Everything you need to know - to build a simple Cmos timer.

COMMENTS SUGGESTIONS	Designing Cmos Timers - Section 1	MORE TIMER CIRCUITS

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Introduction

You can build useful and reliable timer circuits - with a single Cmos inverter. We'll begin by looking at some of the techniques available. And we'll produce a number of simple and useful designs. Then we'll string four of these simple timers together - to create a practical Two-Zone Burglar Alarm.

Although I've divided the material into five separate sections - most of what you need to know is in the first part of the first section. Once you understand how Circuit No.1 works - the rest is easy. The remaining circuits - including the alarm circuit - are just simple variations on Circuit No.1

Circuit No.1

How To Make A Simple Timer
With A Single Cmos Gate

In circuit No.1 - we are only using a single Cmos 4093 gate.

Note that the two input pins are joined together. 

In this configuration - gate 1 is being used as a simple inverter.

When pins 1 & 2 are low - pin 3 will be high.

And when pins 1 & 2 are high - pin 3 will be low.

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If we take pins 1 & 2 low - pin 3 will go high.

While pin 3 is high - it's connected internally to the positive line.

And it supplies base current to Q1 - through R3.

The base current switches the transistor on.

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

And the relay energizes.

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If we take pins 1 & 2 high - pin 3 will go low.

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

And it can no longer supply base current to Q1.

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

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In other words - we can energize and de-energize the relay - by changing the state of pin 3.

And we can change the state of pin 3 - by taking pins 1 & 2 high or low.

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         THE TIMING COMPONENTS

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Pins 1 & 2 are connected to the junction of C1 & R1.

When power is first applied - C1 is in a discharged state.

In that state - it acts like a wire link - joining pins 1 & 2 to ground.

In other words - pins 1 & 2 start off low.

So pin 3 will go high - Q1 will switch on - and the relay will energize.

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At the same time - C1 will begin to charge through R1.

As it does so - the voltage across the capacitor will rise.

This means that the voltage on pins 1 & 2 will rise.

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Eventually - the input pins will change from low to high.

Then pin 3 will switch abruptly from high to low.

The transistor will switch off.

And the relay will de-energize. 

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The voltage level - at which the low input pins become high input pins - is uncertain.

It's somewhere around half the value of your supply voltage.

You have no control over it.

It's fixed at the time of manufacture.

And it varies from one batch of ICs to the next.

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Roughly speaking - with the 12-volt supply - anything below 5-volts is a low.

And anything above 7-volts is a high.

Somewhere between 5v and 7v - the transition will occur.

The low input will becomes a high input.

Pin 3 will switch the transistor off - and the relay will de-energize.


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       THE LENGTH OF THE DELAY 

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There are three main things that determine the length of the delay.

The first is the value of the resistor.

The higher the value of the resistor - the more slowly it will charge the capacitor.

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I've used a variable resistor in the diagram.

It will let you control the speed at which the capacitor charges.

In other words - it will let you set the length of the delay.

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The length of the delay is also determined by the value of C1.

The higher the value of the capacitor - the longer it will take to charge.

The length of the delay is directly proportional to the value of the capacitor.

If you double the value of the capacitor - the delay will last twice as long.

If you halve the value of the capacitor - the delay will last half as long.

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The third factor that determines the length of the delay - is the switching voltage of the gate.

That is - the precise voltage at which the low input becomes a high input.

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We've seen that - with a 12v supply - the switching point of the gate is somewhere between 5v and 7v.

When the capacitor is charging - it will reach 5v before it reaches 7v.

If the switching voltage is as low as 5v - the delay will be shorter.

If the switching voltage is as high as 7v - the delay will be longer.

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Just like the switching voltage of your gate - the values of your capacitor and resistor are subject to manufacturing tolerances.

They're unlikely to correspond exactly to their marked values.

This too will influence the length of the delay.

In other words - don't be surprised if your delay is shorter or longer than mine.

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With the 1 Meg resistor and the 100uF capacitor - my relay de-energized after 70 seconds.

With a 5 Meg pot - it took five times longer.

That's a delay of up to about 6 minutes.

Increase C1 to 1000uF - and the delay will last ten times longer again.

That's a total delay of up to about one hour.




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              THE RESET BUTTON

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Once the capacitor has been charged - it will remain charged.

So it will continue to hold pins 1 & 2 high.

And the relay will remain de-energized.

If you want to run the timer again - you must first discharge C1.

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That's what the reset button does.

It discharges C1 - through R2.

R2 limits the peak discharge current through the switch.

And so prevents damage to the switch contacts.

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The value of R2 is not critical.

But it should be low enough to discharge the capacitor completely - with a relatively brief push on the reset button.

If the value of R2 is too high - the capacitor may not discharge completely.

Consequently - it will not take as long to recharge.

In other words - the preset time period will be shortened.

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I've drawn a momentary action push-to-make switch.

But you can use any type of switch that suits your application.

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When you press the reset button - C1 will discharge - and pins 1 & 2 will go low.

So pin 3 will go high - the transistor will switch on - and the relay will energize.

When you release the reset button - C1 will begin to charge once more.

That is - the timer will restart.

You can press the button - and restart the timer - at any time

Even if it's already running. 


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                THE TRANSISTOR

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There's nothing special about the BC547. 

You can use any small NPN transistor - with a gain (hfe) of at least 100 - and an Ic(max) of at least 100mA.

I think practically every small NPN transistor satisfies these requirements.

But always check the pin configuration of your transistor.

It may be different from that of the BC547.


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                   UNUSED GATES

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The inputs of unused gates should always be connected to one of the supply lines.

If they're left unconnected - the gate will switch back and forth uncontrollably

And this causes substantial and unnecessary current consumption.

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I've connected the inputs of gates 2 & 3 to ground - and the inputs of gate 4 to the positive line.

Which line you connect your inputs to - is simply a matter of convenience.

Connect them to whichever line is the handiest.

Note that unused Cmos output pins should always be left unconnected.


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                      THE DIODE

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

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

Note that the cathode of D1 - the side with the bar - is connected to the positive terminal of the coil. 

The small arrowhead indicates the direction in which the diode will conduct.

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



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                  THE CMOS 4093

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You can make timers with a wide range of Cmos gates, inverters and buffers. 

But you'll get the best performance from the Schmidt-Trigger devices. 

These are the 4093 and the 4106. 

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These ICs have a number of desirable characteristics - one of which is an output pin that's reluctant to change its state.

Whether it's high or low - it holds on till the last possible moment.

And when it finally lets go - it changes state rapidly. 

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This means that for any given values of resistor and capacitor - the delay will be longer.

And when the time comes for the relay to drop out - it will do so quickly and cleanly. 

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With other Cmos ICs - the output pin changes state more slowly.

It goes through a period when it's neither high nor low.

And this can cause relays to rattle for a few moments - before they finally let go.


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                   PRECAUTIONS

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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. 

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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 switch-off the power before inserting or removing the IC.

Also - check that the IC is the right way round - and 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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The small arrows - top and bottom - indicate that pin 14 is connected to the positive line - and pin 7 is connected to the negative line. I've used a SPCO/SPDT relay in the diagram. But you can use a multi-pole relay - if it suits your application.

You Are Here | Section 2 | Section 3 | Section 4 | Section 5

Circuit No.2

How To Make A Simple Timer
With A Single Cmos Inverter

It's assumed that you have read - and understood - the description of Circuit No.1

What follows - concentrates only on those areas where Circuit No.2 is different.

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This circuit is practically identical to Circuit No.1

It just operates the other way round.

The circuit powers-up with the relay de-energized.

And - it only energizes when the time delay expires.

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Note that R1 and C1 have changed places. 

Here - the discharged capacitor begins by taking pins 1 & 2 high.

So - pin 3 will be low - Q1 will be switched off - and the relay will be de-energized.

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C1 is still charged by R1.

But this time R1 is removing positive charge from the capacitor's negative plate

As it does so - the voltage on pins 1 & 2 will fall.

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How quickly it falls depends - as before - on the value of both the capacitor and the resistor.

Eventually - the high input pins will become low input pins.

Then pin 3 will go high - Q1 will switch on - and the relay will energize.

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I got a slightly shorter delay with this configuration.

The 1 Meg resistor and 100uF capacitor gave me about 60 seconds.

This was caused by the relatively high switching voltage of my particular gate.

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Of course - the delay could just as easily be longer.

The length of the delay depends on how far the input voltage has to fall - before it becomes a low.

If the gate's switching voltage were lower - the voltage at the junction of C1 & R1 would have to fall further.

This would require extra time.

And the delay would be longer.

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When you press the reset button - C1 discharges - and pins 1 & 2 are taken high.

So pin 3 will go low - the transistor will switch off - and the relay will de-energize.

When you release the reset button, the timer will restart.

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You can press the button and restart the timer at any time

Even if it's already running.

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The timers will work at anything from 5-volts to 15-volts. All you have to do - is choose a relay with a coil that suits the voltage you want to use. The BC547 will switch 50mA - comfortably. If your relay coil needs more than 50mA - upgrade the transistor to say a BC337.

You Are Here | Section 2 | Section 3 | Section 4 | Section 5

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Designing Cmos Timers - Section 1

Introduction

Circuit No.1

Circuit No.2