

Here you can see that only 1pF of coupling is sufficient to allow strong current pulses through the LED. C1 represents the near-field capacitive coupling to the mains wire. C2 represents the user's body coupling to earth.

Only a few tens of nA are required at the base, thanks to the Darlington triple transistor's high current gain.

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  • \$\begingroup\$ thank you, now I guess I understand what's happening! But I still don't get why the measured value of base-emitter voltage was so low? \$\endgroup\$
    – msmirnov91
    Commented May 26, 2020 at 10:56
  • \$\begingroup\$ It's floating, and it's a effectively a diode so the voltage reading will depend on a lot of factors like the electric fields nearby and how it is held or connected. \$\endgroup\$ Commented May 26, 2020 at 10:58
  • \$\begingroup\$ just to make sure I got everything right: mains voltage is much more than the base-emitter junction (which is essentially a diode) requires to be biased, so current easily flow through the base. But for the reasons like, for example, the capacitance of oscilloscope probe, we are not able to measure it? \$\endgroup\$
    – msmirnov91
    Commented May 26, 2020 at 11:17
  • \$\begingroup\$ Yes. If you had something that was more like an electrometer you could measure the electric field in the room. \$\endgroup\$ Commented May 26, 2020 at 11:18
  • \$\begingroup\$ thank you once again! \$\endgroup\$
    – msmirnov91
    Commented May 26, 2020 at 11:21
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Power lines, electric fields, capacitive coupling and fluorescent tubes (for example): -

enter image description here

Picture from here.

This is what an electric field can do on a larger scale - it passes current to anything capacitively coupled to it. Enough current to light many fluorescent tubes in some cases.

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