When this alarm is activated - the siren will sound for up to 20-minutes. Then it will switch off - and remain off. It will not sound a second time.

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Circuit Description

Click Here For A Photograph Of The Prototype.

The Cmos 4001 has four - independent - two-input  NOR gates.

The four gates are divided into two pairs.

Gates 1 & 2 are wired together to form a type of latching circuit.

When the alarm is triggered - they will latch - and sound the buzzer.

Gates 3 & 4 are wired to form a monostable.

When the entry delay expires - the monostable will sound the siren for a fixed length of time.

Then the siren will switch off - and stay off.

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Cmos 4001 gates only have a high output when both inputs are low.

See the truth table.

When you switch the alarm on - C2 is in a discharged state.

Because it's discharged - the capacitor acts like a wire link connecting pins 2 & 13 to the positive line.

This means that gates 1 & 4 each have one high input.

These high inputs - force the output pins of gates 1 & 4 to remain low.

And while the output pins of gates 1 & 4 are forced to remain low - both the latching circuit and the monostable are disabled.

In other words - the alarm cannot be activated.

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C2 will continue to hold pins 2 & 13 high - until it charges through R3.

This is the exit delay.

Depending on the setting of R3 - it will last for up to a minute or more.

During this period - you may open the loop and leave the building without activating the alarm.

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As C2 charges - the voltage on the negative terminal of the capacitor falls.

When it falls to just below half the supply voltage - it takes pins 2 & 13 low. 

When pins 2 & 13 go low - the outputs of gates 1 & 4 are no longer forced to remain low.

This means that both the latch and the monostable circuits are enabled.

In other words - the exit delay is over - and it's now possible to activate the alarm.

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In standby mode - the normally closed loop connects pin 6 to ground.

And the current through R1 is also diverted to ground through the loop.

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

R1 quickly charges C1 - and the voltage on the capacitor takes pin 6 high.

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Because one of the gate 2 inputs goes high - its output at pin 4 will go low. See the truth table.

When pin 4 goes low - current flows through the emitter-base junction of Q1 and R2 - into pin 4.

This switches Q1 on.

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

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Pin 4 is connected to pin 1.

So when pin 4 goes low - it takes pin 1 low.

You'll recall that R3 is already holding pin 2 low

So when pin 1 goes low - the gate 1 output at pin 3 goes high. 

See the truth table.

When pin 3 goes high - it does two different jobs.

It latches itself on - and it starts the entry delay timer.

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We want pin 3 to remain high - even after the door has been closed and the loop restored.

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

So pin 3 must latch itself on.

It does this by taking pin 5 high.

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For pin 3 to remain high - pin 1 must remain low.

For pin 1 to remain low - pin 4 must remain low.

For pin 4 to remain low - at least one of the gate 2 inputs must remain high.

See the truth table.

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Pin 3 is connected to pin 5.

When pin 3 goes high - it takes pin 5 high.

So pin 3 is holding one of the gate 2 inputs high.

While at least one of the gate 2 inputs is high - pin 4 will remain low.

And while pin 4 holds pin 1 low - pin 3 will remain high.

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In other words - pin 3 has latched itself on.

It no longer needs R1 to hold pin 6 high.

Because pin 3 is holding pin 5 high - pin 1 will remain low - and pin 3 will remain high.

It makes no difference now whether the loop is open or closed.

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Pin 3's second job is to charge C4 through R4.

While pin 3 is low - it holds pin 8 low through R4. 

It also keeps C4 discharged.

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When pin 3 goes high - it begins to charge C4 slowly - through R4.

This is the entry delay.

Depending on the setting of R4 - the entry delay will last for up to about 60-seconds or more.

During this time - you will normally move Sw1 to the off position.

If you fail to do so - when the voltage on C4 reaches just over half the supply voltage - it will take pin 8 high and trigger the siren output.

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Gates 3 & 4 control the relay.

They are wired together to form a monostable.

If you'd like a detailed account of how a monostable works - click on the link below the circuit diagram.

Basically - a monostable will provide an output for a certain length of time - and then switch itself off.

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When C4 takes pin 8 high - pin 10 will go low.

Because C5 is discharged - it acts like a wire link between pins 10 & 12.

So pin 10 takes pin 12 low - through the capacitor.

You'll recall that R3 is already holding pin 13 low.

And - since both of its inputs are now low - the gate 4 output will go high.

It will remain high until C5 charges through R5.

Depending on the setting of R5 - it will remain high for up to about 20-minutes or so.

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When pin 11 goes high - it supplies base current to Q2 through R7.

This switches the transistor on.

Q2 connects the negative side of the relay coil to ground.

This causes the relay to energize

And when the relay contacts close - they supply power to the siren.

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When the monostable triggers - C5 begins to charge through R5

This is the siren cut-off timer.

When the voltage on C5 reaches just over half the supply voltage - it will take pin 12 high.

Since one of its inputs is now high - the output at pin 11 will go low.

See the truth table.

When pin 11 goes low - the Q2 base current ceases.

So the transistor will switch off - the relay will de-energize - and the siren will be silenced.

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The siren will not sound a second time because the output monostable cannot be re-triggered.

The output monostable cannot be re-triggered because C5 is charged.

The monostable will only trigger if C5 is in a discharged state.

If it's not in a discharged state - it can't take pin 12 low - and switch the gate 4 output on.

In other words - while C5 remains charged - the monostable cannot be re-triggered.

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A discharged capacitor has no potential difference across its plates.

You can discharge a capacitor in two different ways.

You can remove the positive charge from the positive plate - or you can apply a positive charge to the negative plate.

Either way - the result is the same.

Both plates of the capacitor are at the same potential.

And therefore - the capacitor is discharged.

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The monostable uses the second method to discharge C5.

When pin 10 goes high - it applies a positive charge to the negative plate.

But - Pin 10 cannot go high until both of the gate 3 inputs go low.

So - while C4 holds pin 8 high - C5 will not discharge.

And C4 will hold pin 8 high indefinitely - because the latching circuit is holding pin 3 high - and pin 3 is keeping C4 charged.

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When you switch the alarm on - the Buzzer will warn you if the loop is open.

If the loop is open - pin 6 is high and pin 4 is low.

Pin 4 connects the base of Q1 to ground.

This switches the transistor on - and the Buzzer sounds.

If the Buzzer sounds when you switch the alarm on - the loop must be open.

Switch off again and check for open doors and windows.

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If the loop is closed - pin 6 is low and pin 4 is high.

Because pin 4 is high - Q1 is switched off - and the Buzzer is silent.

If the Buzzer is silent - the loop is intact - and it's safe to leave the building.

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When you open the door to leave - pin 6 will go high - pin 4 will go low - Q1 will switch on - and the Buzzer will sound.

If you close the door before the exit delay expires - the Buzzer will switch off.

If you don't close the door in time - the Buzzer will continue to sound - the entry delay will begin - and the alarm will proceed to full activation.

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When Sw1 is moved to the "off" position - it discharges C2 rapidly - through R8 and D2.

It also discharges C4 through D3 and R8.

This resets both the exit and the entry delays - and the alarm is ready for immediate re-use.

R8 limits the peak discharge current through Sw1 and the diodes.

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I could simply have connected R8 across the supply rails.

But that would have increased the standby current significantly.

By connecting it to the positive rail through the switch - it's only in circuit when the alarm is off.

That is - while the alarm is disconnected from the battery.

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

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

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D4 & R6 are there to sharpen the point at which the relay drops out.

Although Cmos 4001 gates are digital devices - there are times when they behave like analogue devices.

When timing components are used - the voltage on the inputs rises and falls slowly.

It may hover around the gate's switching point for a significant period of time. 

During this period - the output voltage will rise or fall in an analogue manner. 

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This means that any device connected to the output will switch on or off - slowly.

The effect is particularly noticeable where the timer circuit is set up to provide long delays. 

It can cause relays to rattle - and buzzers to fade in and out slowly.

The monostable may even re-trigger.

To prevent this from happening - we need a little positive feedback.

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As C5 charges through R5 - and pin 12 starts to go high - pin 11 begins to go low.

As pin 11 starts to go low - Q2 begins to switch off.

As Q2 starts to switch off - the voltage on its collector begins to rise.

As the voltage on the Q2 collector rises - it charges C5 through D4 and R6. 

This speeds up the rate at which C5 charges.

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In other words - D4 & R6 create an avalanche effect.

Once Q2 starts to switch off - D4 & R6 push it over the top.

At the last minute - they take over from R5 - and rapidly complete the charge of C5.

This causes the relay to drop out quickly and cleanly.

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D4 is there to create a one-way path.

It allows a high collector voltage to charge C5.

But when Q2 is switched on - it prevents the transistor from taking pin 12 low.

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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 33mA.

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

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There is nothing special about the transistors.

So long as the NPN/PNP requirement is satisfied - any small 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 and BC557.

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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 2 or 3 microamps - i.e. just the current that flows through R1 and the normally-closed loop.

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

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

We can use such a high value resistor because Cmos gates are operated by voltage - and not by current.

Cmos inputs have a very high internal resistance (impedance).

So virtually no current flows into the input pins.

Instead - the mere presence of a voltage at the input pins - is enough to switch the gate.

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

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

The moisture may connect pin 6 to ground - and create a potential-divider with R1.

This may prevent pin 6 from reaching a high enough voltage to trip the alarm. 

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

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The wires going to the doors and windows act like radio antennas.

They pick up interference that could cause false alarms.

C1 short circuits the high frequency signals to ground.

C3 does a similar job for the power leads.

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It's only when the relay is energized - 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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How A Cmos 4001 Monostable Works

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 25 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 Fifteen 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.

Add The Fifteen Wire Links

The next stage is to fit the six diodes, the 5 fixed resistors - and the 3 presets. To fit the presets you may need to enlarge the holes slightly. Pay particular attention to the orientation of the diodes. Both the resistors and diodes are drawn lying flat on the board. However, those connected between close or adjacent tracks are mounted standing upright. See the Photo Of The Prototype.

Six Diodes - Five Resistors - Three Presets

Six Diodes - Five Resistors - Three Presets

Fit the remaining components - the five capacitors - the two transistors - the IC socket - and the relay. The PNP transistor - BC557 - is the one with its emitter arrow coloured red. 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 the two 22uF capacitors are mounted with their negative terminals facing the top of the board. See the Photo Of The Prototype.

Two Transistors
Five Capacitors -
IC Socket and Relay

Two Transistors
Five Capacitors -
IC Socket and Relay

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 6 solder bridges.

Finish off by inserting the Cmos 4001 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

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How To Test The Finished Alarm

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Actual Size Of Pattern

External Connections

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