Build this DIY Burglar Alarm System using cheap off-the-shelf components. It has automatic Exit and Entry delays and a timed bell Cut-off. Use it in your home - small business premises - lock-up - caravan - mobile home, camper van etc.

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Circuit Description

Click Here For A Photograph Of The Prototype

The Automatic Intruder Alarm uses a Cmos 4011.

The 4011 has four two-input NAND gates. 

NAND gates only produce a low output when both inputs are high.

See the truth table. 

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While the alarm is off - Sw1 holds pin 6 low. 

If the loop is opened - and R1 takes pin 5 high - it will make no difference. 

While SW1 holds pin 6 low - the output from gate 2 must remain high.

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While Sw1 is in the off position - it keeps C2 in a discharged state.

So -  when Sw1 is moved to the set position - C2 will continue to hold pin 6 low.

Pin 6 will remain low until C2 charges through R3 and R4. 

This is the Exit delay. 

During this time the alarm cannot be activated.

So it's safe to open the loop and exit the building. 

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Once C2 has charged - the Exit Delay is finished - and R3 holds pin 6 high. 

When you return to the building and open the door - R1 will take Pin 5 high.

Since both gate 2 inputs are now high - pin 4 will go low.

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When pin 4 goes low - two things happen.

C3 charges rapidly - through R8 & D4.

And pin 4 takes pins 1 & 2 low through D4.

So the gate 1 output - at pin 3 - will go high.

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Pin 3 does two jobs. 

The first is to provide base-current for Q1 through R2.

The transistor switches on - connects the negative lead of the buzzer to ground - and the buzzer sounds.

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Its second job is to charge C5 through R6.

C5 & R6 provide the Entry Delay. 

After about 30-seconds - the voltage across C5 will takes both of the gate 3 inputs high. 

So the gate 3 output - at pin 10 - will go low.

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Pin 10 takes the inputs of gate 4 low. 

So the gate 4 output - at pin 11 - will go high. 

Pin 11 provides base-current for Q2.

The transistor connects the negative side of the relay coil to ground.

So the relay energizes - and the siren sounds.

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While at least one of the trigger switches remains open, the siren will continue to sound. 

However - if the normally-closed loop is restored - pin 5 will go low. 

Since one of its inputs is now low - the output of gate 2 will go high.

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Now - pin 4 is no longer holding pins 1 & 2 low.

Instead, pins 1 & 2 are held low by the charge stored in C3.

They will remain low until C3 discharges through R9.

As it does so - the voltage on pins 1 & 2 will rise. 

After about 15-minutes - when it reaches roughly half the supply voltage - the inputs will go high. 

And the gate 1 output - at pin 3 - will go low.

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When pin 3 goes low - the base current to Q1 is cut off. 

So the transistor switches off - and the Buzzer is silenced. 

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Pin 3 also discharges C5. 

During the Entry Delay - D2 forces C5 to charge slowly through R6.

But it allows C5 to discharge rapidly through R7.

Without R7 and D2 - it would take about 30 seconds for C5 to discharge. 

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When C5 discharges - pin 3 takes pins 8 & 9 low - through R7 and D2. 

When pins 8 & 9 go low - pin 10 will go high. 

Pin 10 takes pins 12 & 13 high

So the output of gate 4 - at pin 11 - will go low. 

This cuts off the Q2 base current.

So the transistor switches off - the relay drops out - and the siren is silenced.

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Relay coils and some Sounders produce large reverse voltage spikes that will damage Cmos ICs. 

D1, D5 & D6 short circuit these spikes at source - before they can do any damage. 

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The rest of the diodes are used to create one-way paths.

D4 allows a low pin 4 to take pins 1 & 2 low. 

It also allows a low pin 4 to charge C3.

But it prevents a high pin 4 from taking pins 1 & 2 high. 

It also prevents a high pin 4 from discharging C3.

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D2 prevents C5 from charging rapidly through R7 - instead of slowly through R6.

It takes about 30-seconds for a high pin 3 to charge C5 - through R6.

This provides the Entry Delay.

The purpose of R7 is to reset the Entry Delay.

R7 discharges C5 rapidly - through D2 - into a low pin 3.

Thus - the moment the alarm is switched off - the Entry Delay is reset.

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When the alarm is switched off - C3 discharges rapidly - through R5, the red LED, D3 and R8.

Without D3 - pins 1 & 2 would remain low until C3 discharges through R9. 

With D3 - the moment the alarm is switched off - it's reset - and ready for immediate re-use.

We need to use a diode - and not a wire link - because the diode prevents Sw1 from taking pins 1 & 2 low.

If Sw1 could take Pins 1 & 2 low - the alarm would activate the moment it was set.

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There's nothing special about the BC547s. 

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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While Sw1 is in the set position - current through R5 lights the red LED.

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

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The three 1k resistors are there to limit the peak current when the capacitors charge and discharge. 

The value was kept low so that the capacitors would charge and discharge rapidly. 

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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 is 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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Learn More About The Cmos 4011

Power Supply

The circuit itself should not use more than a few milliamps. It is 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 26 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 fit the 10 resistors and the Five Wire Links. For the 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.

The resistors 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 Ten Resistors -
The Five Wire Links -
The Diodes and Transistors

Add The Ten Resistors -
The Five Wire Links -
The Diodes and Transistors

The next stage is to fit the diodes and transistors. You'll want to mount the LEDs where they can be seen - perhaps on the front of your alarm-panel. They should be connected to the circuit board using a light flexible wire - e.g. the cores from a piece of alarm cable. They may be attached to the circuit board from the front or the rear. Make them long to begin with. You can shorten them later when you assemble your alarm-panel. Twist the wires into pairs to keep them tidy

Fit the remaining seven components (capacitors, relay and 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 electrolytic capacitors. Note that electrolytic capacitors generally have a light coloured stripe down the side next to the negative terminal. C2 is mounted with its positive terminal facing downwards. See The Photo Of The Prototype

Add The Remaining Components
And The Eleven Solder Bridges

Add The Remaining Components
And The Eleven Solder Bridges

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 11 solder bridges.

Finish off by inserting the Cmos 4011 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 Automatic Intruder Alarm

Modification No.1

Add A Normally-Open
Input Trigger To The Circuit

Modification No.2

The following simple modification to the trigger circuit - will cause the Siren cut-off timer to run - the moment the alarm is activated. In other words - the Siren will stop after 15-minutes - regardless of the state of the loop.

Fix The Maximum Output Time

General Information

Test Your Finished Circuit Board

Automatic Intruder Alarm Schematic	The Rest of Ron's Circuits	Write To Ron	More Free-to-Use Circuits	Circuit Exchange International

Automatic Intruder Alarm - Support Material