Entering the code operates a small relay. Build the circuit yourself using cheap off-the-shelf components. Although it was designed to control a Burglar Alarm - it has many other uses.

COMMENTS SUGGESTIONS	Enhanced Alarm Keypad - Support Material	MORE KEYPAD CIRCUITS

Five-Digit Alarm Keypad Schematic	The Rest of Ron's Circuits	Write To Ron	More Free-to-Use Circuits	Circuit Exchange International

Test Your Finished Circuit Board

There is both a Four-Digit and a Five-Digit version of this switch. Although they will have other applications - both are designed to control the Modular Burglar Alarm circuit. Simply use the relay's changeover contacts in place of SW1. The poles and the points marked 'set' and 'off' represent the corresponding connections on both the keypad relay and the alarm switch. Click Here for a Diagram.

Circuit Description
	This circuit uses a Cmos 4081. It has four separate two-input AND gates. AND gates only have a high output - when BOTH of the input pins are high. See the Truth Table. ============== When power is applied - all eight input pins are low. So the four output pins are low also. ============== The alarm is set by using the first four - of your five chosen keys. If A B C & D are pressed in the right order - pin 11 will go high. When pin 11 goes high - it's connected internally to the positive line. And it supplies base current to Q6 - through R11. ============== The base current switches Q6 on. The transistor connects the negative side of the relay coil to ground. And the relay energizes. ============== When the relay energizes - the contacts switch the 12v output from the OFF to the SET terminal. This is the action that arms the Modular Alarm. The relay contacts also do two further jobs. ============== After the turn ON code has been entered - the code entry sequence resets. It has to reset - in order to be ready to receive the turn OFF code. When the sequence resets - pin 11 goes low. And the low pin 11 can no longer supply base current to Q6. ============== Without base current - Q6 would switch off. The relay would de-energize. And the alarm would switch off. ============== The relay contacts themselves - prevent this from happening. When the relay energizes - the contacts supply an alternative base current to Q6 - through R12. So when pin 11 goes low - and current through R11 ceases - the relay will remain energized. ============== The relay contacts also provide the current required to light the green LED. It might perhaps be mounted remotely - along side the keypad switches. Its purpose is to provide visual confirmation that the code entry has been successful. ============== To switch the alarm OFF - you have to enter the entire five-digit code. If A B C D & E are pressed in the right order - pin 10 will go high. When pin 10 goes high - it's connected internally to the positive line. And it supplies base current to Q5 - through R7. ============== When Q5 switches on - it connects the base of Q6 to ground. In effect - it diverts the R12 current to ground. This switches Q6 off - and the relay de-energizes. ============== CODE ENTRY ============== The IC is a Cmos 4081 - with four two-input AND gates. AND gates only have a high output - when both inputs are high. See the Truth Table. ============== To start with - pin 1 is held low by R2. When A is pressed - C1 charges - through R1. This takes pin 1 high. And pin 1 will remain high - until C1 discharges through R2. ============== The remaining four digits must be entered before - pin 1 goes low again. That is - before C1 discharges through R2. In effect - the charge in C1 enables the code entry sequence to proceed. ============== When B is pressed - and pin 2 is taken high by R1 - both of the gate 1 inputs are high. So the gate 1 output - at pin 3 - will go high. ============== This output does two jobs. Firstly - it latches itself on - by holding pin 2 high - through R3. In this way - pin 2 will remain high - after B is released. ============== Secondly - pin 3 enables gate 2 -by taking pin 5 high. In other words - pin 3 is doing for gate 2 - what C2 does for gate 1. ============== Now - in exactly the same way - when C is pressed and pin 6 is taken high - the output from gate 2 will go high. Pin 4 latches itself on through R4 - and it enables gate 3 - by taking pin 12 high. ============== Again - when D is pressed - and pin 13 is taken high - the output from gate 3 will go high. It latches itself on - through R5. And the high pin 11 - provides the necessary base current to switch Q6 on - and energize the relay. ============== THE CODE RESET ============== The end effect of entering the correct code is that pin 11 goes high. The code entry sequence works because each successive gate is enabled by its predecessor. But the whole process relies on C1 - because it enables the first gate in the sequence. ============== Once C1 discharges - pin 1 will go low. So pin 3 will go low. Pin 3 takes pin 5 low. So pin 4 will go low. Pin 4 takes pin 12 low. So pin 11 will go low. ============== Pin 11 will no longer supply base current to Q6. But that doesn't matter - because Q6 is receiving an alternative supply - through R12. So the transistor remains switched on - and the relay remains energized. ============== THE TURN-OFF CODE ============== The turn-off code works in exactly the same way. The only difference is that the sequence doesn't stop at D. ============== When you get to D - pin 11 will once more supply base current to Q6. But this will have no effect - because the transistor is already switched on - and the relay is already energized. ============== But the high pin 11 also takes pin 8 high. And this enables gate 4. So - when E is pressed and pin 9 is taken high - pin 10 will go high. ============== The high pin 10 switches Q5 on. Q5 diverts the Q6 base current to ground. So Q6 switches off - and the relay de-energizes. ============== When the relay de-energizes - the 12v moves from the SET terminal - back to the OFF terminal. This switches the Modular Alarm off. And the green LED is extinguished. ============== THE WRONG KEYS ============== All the keys connected to F are "Wrong" keys. When a wrong key is pressed - current through R9 will switch Q4 on. Q4 discharges C1 through R8. This takes pin 1 low - removing the enable from gate 1. ============== As a result - the attempted code entry will fail. It fails because the support each gate provides for its successor - disappears the moment pin 1 is taken low. When pin 1 goes low - pin 3 goes low - and pin 3 takes pin 5 low. When pin 5 goes low - pin 4 goes low - and pin 4 takes pin 12 low. When pin 12 goes low - pin 11 goes low - and pin 11 takes pin 8 low. And when pin 8 goes low - pin 10 goes low. ============== In other words - when pin 1 goes low - the outputs of all four gates go low. The attempted code entry fails. And the code entry sequence is reset. ============== There is one other situation that must be dealt with - and that is where the "right" keys are pressed - but in the "wrong" order. ============== This eventuality is covered by Q1, Q2 and Q3. If - to begin with - any key other than A is pressed - nothing will happen. Until A is pressed the code entry sequence will not start. ============== Once A is pressed - it must be followed by B. If it's not followed by B then pin 3 will remain low. While pin 3 is low - it's connected internally to the negative rail. ============== The emitter of Q1 is connected to pin 3. So - until B is pressed - the emitter of Q1 is connected to ground. If C is pressed before B - current will flow through R1 - into the base-emitter junction of Q1. ============== So Q1 will switch on. And C1 will discharge - through D1, Q1 and pin 3. And the attempted code entry will fail. ============== In other words - pressing C out of sequence - has the same effect as pressing one of the keys connected to F ============== The emitter of Q2 is connected to pin 4. If D is pressed at any time before C - pin 4 will still be low. So - until C is pressed - the emitter of Q2 is connected to ground. ============== If D is pressed before C - current will flow through R1 - into the base-emitter junction of Q2. So Q2 will switch on. The transistor will discharge C1 - into pin 4. And the attempted code entry will fail. ============== In other words - pressing D out of sequence - has the same effect as pressing one of the keys connected to F ============== The emitter of Q3 is connected to pin 11. If E is pressed at any time before D - pin 11 will still be low. So - until D is pressed - the emitter of Q3 is connected to ground. ============== If E is pressed before D - current will flow through R1 - into the base-emitter junction of Q3. So Q3 will switch on. The transistor will discharge C1 - into pin 11. And the attempted code entry will fail. ============== In other words - pressing E out of sequence - has the same effect as pressing one of the keys connected to F ============== It's only when the keys are pressed in the right order - that pins 3, 4 and 11 will be high at the appropriate time. Then the emitters of Q1, Q2 and Q3 are no longer connected to the negative rail - and the transistors cannot discharge C1. ============== Transistors are generally drawn with a small arrow - indicating the base-emitter junction. In effect - this junction is a diode. It allows current to flow in the direction of the arrow. ============== However - there is a similar junction between the base and the collector. It too could be drawn with an arrow pointing to the collector. And it too - acts like a diode. ============== Under the right conditions - it will allow current to flow from the base terminal - to the collector terminal. Without D1 - this current would tend to charge C1 - or to keep C1 charged. The diode creates a one way path. It blocks off the unwanted charge path - while still keeping the three discharge paths open. ============== The purpose of R1 is to limit the base current through Q1, Q2 and Q3. However - R1 also has to be capable of taking A, B, C, D & E high. In what follows I will take gate 1 as the example. But the same reasoning applies to all four gates. ============== When pin 3 is low - it's connected internally to the negative line. And it's holding pin 2 low - through R3. When B is pressed - R1 has to be able to take the voltage on pin 2 high. That is - it has to be able to take the pin above about half the value of the supply voltage. ============== But when B is pressed - R1 & R3 form a potential divider between the 12-volt supply and the low pin 3. So the voltage at the junction of R1 & R3 - that is the voltage at pin 2 - depends on the values of R1 & R3. If the value of R3 is too low - that is less than that of R1 - pressing B will not take pin 2 high. ============== The value of R3 must be greater than the value of R1. To be on the safe side - it should be several times greater. I picked 27k because I happened to have a lot of 27k resistors. ============== 27k is over five times the value of R1. That is - there is a ratio of about 5 to 1 between the values of R1 & R3. This means that about one-sixth of the supply voltage will appear across R1. And about five-sixths of the supply voltage will appear across R3. In other words - pin 2 will reach about 10-volts. ============== THE ROLE OF C7 ============== There's a potential problem that could arise - if you try to set the alarm too soon after you've switched it off. It takes a few seconds for C1 to discharge through R2. When you switch the alarm off - pin 10 will remain high until C1 discharges. And - while pin 10 is high - Q5 keeps Q6 switched firmly off. ============== If you try to set the alarm again - before C1 discharges - you will only succeed in topping up the charge in C1. And this will prolong the period of time it takes for Q5 to switch off. In fact - you could keep trying to enter the turn-on code - and never succeed in setting the alarm. ============== It's true that in normal use - this problem is not likely to arise. But - just in case it might - it's solved by C7. ============== When Q6 switches off - its collector rises rapidly - from 0v to 12v. And current flows briefly - through the relay coil - the discharged C7 - and R9 - into the base of Q4. This current switches the transistor on - long enough for it to complete the discharge of C1. ============== In other words - C7 has the same effect as pressing a "wrong" key. Q4 discharges C1 rapidly - and resets the code entry sequence immediately. So you can set the alarm again - straight away. ============== C7 charges rapidly through R10 - and the brief reset pulse ends. Then Q4 switches off - and you're free to begin the new code entry sequence. ============== While C7 remains charged - it cannot switch Q4 on again. However - the next time the alarm is set - Q6 will connect the positive plate of C7 to ground. In effect - Q6 connects the capacitor across R10. R10 discharges C7 - and the capacitor is ready to supply a fresh reset pulse - the next time Q6 switches off. ============== THE 100nF CAPACITORS ============== Cmos gates react very quickly to changes in input voltage. They can be triggered by stray electrical or radio interference. This is picked up by the wires that connect the keypad switches to the circuit board. The 100nF capacitors are there to slow down the inputs - and make them less sensitive to stray signals. ============== To high frequencies - 100nF capacitors represent a virtual short circuit. Unwanted incoming signals are shorted to ground - before they reach the input pins. ============== THE RELAY ============== 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 Q6 will be about 45mA. That is:- 12v � 270 ohms. This is well within the limits of a BC547 - which has an IC(max) of 100mA. ============== I used an SPCO relay - but you can use a multi-pole relay if it suits your application. All relay coils produce large reverse voltage spikes that will destroy Cmos ICs. D2 is included to short circuit the spikes at source - before they can do any damage. ==============

Circuit Description

Circuit Diagram For
An Alarm Control Keypad

This circuit uses a Cmos 4081. 

It has four separate two-input AND gates. 

AND gates only have a high output - when BOTH of the input pins are high.

See the Truth Table.

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

When power is applied - all eight input pins are low.

So the four output pins are low also.

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

The alarm is set by using the first four - of your five chosen keys.

If A B C & D are pressed in the right order - pin 11 will go high.

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

And it supplies base current to Q6 - through R11.

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

The base current switches Q6 on.

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

And the relay energizes. 

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

When the relay energizes - the contacts switch the 12v output from the OFF to the SET terminal.

This is the action that arms the Modular Alarm.

The relay contacts also do two further jobs.

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

After the turn ON code has been entered - the code entry sequence resets.

It has to reset - in order to be ready to receive the turn OFF code.

When the sequence resets - pin 11 goes low.

And the low pin 11 can no longer supply base current to Q6.

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

Without base current - Q6 would switch off.

The relay would de-energize.

And the alarm would switch off.

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

The relay contacts themselves - prevent this from happening. 

When the relay energizes - the contacts supply an alternative base current to Q6 - through R12.

So when pin 11 goes low - and current through R11 ceases - the relay will remain energized. 

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

The relay contacts also provide the current required to light the green LED.

It might perhaps be mounted remotely - along side the keypad switches.

Its purpose is to provide visual confirmation that the code entry has been successful.

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

To switch the alarm OFF - you have to enter the entire five-digit code.

If A B C D & E are pressed in the right order - pin 10 will go high.

When pin 10 goes high - it's connected internally to the positive line.

And it supplies base current to Q5 - through R7.

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

When Q5 switches on - it connects the base of Q6 to ground.

In effect - it diverts the R12 current to ground.

This switches Q6 off - and the relay de-energizes.

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

         CODE ENTRY

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

The IC is a Cmos 4081 - with four two-input AND gates. 

AND gates only have a high output - when both inputs are high.

See the Truth Table. 

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

To start with - pin 1 is held low by R2. 

When A is pressed - C1 charges - through R1.

This takes pin 1 high.

And pin 1 will remain high - until C1 discharges through R2.

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

The remaining four digits must be entered before - pin 1 goes low again.

That is - before C1 discharges through R2.

In effect - the charge in C1 enables the code entry sequence to proceed.

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

When B is pressed - and pin 2 is taken high by R1 - both of the gate 1 inputs are high.

So the gate 1 output - at pin 3 - will go high. 

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

This output does two jobs. 

Firstly - it latches itself on - by holding pin 2 high - through R3. 

In this way - pin 2 will remain high - after B is released. 

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

Secondly - pin 3 enables gate 2 -by taking pin 5 high. 

In other words - pin 3 is doing for gate 2 - what C2 does for gate 1. 

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

Now - in exactly the same way - when C is pressed and pin 6 is taken high - the output from gate 2 will go high.

Pin 4 latches itself on through R4 - and it enables gate 3 - by taking pin 12 high. 

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

Again - when D is pressed - and pin 13 is taken high - the output from gate 3 will go high.

It latches itself on - through R5.

And the high pin 11 - provides the necessary base current to switch Q6 on - and energize the relay.

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

      THE CODE RESET

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

The end effect of entering the correct code is that pin 11 goes high.

The code entry sequence works because each successive gate is enabled by its predecessor.

But the whole process relies on C1 - because it enables the first gate in the sequence.

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

Once C1 discharges - pin 1 will go low.

So pin 3 will go low.

Pin 3 takes pin 5 low.

So pin 4 will go low.

Pin 4 takes pin 12 low.

So pin 11 will go low.

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

Pin 11 will no longer supply base current to Q6.

But that doesn't matter - because Q6 is receiving an alternative supply - through R12.

So the transistor remains switched on - and the relay remains energized.

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

  THE TURN-OFF CODE

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

The turn-off code works in exactly the same way.

The only difference is that the sequence doesn't stop at D.

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

When you get to D - pin 11 will once more supply base current to Q6.

But this will have no effect - because the transistor is already switched on - and the relay is already energized.

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

But the high pin 11 also takes pin 8 high.

And this enables gate 4.

So - when E is pressed and pin 9 is taken high - pin 10 will go high.

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

The high pin 10 switches Q5 on.

Q5 diverts the Q6 base current to ground.

So Q6 switches off - and the relay de-energizes.

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

When the relay de-energizes - the 12v moves from the SET terminal - back to the OFF terminal.

This switches the Modular Alarm off.

And the green LED is extinguished.


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

      THE WRONG KEYS

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

All the keys connected to F are "Wrong" keys.

When a wrong key is pressed - current through R9 will switch Q4 on. 

Q4 discharges C1 through R8.

This takes pin 1 low - removing the enable from gate 1.

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

As a result - the attempted code entry will fail.

It fails because the support each gate provides for its successor - disappears the moment pin 1 is taken low.

When pin 1 goes low - pin 3 goes low - and pin 3 takes pin 5 low.

When pin 5 goes low - pin 4 goes low - and pin 4 takes pin 12 low.

When pin 12 goes low - pin 11 goes low - and pin 11 takes pin 8 low.

And when pin 8 goes low - pin 10 goes low.

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

In other words - when pin 1 goes low - the outputs of all four gates go low.

The attempted code entry fails.

And the code entry sequence is reset. 

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

There is one other situation that must be dealt with - and that is where the "right" keys are pressed - but in the "wrong" order.

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

This eventuality is covered by Q1, Q2 and Q3. 

If - to begin with - any key other than A is pressed - nothing will happen.

Until A is pressed the code entry sequence will not start. 

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

Once A is pressed - it must be followed by B. 

If it's not followed by B then pin 3 will remain low. 

While pin 3 is low - it's connected internally to the negative rail. 

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

The emitter of Q1 is connected to pin 3.

So - until B is pressed - the emitter of Q1 is connected to ground.

If C is pressed before B - current will flow through R1 - into the base-emitter junction of Q1.

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

So Q1 will switch on.

And C1 will discharge - through D1, Q1 and pin 3.

And the attempted code entry will fail.

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

In other words - pressing C out of sequence - has the same effect as pressing one of the keys connected to F 

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

The emitter of Q2 is connected to pin 4.

If D is pressed at any time before C - pin 4 will still be low. 

So - until C is pressed - the emitter of Q2 is connected to ground.

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

If D is pressed before C - current will flow through R1 - into the base-emitter junction of Q2.

So Q2 will switch on.

The transistor will discharge C1 - into pin 4.

And the attempted code entry will fail.

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

In other words - pressing D out of sequence - has the same effect as pressing one of the keys connected to F 

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

The emitter of Q3 is connected to pin 11.

If E is pressed at any time before D - pin 11 will still be low. 

So - until D is pressed - the emitter of Q3 is connected to ground.

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

If E is pressed before D - current will flow through R1 - into the base-emitter junction of Q3.

So Q3 will switch on.

The transistor will discharge C1 - into pin 11.

And the attempted code entry will fail.

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

In other words - pressing E out of sequence - has the same effect as pressing one of the keys connected to F 

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

It's only when the keys are pressed in the right order - that pins 3, 4 and 11 will be high at the appropriate time. 

Then the emitters of Q1, Q2 and Q3 are no longer connected to the negative rail - and the transistors cannot discharge C1.

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

Transistors are generally drawn with a small arrow - indicating the base-emitter junction.

In effect - this junction is a diode.

It allows current to flow in the direction of the arrow.

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

However - there is a similar junction between the base and the collector.

It too could be drawn with an arrow pointing to the collector.

And it too - acts like a diode.

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

Under the right conditions - it will allow current to flow from the base terminal - to the collector terminal.

Without D1 - this current would tend to charge C1 - or to keep C1 charged.

The diode creates a one way path.

It blocks off the unwanted charge path - while still keeping the three discharge paths open.

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

The purpose of R1 is to limit the base current through Q1, Q2 and Q3. 

However -  R1 also has to be capable of taking A, B, C, D & E high.

In what follows I will take gate 1 as the example.

But the same reasoning applies to all four gates.

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

When pin 3 is low - it's connected internally to the negative line.

And it's holding pin 2 low - through R3.

When B is pressed - R1 has to be able to take the voltage on pin 2 high. 

That is - it has to be able to take the pin above about half the value of the supply voltage.

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

But when B is pressed - R1 & R3 form a potential divider between the 12-volt supply and the low pin 3.

So the voltage at the junction of R1 & R3 - that is the voltage at pin 2 - depends on the values of R1 & R3.

If the value of R3 is too low - that is less than that of R1 - pressing B will not take pin 2 high. 

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

The value of R3 must be greater than the value of R1. 

To be on the safe side - it should be several times greater.

I picked 27k because I happened to have a lot of 27k resistors.

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

27k is over five times the value of R1.

That is - there is a ratio of about 5 to 1 between the values of R1 & R3.

This means that about one-sixth of the supply voltage will appear across R1.

And about five-sixths of the supply voltage will appear across R3.

In other words - pin 2 will reach about 10-volts.


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

      THE ROLE OF C7

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

There's a potential problem that could arise - if you try to set the alarm too soon after you've switched it off.

It takes a few seconds for C1 to discharge through R2.

When you switch the alarm off - pin 10 will remain high until C1 discharges.

And - while pin 10 is high - Q5 keeps Q6 switched firmly off.

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

If you try to set the alarm again - before C1 discharges - you will only succeed in topping up the charge in C1.

And this will prolong the period of time it takes for Q5 to switch off.

In fact - you could keep trying to enter the turn-on code - and never succeed in setting the alarm.

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

It's true that in normal use - this problem is not likely to arise.

But - just in case it might - it's solved by C7.

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

When Q6 switches off - its collector rises rapidly - from 0v to 12v.

And current flows briefly - through the relay coil - the discharged C7 - and R9 - into the base of Q4.

This current switches the transistor on - long enough for it to complete the discharge of C1.

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

In other words - C7 has the same effect as pressing a "wrong" key.

Q4 discharges C1 rapidly - and resets the code entry sequence immediately.

So you can set the alarm again - straight away.

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

C7 charges rapidly through R10 - and the brief reset pulse ends.

Then Q4 switches off - and you're free to begin the new code entry sequence.

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

While C7 remains charged - it cannot switch Q4 on again.

However - the next time the alarm is set - Q6 will connect the positive plate of C7 to ground.

In effect - Q6 connects the capacitor across R10.

R10 discharges C7 - and the capacitor is ready to supply a fresh reset pulse - the next time Q6 switches off.


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

THE 100nF CAPACITORS

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

Cmos gates react very quickly to changes in input voltage.

They can be triggered by stray electrical or radio interference. 

This is picked up by the wires that connect the keypad switches to the circuit board.

The 100nF capacitors are there to slow down the inputs - and make them less sensitive to stray signals.

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

To high frequencies  - 100nF capacitors represent a virtual short circuit.

Unwanted incoming signals are shorted to ground - before they reach the input pins. 


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

          THE RELAY

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

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 Q6 will be about 45mA.

That is:- 12v � 270 ohms.

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

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

I used an SPCO relay - but you can use a multi-pole relay if it suits your application.

All relay coils produce large reverse voltage spikes that will destroy Cmos ICs. 

D2 is included to short circuit the spikes at source - before they can do any damage.

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

There is nothing special about the transistors. Any small NPN transistors with a gain (hfe) greater than 100 and an IC(max) of at least 100mA should do. But remember that the pin configuration of your transistors may be different from that of the BC547. Just because it looks like a BC547 - do not assume that it has the same pin configuration.

Suppliers	Worldwide RS Components	Uk & Ireland Maplin

Parts List

Construction

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 28 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 make and fit the Thirteen Wire 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 Thirteen Wire Links

The next stage is to fit the 13 resistors - the 8 capacitors - and the IC socket. Using a socket reduces the chances of damaging the IC - and makes it easier to replace if necessary. Pay particular attention to the orientation of the two electrolytic capacitors. The resistors are all shown lying flat on the board. However - those connected between close or adjacent tracks - are mounted standing upright.

Construction Of The
Circuit Board - Continued

Construction Of The
Circuit Board - Continued

Next fit the transistors - diodes - and relay. Pay particular attention to the orientation of the transistors and diodes. Again - D1 is shown lying flat on the board - but it's actually mounted standing upright.

Finally - examine the underside of the board carefully - to make sure that there are no unwanted solder bridges or other connections between the tracks. Again - if you backlight the board - it makes potential problem areas easier to spot.

When you're satisfied that everything is in order - add the 9 solder bridges to the underside of the board. Pay particular attention to the bridge that joins pins 4 & 12 of the IC. When positioning the bridges - it may help to make a copy of the above drawing - and flip it left to right.

Finish off by inserting the Cmos 4081 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

Keypad Layout

One of the advantages of making your own keypad is that you can choose the number of keys you use. For example - with a 16-key pad the number of choices are: 16 for "A" 15 for "B" 14 for "C" 13 for "D" 12 for "E" This gives (16 x 15 x 14 x 13 x 12) = 524 160 possible codes.

General Information

Test Your Finished Circuit Board

Five-Digit Alarm Keypad Schematic	The Rest of Ron's Circuits	Write To Ron	More Free-to-Use Circuits	Circuit Exchange International