Free Details Of How To Build An Anti-Hijack Alarm - Using Basic Readily Available Components.

COMMENTS SUGGESTIONS	Hijack Alarm No.3 - Support Material	MORE ALARM CIRCUITS

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Circuit Description

Click Here For A Photograph Of The Prototype.


The Cmos 4001 has four two-input NOR 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.

               *******

The NOR logic is only relevant with respect to gate 1.

The output of a NOR gate will not go high until both of its inputs are low.

See the bottom line of the truth table.

               *******

By controlling each input separately - we can require two different conditions to be satisfied - before pin 3 will go high.

The first condition is that the ignition should be on.

And the second condition is that a door should be opened.

Until both conditions are satisfied simultaneously - pin 3 will remain low.

               *******

While the ignition is off - R2 holds pins 5 & 6 low. 

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

It in turn holds pin 2 high through R5. 

               *******

If a door is opened - and the door switch takes pin 1 low - it will have no effect. 

The gate 1 output will remain low - because the other input pin is high.

               *******

However - when the ignition is switched on - pins 5 & 6 are taken high through D3.

So the output of gate 2 will go low. 

And it in turn will takes pin 2 low - through R5.

               *******

Now - the primary condition is satisfied.

Because the ignition is on - pin 2 is low.

And - because pin 2 is low - the door switch trigger is enabled.

               *******

While all the doors remain closed - R1 holds pin 1 high - and pin 3 remains low. 

But if a door is opened - the door switch will take pin 1 low. 

This means that both conditions are now satisfied.

Both the gate 1 inputs are low - and the output goes high. 

               *******

When pin 3 goes high - it's connected internally to the positive rail - and acts as a source of current.

It has four different jobs to do - the first of which is to latch itself on.

               *******

If you could defeat the alarm by simply closing the door - and allowing R1 to take pin 1 high again - the device would be useless.

Similarly - if you could take pin 2 high again - by simply turning off the ignition - the alarm wouldn't be very effective.

So once the alarm has triggered - we have to force pins 1 & 2 to remain low - regardless of the state of the doors and/or the ignition.

               *******

When Pin 3 goes high - it takes pins 12 & 13 high through R10.

This means that the gate 4 output will go low. 

When pin 11 goes low - it's connected internally to the negative line.

So pin 11 connects pins 1 & 2 to ground - through D4 & D5.

               ******* 

In effect - pin 11 has taken over the jobs of both the door switch and the ignition switch.

It's holding pins 1 & 2 low.

If you close the door or turn off the ignition - it will have no effect on gate 1. 

The alarm has been activated - pin 3 is high - and it will remain high. 

               ******* 

Pin 3's second job is to provide an early warning signal. 

For a brief period - while C4 charges through R6 - pin 3 supplies base current to Q1. 

This switches the transistor on briefly - and it connects the negative lead of the buzzer to ground.

The buzzer will give a short beep to let you know that the alarm has been activated. 

If the activation has been accidental -  the beep will remind you to press the reset button. 

               ******* 

Thirdly - pin 3 provides current to light the LED. 

The main purpose of the LED is to demonstrate that the reset button has worked. 

When the LED switches off - it provides visual reassurance that the circuit has reset. 

               *******

If the beep is kept short - and the LED is small - they will not attract the attention of a thief.

And even if they do - he will not understand their significance.

               ******* 

Finally - current from pin 3 charges C5 slowly - through R8 & R9.

After about 3 minutes - C5 will take pins 8 & 9 high - and the gate 3 output will go low.

This is the period of time during which the thief puts some distance between himself and you.

               ******* 

When the gate 3 output goes low - it's connected internally to the negative rail.

And current flows through the emitter-base junction of Q2 and R11 - into Pin 10.

This switches Q2 on.

Q2 supplies power to the coil of Ry1 - and the relay energizes.

The relay contacts connect the negative leads of the siren and the buzzer to ground - and they both sound. 

               ******* 

D7 allows the relay to sound the buzzer - by connecting the negative lead of the buzzer to ground.

But D7 prevents Q1 from connecting the negative lead of the siren to ground. 

If Q1 connected the negative lead of the siren to ground - the high current would destroy the transistor. 

               ******* 

Ty1 is an SCR - short for Silicon Controlled Rectifier. 

They're also known as Thyristors.

You can think of an SCR as a simple latching switch. 

It switches on when a small current flows into the gate (g) terminal. 

Typically - the size of  the current required is under 0.2mA.

               *******

In standby mode pin 10 is high. 

That is - pin 10 is connected internally to the positive line.

And it provides gate current to Ty1 - through R12.

This means that Ty1 is switched on.

So the negative side of the relay coil is connected to ground - through the SCR.

               *******

When you turn on the ignition you supply power to the positive side of the coil.

So Ry2 will energize. 

When the relay energizes - it should disconnect YOUR cut-out device - and thus allow the engine to start. 

               *******

When the alarm is activated and C5 takes pins 8 & 9 high, pin 10 will go low. 

Therefore - the Ty1 gate current will disappear.

However - this does not mean that Ty1 will switch off.

               *******

Once an SCR switches on - its internal circuitry latches it in the on position.

When this happens - it no longer needs the gate current.

So even after the gate current disappears - Ty1 will stay on all by itself.

Consequently - Ry2 will remain energized - and the engine will continue to run.

               *******

What keeps the SCR switched on - is the current flowing through it.

That current is coming directly from the relay coil.

And the relay coil gets its current from the ignition switch.

So Ty1 will remain switched on until the ignition is switched off.

               *******

When the ignition is switched off - Ry2 will drop-out. 

The current flowing through Ty1 will cease.

And the SCR will switch off.

               *******

Since pin 10 is low - no gate current is available to switch the SCR on again.

And since the SCR is switched off - the negative side of the relay coil is no longer connected to ground.

So when the ignition is switched on again - the relay will not energize.

In other words - while pin 10 remains low - the engine will not re-start.

And pin 10 will remain low until the alarm is reset.

               *******

The reset button takes pins 12 & 13 low.

This causes pin 11 to go high.

When pin 11 goes high - it no longer holds pins 1 & 2 low.

So - if the ignition is off - and/or the doors are closed - pin 3 will go low.

               *******

When pin 3 goes low - the LED will switch off.

C5 will discharge rapidly through R9 & D6.

Pins 8 & 9 will go low - and pin 10 will go high.

               *******

When pin 10 goes high - the Q2 base-current will cease - and the transistor will switch off.

When Q2 switches off - Ry1 drops out - and both the siren and the buzzer are silenced.

               *******

Since pin 10 is now high - the Ty1 gate current is restored.

The SCR is switched on - and the negative side of the relay coil is once more connected to ground.

So the next time you switch on the ignition - Ry2 will energize - and the engine will start.

               *******

The 100nF ceramic capacitors are there to slow down the response time of the inputs.

They also remove stray signals that might cause damage or a false activation. 

               *******

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

D1, D8, D9 & D10 are included to short circuit the spikes at source - before they can do any damage.

               *******

D4 & D5 allow both of the gate 1 inputs to be connected to the same pin - without actually being connected to each other.

In other words - although pins 1 & 2 are both connected to pin 11 - they can still be controlled independently of one another.

D4 & D5 also perform a second function.

They allow a low pin 11 to take both inputs low.

But they prevent a high pin 11 from taking both inputs high. 

               *******

The remaining diodes are used to provide one-way paths. 

When pin 3 goes high - D6 forces C5 to charge slowly through R8 & R9.

This means that it takes about 3-minutes before the siren sounds.

But when the alarm is reset - D6 allows C5 to discharges rapidly through R9 alone. 

               *******

Without D6 - the capacitor would have to discharge through both R8 & R9.

And that would take about three minutes.

But with D6 - the siren delay timer resets almost immediately.

Within a second or so - Ry1 drops out - and the siren is silenced.

               *******

The alarm may be fitted to a wide variety of vehicles – with different electrical systems.

D2 and D3 are not strictly necessary.  I included them as a precaution.

They restrict - to the minimum necessary - the effects the vehicle’s electrical system can have on the alarm circuit, and vice-versa. 

               *******

It may do no harm for the door switch circuitry to take pin 1 high – but it’s not necessary.
 
It may do no harm for the ignition switch circuitry to take pins 5 & 6 low – but it’s not necessary.

Although 1N4148 diodes might be adequate - I felt that for long-term reliability it would be better to use the more robust 1N4001

               *******

D7 allows the relay contacts to connect the negative lead of the buzzer to ground.

But it prevents Q1 from connecting the negative lead of the siren to ground.

In other words - the relay can sound the buzzer - but Q1 cannot sound the siren.

               *******

As well as protecting the transistor from the siren current - D7 prevents the siren from sounding when the buzzer gives its early warning beep.

Again although a 1N4148 might be adequate - I felt that for long-term reliability it would be better to use the more robust 1N4001. 

               *******

Cmos ICs will work up to about 18-volts - but anything over this will destroy them. 

Z1 is there to clamp the supply rail to a maximum of 16-volts.

This protects the IC from any transient high-voltage spikes produced by the vehicle's electrical system. 

               *******

I specified relays with a minimum coil resistance of 270 ohms. 

Use a higher value if you wish.

With a 12-volt supply - the maximum current through Q2 will be about 45mA.

This is well within the limits of the BC557 - which has an IC(max) of 100mA.

It's also well within the limits of practically every SCR.

               *******

There is nothing special about the transistors.

Provided 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.

               *******

For the SCR - I used a TICP106D. 

It's in a TO92 case - and looks like a BC547 transistor.

However - the current, voltage and power requirements are negligible.

So - practically any type of SCR should work fine. 

But again - check the pin configuration.

It may be different from that of the TICP106D.

               *******

SCRs trigger very easily. 

All you need is about 1-volt at the gate - and a current of less than 0.2mA. 

In fact - simply touching the gate pin is sometimes enough to switch them on.

               *******

R13 and C7 slow down the speed at which the gate operates.

R13 ties the gate to ground - and it keeps C7 discharged.

It also increases the current required to trigger the SCR.

More about this later.

As a result - the SCR won't be triggered by stray electrical or radio frequency interference.

In other words - it won't trigger until pin 10 goes high.

               *******

The reset button takes pins 12 & 13 low. 

If the ignition is off and/or all the doors are closed - pin 3 will go low and the alarm will reset.

But at the very instant the reset button is pressed - pin 3 is still high.

The purpose of R10 is to allow the reset button to take pins 12 & 13 low - without shorting the high pin 3 to ground.

               *******

The value of R10 is not critical.

For reasons I'll explain later - I wanted R1 & R12 to be 4k7.

Since the values of R2, R3, R5, R6, R9, R10 & R11 aren't critical - I made them 4k7 also.

               *******

R2 holds pins 5 & 6 low while the ignition is off.

When the ignition is off - pin 4 is high.

R5 prevents a low pin 11 from shorting a high pin 4 to ground - through D5.

And R9 limits the C5 peak discharge current - when the alarm is reset.

               ******* 

The values of R6 & C4 were chosen to give a short beep on the buzzer.

Although other combinations of values will work just as well - I used 4k7 and 22uF.

R3 keeps Q1 switched firmly off until it's needed

               ******* 

Using a 4k7 resistor for R11 gives a Q2 base current of over 2mA.

Ry1 requires a maximum coil current of about 45mA.

So Q2 only need to have a DC gain of about 20.

In other words - practically any PNP transistor will work fine in this position.

               ******* 

The gate of Ty1 only needs to reach 1-volt for the SCR to trigger.

I wanted C7 to have a large capacity - so that the time it takes to charge to 1-volt is relatively long.

That way - short interference pulses will be unable to switch the SCR on.

               *******

At the same time - I wanted to keep the value of R13 relatively low.

The SCR only needs a gate current of about 0.2mA - to take the gate to 1-volt.

But a 1k resistor needs a current of at least 1mA flowing through it - before the voltage across it will reach 1-volt.

When the resistor is connected in parallel with the gate junction - they require a combined current of 1.2mA in order to reach 1-volt.

So a current of about 1.2 mA is now required to switch the SCR on.

Consequently - the sensitivity of the gate is reduced.

And the SCR is less likely to be triggered by extraneous electrical interference.

               ******* 

R12 needs to be small enough to supply the 1.2mA required to trigger the SCR.

It also has to allow the voltage at the junction of R12 & R13 to reach at least 1-volt.

I chose 4k7 because it will supply over 2mA - more than enough current to operate gate.

And the potential divider it forms with the 1k resistor - would allow the junction of R12 & R13 to reach about 2-volts.

So it won't prevent the gate of the SCR from reaching 1-volt, the level it needs to trigger.

               ******* 

Depending on whether the ignition is on or off - and the doors are open or closed - the charge on C3 can be in either direction.

Sometimes pin 1 will be low and pin 4 will be high.

Sometimes pin 4 will be low and pin 1 will be high.

               ******* 

If a polarized capacitor is biased in the reverse direction - it will conduct direct current.

If it conducts enough current - it will heat up and explode.

Although it's highly unlikely that this will happen in the present circumstances - strictly speaking C3 should be non-polarized. 

               *******

Non-Polarised capacitors are not widely available.

However, you can simulate a non-polarised capacitor by connecting two regular capacitors back-to-back.

Since one of the capacitors is always correctly biased - no DC current can flow through the pair.

               *******

When two capacitors are connected in series - their combined capacitance is reduced.

The formula for capacitors connected in series - is similar to the formula for resistors connected in parallel.

Where just two capacitors are connected in series - the formula may be expressed as : -

   (The Product ÷ The Sum)
                      Or 
       (Ca x Cb) ÷ (Ca + Cb)

So two 22uF capacitors - connected in series - have a combined capacity of : -

         (22 x 22) ÷ 44 
             = 11uF

               *******

When - as here - both capacitors have the same value - there's a simpler calculation.

The combined capacitance is equal to half the value of one capacitor.

So two 22uF capacitors - connected in series - have a combined capacity of : - 

           22 ÷ 2 = 11uF.

               *******

In standby mode C3 is in a discharged state.

When the ignition is switched on - pin 4 goes low.

That is - pin 4 is connected internally to the negative rail.

For a brief moment - pin 1 is also connected to ground through C3, R4 & pin 4.

In other words - pin 4 takes pin 1 low.

This simulates an open door - and the alarm triggers.

               *******

The reason for using a capacitor is that we only want pin 1 to go low briefly.

If R4 were connected directly to pin 1 - pin 4 would take pin 1 permanently low.

So the alarm could not be reset.

But the introduction of C3 means that pin 4 can only take pin 1 briefly low.

Pin 4 will only take pin 1 low for the length of time it takes R1 to charge C3.

When R1 charges C3 - pin 1 will go high again.

               *******

A 1uF capacitor gave a long enough pulse to trigger the prototype.

So the specified 10uF capacitor is more than large enough to trigger the alarm reliably.

If you don't want the alarm to trigger every time the ignition is switched on - leave out C3 and R4.

               *******

R4 is needed to prevent the opposite from happening.

Without R4 - every time the door is opened the alarm will trigger - even when the ignition is off.

The reason for this is that the door switch takes pin 2 low - through C3 and R5.

The cause of the problem is the impedance - or internal resistance - of the gate 2 output pin.

               *******

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

It's not connected directly to the positive line.

It's connected through a metal-oxide transistor.

And the transistor offers some resistance to the passage of current.

Typically - it's the equivalent of about 200 ohms.

               *******

So you can think of a Cmos gate as having an internal 200 ohm resistor in series with its output pin.

Without R4 - when the door switch is closed - virtually the whole of the supply voltage will be dropped across the internal resistance.

In other words - pin 4 will go briefly low.

Pin 4 will take pin 2 low - through R5.

So both the gate 1 inputs will be low - and the alarm will activate.

               *******

The purpose of R4 is to create a potential divider with the internal resistor.

When the door switch takes R4 low - through C3 - the voltage will be dropped across both R4 and the internal resistance.

Roughly speaking - over 80% of the voltage will drop across R4 - and less than 20% will drop across the internal resistance.

This means that pin 4 will only drop to about 10-volts.

Consequently - it will not take pin 2 low - and the alarm will not activate.

               *******

Since we have introduced a resistor in series with pin 4 - we have to consider its effect in both directions.

We want pin 4 to be able to take pin 1 low - through C3.

That's what makes the alarm activate when you turn on the ignition.

               *******

R1 can't have a very high resistance.

If it did - condensation or other moisture on the door switches might succeed in taking pin 1 low - and cause a false alarm.

But the value of R1 can't be too low either.

R4 and the pin 4 internal resistance - form a potential divider with R1.

So the value of R1 has to be such that over half the supply voltage will drop across it.

Otherwise - pin 4 can't take pin 1 low.

               *******

4k7 is about four times the value of R4 and the internal resistance combined.

This means that about 80% of the supply voltage will drop across R1.

And the remaining 20% will drop across R4 and the pin 4 resistance.

In other words - because R1 is 4k7 - when pin 4 goes low - R4 will take pin 1 down to just over 2-volts.

This is more than low enough to activate the alarm.

               *******

Click Here To Learn More About The Cmos 4001

Parts List

Suppliers

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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 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 - 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 Eleven 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 Eleven Wire Links

Now fit the 13 resistors. 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 Thirteen
Resistors

Next fit the 2 transistors, the 10 diodes, the 16v zener and the thyristor. The pnp transistor - BC557 - is the one with the emitter arrow coloured red. If you are using transistors other than the BC547 and BC557 - check their pin configuration before soldering them in place. Just because they look the same - don't assume that they have the same pin configuration.

Pay particular attention to the orientation of the diodes. Again - they're 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.

The positions of the kathode, anode and gate connections are marked on the stripboard layout. I used a TICP106D in the prototype because it's widely available. You'll probably find that any SCR - 30-volts or better - will work fine. But remember that the pin configuration of your SCR may be different from that of the TICP106D.

Add The Semiconductors

Fit the capacitors, the IC socket and the two relays. Pay particular attention to the orientation of the electrolytic capacitors. Note that electrolytic capacitors generally have a stripe down the side next to the negative terminal. See the Photograph Of The Prototype.

Add The Remaining Components

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 12 solder bridges.

Add The Twelve 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

Veroboard Layout	The Rest of Ron's Circuits	Write To Ron	More Free-to-Use Circuits	Circuit Exchange International

Hijack Alarm No.3 - Support Material

Construction Guide

Actual Size Of Pattern