Use this DIY Alarm in your home - small business premises - shed - garage - lock-up etc. It can be battery powered and it has provision for normally-closed input devices such as - Magnetic-Reed Contacts, Foil Tape, Movement Detectors, PIRs etc.

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Circuit Description

Click Here For A Photograph Of The Prototype.

Circuit Diagram For A
Cmos 4093 Burglar Alarm

The Cmos 4093 is the Schmidt-Trigger version of the 4011. 

While its logic is identical to that of the 4011 - what makes it different is a property known as "Hysteresis".

The circuit will work with a 4011. But - because the 4011 does not posses hysteresis - it will not work as well. 

We'll come back to this later.

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Like the 4011 - the 4093 has four two-input NAND gates. 

These gates only produce a low output when both inputs are high - see the truth table.

While the alarm is switched off - C2 remains in a discharged state.

Because C2 is discharged - when you move Sw1 to the "set" position -the capacitor continues to hold pin 1 low. 

It will remain low until C2 charges through the relay coil, R7 and D4. This is the Exit delay.

C2 also receives some charge through R6 - but its effect is insignificant - and it may be ignored. 

So - for practical purposes - the length of the Exit delay is controlled by the value of R7 alone.

During this period - you can open the door and leave the building. 

As you do so - R3 will take pin 2 high. 

But - because C2 is still holding pin 1 low - the output of gate 1 will not change.

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After the Exit delay has expired - the charge stored in C2 will hold pin 1 high. 

When you return to the building and open the door - R3 will take pin 2 high.
 
Now that both the inputs of gate 1 are high - its output will go low.

Pin 3 takes the inputs of gate 2 low. 

So the output of gate 2 - at pin 4 - will go high. 

This output does 3 jobs.

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Firstly - it latches itself on by taking pin 2 high through D2. 

This means that when you close the door after activation - the normally-closed loop cannot take pin 2 low again.

In other words - you cannot defeat the alarm by simply closing the door behind you.

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D2 creates a one-way path.

It allows a high pin 4 to take pin 2 high.

But it prevents a low pin 4 from clamping pin 2 low.

If this were to happen - opening the loop would not take pin 2 high.

So the alarm would not activate.

R4 prevents the normally-closed loop from shorting a high pin 4 to ground - through D2.

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Secondly - pin 4 provides base current for Q1 - through R2.

When pin 4 goes high - it switches Q1 on.

The transistor connects the negative side of the Buzzer to ground - and the Buzzer sounds.

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The third task performed by pin 4 is to charge C4 through R5.

This is the Entry delay. 

During this period you will normally switch the alarm off.

If Sw1 is not switched off within about 30-seconds - C4 will take pins 8 & 9 high.

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

Pin 10 takes pins 12 & 13 low. 

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

Pin 11 provides base current for Q2 - through R8. 

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

This causes the relay to energize - and the siren to sound.

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Since the negative side of the relay coil is now connected to ground - the C2 charge path through the coil, R7 and D4 no longer exists.

Instead - a new discharge path has been opened - through Q2.

D4 prevents C2 from discharging rapidly through R7.

Instead - the capacitor discharges slowly through R6.

This slow discharge controls the length of the Siren cut-off and Reset time.

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As soon as the relay energizes - the reset time begins.

It takes about 10-minutes for the voltage on pin 1 to go low.

Since one of its inputs is now low - the gate 1 output goes high.

It takes pins 5 & 6 high - and pin 4 goes low.

At this point - three things happen.

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Firstly - the Buzzer switches off.

Secondly - since D2 is no longer holding pin 2 high - the loop can take pin 2 low.

And thirdly, C4 discharges into pin 4 - through R5.

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When pins 8 & 9 go low - pins 10,12 & 13 go high.

This causes pin 11 to go low - Q2 to switch off - and the relay to drop out.

Thus, the C2 charge path - through the relay coil, R7 and D4 - is restored.

And after a short delay - pin 1 will go high - and the alarm will return to standby mode.

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

D1, D6 & D7 short circuit these spikes at source - before they can do any damage.

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I specified a relay with a minimum coil resistance of 270 ohms. Use a higher value if you wish.

With a 9-volt supply - the maximum current through Q2 will be about 35mA.

This is well within the limits of a BC547 - which has an IC(max) of 100mA.

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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 low standby current makes battery power a genuine alternative to a mains supply.

The current drawn by the circuit is only about 3 or 4 microamps - i.e. just the current that flows through R3, R4 and the normally-closed loop.

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To keep this current to a minimum - I used a 4.7 meg resistor.

The downside of using such a high value resistor is that the trigger circuit is very sensitive to condensation.

The moisture connects pin 2 to ground. 

This creates a potential-divider with R3 - and may prevent pin 2 from reaching a voltage high enough to trip the alarm. 

It's important that both the circuit board and trigger switches be protected from moisture.

For the same reason - it's also important that C1 should not leak. If it does - pin 2 may not reach a high enough voltage to trigger the alarm.

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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. 

C3 does a similar job for the supply.

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When Sw1 is moved to the "off" position - it connects R1 to the positive rail.

Since the power is switched off - the resistor is not drawing any current from the supply.

Instead - it provides a rapid discharge path for C2 and C4 - through D3 and D5.

This means that the instant the alarm is switched off - the Exit and Entry delays are reset - and the circuit is ready for immediate re-use.

R1 limits the maximum current surge through D3, D5 and the contacts of Sw1.

I could simply have connected R1 across the supply rails - but that would have increased the standby current significantly.

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I set the Entry and Exit delays at about 30-seconds.

If you think this is too long - reduce the value of R5 and/or R7. 

If you think it's too short - increase the values.

The reset time is controlled by R6. 

You can also change its value to suit your requirements.

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In standby mode - it takes a minute or so for C2 to charge completely.

Once it has charged completely - it will provide an initial reset time of about 10 minutes.

At the end of the reset time - if the loop is still open - the alarm will reactivate.

           ============

In this case - the alarm doesn't actually enter standby mode.

So C2 doesn't have the extra minute or so it requires to complete its charge.

Consequently - subsequent reset times are somewhat shorter.

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In other words - the initial reset time is about ten minutes.

After that - the alarm will attempt to reset itself about every five minutes or so.

It will go on doing this until the loop is closed - or the alarm is switched off. 

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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.

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I've drawn the circuit with a 9-volt supply - but the alarm will work at anything from 5 to 15-volts.

All you have to do is choose a relay, siren and buzzer that will work at the supply voltage you are using.

           ============

So - back to "Hysteresis".

Hysteresis is simply a delay between an input and output - between a cause and its effect. 

Sometimes the delay is useful - and sometimes it's a nuisance.

In the case of a thermostat - hysteresis is a desirable attribute - and it's built in deliberately.

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A thermostat doesn't try to maintain a constant temperature. 

In order to do so - it would have to keep switching on and off every few seconds.

Instead - it keeps the temperature within a specific range.

When the preset temperature has been reached - it switches off. 

And it only switches on again - when there has been a significant change in temperature.

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The difference between the temperature at which the thermostat switches off - and the temperature at which it switches on again - is the hysteresis.

Without this hysteresis - your central heating and your refrigerator etc. would keep switching on and off every few seconds.

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Like the thermostat - the gates of a Cmos 4093 have hysteresis.

The input voltage that will turn the output on - is different from the input voltage that will turn it off again.

The gate is - as it were - reluctant to change the state of it's output.

It waits for a significant fall in input voltage - before it takes its output high.

And it waits for a significant rise in input voltage - before it takes its output low again.

The difference between these two input voltages - is called the hysteresis voltage.

Its value depends on the supply voltage. 

           ============

With a 9-volt supply - the hysteresis voltage of a 4093 gate is typically about 2-volts.

This means that the input will have to fall to at least 3v5 - before the output will go high.

And it will have to rise to at least 5v5 - before the output will go low again.

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So - what's this got to do with the alarm circuit?

Well - it's only important with regard to gate 1 - and the automatic reset timer.

Let's assume that we are using a Cmos 4011.

When the alarm is activated - it will take about 10 minutes for C2 to discharge through R6 and Q2.

This is because C2 is fully charged when the reset timer begins.

However - if the loop is still open - the attempted reset will fail.

The alarm will not enter standby mode.

And C2 will not have enough time to complete its charge.

           ============

Because a 4011 doesn't have hysteresis - the charge on C2 will remain close to the switching point of the gate - say 4v5.

In a matter of seconds - R7 will take it just above 4v5 - and re-trigger the alarm.

The relay will re-energizes - and the capacitor will start to discharge again - through R6.

So in a matter of seconds - R6 will take it just below 4v5 - and the alarm will attempt a reset.

In other words - after the initial 10-minute reset period - the gap between subsequent reset attempts is very short.

           ============

But - with a 9-volt supply - the 4093 has about 2-volts of hysteresis.

This means that the voltage on pin 1 will have to fall to at least 3v5 before the reset is attempted.

And it will have to rise to at least 5v5 - before the open loop can re-activates the alarm.

Actually - because of the influence of the Entry Delay - the relay doesn't re-energize immediately.

Until it does - the voltage on C2 goes on rising. So it will end up somewhat higher than 5v5. 

And it will take a significant time before it falls back to 3v5 - and another reset is attempted.

With a 4093 - the reset attempts occur about every 5-minutes or more.

And they will continue to do so until - the loop is restored - or the alarm is switched off.

Learn More About The Cmos 4011 And 4093

Parts List

Suppliers

Worldwide RS Components

UK & Ireland Maplin

Construction Guide

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 19 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 8 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 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 Eight Resistors -
The Five Wire Links -
The Seven Diodes -
And Two Transistors

Add The Eight Resistors -
The Five Wire Links -
The Seven Diodes -
And Two Transistors

The next stage is to fit the diodes and transistors. Pay particular attention to the orientation of the diodes. Note that one diode - D3 - is pointing down.

Fit the remaining six components (capacitors, relay and IC socket), Using a socket reduces the chances of damaging the IC - and makes it easier to replace if necessary.

Add The Remaining Components
And The Nine Solder Bridges

Add The Remaining Components
And The Nine Solder Bridges

Turn the board over and examine the underside 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 9 solder bridges.

Finish off by inserting the Cmos 4093 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

External Connections

Connecting External Devices
To The Enhanced Alarm

Test Your Finished Circuit Board

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Enhanced Alarm - Support Material

Circuit Description

Parts List

Construction Guide

Actual Size Of Pattern

External Connections