A Beginner's Guide To The Cmos 4011

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

Circuit Diagram For
A Simple Fridge Alarm

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This article starts with a detailed look at the operation of the Cmos 4011. 

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 4011 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 4011 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 4011 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 4011 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 High.

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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 4011 are known as "NAND" gates. 

This is short for "NOT AND". 

It means that they are "AND" 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 NAND gate may be described as follows: -

If input "a" is High

AND  input "b" is High

Then the output will NOT be High.

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Or to put it more clearly - while at least one of the inputs is Low - the output will be High. 

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 NAND gate by the way its symbol is drawn.

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

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

And an AND gate with an inverted output - is a NAND 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 fire-door alarm circuit. 


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In the circuit on the left - gates 3 and 4 are connected to form an astable oscillator.

The essence of an oscillator is that it's constantly switching back and forth from one state to another. 

Gate 3 and 4  take turns at switching each other "ON" and "OFF". 
And the speed at which they do this is controlled by R2 and C2. 

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Every time 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 gives a short beep.

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The oscillator 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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It's very important that fire-doors are kept closed at all times. 

If they're left propped open they're no longer doing their job. 

If SW1 is connected to the fire-door - this circuit will allow you to open and close the door without sounding the buzzer. 

But if the door is left open for more than about 30 seconds -the buzzer will start to give a series of short warning beeps.

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The length and speed of the beeps is set by R2 and C2. 

With the values shown - the buzzer will switch on and off at about half-second intervals. 

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In standby mode SW1 holds pin 8 Low. 
This forces the gate 3 output - at pin 10 - to remain High. 

And while one of the gates is prevented from changing the state of its output - the oscillator can't run.

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The High pin 10 holds pins 12 and 13 High.

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

The Low pin 11 is unable to supply base current to the transistor. 

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

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Note that R2 and C2 are connected in series between pins 10 and 11. 

Note also that pin 9 is connected to the junction between the resistor and capacitor. 

It's the voltage at this junction that controls the voltage on pin 9.

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In standby mode - current from pin 10 charges C2 through R2.
 So the voltage on the junction of C2 and R2 is High. 

Consequently - the voltage on pin 9 is also High 

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When SW1 is opened - C1 starts to charge slowly - through R1 and R4. 

And - as it does so - the voltage on pin 8 will begin to rise.

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If the door is re-closed immediately - C1 will discharge rapidly - through R4 and SW1

SW1 will take pin 8 Low. 

And the circuit will reset.

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But - if the door is left open - C1 will continue to charge.

And the voltage on pin 8 will continue to rise.

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

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You'll recall that pin 9 is already High.

So now - both of the gate 3 inputs are High.

And the gate 3 output - at pin 10 - will go Low. 

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Pin 10 takes pins 12 and 13 Low.

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

The High pin 11 provides base current for Q1. 

This switches the transistor on.

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

And the buzzer sounds.

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Note that the polarity of the voltage across R2 and C2 had reversed. 

Pin 11 is now High - and pin 10 is Low 

C2 begins to charge once more - but this time it's in the opposite direction.

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Since pin 11 is High - the junction of R2 and C2 represents the negative side of the capacitor. 

And as C2 charges - the voltage on this junction will fall. 

This means that the voltage on pin 9 will fall. 

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When it falls to just under half the supply voltage - about 4-volts - the pin 9 changes from High to Low. 

And now that one of its inputs is Low - the output of gate 3 will go High.

See the Truth Table.

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The High pin 10 takes pins 12 and 13 High.

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

The Low pin 11 can no longer supply Q1 with base current. 

So the transistor switches off.

And the buzzer is silenced.

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Note that the polarity of the voltage across R2 and C2 had reversed once more.

Now pin 10 is High - and pin 11 is Low. 

C2 begins to charge once more - but this time it's in the opposite direction.

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Since pin 10 is High - the junction of R2 and C2 once again represents the positive side of the capacitor. 

And as the capacitor charges - the voltage on this junction will rise. 

This means that the voltage on pin 9 will rise.

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

And now that both of its inputs are High - the output of gate 3 will go Low.

See the Truth Table.

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While SW1 remains open - pin 8 will remain High.

And while pin 8 is High - the oscillator will continue to run.

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The oscillator runs because the polarity of the charge on C2 is constantly reversing. 

As a result - the junction of R2 and C2 keeps changing back and forth between High and Low.

So pin 9 keeps changing back and forth between High and Low. 

And because pin 9 keeps changing back and forth between High and Low - the polarity of the charge on C2 keeps reversing.

It's a sort of controlled feedback - where the speed at which the buzzer switches on and off is controlled by the values of R2 and C2 

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While SW1 remains open - and pin 8 is High - the transistor will continue to turn the buzzer on and off.

But once SW1 is re-closed - C1 will discharge rapidly - through R4 and SW1.

The initial delay will reset - pin 8 will be taken Low - and the oscillator will be disabled. 


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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 a "false alarm".

As well as providing the initial delay - C1 tunes-out interference picked up by the wires going to SW1. 

It makes the circuit more stable - by slowing down the response time of the trigger. 

C3 does the similar job of removing stray 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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The purpose of R4 is to protect SW1 from the surge current as C1 discharges. 

It's true that with a 10uF capacitor the resistor is probably not necessary. 

But its presence allows you to extend the initial delay - by increasing C1 to any value you like. 

The value of R4 is not critical. 

So long as it discharges C1 reasonably quickly - any value up to 100k will work fine. 

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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.

If you have more than one door - you can use more than one switch. 

Just wire all of your switches 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 - both the length of the beeps and the length of the pauses between the beeps - are proportional to the values of R2 and C2. 

In other words - if you double the value of either R2 or C2 - both the beeps and pauses will be twice as long. 

If you halve the value of either R2 or C2 - both the beeps and pauses will be half as long. 

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If the time period you want is long - and C2 is an electrolytic capacitor - it should be "non-polarized" - because the charge on C2 constantly reverses polarity. 

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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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Although a Cmos 4011 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.

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One way to reduce this effect - is to replace the Cmos 4011 - with a Cmos 4093. 

The 4093 is a "Schmidt Trigger" version of the 4011. 

Its gates have an additional circuit element that speeds up the rate at which the output changes state. 

In other words - once the gate starts to switch - it does so very quickly. 


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An Introduction To The Cmos 4001
Fire-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 4011

Simple Fire-Door Alarm Circuit
