Circuit Description

Garage / Shed Alarm - Circuit Simulation

Circuit Diagram For A
Battery Powered Burglar Alarm

The Cmos 4011 has four two-input NAND gates. 

Gates 2, 3 & 4 each have their two inputs connected together.

In this configuration - the three gates are being used as simple inverters.

If the two inputs are high - the output will be low.

If the two inputs are low - the output will be high.

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The NAND logic is only relevant with respect to gate 1.

The output of a NAND gate will only be low - if both of its inputs are high.

See the top line of the truth table.

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

In this condition - it acts like a wire link connecting pin 1 to ground.

That is - it's holding pin 1 low.

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When you move Sw1 to the set position - C2 continues to hold pin 1 low. 

Pin 1 will remain low until C2 charges through R5. 

This is the exit delay.

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

As you do so - R2 will take pin 2 high. 

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

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After 10 seconds or so - the charge on C2 will take pin 1 high.

This means that the exit delay has expired - and the trigger is enabled.

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When you return to the building and open the door - R2 will take pin 2 high. 

Now - both the gate 1 inputs are high - so the output will go low.

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Pin 3 takes the inputs of gate 2 low. 

So the output of gate 2 will go high. 

When Pin 4 goes high - it does 3 different jobs.

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Firstly, it must latch itself on.

Otherwise - you could defeat the alarm by simply closing the door behind you.

Pin 2 would go low - pins 3, 5 & 6 would go high - and pin 4 would go low.

In other words - the alarm would return to standby mode.

And the activation would fail.

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Pin 4 latches itself on by taking pin 2 high through D1.

This forces pin 2 to remain high - even after the loop is restored.

Now - it makes no difference whether the door is open or closed.

Pin 4 is high - and it will stay high.

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When pin 4 goes high it's connected internally to the positive line.

Its second job is to supply the current needed to light the green LED.

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The LED is not necessary to the operation of the circuit. 

It's there to re-assure you that the alarm is working.

It's doing the same job as a buzzer.

When you enter the building - it lets you know that the alarm has been activated - and that the siren is about to sound.

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Pin 4's third job is to charge C4 through R6.

This is the entry delay. 

During this period you will normally switch the alarm off.

If Sw1 is not moved to the off position in time - 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 will go high.

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When Pin 11 goes high - it's connected internally to the positive line.

And it supplies base-current for the transistor - through R7. 

This switches the transistor on.

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

This causes the relay to energize.

The relay contacts deliver power to the siren.

And the siren sounds.

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        COMPONENTS

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

D4 & D5 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 the transistor will be about 33mA.

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 transistor.

Any small NPN transistor 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 transistor may be different from that of the BC547.

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D1 creates a one-way path.

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

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

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In standby mode - if pin 4 were to hold pin 2 low - opening the loop would not take pin 2 high.

So the alarm would not activate.

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R3 prevents the normally-closed loop from shorting a high pin 4 to ground - through D1

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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 micro-amps.

That is - the current flowing to ground through R2, R3 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.

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If moisture connected pin 2 to ground - it would create a potential-divider with R2.

This could prevent pin 2 from reaching a voltage high enough to trip the alarm. 

So it's important that both the circuit board and trigger switches are protected from moisture.

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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 you turn off the alarm - it's important that C2 and C4 should discharge completely.

This resets the exit and entry timers.

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

R1 provides a rapid discharge path for C2 and C4 - through D2 and D3.

This means that both timers are reset the instant the alarm is switched off.

And the circuit is ready for immediate re-use.

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R1 limits the maximum current surge through D2, D3 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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Because R1 is connected to the off terminal - it's only in circuit when the alarm is off.

That is - when the battery is disconnected.

So R1 has no adverse effect on the level of standby current.

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On the basis that this alarm would be used in a garage or shed - I set the Exit and Entry delays at a short 10 to 15 seconds.

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If you think this is too short a time - increase the value of R5 and/or R6. 

If you think the delay is too long - decrease the values.

Roughly speaking - if you double the value of the resistors - you'll double the time.

If you halve the value - you'll halve the time.

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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 typical electronic siren will require less than 300mA.

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The drawing shows 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 and siren that will work at the supply voltage you want to use.

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The components used in the circuit should be widely available. However, none of them are critical. So - if you can't find the specified parts - You're certain to find something that will do just as well. And - if you prefer to make your own PCB - The Stripboard Layout Is Easily Converted.

Stripboard or Veroboard is a board drilled with a matrix of 1mm holes spaced approximately 2.5 mm apart and joined in rows by copper strips. The piece used for the prototype has 15 rows with 26 holes in each - and measures about 7 cm by 4 cm (2.6" by 1.5"). The drawing shows the board with PCB mounting terminal blocks. But to save money - and/or space - the wires may be soldered to veropins - or directly to the board itself.

Parts List

Click Here For A Photograph Of The Prototype.

Component Suppliers	Worldwide RS Components	Uk & Ireland Maplin	How To Choose The Right Parts

Construction Notes

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 20 places where the tracks are to be cut. Before you cut the tracks, use the "actual size" drawing to Check The Pattern is correctly marked .

Actual Size

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 - 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

Next fit the 7 resistors and the Six 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 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 Photo Of The Prototype.

Add The Seven Resistors -
The Six Wire Links -
The Diodes and Transistor

The next stage is to fit the diodes and the transistor. Pay particular attention to the orientation of the diodes. Note that D2 faces downwards. Again - they are all drawn lying flat on the board. But those connected between close or adjacent tracks are mounted standing upright. See The Photo

You'll want to mount the LED where it can be seen - perhaps on the front of your alarm-panel. It should be connected to the circuit board using two light flexible wires - e.g. the cores from a piece of alarm cable. The wires 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 together to keep them tidy.

Fit the remaining six components (capacitors, relay and IC socket). Pay particular attention to the orientation of the electrolytic capacitors. They generally have a stripe down the side next to the Negative terminal. Note that C2 & C4 face in opposite directions. See The Photo Of The Prototype. .

Add The Remaining Components
And The Ten 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 10 solder bridges to the underside of the board. The bridges are just small blobs of solder - used to connect adjacent tracks.

Finish off by inserting the Cmos 4011 into the socket. Using a socket reduces the chances of damaging the IC - and makes it easier to replace if necessary. 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 Are Now Ready To Test Your Alarm

Garage / Shed Alarm - Support Material

Circuit Description

Parts List

Construction Notes

Actual Size

External Connections