This is an alarm with an adjustable bell cut-off timer. The bell will sound for up to about 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.
	OVERVIEW =========== The Cmos 4060 is a 14-bit binary counter. So counting is central to the way the circuit operates. ***** The sequence begins when you power up the alarm. First of all - the 4060 counts up to eight. While it's counting up to eight - the alarm cannot be activated. Therefore - the time it takes for the count to reach eight - is the exit delay. *** When the count reaches eight - it pauses. At this stage - the exit delay is over - and the system is armed. *** When the loop is opened - the alarm is activated. The Buzzer sounds - and the 4060 continues the count. The time it takes to count on from eight to sixteen - is the entry delay. Before the count reaches sixteen - you will normally switch the alarm off. *** If the alarm is not switched off before the count reaches sixteen - the relay will energize and the Siren will sound. *** While the Siren is sounding - the 4060 continues the count beyond sixteen. It carries on counting until it reaches a total of 256. The time it takes to reach 256 - is the Siren Cut-Off time. *** When the 4060 reaches 256 - the count will stop. At the same time - the relay will drop out - and the Siren will be silenced. *** The alarm cannot be re-activated until the counter is reset. And the counter will not reset until the alarm has first been switched off - and then switched on again. ABOUT THE 4060 =============== The 4060 is a basically a 14-bit binary counter. We are only using it to count up to 256. But it's capable of counting up to 16 384. That is: 2 raised to the power 14. *** The various bits of the counter are connected internally to the output pins. You'll see from the diagram that bit-12 (Q12) is connected to pin 1. Bit-13 (Q13) is connected to pin 2 etc. Although the 4060 is a 14-bit counter - only 10 of those bits are available as outputs. Q1, Q2, Q3 & Q11 do exist - but they are not connected to output pins. *** Because the counter has to have something to count - the Cmos 4060 comes with two built-in inverters. They are connected in series across pins 11, 10 & 9. Together with R5, R6, R7 & C4 - the inverters form a simple oscillator. *** While the oscillator is running - pin 9 switches continuously from high to low and vice-versa. It's rather like the swing of a pendulum - or the ticking of a clock. Pin 9 is connected internally to the binary counter - and the counter counts the ticks. THE EXIT DELAY =============== The exit delay starts automatically when you turn the power on. It lasts until the count reaches eight. That is - it lasts until pin 7 goes high. *** During this period - the oscillator is running. And every time pin 9 goes low - it's connected internally to the negative supply line. *** This means that current will flow to ground - through the emitter-base junction of Q2, R10 and pin 9. In other words - every time pin 9 goes low - Q2 will switch on. When Q2 switches on - it connects the negative lead of the Buzzer to ground - and the Buzzer sounds. *** The Buzzer will sound eight times - and then stop. That's the number of oscillations required to take pin 7 high. *** When pin 7 goes high it does two jobs - the first of which is to stop the oscillator. It stops the oscillator by taking pin 11 high - through R3 and D8. Stopping the oscillator stops the count. It also stops the Buzzer. *** Pin 7's second job is to act as a source of trigger current for Ty1. Until pin 7 goes high - no trigger current is available for the SCR. So - you can open the loop without activating the alarm. *** It's only after pin 7 goes high that trigger current becomes available - through R4 and R8. Consequently - the length of the exit delay - is the length of time it takes for pin 7 to go high. In other words - it's the length of time it takes for the counter to count up to eight. *** R6 controls the speed of the oscillator. Therefore - it controls the speed at which the count progresses. Consequently - it controls the length of the exit delay. *** When pin 7 goes high the count is paused at eight. The exit delay has expired - and the alarm is in standby mode. If the loop is opened - pin 7 will supply the current necessary to trigger Ty1. THE SCR ======== Ty1 is an SCR - short for Silicon Controlled Rectifier. They are 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 about 0.2mA. *** When the SCR switches on - its internal circuitry latches it in the on position. It no longer needs the gate current. Even if the gate current ceases - the SCR will stay on all by itself. THE TRIGGER CIRCUIT ==================== The alarm's trigger circuit consists of R4, R8 and the normally-closed loop. While the loop remains closed - it diverts the R4 current to ground. Consequently - no current will flow into the gate terminal of Ty1 - through R8. *** If the loop is opened - even for an instant - the R4 current will enter the gate terminal - through R8. This will switch the SCR on. It doesn't matter now if the loop is restored and the gate current ceases. The SCR is switched on - and it will stay on all by itself. *** Ty1 does two jobs. Firstly - it connects the negative lead of the Buzzer to ground - through D10. So the Buzzer sounds. *** The second job is to connect the junction of R3 and D8 to ground - through D9. When it takes this junction low - the SCR destroys pin 7's ability to hold pin 11 high. So pin 7 can no longer pause the oscillator. In other words the - count starts up again from where it left off. THE ENTRY DELAY ================ The entry delay lasts until pin 5 goes high. This requires a further eight oscillations. Since the speed of the oscillator is controlled by R6 - the entry delay is more or less the same length as the exit delay. *** During the entry delay - pin 9 will continue to switch Q2 on and off - as it did during the exit delay. But this will have no effect on the Buzzer. Because Ty1 has provided it with an alternative connection to ground - through D10 - the Buzzer is already sounding continuously. THE SIREN OUTPUT ================= When pin 5 goes high - it's connected internally to the positive line. And it supplies base-current to Q1 - through D4 and R2. This switches the transistor on. *** Q1 connects the negative side of the relay coil to ground. This causes the relay to energize. And the relay contacts deliver power to the Siren. *** It has taken a total count of 16 for pin 5 to go high. Pin 5 is a binary bit. So - after a further count of 16 - pin 5 will go low again. This would cause the relay to drop out. And the Siren Cut-Off time would be too short. In other words - we cannot rely on pin 5 alone. *** That's why pins 4, 5, 6 and 14 are all connected to the base of Q1 - through R2. Until the count reaches 256 - at any given time at least one of these four pins will be high. So at least one of the four pins will be supplying base-current to Q1. In other words - until the count reaches 256 - the relay will remain energized and the Siren will continue to sound. *** If you've set the exit delay at about 30-seconds - this means that the counter will take 30-seconds to count up to eight. Since the entry delay is also an eight count - it too will last about 30 seconds. And the subsequent count from 16 to 256 will take about ((256-16) ÷ 8) x 30 sec = 15 min. *** At this point - pins 4, 5, 6 and 14 will all be low. And none will be supplying base-current to Q1. So the transistor will switch off - the relay will drop out - and the Siren will be silenced. THE RESET =========== When the count reaches 256 - pin 13 goes high. Pin 13 is connected to pin 11 through D7. Just as pin 7 stopped the count temporarily at eight - pin 13 stops it permanently at 256. *** The Buzzer will continue to sound because it's connected to ground through D10 and Ty1. But Q1 is switched off and the relay is de-energized - so the Siren is silent. It can't sound again until the next time the alarm is switched on - and the counter is reset. *** To reset the counter to zero - pin 12 must first be taken high - and then taken low. At the exact moment that pin 12 goes low - the counter resets. That is - all the outputs go low - and the counter begins to count up from zero. *** The purpose of C1 and R1 is to perform an automatic reset when power is applied. That way - we can be sure that count will always start at zero. Initially - at power-up - C1 is discharged. In a discharged state it acts like a wire link connecting pin 12 to the positive rail. That is - it takes pin 12 high. Almost immediately - C1 charges through R1 - and the resistor takes pin 12 low. This causes all of the outputs to go low - and the count begins at zero. In other words - the exit delay begins. THE OSCILLATOR =============== Note that R5, R6 & C4 are connected in series between the outputs of the two inverters. If the cycle starts with pin 11 low - then pin 10 will be high - and pin 9 will be low. So the capacitor will begin to charge from pin 10 - through R5 & R6. The speed at which it charges depends on the value of the capacitor - and the setting of R6. As it charges - the voltage at the junction of C4 & R6 will rise. When it reaches just over half the supply voltage - it takes pin 11 high - through R7. *** Now everything reverses. Pin 11 is high - so pin 10 is low - and pin 9 is high. C4 begins to charge again - only this time it's in the opposite direction. As Pin 9 charges the capacitor - the voltage at the junction of C4 & R6 will fall. The speed at which it falls depends on the value of the capacitor - and the setting of R6. When it falls to just below half the supply voltage - it takes pin 11 low - through R7. With pin 11 low - the cycle is complete -and the next cycle can begin. *** The oscillator runs because the polarity of the charge on C4 keeps reversing. As a result - the junction of C4 & R6 keeps taking pin 11 back and forth between high and low. And since pin 11 keeps changing back and forth between high and low - the polarity of the charge on C4 keeps reversing. It's a sort of controlled feedback - where the frequency generated is determined by the values of R5, R6 & C4. *** The main purpose of R5 is to set an upper limit on the frequency. Its value was chosen to provide a minimum exit/entry period of about 10 seconds. The value of R6 was chosen to provide a maximum exit/entry period of about a minute. *** The oscillator stops when D7 or D8 take pin 11 high. The diodes allows a high output pin to take pin 11 high. But - prevent a low output pin from holding pin 11 low. By clamping pin 11 high - the diodes forces pin 10 to remain low. So pin 9 has to remain high. *** Since pin 11 can no longer change its state - the polarity of the charge on C4 cannot change. So the oscillator cannot run. *** Note that R5, R6 & R7 are connected in series between pins 10 and 11. They join the output of the first inverter to its own input. The value of R7 needs to be very high. That way - any influence the output of the first inverter has on its own input will be very small. It will be swamped by the influence of the changing charges on C4. *** The data sheet for the 4060 says that the value of R7 should be at least ten times the value of R5 + R6. That would make it about 3M3. The value of 4M7 was chosen because it's widely available. It's possible to use such high value resistors because Cmos inputs are operated by voltage - and not by current. Because no current flows through the 4M7 resistor - there's no voltage drop across it. *** When polarized capacitors are biased in the reverse direction - they will conduct direct current. Since the charge on C4 is constantly reversing in polarity - the capacitor needs to be non-polarized. If you use a polarised capacitor in this position - the oscillator may not run - or it may not run reliably. *** 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. THE COMPONENTS =============== 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 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 0.2mA. In fact - simply touching the gate pin is sometimes enough to switch them on. *** 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'. R8 and C3 slow down the speed at which the gate operates. So it won't be triggered by electrical interference. *** To remain switched on - the SCR must have a certain minimum current flowing through it. This is known as the holding current. Some Buzzers require a very small current that may be insufficient to keep the SCR switched on. And - even if it is sufficient - Q2 repeatedly diverts that current to ground. So - it's not constant - and it cannot be used to keep the SCR switched on. *** The purpose of R9 is to provide an independent - and constant - source of holding current. The current through R9 guarantees that - once the SCR has switched on - it will stay on. If you're using a lower supply voltage - and you find that your SCR is not latching - reduce the value of R9. *** I specified a relay with a minimum coil resistance of 270 ohms. Use a higher value if you wish. With a 12-volt supply - the maximum current through Q1 will be about 45mA. This is well within the limits of the BC547 - which has an IC(max) of 100mA. *** 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. *** Relay coils and some Sounders produce large reverse voltage spikes that will destroy Cmos ICs. D1, D2 & D11 are included to short circuit the spikes at source - before they can do any damage. *** The remaining diodes are used to create one-way paths. D3, D4, D5 & D6 allow any and all of the four output pins to supply base current to Q1. At any given time - some of these pins will be high - and some will be low. The four diodes prevent the high and low outputs from shorting together. *** D7 allows a high pin 13 to take pin 11 high - but it prevents a low pin 13 from holding pin 11 low. D8 allows a high pin 7 to take pin 11 high - but it prevents a low pin 7 from holding pin 11 low. *** When the SCR is triggered - D9 allows it to take the junction of R3 & D8 low. But while the SCR is off - D9 prevents R9 from taking pin 11 high - through D8. *** D10 allows Ty1 to connect the negative lead of the Buzzer to ground. But it prevents Q2 from connecting the anode of Ty1 to ground. If Q2 connected the anode of Ty1 to ground - the SCR would unlatch. *** D10 has to carry the Buzzer current. A 1N4148 would probably be fine in this position. But - in the interest of long term reliability - I specified the more robust 1N4001. *****

Circuit Description

Click Here For A Photograph Of The Prototype.

Circuit Diagram For A
Cmos 4060 Burglar Alarm


                 OVERVIEW
                ===========

The Cmos 4060 is a 14-bit binary counter. 

So counting is central to the way the circuit operates.

                     *******
The sequence begins when you power up the alarm.

First of all - the 4060 counts up to eight.

While it's counting up to eight - the alarm cannot be activated.

Therefore - the time it takes for the count to reach eight - is the exit delay.

                     *******

When the count reaches eight - it pauses.

At this stage - the exit delay is over - and the system is armed.

                     *******

When the loop is opened - the alarm is activated.

The Buzzer sounds - and the 4060 continues the count.

The time it takes to count on from eight to sixteen - is the entry delay.

Before the count reaches sixteen - you will normally switch the alarm off. 

                     *******

If the alarm is not switched off before the count reaches sixteen - the relay will energize and the Siren will sound.

                     *******

While the Siren is sounding - the 4060 continues the count beyond sixteen.

It carries on counting until it reaches a total of 256.

The time it takes to reach 256 - is the Siren Cut-Off time.

                     *******

When the 4060 reaches 256 - the count will stop.

At the same time - the relay will drop out - and the Siren will be silenced.

                     *******

The alarm cannot be re-activated until the counter is reset.

And the counter will not reset until the alarm has first been switched off - and then switched on again.


            ABOUT THE 4060
           ===============

The 4060 is a basically a 14-bit binary counter.

We are only using it to count up to 256.

But it's capable of counting up to 16 384.

That is: 2 raised to the power 14.

                     *******

The various bits of the counter are connected internally to the output pins.

You'll see from the diagram that bit-12 (Q12) is connected to pin 1. Bit-13 (Q13) is connected to pin 2 etc.

Although the 4060 is a 14-bit counter - only 10 of those bits are available as outputs.

Q1, Q2, Q3 & Q11 do exist - but they are not connected to output pins.

                     *******

Because the counter has to have something to count - the Cmos 4060 comes with two built-in inverters.

They are connected in series across pins 11, 10 & 9.

Together with R5, R6, R7 & C4 - the inverters form a simple oscillator.

                     *******

While the oscillator is running - pin 9 switches continuously from high to low and vice-versa. 

It's rather like the swing of a pendulum - or the ticking of a clock.

Pin 9 is connected internally to the binary counter - and the counter counts the ticks. 


             THE EXIT DELAY
            ===============

The exit delay starts automatically when you turn the power on.

It lasts until the count reaches eight.

That is - it lasts until pin 7 goes high.

                     *******

During this period - the oscillator is running.

And every time pin 9 goes low -  it's connected internally to the negative supply line.

                     *******

This means that current will flow to ground - through the emitter-base junction of Q2, R10 and pin 9.

In other words - every time pin 9 goes low - Q2 will switch on.

When Q2 switches on - it connects the negative lead of the Buzzer to ground - and the Buzzer sounds.

                     *******

The Buzzer will sound eight times - and then stop.

That's the number of oscillations required to take pin 7 high.

                     *******

When pin 7 goes high it does two jobs - the first of which is to stop the oscillator.

It stops the oscillator by taking pin 11 high - through R3 and D8.

Stopping the oscillator stops the count.

It also stops the Buzzer.

                     *******

Pin 7's second job is to act as a source of trigger current for Ty1. 

Until pin 7 goes high - no trigger current is available for the SCR.

So - you can open the loop without activating the alarm.

                     *******

It's only after pin 7 goes high that trigger current becomes available - through R4 and R8.

Consequently - the length of the exit delay - is the length of time it takes for pin 7 to go high.

In other words - it's the length of time it takes for the counter to count up to eight.

                     *******

R6 controls the speed of the oscillator.

Therefore - it controls the speed at which the count progresses.

Consequently - it controls the length of the exit delay.

                     *******

When pin 7 goes high the count is paused at eight.

The exit delay has expired - and the alarm is in standby mode.

If the loop is opened  - pin 7 will supply the current necessary to trigger Ty1.

 
                  THE SCR
                  ========

Ty1 is an SCR - short for Silicon Controlled Rectifier. 

They are 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 about 0.2mA.

                     *******

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

It no longer needs the gate current.

Even if the gate current ceases - the SCR will stay on all by itself.


       THE TRIGGER CIRCUIT
       ====================

The alarm's trigger circuit consists of R4, R8 and the normally-closed loop.

While the loop remains closed - it diverts the R4 current to ground.

Consequently - no current will flow into the gate terminal of Ty1 - through R8.

                     *******

If the loop is opened - even for an instant - the R4 current will enter the gate terminal - through R8.

This will switch the SCR on.

It doesn't matter now if the loop is restored and the gate current ceases.

The SCR is switched on - and it will stay on all by itself.

                     *******

Ty1 does two jobs.

Firstly - it connects the negative lead of the Buzzer to ground - through D10.

So the Buzzer sounds.

                     *******

The second job is to connect the junction of R3 and D8 to ground - through D9.

When it takes this junction low - the SCR destroys pin 7's ability to hold pin 11 high.

So pin 7 can no longer pause the oscillator.

In other words the - count starts up again from where it left off.


          THE ENTRY DELAY
          ================

The entry delay lasts until pin 5 goes high.

This requires a further eight oscillations.

Since the speed of the oscillator is controlled by R6 - the entry delay is more or less the same length as the exit delay.

                     *******

During the entry delay - pin 9 will continue to switch Q2 on and off - as it did during the exit delay.

But this will have no effect on the Buzzer.

Because Ty1 has provided it with an alternative connection to ground - through D10 - the Buzzer is already sounding continuously.


          THE SIREN OUTPUT
          =================

When pin 5 goes high - it's connected internally to the positive line.

And it supplies base-current to Q1 - through D4 and R2.

This switches the transistor on.

                     *******

Q1 connects the negative side of the relay coil to ground.

This causes the relay to energize.

And the relay contacts deliver power to the Siren.

                     *******

It has taken a total count of 16 for pin 5 to go high. 

Pin 5 is a binary bit.

So - after a further count of 16 - pin 5 will go low again.

This would cause the relay to drop out.

And the Siren Cut-Off time would be too short.

In other words - we cannot rely on pin 5 alone.

                     *******

That's why pins 4, 5, 6 and 14 are all connected to the base of Q1 - through R2.

Until the count reaches 256 - at any given time at least one of these four pins will be high.

So at least one of the four pins will be supplying base-current to Q1.

In other words - until the count reaches 256 - the relay will remain energized and the Siren will continue to sound.

                     *******

If you've set the exit delay at about 30-seconds - this means that the counter will take 30-seconds to count up to eight.

Since the entry delay is also an eight count - it too will last about 30 seconds.

And the subsequent count from 16 to 256 will take about 

         ((256-16) ÷ 8) x 30 sec 
                 = 15 min.

                     *******

At this point - pins 4, 5, 6 and 14 will all be low.

And none will be supplying base-current to Q1.

So the transistor will switch off - the relay will drop out - and the Siren will be silenced.


                THE RESET
              ===========

When the count reaches 256 - pin 13 goes high.

Pin 13 is connected to pin 11 through D7.

Just as pin 7 stopped the count temporarily at eight - pin 13 stops it permanently at 256.

                     *******

The Buzzer will continue to sound because it's connected to ground through D10 and Ty1.

But Q1 is switched off and the relay is de-energized - so the Siren is silent.

It can't sound again until the next time the alarm is switched on - and the counter is reset.

                     *******

To reset the counter to zero - pin 12 must first be taken high - and then taken low.

At the exact moment that pin 12 goes low - the counter resets.

That is - all the outputs go low - and the counter begins to count up from zero. 

                     *******

The purpose of C1 and R1 is to perform an automatic reset when power is applied.

That way - we can be sure that count will always start at zero.

Initially - at power-up - C1 is discharged. 

In a discharged state it acts like a wire link connecting pin 12 to the positive rail.

That is - it takes pin 12 high. 

Almost immediately - C1 charges through R1 - and the resistor takes pin 12 low. 

This causes all of the outputs to go low - and the count begins at zero.

In other words - the exit delay begins.


        THE OSCILLATOR
        =============== 

Note that R5, R6 & C4 are connected in series between the outputs of the two inverters. 

If the cycle starts with pin 11 low - then pin 10 will be high - and pin 9 will be low.

So the capacitor will begin to charge from pin 10 - through R5 & R6. 

The speed at which it charges depends on the value of the capacitor - and the setting of R6. 

As it charges - the voltage at the junction of C4 & R6 will rise. 

When it reaches just over half the supply voltage - it takes pin 11 high - through R7.

                     *******

Now everything reverses. 

Pin 11 is high - so pin 10 is low - and pin 9 is high. 

C4 begins to charge again - only this time it's in the opposite direction.

As Pin 9 charges the capacitor - the voltage at the junction of C4 & R6 will fall. 

The speed at which it falls depends on the value of the capacitor - and the setting of R6. 

When it falls to just below half the supply voltage - it takes pin 11 low - through R7. 

With pin 11 low - the cycle is complete -and the next cycle can begin.


                     *******

The oscillator runs because the polarity of the charge on C4 keeps reversing. 

As a result - the junction of C4 & R6 keeps taking pin 11 back and forth between high and low. 

And since pin 11 keeps changing back and forth between high and low - the polarity of the charge on C4 keeps reversing. 

It's a sort of controlled feedback - where the frequency generated is determined by the values of R5, R6 & C4.

                     *******

The main purpose of R5 is to set an upper limit on the frequency. 

Its value was chosen to provide a minimum exit/entry period of about 10 seconds. 

The value of R6 was chosen to provide a maximum exit/entry period of about a minute.

                     *******

The oscillator stops when D7 or D8 take pin 11 high. 

The diodes allows a high output pin to take pin 11 high.

But - prevent a low output pin from holding pin 11 low. 

By clamping pin 11 high - the diodes forces pin 10 to remain low.

So pin 9 has to remain high. 

                     *******

Since pin 11 can no longer change its state - the polarity of the charge on C4 cannot change. 

So the oscillator cannot run. 

                     *******

Note that R5, R6 & R7 are connected in series between pins 10 and 11. 

They join the output of the first inverter to its own input. 

The value of R7 needs to be very high. 

That way - any influence the output of the first inverter has on its own input will be very small. 

It will be swamped by the influence of the changing charges on C4. 

                     *******

The data sheet for the 4060 says that the value of R7 should be at least ten times the value of R5 + R6. 

That would make it about 3M3. 

The value of 4M7 was chosen because it's widely available. 

It's possible to use such high value resistors because Cmos inputs are operated by voltage - and not by current. 

Because no current flows through the 4M7 resistor - there's no voltage drop across it.

                     *******

When polarized capacitors are biased in the reverse direction - they will conduct direct current.

Since the charge on C4 is constantly reversing in polarity - the capacitor needs to be non-polarized. 

If you use a polarised capacitor in this position - the oscillator may not run - or it may not run reliably.

                     *******

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.



       THE COMPONENTS
        ===============

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 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 0.2mA. 

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

                     *******

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'.

R8 and C3 slow down the speed at which the gate operates.

So it won't be triggered by electrical interference.

                     *******

To remain switched on - the SCR must have a certain minimum current flowing through it. 

This is known as the holding current.

Some Buzzers require a very small current that may be insufficient to keep the SCR switched on.

And - even if it is sufficient - Q2 repeatedly diverts that current to ground.

So - it's not constant - and it cannot be used to keep the SCR switched on.

                     *******

The purpose of R9 is to provide an independent - and constant - source of holding current.

The current through R9 guarantees that - once the SCR has switched on - it will stay on.

If you're using a lower supply voltage - and you find that your SCR is not latching - reduce the value of R9.

                     *******

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

Use a higher value if you wish.

With a 12-volt supply - the maximum current through Q1 will be about 45mA.

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

                     *******

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

D1, D2 & D11 are included to short circuit the spikes at source - before they can do any damage.

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

D3, D4, D5 & D6 allow any and all of the four output pins to supply base current to Q1.

At any given time - some of these pins will be high - and some will be low.

The four diodes prevent the high and low outputs from shorting together.

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D7 allows a high pin 13 to take pin 11 high - but it prevents a low pin 13 from holding pin 11 low.

D8 allows a high pin 7 to take pin 11 high - but it prevents a low pin 7 from holding pin 11 low.

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When the SCR is triggered - D9 allows it to take the junction of R3 & D8 low.

But while the SCR is off - D9 prevents R9 from taking pin 11 high - through D8.

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D10 allows Ty1 to connect the negative lead of the Buzzer to ground.

But it prevents Q2 from connecting the anode of Ty1 to ground.

If Q2 connected the anode of Ty1 to ground - the SCR would unlatch.

                     *******

D10 has to carry the Buzzer current.

A 1N4148 would probably be fine in this position.

But - in the interest of long term reliability - I specified the more robust 1N4001.


                     *******

How To Use The Cmos 4060 As A Timer

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 24 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 Five Wire Links - the nine fixed resistors - and the preset. To fit the preset you may need to enlarge the holes slightly. 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. All the resistors 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.

Add The Five Wire Links
The Nine Fixed Resistors
And The Preset.

Add The Five Wire Links
The Nine Fixed Resistors
And The Preset.

The next stage is to fit the eleven diodes - the two transistors - and the SCR. 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.

Pay particular attention to the orientation of the diodes. Note that one of the diodes (D7 - right in the middle) faces downwards. All the 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. The PNP transistor (BC557) is the one drawn with its emitter arrow coloured red.

Eleven Diodes - Two Transistors - And Thyristor

Eleven Diodes - Two Transistors - And Thyristor

Fit the remaining components - the five capacitors - the IC socket - and the relay. 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 connected together. See the Photo Of The Prototype.

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 5 solder bridges.

Finish off by inserting the Cmos 4060 into the socket. Pin 1 of the IC should be in the top left-hand corner. Check that all 16 pins have entered the socket. Sometimes - instead of entering the socket - a pin will curl up under the IC.

External Connections

Connecting External Devices
To The Cmos 4060 Alarm

How To Test The Finished Alarm

Cmos 4060 Alarm Veroboard Layout	The Rest of Ron's Circuits	Write To Ron	More Free-to-Use Circuits	Circuit Exchange International

Cmos 4060 Alarm - Support Material

Construction Guide

Actual Size Of Pattern

External Connections