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I am wondering if it is possible to make an alarm that normally operates/sounds when electricity is flowing through a wire to do the complete opposite. The alarm sounds when there is a constant flow of electricity. When the flow of electricity is interrupted, the device does not sound. I would like the alarm to sound only when there is an interruption or holt of electricity as well as be silent when there is a constant flow of electricity.

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Please reread and edit your question, as it is confusing. I.e.: clearly define sentence subjects, such as "the alarm" and "the device", to remove ambiguity. (To which "alarm" or "device" are you referencing in each sentence? The one you want, the one "that normally operates...", or another?) – tyblu Jun 9 '12 at 23:08
I answered your question because I think I understand what you need. But I had to read it three times, so -1. Like tyblu says, reread and edit. – stevenvh Jun 10 '12 at 10:21

3 Answers

up vote 2 down vote

enter image description here

This is an alarm that will not sound in rest, i.e. when contacts SW1 and SW2 are closed. When both contacts are closed they will connect the transistor's base to ground, and there will flow no current into the base. So no collector current and the relay will not activate. The relay's contact will remain open, and the horn silent.
When either or both of the switches are opened the transistor will get base current from +12V through R1, the relay will switch on, and the alarm will sound.

So the alarm will sound if there's an interruption in the current through SW1 and SW2, and be silent if there's current through them.

Some will use a MOSFET instead of the BC547C BJT, because it allows them to use a much higher value for R1, so that you have a much lower idle current than the 1mA you have now. That may not be a good idea. Most electromechanical contacts require some current through them when switching to keep the contacts clean. And 1mA at 12V is only 12mW, that's 0.1kWh per year, or less than 2 cents.
You can increase switching current through SW1 and SW2 by placing a 100nF capacitor across each contact.

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1  
stevenvh has discovered CircuitLab – Federico Russo Jun 10 '12 at 9:55
@Federico - For complaints you have to go to clabacchio, it's all his fault! :-) – stevenvh Jun 10 '12 at 10:24
@Federico - First thing I don't like (and I think Olin will agree) is how connections are shown. In this answer I had to paint the dots with India ink for clarity. :-( – stevenvh Jun 10 '12 at 17:36
@stevenvh - I agree, it should at least have option for using junction dots. I haven't used it but I'm assumed it used the "two lines crossing are never connected" rule and disallows connections like this (Altium does this) but looking at your link it seems e.g. the R2, R3, D1, RLY1 junction is allowed, which I'm not keen on. – Oli Glaser Jun 10 '12 at 18:21
@Oli - Yes, it's that kind of junctions that worry me (and makes Olin utter grumpy comments). I try to avoid them, and always have them at a "T", but here I knew I had to place the dots anyway. – stevenvh Jun 10 '12 at 18:27
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up vote 1 down vote

at first.

a good alarm sounds when there is a open circuit. and it is quiet when it is short
but it can be active high or active low, which means:

active low: (actually inactive low; active open): the alarm is quiet when sense wire is shorted to ground. (you measure 0V) and sound when open (you measure VCC (e.g. 5V or 12V);

active high: (actually inactive high; active open): the alarm is quiet when sense line ( wire) is shorted to VCC. you measure between wires 0V, between sense and ground: VCC and sounds when open: (you measure 0V between sense and GND, ans -VCC between sense and Vcc)

the reason is: if somebody cuts the alarm cable, he directly activates the alarm, in stead of disabling it.

second:

your voltage monitor: is it on High voltage AC (120V~ or 230V~);
or on low voltage DC (12V= or 5V= or 24V=,...)

low voltage is simple to build your own;

enter image description here

for the values of the resistors and transistors i need to know what voltage and which current your alarm consumes, the diode can be a regular 1N4007

the L is the sense line ( the voltage to monitor), and the big black blob is the output.

if you want to build something to monitor High Voltage, you first needs a lot of experience with electronics, because it is very dangerous to play with it without exactly knowing what you are doing.

if you would build a circuit to convert a alarm you already have bought: please post the manufacturer and the type of the device

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up vote 0 down vote

Choose the momentary button and mechanical relay with coil, normally open and normally closed contacts. Total 6 pins. Connect relay coil to its own normally open contacts. Connect a button in parallel to normally open contacts. Power the relay. Push the button. The alarm is ready.

If intruder will break the wire, the relay will return to unpowered state and close normally closed contacts. Even if intruder will restore power quickly, the relay will wait for secret button to push.

This normally closed contacts will be the alarm signal.

enter image description here

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