Build this DIY Burglar Alarm System using cheap off-the-shelf components. It has automatic Exit and Entry delays, a timed Bell Cut-off. Use it in your home - small business premises - lock-up etc. It has provision for normally-open and normally-closed switches and will accommodate the usual input devices (Pressure Mats, Magnetic-Reed contacts, Foil Tape, Movement Detectors and Inertia Sensors).

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Circuit Description

Click Here For A Photograph Of The Prototype

The Cmos 4081 has four separate two-input AND gates. 

AND gates only produce a high output when both inputs are high. See the truth table. 

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While the alarm is off - Sw1 holds pins 2, 6, 9, 12 & 13 low. 

Because gate 4 is not used - we can ignore pins 12 & 13.

Thus - Sw1 holds pins 2, 6 & 9 low.

This means that each of the remaining three gates has one low input.

Therefore - all three outputs are low.

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While it's in the off position - Sw1 keeps C6 in a discharged state.

Initially - when Sw1 is moved to the set position - C6 will continue to hold pins 2, 6 & 9 low.

The time it takes for C6 to charge - and take pins 2, 6 & 9 high - is the exit delay.

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The C6 charge path is through R10 - the normally-closed relay contacts - D7 and R12.

It's true that current through R11 also makes a small contribution to the charge.

But - the length of time it takes for C6 to charge is effectively controlled by the value of R12 alone.

Until the voltage on C6 takes pins 2, 6 & 9 high - the alarm cannot be activated.

With the component values shown - you have about 30 to 40 seconds to leave the building.
 
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As you leave the building - R2 will take pin 1 of the 4081 high.

And current from R2 - flowing in the reverse direction through the 8v2 zener diode - will turn Q2 on.

Q2 connects the negative lead of the buzzer to ground - through D5 and Sw1.

This causes the buzzer to sound.

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When you close the door behind you - pin 1 will go low again - the current through Z1 will cease - and the buzzer will switch off. 

This indicates that the trigger circuit has been successfully restored within the time allowed.

If you don't want this feature - just leave out the zener diode.

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Z1 permits current flowing through R2 to reach the base of Q2.

But the voltage drop across the zener diode prevents the base-emitter junction of Q2 from holding pin 1 permanently low.

If the base of Q2 were connected directly to pin 1 - R2 could never take pin 1 high - so the alarm would never trigger.

The buzzer's path to ground has to go through Sw1 - otherwise it would sound every time the door is opened - even when the alarm is off.

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When Sw1 is moved to the set position - current through R9 lights the red LED.

The red LED provides a visual indication that the alarm is set.

I found that while the alarm was off - the red LED was glowing dimly.

It was finding some sort of path to ground through Q2.

D5 closes this path. 

The diode is required to carry the current drawn by the buzzer.

A 1N4148 would probably be fine in this position.

However, I felt that for long-term reliability - it would be better to use the more robust 1N4001.
 
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30 to 40 seconds after you move Sw1 to the set position - the voltage on C6 will take pins 2, 6 & 9 high.

This will enable gates 1, 2 & 3.

While pins 2, 6 & 9 were held low - we couldn't change the outputs of gates 1, 2 & 3.

That is - we couldn't activate the alarm.

But once the exit delay has expired - and pins 2, 6 & 9 are high - we can change the outputs.

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The trigger circuit is coloured blue in the diagram. 

In standby mode - R1 supplies base-current to Q1. 

This keeps the transistor switched on.

You can think of Q1 as an extra normally-closed switch - connected in series with the normally-closed loop

Current flows to ground through R2, the green LED, Q1 and the normally-closed loop.

The green LED confirms that the trigger switches are intact - and that pin 1 is low.

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When you return and open the door - the normally-closed loop is broken.

The R2 current can no longer flow to ground through the loop.

So - the green LED switches off; and R2 takes pin 1 high.

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The same thing will happen if a normally-open switch is closed.

If the base of Q1 is connected to ground - the transistor will switch off.

The R2 current can no longer flow to ground through the transistor and the normally-closed loop.

So the green LED will switch off - and R2 will take pin 1 high.

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You'll recall that C6 is holding pin 2 high.

When R2 takes pin 1 high - both the gate 1 inputs are high.

So - the gate 1 output will go high also.

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When the gate 1 output goes high - current from pin 3 flows into C3 - through D2 & R4.

C3 charges almost immediately - and takes pin 5 high. 

Because C6 is holding pin 6 high - when pin 5 goes high - pin 4 will go high also.

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In other words - when you activate the alarm - the immediate result is that pin 4 goes high.

When pin 4 goes high it does three jobs. 

It latches itself on - sounds the buzzer - and begins the entry delay.

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It latches itself on by taking its own input high - through D3 and R6. 

It doesn't matter now - if you close the door behind you - and pin 3 goes low.

Pin 4 has taken over from pin 3 - the job of keeping C3 charged - and thereby holding pin 5 high.

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Pin 4 sounds the buzzer by providing base current for Q2 through R3. 

This switches the transistor on.

Q2 connects the negative lead of the buzzer to ground - through D5 and Sw1 - and the buzzer sounds.

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The entry delay begins when current from pin 4 starts to charge C5 - through R8.

After 30-seconds or so - C5 will take pin 8 high.

Because C6 is already holding pin 9 high - when pin 8 goes high - pin 10 will go high also.

Pin 10 supplies base current to Q4 - through R15.

The transistor switches on - and connects the negative side of the relay coil to ground.

So the relay will energize - and the siren will sound.

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When the relay energizes - the C6 charge path through the normally-closed contacts is cut.

And a new discharge path to ground  - through R11 and the relay contacts - is opened. 

This is a much slower path. It provides the reset time.

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After 15-minutes or so - the voltage on C6 will take pins 2, 6 & 9 low.

Since one input from each gate is now low - all the outputs go low. 

This means that the base-current supplied to Q2 by pin 4 - is cut off. 

So the transistor will switch off - and the buzzer will be silenced. 

The base-current supplied to Q4 by pin 10 - is also cut off. 

So the transistor will switch off - the relay will drop out - and the siren will be  silenced. 

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Pins 2, 6 & 9 go low when the voltage on C6 falls to just below half the supply voltage.

This means that when pins 2, 6 & 9 go low - C6 is only partially discharged.

There's still almost 6-volts across it.

This is not necessarily a problem.

But - Q3 and its associated components - coloured red in the diagram - finish the discharge of C6 - and perform a complete reset.

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While R11 is partially discharging C6 - it also partially discharges C7 - through D7 and R12.

Because C7 is partially discharged - when the relay drops out - current will flow briefly through R10 & C7 - into the base of Q3. 

This turns Q3 on for just long enough to complete the discharge of C6 - and thus reset the exit delay. 

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In effect - Q3 is doing the same job as Sw1. 

It switches the alarm off for a moment - by connecting pins 2, 6 & 9 to ground. 

R14 keeps Q3 switched firmly off when it's not needed.

And R13 limits the surge current through Q3 as it discharges C6.

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If the trigger circuit has been restored - the alarm will reset and return to standby mode. 

If the trigger circuit has not been restored then - when the exit and the entry delays expire - the siren will sound again. 

The alarm will continue trying to reset itself every 15 to 20 minutes until the trigger circuit is restored - or until the alarm is switched off.  

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Relay coils and some sounders can produce high reverse voltage spikes that will destroy Cmos ICs. 

D1, D6 and D8 are there to short-circuit the spikes before they do any damage.

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The remaining diodes create one-way paths.

D2 allows a high pin 3 to charge C3 - through R4.

But it prevents a low pin 3 from discharging C3 - through R4. 

Instead - C3 has to discharge slowly - through R5.

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If you've fitted inertia sensors - a long series of minor vibrations could build up the charge on C3 to the point where it takes pin 5 high - and causes a false alarm.

In standby mode - R5 prevents this from happening by keeping C3 discharged.

However - the R5 discharge path is slow enough to ensure that a sustained - but low level - attack will gradually charge C3 - and so cause an activation.

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D3 allows a high pin 4 to take pin 5 high through R6 - but it prevents a low pin 4 from holding pin 5 low.

If pin 4 were to hold pin 5 low - pin 3 wouldn't be able to take pin 5 high.

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When pin 4 goes high - D4 prevents C5 from charging rapidly through R7.

Instead - it has to charge slowly through R8.

However - when pin 4 goes low - D4 allows C5 to discharge rapidly through R7.

This rapid discharge resets the entry timer - the moment the alarm is switched off.

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D7 allows C6 to charge through R12 - but removes any tendency it might have to discharge through R12, C7, Q4, R13 & R14. 

D7 may not be strictly necessary - but it means that the siren cut-off timer is controlled exclusively by R11.

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There is nothing special about the transistors. 

Any small NPN transistors with a gain (hfe) greater than 100 and an Ic(max) of at least 100mA should do.

But remember that the pin configuration of your transistors may be different from that of the BC547.  
 
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The wires going to the switches on the doors, windows, etc. can act like a radio antenna.

They pick up stray signals that can cause false alarms.

C1 and C2 are there to de-tune the wires - and short-circuit any high frequency signals to ground. 

C4 does a similar job for the supply.

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How Inertia Sensors Work

Inertia Sensors are devices that react to vibrations. They do not react to constant low level vibrations of the sort you would get from loud music or heavy road traffic. Instead - they have a certain amount of inertia that must be overcome before they react. In other words - they need a sharp blow - such as that required to break a pane of glass.

How Inertia Sensors Work

Essentially, they are small normally closed switches - the contacts of which are joined by a specially weighted bar. It's the weights on the bar that provides the inertia. When a blow is struck - the bar bounces - and the switch opens briefly. The length of time the switch is open - i.e. how high the bar bounces - depends on the weight of the bar and the strength of the blow.

The alarm circuit sensitivity is adjusted by altering the speed at which it reacts - i.e. the speed at which C3 charges through R4. At its most sensitive - the value of R4 is zero. So C3 will charge very quickly. The alarm will react to a very short opening of the circuit - so a light tap on the sensor is all that's required.

Because a large number of small insignificant impulses might build up the charge on C3 to the point where it would cause a false alarm - R5 is included to keep C3 discharged in standby mode.

Power Supply

In standby mode - the circuit should not use more than a few milliamps. It's only when the relay is energised and the siren is sounding that any real current is required. The actual current consumption depends on the sounder you use. A conventional bell uses up to about 400ma. An electronic siren generally uses less. In any case - a 1-amp power supply should be sufficient. The Alarm Power Supply is ideal.

Parts List

Suppliers

Worldwide RS Components

UK & Ireland Maplin

Construction

Click here if you're new to constructing stripboard projects.

The terminals are a good set of reference points. To fit them, you may need to enlarge the holes slightly. Then turn the board over and use a felt-tip pen to mark the 34 places where the tracks are to be cut. Before you cut the tracks, use the "actual size" drawing to Check That The Pattern is Correctly Marked .

When you're satisfied that the pattern is right - cut the tracks. Make sure that the copper is cut all the way through. Sometimes a small strand of copper remains at the side of the cut and this will cause malfunction. Use a magnifying glass - and backlight the board. It only takes the smallest strand of copper to cause a problem. If you don't have the proper track-cutting tool - then a 6 to 8mm drill-bit will do. Just use the drill-bit as a hand tool - there's no need for a drilling machine.

Pattern for Cutting
The Tracks on the
Underside of the Board

Pattern for Cutting
The Tracks on the
Underside of the Board

Actual Size Of Pattern

Pattern for Cutting
The Tracks on the
Underside of the Board
ACTUAL SIZE

Next make and fit the Nine Wire Links. I used bare copper wire on the component side of the board. Telephone cable is suitable - the single stranded variety used indoors to wire telephone sockets. Stretching the core slightly will straighten it - and also allow the insulation to slip off.

Add The Nine Wire Links

The next stage is to fit the 14 resistors - the preset - most of the diodes - and the IC socket. Using a socket reduces the chances of damaging the IC - and makes it easier to replace if necessary. Pay particular attention to the orientation of the diodes. The resistors and diodes are all shown lying flat on the board. However, those connected between close or adjacent tracks are mounted standing upright. See the Photograph Of The Prototype

Add The Components

Next - fit the four transistors - the remaining diode - the two LEDs - the seven capacitors - and the relay. Pay particular attention to the orientation of the electrolytic capacitors. They generally have a light coloured stripe down the side next to the negative terminal. Note that C6 and C7 have their positive terminals facing the bottom of the board. See the Photo Of The Prototype

Add The Ten Solder Bridges

Finally, examine the underside of the board carefully - to make sure that there are no unwanted solder bridges or other connections between the tracks. If you backlight the board during the examination - it makes potential problem areas easier to spot.

When you're satisfied that everything is in order - add the ten solder bridges to the underside of the board. The tracks are cut to accommodate two sizes of preset. If you've used the smaller preset - like the one in The Prototype - you'll need to add an extra solder bridge to connect the wiper to the track above.

Finish off by inserting the Cmos 4081 into the socket. Pin 1 of the IC should be in the top left-hand corner. Check that all 14 pins have entered the socket. Sometimes - instead of entering the socket - a pin will curl up under the IC.

You're Now Ready To Test Your Circuit

Connections

Connecting External Devices
To The Cmos Single Zone Alarm

General Information