The copper wire is part of the existing circuit. So when there is a open copper wire, the latch relay should light up the LED and remain so until it been reset. As for the voltage/current i not sure and the distance can be like 100 to 200 meter away. I know if there is an open cooper wire, the resistance is around 50M ohm and if is not open, is around 50ohm or so. Please advise on a simple design. Thanks.We need a lot more info. Is the copper wire part of an existing circuit? What voltage/current/distance/AWG are involved? Give us the big picture.
Ken
Sorry then, can't help.sorry, i do not have a schematic or diagram. =(
The problem is: the open-detection circuit has to deal with the voltage/current that is normally present on the copper wire. That's what I'm looking for...a schematic for what is normally on both ends of the wire.The copper wire is part of the existing circuit. So when there is a open copper wire, the latch relay should light up the LED and remain so until it been reset. As for the voltage/current i not sure and the distance can be like 100 to 200 meter away.
ok, i will take measurement reading on that. sorry, i cant give you the schematic on the both end of wire. It is just purely just amplifier on the other end of wire.The problem is: the open-detection circuit has to deal with the voltage/current that is normally present on the copper wire. That's what I'm looking for...a schematic for what is normally on both ends of the wire.
Ken
...sound so complicated. I just wan to monitor the alarm unit through LEDs.. sound easy but implement is hard...There are problems with adding components to an existing alarm system. If you change the resistance of the alarm wire, it may trip continuously.
I really haven't worked with home alarm systems much, and it was a very long time ago. However, if the resistance in the loop is either too low or too high, the alarm will go off.
So, adding in something like the circuit Oclaf proposed could possibly cause the alarm to trip continuously, or perhaps not function at all when there was a fault condition.
should be positive voltage. The power supply should be 5VIf you measure the voltage compared to the negative lead on the battery, is it positive voltage or no voltage. I have a circuit in mind that may work. Basically I'm looking for a positive voltage to work with.
What is the power supply voltage I have to work with?
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