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The following is the schematic of a security system designed to work on the break-wire principle and intended to be used on a perimeter fence. The input 2 of logic gate U-3.1 is kept low via a long thin (60 ft) bifilar wire. Upon breakage of the wire, R4 makes the aforementioned input 1 high, which turns Q2 on. Although I have dedicated filter cap C2 (100nF ceramic) and ferrite bead on the break-wire near the circuit, Do you think that would be enough? Please post your suggestion to improve this filtering and to make this circuit more immune to false alarm induced by RF, EMI etc. The whole circuit board will be placed in an aluminium box except for the break-wire. The power source is 9v battery. enter image description here Kindest regards

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    \$\begingroup\$ With such high impedances CMOS IC, 3.3M pull-up you have created an antenna. IMO this circuit is designed to pick up noise rather than to reject it. \$\endgroup\$ – Marko Buršič Mar 10 '17 at 14:44
  • \$\begingroup\$ The CMOS was used because the circuit needs to operate on 9v battery for a long time. \$\endgroup\$ – Baphomet Mar 10 '17 at 14:47
  • \$\begingroup\$ I agree 3.3 M ohm on that line is asking for trouble, very likely it will pick up plenty of 50 or 60 Hz from the mains making the gates flip in random states. I'd start with lowering that 3 M ohm to 10 kohm or even 1 kohm ig you can live with the current which will flow. \$\endgroup\$ – Bimpelrekkie Mar 10 '17 at 14:48
  • \$\begingroup\$ The problem is the battery life would decrease significantly. Is there any way to remove these noise before reaching the input? \$\endgroup\$ – Baphomet Mar 10 '17 at 14:50
  • \$\begingroup\$ OK, when battery fed you the supply is floating, that helps. You might get away with that 3.3 M ohm then. I'd filter the node going to all the gates though but not like you suggest. Let me draw a circuit... \$\endgroup\$ – Bimpelrekkie Mar 10 '17 at 14:52
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I would make the tripwire input like this:

schematic

simulate this circuit – Schematic created using CircuitLab

This way you keep the low quiescent current but improve the filtering. if the wire breaks.

Also: you can leave out the ferrite bead, it will not help to improve anything.

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  • \$\begingroup\$ Could you please explain theory behind adding the extra resistor? \$\endgroup\$ – Baphomet Mar 10 '17 at 14:57
  • \$\begingroup\$ R2 and C1 make a lowpass filter. Previously you did not have R2 so anything picked up by the wire goes straight to the gates. Now the resistor provides a high impedance and C1 will provide a low impedance for high frequency signals. It is not that different from what you had only a bit more robust because of the extra R2. \$\endgroup\$ – Bimpelrekkie Mar 10 '17 at 15:00
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    \$\begingroup\$ Nearby lightning strikes might induce an incredibly large transient. Perhaps a 9v avalanche diode from junction of R1 & R2 to ground to absorb such a transient? \$\endgroup\$ – glen_geek Mar 10 '17 at 15:24
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    \$\begingroup\$ does the value of R2 need to be same as R2 No, another "best guess" since you had a 3.3 M ohm I added another one. If R2 is 1 M ohm or 10 Mohm the circuit will also work. \$\endgroup\$ – Bimpelrekkie Mar 10 '17 at 15:31
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    \$\begingroup\$ No actually the box is not grounded. It just to provide more shielding for EMI. \$\endgroup\$ – Baphomet Mar 10 '17 at 16:16
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3.3 MOhms - can't be serious! A drop of water bypasses your wire and nothing is detected. Change the pullup resistor to 500 Ohm. If that's too much in current consumption, consider to have pulsed operation - say 1 ms in every half second.

No test? Again: Cant't be serious! You should have a relay that makes a test breakage somewhere, preferably in the middle of the loop and a checking plan, too, preferably automatic.

A lowpass RC filter at the input of your nand gate kills the RF, have a time constant 1...100ms.

Consider more specific detection. ON-OFF does not see the bypassing. At least make total bypassing difficult by keeping the outside ends of the loop unreachable at the same time.

ADDENDUM:

The test relay: A normally closed switch is inserted as a part of your loop. The switch is remotely controlled.The simplest form of it is a relay. If opening the switch does not cause alarm, the current has a bypass way somewhere (=a bypass circuit is inserted, broken insulations and the vire is in the water)

Before any further circuit analysis a threat analysis should be performed. The most important things after deciding that the intrusion trial is probable and a tripwire is useful:

  • does the intruder know beforehand that there's a tripwire (=by having some inside knowledge)
  • can the intruder survey freely the systems as long as he wants (=no dogs, no cameras, no motion sensors)
  • if no foreknowledge, can the intruder see the tripwire before it's too late
  • what's the workaround skill level of the intruder

I spotted word "BIFILAR" in your text. Your loop seems to have 2 parallel wires. Thus it can have 4 wire ends, all available where your electronics reside. See the illustration:

enter image description here

The bifilarity (=2 parallel identical wires) gives a lot of new possiblities that you should explore

  • In heavy disturbing fields (=powerline, radio transmitters, high current works around) your system can feed the test voltage between the 2 wires. Any disturbing signal that affects equally to both wires can be cancelled
  • If there appears a resistance difference between the wires, something is added to the circuit
  • if there appears a connection between the wires, something is added to the circuit
  • a voltage or intentional signal between the wires can be used to do something useful. The wires can be at the same time for power supply, signal transmission and your original tripwire loop
  • by measuring pulse reflections the place of the breakage can be deduced (not useful if the max distances are only a few meters, but very useful if the distance is beyond the field of the vision)

I have not written about how to protect against the lightning (not direct hit, but nearby). For that the methods can be borrowed from the landline telephony.

A way to reduce induced voltages - not much better than a twisted pair as a single wire loop and makes bypassing easy. Gathers common mode (=against the earth) voltages as easily as a single wire loop => NOT RECOMMENDED. To utilize properly the 4 end loop, you must have more complex circuit.

enter image description here

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  • \$\begingroup\$ Thanks for your comment, the solutions has been posted by Trevor and fakemoustache. Could you explain more on using relay? \$\endgroup\$ – Baphomet Mar 10 '17 at 15:54
  • \$\begingroup\$ All good points. Not sure the water drop is an issue though, unless it's right where the wires come together. I was assuming the wire was insulated out to some distance apart. \$\endgroup\$ – Trevor_G Mar 10 '17 at 16:02
  • \$\begingroup\$ BTW : There is also a method to use the wire single or double ended fed with either a pulse or continuous AC waveform / signal, and tracking the capacitance or inductance of the line. That type is very hard to defeat. \$\endgroup\$ – Trevor_G Mar 10 '17 at 16:05
  • \$\begingroup\$ Do you have any schematics for this type? I think it is also known as 'ground sensor". \$\endgroup\$ – Baphomet Mar 10 '17 at 16:14
  • \$\begingroup\$ Not off hand no. Would probably involve a signal generator and tuned receiver that integrates the average return signal level and trips on a variance on that level. \$\endgroup\$ – Trevor_G Mar 10 '17 at 16:21
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Use a large value resistor tree, and put the capacitor on the same side as the pull-up not the ground side. It will take a little longer to switch, but not long enough to care about.

I'd also add series resistors and a protection diode as shown to give you some protection from static and maybe even a Transorb go give you some protection from static discharge, though it won't protect you from a lightning strike directly on the fence.

Both the ground side AND the signal side need to go through the bead with a turn or four. Though you may not need it.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Thank you Trevor, the problem is circuit is operated by 9v battery and therefore need to have very low quiescent current. \$\endgroup\$ – Baphomet Mar 10 '17 at 15:08
  • \$\begingroup\$ Ok.. then combine my answer with @FakeMoustache's \$\endgroup\$ – Trevor_G Mar 10 '17 at 15:12
  • \$\begingroup\$ Thank you, may I ask where the other ends of R2 and R3 are connected to? \$\endgroup\$ – Baphomet Mar 10 '17 at 15:16
  • \$\begingroup\$ That's the ends of your trip wire, or connector for same. \$\endgroup\$ – Trevor_G Mar 10 '17 at 15:17
  • \$\begingroup\$ Yes you re right, may I ask what is the value of zener diode? considering the power supply is 9v. \$\endgroup\$ – Baphomet Mar 10 '17 at 15:18

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