# Increasing 0.15v to +1v?

Discussion in 'The Projects Forum' started by stevejnr, Sep 8, 2011.

1. ### stevejnr Thread Starter New Member

Sep 8, 2011
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Hi Folks

I'm quite a novice when it comes to circuits.. So hopefully someone could please help with the following problem I'm facing..

I've got a vehicle tracking system that logs when a vehicle door is unlocked. It is negative triggered and as such when the door is locked a voltage of 0.14-0.16V passes over the central locking unlock wire. When I unlock it drops to 0.

The issue is my tracking devices only pick up the door as locked if the voltage across the central locking wire is >1V and as such interperets the 0.14-0.16V as unlocked, even though it is not.

I have tapped in to the door wire and I'm running off a seperate feed, I also have a 12V DC supply if required. The input for my tracker can take up to 45V, 500mA max. Could someone kindly recommend a work around to get me from 0.14-0.16V input to >1V output while still allowing for the voltage to drop back to 0 when unlocked?

Steve

Last edited: Sep 8, 2011

Nov 16, 2007
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3. ### stevejnr Thread Starter New Member

Sep 8, 2011
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Ahh ok, sorry..

4. ### wayneh Expert

Sep 9, 2010
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So you want to detect a break in a circuit that occurs when the door is opened. And you want to use the existing signal in that circuit, which is ~0.15V or "open" (no voltage, but not grounded, or is it?). Is this right?

One way would be a comparator circuit, for instance using the widely available LM339 quad comparator. You could power it with your 12V supply, and compare the door circuit voltage to ground as the two inputs to the comparator. When the comparator sees a tiny voltage in the circuit, as compared to the ground reference, it will output a voltage of either 12v (supplied by a pull-up resistor of ~3KΩ) or 0V, your choice, depending how you wire it. In either case the current output is small, less than 5mA, so you may need to use a transistor to amplify the current output. Depends what your tracker needs.

Here's an example that would light the LED when the voltage is detected.

If you think this approach will work for you, there are some additional details that may be needed to interface it with your tracker.

Sep 9, 2010
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