How can I pass voltage through a wire only when it goes over 7 volts?

I'm trying to hook up a Z-Wave contact sensor to my home alarms wired contact sensor. The alarm panel seems to provide a constant voltage of 6.8 when the door contact is closed and 13.5 when it opens. 

The z-wave module I am connecting however looks for no voltage in order for it to be closed and any voltage for it to be open.

So how can I only pass the voltage to the z-wave sensor when it goes over 6.8 volts? I have a ground and power wire running to the respective terminals on my alarm panel.

I'm thinking I may be able to wire in a resistor but am not sure if that is what I need or what size.

Thank you

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rickharris2 years ago

i know nothing about the system you are using, but long wires often pick up stray voltage from other sources. there will be little to no current available.

In general this is often a result of measuring voltage with a modern high input resistance digital multi meter.

A resistor - perhaps 10 k or so from the wire to ground should sort out your odd 6.8 volts.

Mweston011 (author) 2 years ago

Thanks for the responses, some were a bit complicated but after a closer look all I needed was a simple resistor to the ground wire. I found an extra one laying in the bottom of the alarm box that I guess they were using when installing it. I'm not sure what type it is but it worked perfectly.


Looks like a 3.3K 1/4W

iceng2 years ago

Here is a simple op-amp circuit solution that should work on any voltage from 10-30 volts. Here it is set up for 15 volts.

Works very simply, op-amp 'B' divides 15 volts to a 7.5 volt Reference.

This reference goes to the --pin2 of op-amp 'A' , the alarm goes to the +pin3 of op-amp 'A'.

Now, IF the alarm is below the7.5 v reference THEN pin 1 of U4A will be 0.0

But IF the alarm is above the 7.5 v reference THEN pin 1 will go to 15 volts.

Just for fun I added a resistor R3 and a zener diode which will give you the 5v digital signal when the door is open....

Door CLOSED = 6.8v ===> OUTPUT = 0.0v

Door OPENED = 13.5v ==> OUTPUT = 5.1v

Attached is the PDF if you want more info on the Quad op-amp.

BTW you should tie or ground unused input pins 9,10,11,12,13 together.

LP324.pdf364 KB
iceng iceng2 years ago

BTW, IF two op-amps are stupefying, get rid of op-amp 'B' and replace it with a 7.n Zener diode also get rid of R2.

The zener Anode goes to ground, Cathode to pin2 and R1.

If you don't want the 5.1v output then you can get rid of the 5.1v zener and R3 too.

iceng iceng2 years ago

Hey weston are you following our discourse ?

We really like to help you !

We also like you to select a best answer after there are no more ideas presented.

Can you understand schematic drawings like the one here ?

Do you know how to solder or use a plastic breadboard ?

You can talk to us in this question thread



I'd put a 10pF cap across the opamp from output to - it makes them quieter. And ideally you want some sort of hysteresis too.

Off the subject......

Attended my first Comic Con and had this slow down fun peeking over curtains and spotting grand children making a selfie

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Yes, excellent feature, now you must explain the advantage of

hysteresis to weston :) who probably isn't reading our output anyway.

Though there are almost 50 views !


You'd need some additional electronics, like a comparator circuit, to isolate the alarm signal from the Z thing.

Wired_Mist2 years ago

I wouldn't mess with it. The reason you read the two voltages is because it is used to tell if...

A 13.5v Door is Open

B 6.8v Door is closed

C 0V sensor is malfunctioning, and is flagged by your home alarm company.

Chances are they will already have a product you can use. they will probably charge for the part + installing it. It's really not worth damaging your system, that could lead to pricy repairs before they will continue your service. They Will know if you have messed with it. Hope this helps !