How To Build A Simple Keypad-Operated Control Switch - With A Difference.

COMMENTS SUGGESTIONS	Parallel Keypad Switch - Support Material	MORE KEYPAD CIRCUITS

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Circuit Description

Click Here For A Photograph Of The Prototype.

Circuit Diagram For A
Keypad-Operated Switch

This circuit uses a Cmos 4081.

The 4081 contains four separate two-input AND gates.

These are labelled 1 to 4 in the diagram.

The small numbers - next to each gate - identify the IC pins to which that gate is connected.

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The small arrows - top and bottom - indicate that the positive line is connected to pin 14 of the IC - and the negative line is connected to pin 7.

All four gates share the same two connections to the power supply. 

However - apart from that - they operate entirely independently of one another. 

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Each gate has two input pins and one output pin. 

The input pins are on the left of the gate symbol - 1, 2, 5, 6, etc.

And the output pin is on the right of the gate symbol - 3, 4, etc.

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The two inputs of a Cmos 4081 gate are identical to one another.

They operate in exactly the same way. 

So from a circuit design point of view - they are interchangeable. 

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The output of an AND gate will only be high - when both of its inputs are high.

See the truth table.

At power-up - the four inputs of gates 1 & 2 - are held low by the four 100k resistors.

Consequently - pins 3 & 4 will be low also.

And - since these two pins are connected to the gate 3 inputs - pin 11 will be low as well.

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Pin 11 is connected to pin 9.

This means that one of the gate 4 inputs is low.

And while at least one of its inputs is low - the gate 4 output is also low.

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When a Cmos output pin is low - it's connected internally to the negative line.

This means that the Q1 base-current will flow to ground - through R6 and pin 10.

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The direction of the base current - through the transistor - is indicated by the small arrow.

The arrow is always drawn on the emitter.

And it always points towards an "N" terminal.

The BC557 is a PNP transistor - so the arrow points to the base.

And the direction of current flow - is from the emitter - to the base.

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When the Q1 base-current flows to ground - through pin 10 - it switches the transistor on.

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

And the relay energizes.

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To de-energize the relay - you must take the inputs of gates 1 & 2 - high.

The four buttons that represent your code - are connected to A, B, C & D respectively.

When you press "A & B" together - pins 1 & 2 will be taken high by R7.

This means that pin 3 will go high.

And pin 3 will take one of the gate 3 inputs high.

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If you add "C & D" - R7 will take pins 5 & 6 high.

This means that pin 4 will go high.

And pin 4 will take the second gate 3 input high.

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Since both the gate 3 inputs are now high - pin 11 will go high as well.

And pin 11 will take pin 9 high.

Note that pin 8 is already being held high - by R9.

So - when pin 11 takes pin 9 high - the gate 4 output will go high.

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When a Cmos output pin is high - it's connected internally to the positive rail.

This means that the Q1 base current can no longer flow to ground - through pin 10.

As a result - the transistor will switch off.

And the relay will de-energize.

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The alternative NPN transistor - a BC547 - works in more or less the same way.

The only difference is that the NPN base-current flows in the opposite direction - from the base to the emitter.

So a high pin 10 will switch the transistor ON.

And a low pin 10 will switch it OFF.

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This means that when the circuit powers-up - and pin 10 is low - the BC547 will be off - and the relay will be de-energized.

When you enter the code - and pin 10 goes high - the BC547 will switch on - and the relay will energize.

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Note that the emitter of the PNP transistor is connected to the relay coil.

However - the emitter of the NPN transistor - should be connected to ground.

Basically - on the veroboard layout -the two transistors face in opposite directions.

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When pin 10 goes high - it does a second job.

It takes the inputs of gates 1 & 2 high - through R5 - and through D1, D2, D3 & D4 respectively.

In effect - pin 10 latches itself on.

When you release the buttons - the four inputs will remain high.

And the relay will remain de-energized

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It will remain de-energized - until one of the keys connected to "E" is pressed.

It doesn't matter which one.

The result is to connect the common terminal to ground.

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When the common terminal is connected to ground - it takes pin 1 low - through D5.

Since one of its inputs is low - the gate 1 output will go low.

Gate 1 takes one of the gate 3 inputs low.

And gate 3 takes one of the gate 4 inputs low.

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This causes pin 10 to go low.

And the low pin 10 switches the transistor on - and energizes the relay.

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Note that pin 10 is no longer holding the four input pins high.

Once more - they are being held low by the four 100k resistors.

And once more - the security code is required - to de-energize the relay. 

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          THE TIME LOCK

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At power-up - pin 8 is held high by R9.  

And - when the correct code is entered - pin 9 goes high.

Since both inputs are now high - pin 10 will go high as well.

And the relay will de-energize.

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Note that both plates of C6 are at 12 volts.

The positive plate is connected to the 12v line.

And the negative plate is connected to pin 8 - which is also 12v.

Since there's no difference in potential across the plates - the capacitor is in a discharged state.

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If you press an incorrect key - it connects the common terminal to ground.

It also connects the negative plate of C6 to ground - through R8 & D6.

This begins to remove the positive charge - from the negative plate of the capacitor.

In other words - C6 begins to charge.

Every time you press an incorrect key - C6 will charge a little bit more.

And - as C6 charges - the voltage on pin 8 falls.

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When it falls below about half the value of the supply - roughly 6v - the input pin becomes low. 

While pin 8 is low - pin 10 is also low.

Now - even if you enter the correct code - it will be unsuccessful.

While pin 8 remains low - the keypad is locked.

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To unlock the keypad - you must wait for C6 to discharge - through R9.

How long this takes depends on the value of R9 - and on how much charge there is in C6.

The amount of charge in C6 depends on the value of R8.

It also depends on the number of times the wrong key was pressed - and on how long it was held down.

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As it stands - if you press at least three or four wrong keys in succession - the timer will lock you out for about a minute.

This should allow you to make a couple of mistakes - and still have time to enter the correct code.

If you need more time - increase the value of R8.

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              THE DIODES

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The diodes create one-way paths.

D1, D2, D3 & D4 allow a high pin 10 to take the four input pins high.

But they prevent a low pin 10 from taking the four input pins low.

The diodes also allow all four inputs to be connected to R5 - without actually being connected to each other.

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D5 allows a wrong key to take pin 1 low - through the common terminal.

But - it prevents R7 from taking pin 1 high - until "A" is pressed.

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D6 allows a wrong key to charge C6 - through R8. 

But - it prevents C6 from discharging rapidly through R7 & R8.

Instead - the capacitor must discharge slowly - through R9.

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When a wrong key connects "A" to ground - through D5 - R5 prevents pin 10 from shorting to ground - through D1.

R6 limits the amount of base current flowing through Q1.

It's roughly 1mA - more than enough to switch the transistor on.

When a wrong key connects the common terminal to ground - R7 limits the size of the current flowing through the keypad-switch.

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The 100nF capacitors are there to slow down the inputs of the Cmos 4081 - and to make them less sensitive to stray high-frequency signals. 

The capacitors allow you to mount the actual keypad - at some distance from the circuit board.

To a high frequency signal - a 100nF capacitor looks like a wire link.

Any stray RF signals - picked up by the long wires - will simply short to ground through the capacitors. 

C5 does a similar job for the supply lines.

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I've drawn a SPCO/SPDT relay. But you can use a multi-pole relay - if it suits your application. 

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

D7 short-circuits these spikes at source - before they can do any damage.

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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 Q1 will be about 45mA. (Ohms Law)

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

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There is nothing special about the transistor.

Depending on the version of the circuit you're building - Q1 has to be a PNP - or an NPN - transistor

Apart from this requirement - any small transistor - with a gain (hfe) greater than 100 - and an IC(max) of at least 100mA - should work fine.

But check the pin configuration. It may be different from that of the BC547 / BC557.

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Do not use the "on-board" relay to switch mains voltage. The board's layout does not offer sufficient isolation between the relay contacts and the low-voltage components. If you want to switch mains voltage - mount a suitably rated relay - somewhere safe - Away From The Board.

Parts List

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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 21 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 .

Actual Size Of Pattern

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

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.

Add The Ten Wire Links And Nine Resistors

Add The Ten Wire Links And Nine Resistors

Next fit the Ten Wire Links and the nine resistors. 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.

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.

Next - fit the seven diodes - and the six capacitors. Pay particular attention to the orientation of the diodes and the electrolytic capacitor. Note that D6 and the electrolytic capacitor both have their positive terminals facing the bottom of the board. See The Photograph Of The Prototype.

Add The Remaining Components

Fit the transistor - the IC socket - and the relay. If you want the circuit to power-up with the relay energized - fit the PNP transistor shown above. It's the one with its emitter arrow coloured red. It will allow your security code to de-energize the relay.

If you want the circuit to power-up with the relay de-energized - fit the NPN transistor shown below. Note that it faces in the opposite direction. It will allow your security code to energize the relay.

Fit The Alternative NPN Transistor

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. If you flip the above drawing horizontally - it will help you place them accurately.

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 Are 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 in the pad. The more keys you use - the more secure will be the code. For example - a 12-key pad has 495 different groups of four keys. But a 16-key pad offers 1820 different groups.

Construction
Of Keypad

You Are Now Ready To Test Your Circuit

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

Parallel Keypad Switch - Support Material

Parts List

Construction

Actual Size Of Pattern

Keypad Layout