Help to improve this filtering circuit

[DDT]

The following shows the diagram for a sensing and filtering input of a break wire alarm. The logic input is held low by the break wire and goes high once the wire is broken. The triggering voltage for the logic high is around 6V when Vcc is 10v. R2 needs to be very high ( min 1M ) to minimise the current consumption, the R1 makes a low pass filter combined with C1. Once the wire is broken, the voltage on C1 needs to be as high as possible ( over VT+ 6v) to maximise reliability. Considering TVS has 1uA leakage, how would you choose R1 and R2? Feel free to suggest a more effective alternative circuit. Regards

IC is CD4093BE with Schmitt trigger.

[rstofer]

Your circuit allows the bad guy to simply short-circuit the wire to disable the alarm.  Usually, there is some kind of balanced circuit that detects both shorts and opens with the circuit terminated at the far end in a specific resistor value called an end-of-line resistor.

I did a brief search and didn't turn up anything useful.

[DDT]

That bifilar wire is very thin (0.1mm) and it's insulated like a single wire. The wire is 20cm above the ground in the grass, extremely hard to see. I am mostly concern about the TVS leakage causing an undesirable voltage drop so the Vcc to input of the logic gates wouldn't function reliably. Regards

[Kire Pûdsje]

Assuming you want to keep the sense wire short circuited, the leakage current of the TVS does not matters, since the voltage across this diode is essentially 0V.
From an RFI perspective, please place a small capacitor with very short leads, (or preferably SMT) across the TVS diode, to prevent RF detection of eg cell-phones. This should be oom 100pF ... 1nF. Do not expect too much from ferrites at GHz frequencies.
R1 will help a little for ESD, but does almost nothing for the time constant as this will be determined by R2*C.

[moffy]

The TVS leakage is a killer. Why not use a diode(1N4148) with the kathode connected to the +9v and the anode connected to the input? Basically in parallel with the pull up resistor. But then again the TVS is not necessary because of D1. Replace the TVS with a 1N4148 diode.

[DDT]

From an RFI perspective, please place a small capacitor with very short leads, (or preferably SMT) across the TVS diode

Ok thank for the suggestion. Wouldn't C1 serve the same function though?

Replace the TVS with a 1N4148 diode.

If I replace TVS with a normal diode, wouldn't the leakage still be an issue? TVS is the there to dump any voltage induced in the wire over 12v to ground. Could you please explain why you don't think it's necessary?

It has been suggested the best option is to reduce R2. The following shows the calculation made by MikeML from electro.tech.online.com for R2= 1M.  As you can see the voltage developed on the node CMOSIN is 6.9v when the wire is broken. The specification of CD4093BE specifies the trigger voltage 5.9V at VCC=10V. Would you say the R2=1M is good or I still should go lower. Regards

[moffy]

"If I replace TVS with a normal diode, wouldn't the leakage still be an issue? TVS is the there to dump any voltage induced in the wire over 12v to ground. Could you please explain why you don't think it's necessary?"

You have all theses protections:
1. Reverse breakdown voltage of D1.
2. Current limiting resistors R3, R1.
3. Surge absorbing capacitor C1

That's a reasonable amount of protection already. Replace the TVS with a 1N4148 and your leakage should be around 1na or less, except at extreme temps like 125C.

[DDT]

Is the position of the new diode correct? I've also added 100pF ceramic for high frequency suppression. Regards

[moffy]

No, not quite, literally replace the TVS with a diode in exactly the same position and orientation. It's job is to clamp negative voltages.