Circuit Description

To see the difference between the two circuits move the mouse pointer over the schematic. In the first circuit - the NPN transistor switches the negative side of the coil. In the second circuit - the PNP transistor switches the positive side of the coil. The two circuits are otherwise identical. In what follows - I've concentrated on describing Timer No.1. However - where appropriate - I've included an account of how Timer No.2 differs.

Breadboard No.1		Test Procedure		Breadboard No.2

Image Effect Courtesy Of DynamicDrive.Com

The Cmos 4060 contains a 14-bit binary counter.

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 contains 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.

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The Cmos 4060 also has two built-in inverters.

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

You can tell they're inverters because the symbols are drawn with a small circle at the output.

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As the name suggests - the output of an inverter is always the opposite of its input.

If the input pin is high - the output pin will be low.

If the input pin is low - the output pin will be high. 

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Together with R3, R4, R5 & C3, the inverters form a simple oscillator.

The oscillator gives the 14-bit counter something to count.

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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. 

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As the count progresses - its state is reflected by the output pins. 

For example - after eight oscillations - Q4 (pin 7) will go high.

After sixteen oscillations - Q5 (pin 5) will go high.

After thirty-two oscillations - Q6 will go high - and so on.

It takes 8192 oscillations for Q14 to go high. 

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R4 controls the frequency of the oscillator.

That is - the speed at which the clock ticks. 

By altering the frequency of the oscillator - you can control the speed at which the count progresses.

In other words - you can control the length of time it takes for any output pin to go high.

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

So it will supply base current to the transistor - through R7.

The transistor will switch on - connect the negative side of the relay coil to ground - and the relay will energize.

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When the output pin goes high - it does a second job.

It takes pin 11 high - through D1.

This stops the oscillator - so the count cannot proceed.

In other words - the timer stops when the relay energizes.

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If the oscillator were to continue to run - the counter would go on counting.

And - after a similar time interval - the output pin would go low again. 

So - the transistor would switch off - and relay would de-energize. 

In fact - the relay would keep energizing and de-energizing at the same regular time intervals. 

If you have an application where this would be useful - simply leave out D1.

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The basic difference between the two circuits is: -

The first uses an NPN transistor to connect the relay coil to ground.

And the second uses a PNP transistor to connect the relay coil to the positive line.

In Circuit No.1 - when the output goes high - the transistor switches on.

In Circuit No.2 - when the output goes high - the transistor switches off.

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While the output pin is low - it's connected internally to the negative line. 

In Circuit No1- it connects the base of the NPN transistor to ground through R7.

So the transistor is switched off - and the relay is de-energized.

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

So - it supplies base current to the transistor - through R7.

This causes the transistor to switch on.

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

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In Circuit No2 - while the output pin is low - it connects the base of the PNP transistor to ground through R7.

So current is flowing through the emitter-base junction - and the transistor is switched on.

The transistor connects the positive side of the relay coil to the positive supply line - so the relay is energized.

But when the output pin goes high - it's connected internally to the positive line.

So - it connects the base to the transistor to the positive line - through R7.

Without a path to ground for its base current - the PNP transistor switches off.

This cuts the power to the relay coil - and the relay de-energizes.

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The reason for the two versions of the same circuit is to help keep overall current consumption to a minimum. 

The first circuit uses less power while the timer is running.

The second uses less power after the timer has stopped.

You choose the one that best suits your application.

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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. 

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The purpose of C2 and R6 is to perform an automatic reset when power is applied.

Initially - at power-up - C2 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 - C2 charges through R6 - and the resistor takes pin 12 low. 

This causes all of the outputs to go low - and the count begins at zero. 
 
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When you're operating the timer you can use the optional reset button. 

It takes pin 12 High and discharges C2. 

When you release the button - C2 will charge through R6 - and the resistor will take pin 12 low. 

The moment pin 12 goes low - all of the outputs go low - and the counter begins to count up from zero. 

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However - when you are setting up the timer - you must reset the circuit by interrupting the power supply. 

The reset button only resets the counter - it does not reset the oscillator. 

In fact it stops the oscillator right at the point where it's about to deliver the first pulse. 

So the first cycle is shorter than normal. 

Consequently - the length of time - between the release of the reset button - and the arrival of the first pulse to be counted - is too short. 

In everyday operation this amount of inaccuracy is insignificant. 

However - during setup - the error is magnified by the factors in the setup table.

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Interrupting the power for a few seconds not only resets the counter - it also resets the oscillator. 

This makes the time it takes for the yellow LED to light - much more predictable.

And makes the fine adjustment of the timer much easier.



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  HOW THE OSCILLATOR WORKS

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Note that R3, R4 & C3 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 R3 & R4. 

The speed at which it charges depends on the value of the capacitor - and the setting of R4. 

As it charges - the voltage at the junction of C3 & R4 will rise. 

When it reaches just over half the supply voltage - it takes pin 11 high - through R5.

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Now everything reverses. 

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

C3 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 C3 & R4 will fall. 

The speed at which it falls depends on the value of the capacitor - and the setting of R4. 

When it falls to just below half the supply voltage - it takes pin 11 low - through R5. 

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

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The oscillator runs because the polarity of the charge on C3 keeps reversing. 

As a result - the junction of C3 & R4 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 C3 keeps reversing. 

It's a sort of controlled feedback - where the frequency generated is determined by the values of R3, R4 & C3.

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The main purpose of R3 is to set an upper limit on the frequency. 

Its value was chosen to provide a minimum period of about 30 seconds at pin 7. 

The value of R4 was chosen to provide up to about 24-hours at pin 3. 

Thus - R4 will adjust the timer over the range from 30-seconds to 24-hours.

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

Since the charge on C3 is constantly reversing in polarity - ideally the capacitor should be non-polarized. 

Modern electrolytic capacitors generally have a very low leakage current - even in the reverse direction. 

So you can usually get away with using a regular polarized "aluminium" electrolytic capacitor.
 
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Because the minimum value of R3 + R4 is 150k - very little current will flow through the resistors. 

So - even if C3 does leak in the reverse direction - the worst that can happen is that the oscillator won't run. 

It's easy to tell if the oscillator is working. 

Set R4 at about mid-point. 

If the green LED is turning on and off at about 5-second intervals - the oscillator is running properly.

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By connecting two regular capacitors back-to-back - you can simulate a non-polarised capacitor.

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

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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

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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.

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The oscillator stops when D1 takes pin 11 high. 

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

But - it prevents a low output pin from holding pin 11 low. 

By clamping pin 11 high - D1 forces pin 10 to remain low.

So pin 9 has to remain high. 

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Since pin 11 can no longer change its state - the polarity of the charge on C3 cannot change. 

So the oscillator cannot run. 

In effect - the counter is frozen at the moment your chosen output pin goes high.

It will remain frozen until the circuit is reset. 

That is - until pin 12 is taken high and then low again.

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Note that R3, R4 & R5 are connected in series between pins 10 and 11. 

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

The value of R5 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 C3. 

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The data sheet for the 4060 says that the value of R5 should be at least ten times the value of R3 + R4. 

That would make it about 6M8. 

But a 4M7 resistor seems to work well enough.

The value of 4M7 was chosen because it's widely available - but you can fit a higher value resistor if you wish. 

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. 

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Coils of all sorts give off high reverse-voltage spikes that will destroy Cmos ICs.

Always connect diodes across relay coils - and across any other device that might contain a coil. 

Note that the cathode of D2 - the side with the bar - is connected to the positive terminal of the coil. 

Any reverse-voltage spikes will be short-circuited at source - before they can do any damage.

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I've used a SPCO/SPDT relay in the drawing. 

But you can use a multi-pole relay - if it suits your application. 

I specified a minimum coil resistance of 270 ohms. 

A higher value will improve battery life.

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With a 12-volt supply - the maximum current through the transistor will be 
(12v / 270 ohms) about 45mA. 

This is well within the limits of the BC547 and BC557 - which have an IC(max) of 100mA.

There is nothing special about the transistor. 

Provided the NPN/PNP requirement is satisfied - any small transistor 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 transistor may be different from that of the BC547 and BC557. 

Just because your transistor looks the same - don't assume that its pins are arranged in the same order.

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Although static electricity can destroy Cmos ICs - they are extremely reliable in operation. 

Most modern versions will survive a fair degree of abuse.

But it's worth taking a couple of basic precautions. 

Avoid touching the pins - and don't overheat them with the soldering iron. 

A socket will reduce the chances of damaging the IC - and will make it easier to replace if necessary. 

If you're using a socket - always turn-off the power - before inserting or removing the IC. 

Also - check that the IC is the right way round. Pin 1 should be in the top left-hand corner.

And check that all of the pins are correctly inserted into the socket. 

Sometimes - instead of entering the socket - a pin will curl up underneath the IC. 

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Construction Guide

Circuit No.1

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Construction Notes

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

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 8 mm drill-bit will do. Just use the drill-bit as a hand tool - there's no need for a drilling machine.

Next make and fit the Five 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.

Then fit the six resistors and the preset. The resistors are all shown lying flat on the board. But those connected between close or adjacent tracks are actually mounted standing upright. See The Photo Of The Prototype .

Now fit the transistor - the two diodes - the IC socket - and the relay. Pay particular attention to the orientation of the diodes. Note that both are facing in the same direction. Again - both diodes are shown lying flat on the board - but D2 is actually mounted standing upright.

Next, fit the remaining components - the 3 capacitors and 2 LEDs. Then examine the underside of the board 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 five solder bridges to the underside of the board. These are just small blobs of solder. I've used them to connect adjacent tracks. They are a simple and convenient alternative to wire links.

Finally - insert the Cmos 4060 into the socket. Make sure that pin 1 is in the top left-hand corner - and check very carefully that all of the pins are correctly inserted into the socket. Sometimes - instead of entering the socket - a pin will curl up underneath the IC.

You Are Now Ready To Test Your Timer

Construction Guide

Circuit No.2

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

Next, fit the remaining components - the 3 capacitors and 2 LEDs. Then examine the underside of the board 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 four solder bridges to the underside of the board. These are just small blobs of solder. I've used them to connect adjacent tracks. They are a simple and convenient alternative to wire links.

Finally - insert the Cmos 4060 into the socket. Make sure that pin 1 is in the top left-hand corner - and check very carefully that all of the pins are correctly inserted into the socket. Sometimes - instead of entering the socket - a pin will curl up underneath the IC.

You Are Now Ready To Test Your Timer

General Information

Do not use the "on-board" relays to switch mains voltage. The board layouts do 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.

24-Hour Timers - Support Material

Circuit Description

Construction Guide

Circuit No.1

Construction Notes

Actual Size

Construction Guide

Circuit No.2

Actual Size

General Information