# How can I design a circuit that will turn on when a wire is cut?

I'm trying to build my own cable lock alarm (for science!) and am having trouble figuring out a way to detect that a wire has been cut.

The circuit has to

1. Use no power at idle (or at least, very very little so that the batteries don't have to be changed every hour)
2. Sound a speaker when a specific wire is cut

My knowledge of electronics is minimal.. I know what capacitors, diodes, resistors and other basic stuff are and do, but I don't have a good grasp on how electricity flows in anything other than a single loop.

I seem to remember once making a circuit that was something like this.. (and oh geez, I don't even know how to do a proper diagram so forgive me folks)

/----------[battery]-------\
|                          |
|--------[light bulb]------|
|                          |
\-----[wire to be cut]-----/


And the bulb would only light up if the wire below was cut, because electricity always takes the path of least resistance.

Anyway, this is going to be a battery operated circuit and I'm pretty sure that diagram up there is a short. I think there was a resistor involved but I don't remember where it went.

If anyone can give me some pointers that'd be great!

• I understand that you're eager to get started, but I would suggest to wait a little longer before accepting an answer. Others may have a good solution to, and may not feel like posting if there's already an accepted answer. Waiting a day or so won't hurt. – stevenvh Jul 30 '12 at 6:02
• fair enough -- i'll wait a couple days before selecting an answer – Ben Jul 31 '12 at 18:17

A simple circuit to do this would use a single transistor, one resistor and a buzzer. Connect two 1.5 volt batteries in series to get 3 volts. Connect one end of a 10 kilohm resistor to the positive terminal of the batteries and the other end to the base of a general purpose NPN transistor (2N2222, 2N3904, etc.). Connect the negative end of the battery to the emitter of the transistor. Connect one wire of the buzzer to the positive end of the battery and the other wire to the collector of the transistor. If the buzzer has polarity markings, follow them: positive to positive end of battery, negative to transistor collector. Connect your sensing wire from the base to the emitter of the transistor. As long as the wire is connected , it will short the base to the emitter and prevent the transistor from turning on. When it is cut, the batteries will send current to the base of the transistor through the resistor. This will turn on the transistor which means the collector to emitter voltage will be very small and most of the battery voltage will be across the buzzer which will then turn on. With the sense wire connected, the batteries only have to provide current through the resistor which will be about 3 volts divided by 10 kilohms or 0.3 ma. Two AA batteries can provide this much current for hundreds of hours. If necessary you can use C or D batteries for even greater lifetime. This circuit is simply and can be easily modified to handle other sound sources if needed.

• wow. alright, i know all these parts, i just need to sit down and draw this out so i can make sense of it. will be back to approve! – Ben Jul 30 '12 at 0:08
• 10k base resistor will draw about 0.25 mA as notes. AA Alkalines can supply that for perhaps 1 year. BUT if transistor has high current gain (say BC337-40) you can use a 100 k resistor for 10 year battery life (ie shelf life) and switch about 10 mA. Drive a second transistor stage for more load current. Or use a darlington transistor for more load current. Or use a suitable spec MOSFET with say 1 megohm resistor and as much load current as desired. – Russell McMahon Jul 30 '12 at 2:12
• This looks good Barry! I'm going to start out small and make a test circuit with 2 1.5v AAs, but the end goal is to hook this up to a 95 Wh, 10.95 V battery. I assume this means that I need to up the resistor (to what? -- this depends on what transistor i use, right?) this is great, i haven't been so excited about electronics in years. – Ben Jul 30 '12 at 3:40
• @Ben: that sounds like a rechargable battery. Its worth noting that the battery's self-discharge will BY FAR consume more of its charge than this circuit will. The MOSFET is actually a neat trick and can probably run for years off a coin cell. – Bryan Boettcher Jul 31 '12 at 17:24
• A schematic would be more comfortable. – Antonio Aug 9 '16 at 9:29

I would use an standard nMOSFET (such as the SI2316BDS-T1-GE3) with a 10M resistor connected between gate and the + of the battery. The buzzer should be connected with the - to the transistor drain and the + to the + of the battery. Connect one end of your sensing wire to the gate and the other to the transistor source toghether with the - of the battery and you are done! Make sure that the battery is the last component that you connect because you can potentially damage components if you insert them into the circuit with power applied. If you require more info I could email you a diagram. Gregory

simulate this circuit – Schematic created using CircuitLab

• Please post a diagram in your answer :) It will help others who have a similar problem. I like this solution as well, a 10M resistor will keep the battery alive for a long time. – Bryan Boettcher Jul 31 '12 at 17:28
• I don't understand how this is hooked up, specifically this part: "Connect one end of your sensing wire to the gate and the other to the transistor source toghether with the - of the battery" A diagram would be really helpful! – Ben Jul 31 '12 at 18:19
• "E-MOS"? You probably mean enhancement, but "P-MOS" or "N-MOS" should come first, and you don't say anything about that. And unless otherwise noted a MOSFET is an enhancement FET. – stevenvh Jul 31 '12 at 18:24
• Sorry guys. I was referring to a MOSFET, N-Channel. A practical example is SI2316BDS-T1-GE3. You can buy it at Digi-Key.com – Gregory Ion Aug 2 '12 at 3:58
• I added a diagram, hope it helps. Please check (it's working for me in the real world, and in the simulation, but I might have transcribed it wrong). – stib Feb 10 '14 at 12:18

If you can live with battery life measured in days rather than weeks, you can also do this with an SPDT relay or a normally closed (NC) reed relay. It's not nearly as power efficient as Barry's proposed solution (his will last ~50x longer), but if you're not comfortable with discrete electronic components, it might be easier to build and understand.

Using a low power relay like this one, you can get about 5 days out of a pair of AA batteries or 19 days out of a pair of C cells.

Connect the battery to the relay coil terminals with one leg of the connection representing your "sense wire" (battery negative to one side of the coil, battery positive to one end of your sense wire, and the other end of your sense wire to the other side of the relay coil. Polarity doesn't matter for most relays (unless there is an integral snubber diode))

You'll have a common contact (C), and a normally closed contact (NC) that you're interested in. Connect the battery positive to the common terminal, the NC contact to the positive lead of your buzzer, and the negative lead of your buzzer to the negative battery terminal. Make sure that cutting your "sense wire" will only remove power from the relay coil and will not remove power flowing to the buzzer.

With the sense wire being intact, the relay will be energized, holding the NC contacts open (not connected). When power is removed, the contacts will close, powering your buzzer.

• Upvoting because i learned something new! I'm probably going to go with Barry's solution for the battery life. I'm comfortable with discrete electronic components and all the soldering that goes on. I'm a follow-the-manual kind of guy who's finally made up his mind to learn the theory and start creating stuff instead. – Ben Jul 30 '12 at 3:30