A Beginner's Guide To The Cmos 4001

A Simple Explanation Of How The Cmos 4001 Works.
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Simple Door Alarm Circuit


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This article starts with a detailed look at the operation of the Cmos 4001.

It then considers the practical use of the IC in the small alarm circuit on the left. 

And it provides links to some more examples of Cmos based circuits. 

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The 4001 is a 14 pin - DIL - Integrated Circuit. 

DIL refers to the layout of the pins. 

It stands for "Dual-in-Line".

It simply means that the 14 pins are arranged in two parallel rows.

i.e. There are 7 pins on each side.

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The top of the IC is usually marked with a small notch - coloured yellow in the diagram.

When viewed from above - pin 1 is on the left of the notch - and pin 14 is on the right.

In addition - pin 1 is often identified by a small dot.

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Although static electricity can destroy Cmos ICs - they're 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.

It also makes the IC much easy to replace - if necessary.

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

Always be careful to insert it the right way round.

And always check that all of the pins have entered the socket. 

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

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A basic 4000 Series Cmos IC will work from a minimum of 3-volts up to a maximum of 15-volts.

However, some versions of the 4000 Series will operate up to a maximum of 18-volts.


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One of the most useful features of Cmos ICs is that they use very little current.

This makes them ideal for battery-powered circuits.

The circuit on the left is designed to be powered by a small 9-volt battery.

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


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The Cmos 4001 contains four distinct circuits known as "gates". 

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. 

For example - pins 1 & 2 are the Inputs of gate 1 - and pin 3 is the output .

By altering the voltages on the two input pins - you can turn the output pin on and off.

And - if you can turn the output pin on and off - you can use it - to turn other devices on and off.

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When the output is on - it's connected internally to the positive rail - and it will supply a current of up to about 10mA. 

When the output is off - it's connected internally to the negative rail - and it will "sink" a current of up to about 10mA.  

"Sink" simply means that it will allow a current of up to about 10mA - to flow into the output pin.

When the output is on - it's said to be "High".

And when the output is off - it's said to be "Low". 

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The two inputs of a Cmos 4001 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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Each input has a very high internal resistance (impedance). 

So almost no current flows into the input pins. 

Instead - the mere presence of a voltage at the inputs - is all that's required to make the gate operate.

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In the circuit on the left - both the gate 1 inputs are connected directly to the   
9-volt rail. 

There's no need for an external resistor. 

The high impedance of pins 1 & 2  allows you to connect them to the positive line. 

In this condition - the inputs are said to be "High"

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Cmos outputs do have some internal resistance.

However, it's not very large - and it shouldn't be relied upon to limit the output current to a safe level.

Whenever you draw current from - or sink current into - a Cmos output pin - you should always use an external resistor - to keep the current below the permitted 10mA.

Although an output current of 10mA may seem small, it's enough to operate a transistor switch or an optical-isolator.

So you can control almost any type of device - with the output from a Cmos gate.


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The four gates in a Cmos 4001 IC are sometimes called "Inverters". 

As the name suggests - the state of the gate's output will be the opposite (or inverse) of its inputs.

In the circuit on the left - the inputs of gate 1 are connected to the positive rail -so they are "High". 

And since gate 1 is an inverter - this means that the output - at pin 3 - is "Low".

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On the other hand - the gate 2 inputs are connected to the negative line. 

Actually they're connected to pin 7 - which you'll recall is connected to the negative line.

Their state is said to be "Low".

And since gate 2 is an inverter - this means that the output - at pin 4 - is "High".

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You can tell the gates are inverters from the diagram - because the symbol for each gate is drawn with a small circle at the output. 

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Note that gates 1 and 2 are not used in the circuit. 

When a gate is not required - its inputs should always be connected to one of the supply rails. 

It doesn't matter which.

If the inputs are not tied to one of the supply rails - the output will constantly switch from one state to the other.

And this causes unnecessary current consumption.

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When they are not actually switching - Cmos ICs draw virtually no current.

However - at the exact moment that the output changes state - from High to Low or vice-versa - a significant current flows. 

That's why an unused gate - allowed to switch back and forth between these two states - will cause an increase in current consumption.

It should be noted that the outputs of unused gates MUST always be left unconnected.


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The box filled with Zeros and Ones is known as a "Truth Table".

The table lists the four possible combinations of input and output states.

A "One" represents a High - and a "Zero" represents a Low.

The first two columns represent the state of the gate's two inputs.

And the third column represents the corresponding state of the gate's output.

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You already know that when the two inputs are High - the output will be Low. 

That is the condition of gate 1. 

It's represented by the top row of the table - where the entry is

                  "1 + 1 = 0".

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The condition of gate 2 is represented by the bottom row of the table.

Both inputs are Low - so the output is High. 

The table reads 

                  "0 + 0 = 1".

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There are two more possible situations. 

These arise when one of the inputs is High - and the other is Low. 

If we call the two inputs "a" and "b" - we could make "a" High and "b" Low - or "a" Low and "b" High.

These two configurations are covered by rows two and three of the table. 

In both cases - the output is Low.

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Once you understand the truth table - you can predict the type of output any combination of inputs will produce.


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The gates in a Cmos 4001 are known as "NOR" gates. 

This is short for "NOT OR". 

It means that they are "OR" gates with an "Inverted Output".

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Stay with me. 

It's not as complicated as it sounds.

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The logic of the NOR gate may be described as follows: -

If input "a" is High

OR  input "b" is High

OR both inputs "a" and "b" are High

Then the output will NOT be High.

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In other words - while at least one of the inputs is High - the output will be Low.

That - in a nutshell - is all you need to remember.

It summarizes neatly - the contents of the Truth table.

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You can tell a NOR gate by the way its symbol is drawn.

The concaved shape of the input side of the symbol tells you that it's basically an "OR" gate.

While the small circle at the output - tells you that it's also an Inverter.

And an OR gate with an inverted output - is a NOR gate.


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There's only one more thing we need to examine.

And that is the meaning of High and Low in the present context.

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Depending on whether it's High or Low - the output pin is connected internally to either the positive or the negative line. 

So for practical purposes - you can think of a High as equal to the supply voltage - and a Low as equal to zero volts.

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However - in the case of the input pins - High and Low are relative terms. 

An input pin is High - when it's more than half of the supply voltage.

And an input pin is Low - when it's less than half of the supply voltage.

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Roughly speaking - with the 9-volt supply - anything over 5-volts is a High.

And anything below 4-volts is a Low.

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Somewhere between 4 and 5 volts - is the crossover or switching point of the gate.

Depending on whether the voltage is rising or falling - when it crosses this point - the output pin will switch from High to Low - or vice-versa.

Exactly how this works will become clearer when we consider the operation of the door alarm circuit. 


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In the circuit on the left - gates 3 and 4 are connected to form a monostable.

A monostable is a circuit that - when triggered - will provide an output for a certain length of time - and then it will switch itself off.

Our monostable is triggered by SW1.

And the output - at pin 11 - activates a transistor switch.

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When pin 11 goes high - it supplies base current to Q1 - through R3.

This switches the transistor on.

The transistor connects the negative lead of the buzzer to ground.

And the buzzer sounds.

How long the buzzer sounds - depends on the values of R2 and C2.

With the values shown - it will sound for somewhere between 3 and 5 seconds.

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The monostable circuit is a basic building block with a wide range of uses.

However - if we place our circuit in a practical context - it will help to make its operation easier to understand.

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If SW1 is fitted to a door - every time the door opens - the buzzer will give a short beep.

In an unattended shop - or reception area - the sound of the beep will alert you to the fact that you have a customer.


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In standby mode - R2 holds pins 12 and 13 High. 

So the output of gate 4 - at pin 11 - is Low.

Because pin 11 is Low - it cannot supply base current to Q1.

So the transistor is switched off - and the buzzer is silent.

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The normally-closed switch - SW1 - is holding pin 8 Low.

And pin 11 is holding pin 9 Low.

Since both the gate 3 inputs are Low - the output at pin 10 - is High.

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Note that in this condition both plates of C2 are High.

R2 is holding the positive plate High

And pin 10 is holding the negative plate High. 

Since there is no difference in potential across C2 - the capacitor is in a discharged state.

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Opening SW1 will trigger the alarm.

C1 has a relatively low value.

It will charge quite quickly through R1. 

As it does so - the voltage on pin 8 rises.

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When it reaches just over half the supply voltage - about 5-volts - the state of the input pin changes from Low to High. 

Now that one of its input pins is High - the output of gate 3 will go Low. 

See the Truth Table.

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Because C2 is in a discharged state - it acts like a wire link - between pin 10 and the inputs of gate 4.

In other words - when pin 10 goes low - it takes pins 12 and 13 Low as well 

So the output of gate 4 - at pin 11 - goes High. 

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The High pin 11 provides base current for Q1. 

This switches the transistor on - and the buzzer sounds

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This next bit - is the important bit.

It's pin 8 that starts the timer.

But it's pin 9 that keeps it going.

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When pin 11 goes High - it takes pin 9 High.

By taking pin 9 High - pin 11 guarantees that pin 10 will remain low.

In effect - pin 11 uses pin 9 to latch the gate 3 output Low.

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It doesn't matter now - if you close the door - and take pin 8 low again.

It will have no effect on the duration of the output period.

Pin 11 has taken control of gate 3.

And while pin 11 holds pin 9 High - 
pin 10 will remain Low.

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At this stage - C2 begins to charge slowly through R2.

And the voltage on pins 12 and 13 starts to rise.

How quickly it rises - is determined by the values of R2 and C2.

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When pins 12 and 13 reach just over half the supply voltage  - about 5-volts - the inputs change state from Low to High.

This means that the gate 4 output - at pin 11 - will go Low. 

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When pin 11 goes Low - it can no longer supply base current to Q1.

So the transistor switches off - and the buzzer is silenced.

The Low pin 11 - also takes pin 9 Low.


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At this point there are two possible situations. 

Either SW1 has already been closed - or it's still open

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If Sw1 is already closed - both of the gate 3 inputs are Low

So pin 10 will go High. 

The High pin 10 will discharge C2 - by applying a positive charge to its negative plate. 

And - with C2 discharged - the circuit is reset - and the monostable is ready to be re-triggered.

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However - if SW1 is still open - pin 8 is still High.

And - since at least one of the gate 3 input pins is High - the output pin will remain Low

This means that C2 will remain charged.

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To reset the monostable - you must first close SW1. 

Then both of the gate 3 input pins will be Low.

Pin 10 will go High - C2 will discharge - and the circuit will reset.

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Put more simply - the alarm is triggered by opening SW1.

But you can't open SW1 - if it's already open.

Therefore - you'll have to close it first - before you can re-trigger the alarm.


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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. 

I put one across the buzzer - just in case. 

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Note that the cathode - or plus side - of the diode is always 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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In the circuit - R3 limits the current taken from gate 4 - to about 2 mA.

This 2 mA is the base-current that operates the transistor switch.

And the transistor in turn - sounds the buzzer. 

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I used a small piezo type buzzer that draws a current of about 10 mA.

But the BC547 will reliably cope with a current of up to 50mA.

That's about half its "Ic max".

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So you can replace the buzzer with a relay coil.

And use the relay contacts to operate practically any type of output device you like.

You just need to make sure that the coil of your relay does not draw more than about 50mA.

Otherwise the transistor might be overloaded.


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Cmos gates react very fast. 

High frequency electrical interference could cause the monostable to trigger - causing a "false alarm".

C1 makes the circuit more stable - because it slows down the response time of the trigger.

To high frequency signals - a 100nF capacitor looks like a short circuit.

C1 simply shorts the interference signals to ground.

Similarly - C3 removes interference from the supply line.

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Note that while SW1 is closed, current flows to ground through R1.

The value of 4M7 was chosen to keep the standby current as low as possible.

We can use such a high value resistor - because we don't need current to operate the input pin. 

The mere presence of the voltage at pin 8  - is all that's required to switch the gate. 

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SW1 is drawn as a magnetic reed switch.

But you can use any type of switch - or other device - that suits your application.

For example - if you use a PIR - the buzzer will tell you when someone has entered a restricted area.

If you have more than one door/area to protect - you can use more than one trigger device. 

Just wire all of your switches, PIRs etc. - in series.


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The length of time the buzzer sounds - is controlled by the values of R2 and C2.

There's little point in trying to calculate which values of resistor and capacitor - will give you the output time you require. 

Manufacturing tolerances are such that the results wouldn't be very accurate.

It's best to use trial and error when selecting components.

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Generally speaking - the length of the output time is proportional to the values of R2 and C2.

In other words - if you double the value of either R2 or C2 - you'll double the output time.

And - if you halve the value of either R2 or C2 - you'll halve the output time.

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For example - if you replace R2 with a 4M7 resistor - you'll increase the output time by a factor of about 5.

If you replace C2 with a 470uF capacitor - you'll increase the output time by a factor of about 100.

If you use a 4M7 resistor AND a 470uF capacitor together - you'll increase the time by a factor of about 

               5 X 100 = 500. 

This should give you an output of around half-an-hour or more.

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If you require an accurate output time - you can replace R2 with a variable-resistor.

And adjust its value to suit your needs.


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It should be clear that the output time - is determined by the values of R2 & C2 alone. 

The length of time the trigger switch is open - is irrelevant.

Indeed - while the output timer is running - you can open and close SW1 as many times as you like. 

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The reason is simple.

Pin 11 is now in control of gate 3.

Thanks to the link between pins 11 & 9 - SW1 cannot change the state of pin 10.

So it can't effect C2's state of charge.

And consequently - it can't effect the duration of the output.

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Whether SW1 is opened for a fraction of a second - or for an hour - the length of the output pulse will be the same.

Monostables are often used to lengthen a short input pulse - or to shorten a long input pulse.


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Although a Cmos 4001 gate is a digital device - there are times when its output is more akin to analogue.

When timing components are used - the voltage on the input pins rises and falls slowly.

So it may hover around the gate's switching point - for a significant period of time.

And during this period - the output voltage will rise or fall in an analogue manner.

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It's as though the gate cannot decide whether the input pins are high or low.

As a result - the output pin goes through a transition phase - during which it is neither high nor low.

Because the output pin is changing state slowly - any device connected to it - will switch on or off slowly.

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The effect is particularly noticeable where the timer circuit is set up to provide long delays.

The transition phase can last a significant length of time.

And it can cause relays to rattle - and buzzers to fade in and out slowly.

However - it's usually possible to speed up the transition.

You just need a couple of additional components - to provide the right kind of feedback.

In other words - once it starts to go - you give it a little push to help it on its way.


 
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An Introduction To The Cmos 4011
Door Alarm Construction Guide	The Rest of Ron's Circuits	Write To Ron	More Free-to-Use Circuits	Circuit Exchange International
An Introduction To The Cmos 4001

Simple Door Alarm Circuit
