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I recently built a unisolder soldering station and am struggling to understand how the detection of the soldering iron within its holder is implemented. The unisolder board itself provides a 3-pin connector with +3.3V, GND (floating / not referenced to PE/chassis ground), and a signal pin that's weakly pulled high to 3.3V. The iron is detected as being in its holder when the signal pin is externally pulled to GND.

The following circuit is what's connected to that 3-pin connector via U2:

sensor schematic

J1 is connected to a conductive part of the holder and makes contact with with a part of the soldering iron handle that is directly connected to protective earth. The circuit works in practice, but I don't yet understand why.

Here's what I think I understand so far:

  • C3 is a decoupling capacitor for Vcc
  • C2 is there to filter high-frequency noise on the signal (P1)
  • BSS138 is used to pull P1 low when the iron is present
  • R2 limits the gate current and also forms a low-pass filter together with C4
  • U1 is a quad NOR gate, and only the J output and A+B inputs are actively used
  • the one used NOR gate is configured to invert the logical signal on the anode of D1
  • R1 weakly pulls J1 to 3.3V and forms another filter together with C1

What I'm still really confused by is the role of D1. How does it provide the right input on the inverter to detect whether or not J1 is connected to protective earth? Why is it OK to effectively leave the inverter input floating when the diode is reverse-biased? Why doesn't connecting 3.3V to PE through R1 cause a current to flow that'll trip a ground fault circuit interrupter? Are there perhaps other things about the circuit I've misunderstood?

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  • \$\begingroup\$ Note in the schematic that U1 pins 12 and 13 are shown as not connected. These must also be tied to Vcc (all digital inputs must be kept at a valid digital level.) \$\endgroup\$
    – rdtsc
    Commented Jun 17, 2021 at 18:17
  • \$\begingroup\$ Is it possible that the CD4001 comprises an oscillator and the output changes due to capacitive coupling of J1 to GND when it's in or near the holder? Similar to how a proximity detector operates. \$\endgroup\$
    – PStechPaul
    Commented Dec 1, 2022 at 6:09

1 Answer 1

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Given the schematics, the explanation is that when J1 is connected to GND, probably through the holder of the iron, D1 conduct and the A-B pin is pulled low.

When "open", the line is pulled high by R1, given D1 has some leakage current (5uA), the A-B pin will slowly go high after C1 is charged.

However, CD4001 specify an input leakage of 10uA but it is not clear if it's a leakage to VCC or GND, given that, there is no way to ensure it really works unless testing it and might be really dependent to parts (specific die and manufacturer).

Are you sure this is the correct schematic?

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  • \$\begingroup\$ Based on visually inspecting the circuit and checking continuity between different pins, the schematic does seem to be correct as far as I can tell. It's been provided to me by the person manufacturing and selling them on tindie. \$\endgroup\$ Commented Jun 16, 2021 at 7:44
  • \$\begingroup\$ Is J1 really being connected to GND when the soldering iron is inserted into the holder? I was under the impression that GND in the schematic is floating with respect to protective earth, which is what the part of the handle that makes contact with the stand is connected to. \$\endgroup\$ Commented Jun 16, 2021 at 7:48
  • \$\begingroup\$ That really depends how it's wired, it can be floating if the holder is connected to the GND of the supply. \$\endgroup\$
    – Damien
    Commented Jun 16, 2021 at 9:47
  • \$\begingroup\$ the holder is only connected to protective earth, and there's no connection between the controller's GND and protective earth. \$\endgroup\$ Commented Jun 16, 2021 at 16:32

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