Build this Four-Digit Keypad Operated Switch using off-the-shelf components. Although it was designed to control a Burglar Alarm - it has other uses.

COMMENTS SUGGESTIONS	Four-Digit Keypad - Support Material	MORE KEYPAD CIRCUITS

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

Add A "Beep" To The Keystrokes	Three More Ways To Modify The Circuit

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

Click Here For A Photograph Of The Prototype

Circuit Diagram For
An Alarm Control Keypad

The key connected to E is used to energize the relay. 

The four keys connected to 
A, B, C & D are used to 
de-energize the relay. 

All the other keys on the keypad are connected to F. 

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When E is pressed - current from the Common terminal flows through D2 and R9 into the base of Q5. 

This switches Q5 on.

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

And the relay energizes. 

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The relay contacts switch the 12-volt output from the "off" terminal to the "set" terminal - and the green LED lights. 

The LED is included to give a visual indication that the alarm is set.

You could simply connect it to the negative line.

I connected it to Q5 because it was more convenient for the veroboard layout.

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The relay latches itself on by providing an alternative base current for Q5 - through R10. 

So after E is released - and the current through D2 & R9 ceases - the relay will remain energized. 

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To de-energize the relay it's necessary to press A, B, C & D in the right order. 

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

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

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At the start - pin 1 is held high by R5. 

This enables gate 1. 

When A is pressed - and pin 2 is taken high - both the gate 1 inputs are high.

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

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This output does two jobs. 

Firstly, it latches itself on by holding pin 2 high through R2. 

In this way - pin 2 will stay high after A is released. 

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Secondly, Pin 3 enables gate 2 by taking pin 5 high. 

In other words - pin 3 is doing for gate 2 - what R5 does for gate 1. 

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

Pin 4 latches itself on through R3 - and it enables gate 3 by taking pin 12 high. 

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Again - when C is pressed and pin 13 is taken high - the output from gate 3 goes high.

Pin 11 latches itself on through R4 - and it enables gate 4 by taking pin 8 high. 

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Now - when D is pressed and pin 9 is taken high - the output from gate 4 goes high.

Pin 10 latches itself on through R6. 

But this time it also switches Q4 on - through R8.

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Q4 connects the base of Q5 to ground. 

This diverts the Q5 base current to ground. 

Without its base current - Q5 switches off and the relay drops out. 

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The 12-volt output moves from the "set" terminal - back to the "off" terminal.

And the green LED switches off.


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      THE WRONG KEYS

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The end effect of entering the correct code is that pin 10 goes high.

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

But the whole process relies on R5 - because it enables the first gate in the sequence.

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

When a wrong key is pressed - current through R7 will switch Q3 on. 

Q3 takes pin 1 low - and removes the R5 enable from gate 1.

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This causes the attempted code entry to 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.

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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. 

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D1 lets the key connected to E act as though it's also connected to F. 

But it prevents keys connected to F from behaving as though they are connected to E. 

In other words - E is able to switch Q3 on - but F cannot energize the relay.
 
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This is important for two reasons. 

Firstly - after the correct code has been entered - and the alarm has switched off - pin 10 is high. 

Therefore - Q4 is switched on - and the base of Q5 is connected to ground. 

Instead of entering the base of Q5 - current through D2 and R9 flows to ground through Q4.

In other words - E cannot switch Q5 on.

So - E cannot energize the relay.

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But D1 allows E to behave as though it's connected to F. 

Therefore - it can switch Q3 on - and take pin 1 low.

And when pin 1 goes low - all the outputs go low - including pin 10.

So D1 allows E to turn Q4 off.

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In other words - when E is pressed - it begins by resetting the code entry procedure - and then it energizes the relay. 

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The second reason for the presence of D1 is that E is a "wrong" key.

D1 allows it to act as such - by permitting it to switch Q3 on.

But the diode prevents the remaining "wrong" keys from energizing the relay.

Current from F cannot reach the base of Q5.

So keys connected to F cannot switch Q5 on.

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When the relay is energized - D2 prevents current through R10 and R9 - from reaching the base of Q3 - through D1 and R7. 

If you work out the voltages involved - it's not strictly necessary.

It's included as a precaution - to avoid the remote possibility.

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

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This eventuality is covered by Q1 and Q2. 

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. 

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

If it's not followed by B then pin 4 will remain low. 

While pin 4 is low - it's connected internally to the negative rail. 

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The emitter of Q1 is connected to pin 4.

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.

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So Q1 will switch on.

The transistor will connect pin 1 to ground - through pin 4.

And the attempted code entry will fail.

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

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The emitter of Q2 is connected to pin 11.

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

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

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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 connect pin 1 to ground - through pin 11.

And the attempted code entry will fail.

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

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

Then the emitters of Q1 and Q2 are no longer connected to the negative rail - and the transistors cannot take pin 1 low. 

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

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

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

But the same reasoning applies to all four gates.

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

And it's holding pin 2 low - through R2.

When A 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.

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But when A is pressed - R1 & R2 form a potential divider between the 12-volt supply and the low pin 3.

So the voltage at the junction of R1 & R2 - that is the voltage at pin 1 - depends on the values of R1 & R2.

If the value of R2 is too low - that is less than that of R1 - pressing A will not take pin 2 high. 

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The value of R2 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.

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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 & R2.

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 R2.

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

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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 to the circuit board.

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

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

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

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I specified a relay with a minimum coil resistance of 270 ohms. Use a higher value if you wish.

With a 12-volt supply - the maximum current through Q5 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.

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Relay coils can produce large reverse voltage spikes that will destroy Cmos ICs. 

D3 is included to short circuit the spikes at source - before they can do any damage.

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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.

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The number of different codes available depends on the number of keys in your keypad.

The more keys you have - the more choices you have,

With a 12-key pad there are:

12 possible choices for A 
11 possible choices for B
10 possible choices for C
 9  possible choices for D

This means that there are:

12 x 11 x 10 x 9 = 11 880 

different codes available. 

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Using a bigger keypad - with more "wrong" keys - will increase security - by making an even larger number of codes available.

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Parts List

Suppliers	Worldwide RS Components	Uk & Ireland Maplin

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 29 places where the tracks are to be cut. Before you cut the tracks, use the "actual size" drawing to Check That The Pattern is Correctly Marked .

When you're satisfied that the pattern is right - cut the tracks. Make sure that the copper is cut all the way through. Sometimes a small strand of copper remains at the side of the cut and this will cause malfunction. Use a magnifying glass - and backlight the board. It only takes the smallest strand of copper to cause a problem. If you don't have the proper track-cutting tool, then a 6 to 8mm drill-bit will do. Just use the drill-bit as a hand tool - there's no need for a drilling machine.

Pattern for Cutting
The Tracks on the
Underside of the Board

Pattern for Cutting
The Tracks on the
Underside of the Board

Actual Size Of Pattern

Pattern for Cutting
The Tracks on the
Underside of the Board
ACTUAL SIZE

Next make and fit the Fifteen Wire Links. I used bare copper wire on the component side of the board. Telephone cable is suitable; the single stranded variety used indoors to wire telephone sockets. Stretching the core slightly will straighten it; and also allow the insulation to slip off.

Add The Fifteen Wire Links

The next stage is to fit the 11 resistors, the 5 capacitors and the IC socket. Using a socket reduces the chances of damaging the IC; and makes it easier to replace if necessary. The components 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

Now fit the transistors, diodes and relay. Again, the diodes are all shown lying flat on the board. However, those connected between close or adjacent tracks are mounted standing upright.

Next - 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 7 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" This gives (16 x 15 x 14 x 13) = 43 680 possible codes.

General Information

Add A "Beep" To The Keystrokes	Three More Ways To Modify The Circuit

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