Digital Teaching Aid (DED Philippinen, 86 p.)

Digital Teaching Aid (DED Philippinen, 86 p.) Fundamental Logic Operations - Lesson 1 Lesson Plan (introduction...) Introduction TTL circuits Logic operations Handout No. 1 Worksheet No. 1

Digital Teaching Aid (DED Philippinen, 86 p.)

Fundamental Logic Operations - Lesson 1

Lesson Plan

Titel: Fundamental Logic Operations

Objectives:

- Understand the principle of NOT, AND, OR operations
- Able to convert binary and decimal numbers

Introduction

Fundamental Logic Operations

Digital Electronics

The world of electronics is divided into two areas: analog and digital. Analog circuits consist mainly of amplifiers for voltage or current variations that are smooth and continuous. Digital circuits provide electronic switching of voltage pulses. A pulse has abrupt changes between two extreme amplitude levels (i.e.: 5 volt = high level and 0 volt = low level).

Since the digital signal has only two significant levels, either high or low, it is useful to represent the pulses in a binary number system with the digits 1 and 0. (see also Worksheet 1, Binary number system)

Fig. 1-1: Analog versus digital signals

Transistor in digital electronic

A transistor can be used in analog and digital electronic. In digital electronic the transistor operates usually in hard saturation. (see also Amplifier Teaching Aid, Transistor as switch)

Fig. 1-2: Load line and circuit for a transistor switch

Hard Saturation

To get hard saturation, a designer makes I_C approximately 10 times the value of I_B.

When the max. Vin equals to the supply voltage you can get hard saturation by using a ratio of:

10:1 for R_B/R_C

Inverter Function

A transistor switch (circuit as above) can be used to build the first device in digital electronic, the inverter:

Fig. 1-3: Inverter symbol and truth table

Because in digital electronics mainly ready made devices are used (Integrated circuits IC), we don't care any more how the single device is built up. From now on we will use only these ready made devices.

TTL circuits

1964 TEXAS Instruments introduced a family of digital devices which became standard elements in digital electronic: The TTL (Transistor Transistor Logic) circuits. The High (1) and Low (1) state is represented by voltage levels.

Fig. 1-4: Voltage levels for high and low state in TTL IC's

Logic operations

All decisions and operations can be realized by means of the three basic operations:

NOT, AND, OR

NOT (Inverter)

Fig. 1-5: NOT (Inverter) symbol and truth table

Circuit example: 7404 TTL device (see Handout No. 1)

HO: A 1KHz square wave drives pin 1 of a 7404. What does the voltage waveform at pin 2 look like?

Solution:

Fig. 1-6: Timing diagram, Inverter input (pin 1) and output (pin 2)

AND gates

AND operation: The output supplies an high (1) signal if to all inputs high (1) signals are applied.

Fig. 1-6: AND symbol and truth table

AND application example: An elevator motor may only start to work if the doors are shut AND an operation occured.

Circuit example: TTL device 7408, two input AND gate

Ex: Develop the truth table for the following logic circuit.

Fig. 1-8: Logic circuit with truth table

OR gates

OR operation: An OR operation supplies a high signal at the output if to one or more inputs high signals are applied.

Fig. 1-9: OR gate, symbol and truth table

OR application example: A pump has to be switched on when the water level has fallen to a certain level OR too much water is taken out of the container.

Circuit example: TTL device 7432, two input OR gate

Ex: Develop the truth table for the following logic circuit.

Fig. 1-10: Logic circuit with truth table

Handout No. 1

The 7400 Family of TTL Devices (Sample List)

Figure

Worksheet No. 1

Binary number system

All number systems have a base, which specifies how many digits can be used in each place count. For binary numbers the base is 2, with 0 and 1 as the only two digits. In the decimal system, the base is 10.

Decimal digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

Binary digits: 0, 1

Decimal to Binary Conversion

EX: Convert a 4 bit binary number (100 1) into decimal.

Each digit position has specified weight, for binary numbers the position represents a power of two.

EX: Convert a 8 bit number (10010101) into decimal

HO: Convert the following binary numbers into decimal:

101, 1100, 1110, 1000, 0110, 11001010, 01101111, 10001100

Binary to decimal conversion

EX: Convert the decimal number 138 into a binary number

HO: Convert the following decimal numbers into binary numbers:

12, 8, 127, 247, 139, 255

Digital Teaching Aid (DED Philippinen, 86 p.) Introduction Boolean Algebra - Lesson 2 Lesson Plan Introduction Boolean algebra Boolean equations of logic circuits NOR and NAN gates De Morgan's theorems Worksheet No. 2

Digital Teaching Aid (DED Philippinen, 86 p.)

Introduction Boolean Algebra - Lesson 2

Lesson Plan

Introduction

Titel: Introduction Boolean Algebra

Objectives:

- Able to express the basic operations in Boolean algebra
- Know how to describe a logic circuit in a Boolean equation
- Understand the basic Boolean theorems.

Boolean algebra

George Boole (1854) invented a new kind of algebra that could be used to analyse and design digital and computer circuits.

NOT operation

Fig. 2-1: Inverter symbol and Boolean notation

Ex: If A is 0 (low) ® X = NOT 0 = 1

In Boolean algebra the overbar stands for NOT operation.

OR operation

Fig. 2-2: OR symbol and Boolean notation

Ex: If A = 0, B = 1 ® X = A or B = 0 or 1 = l

In Boolean algebra the + sign stans for the OR operation:

X = A + B

Ex: If A = 1, B = 0 ® X = A + B = 1 + 0 = 1

AND operation

Fig. 2-3: AND symbol and Boolean notation

In Boolean algebra the multiplication sign stands for the AND operation:

X = A · B

or simply:

X = A B

Ex: If A = 1, B = 0 ® X = A B = 1 · 0 = 0

Boolean equations of logic circuits

You can use Boolean algebra as a shorthand notation for digital circuits.

Ex:

Fig. 2-4: Digital circuit

Output of the first gate:

X3 = A + B

Output of the second gate:

X6 = X3 + C = A + B + C

The final output is:

X8 = X6 + D = A + B + C + D = X

HO: Find the Boolean equation for the following circuit.

Fig. 2-5: Logic circuit

Solution:

X = (A + B) (C + D)

NOR and NAN gates

NOR gate

Based on the three fundamental logic operations it is possible to design additionel logic devices.

Fig. 2-6: NOR gate, symbol and truth table

Formula:

NAND gate

Fig. 2-7: NAND grate, symbol and truth table

Formula:

De Morgan's theorems

Augustus De Morgan was the first who found the link between logic and mathematics.

Fig. 2-8: Logic circuits with the same output

Although the circuits are different but the output, as we can see in the truth table, is equal. Therefore we can write:

De Morgan's First Theorem

Also if we compare the two circuits on the next page we can easily see that the output is the same, although the circuits are different.

(see Fig. 2-9)

Fig. 2-9: Logic circuits with the same output

So we can write:

De Morgan's Second Theorem

HO: How can you connect a NAND gate to get an inverter

Solution:

Figure

HO: How can you connect NAND gates to get an OR gate?

Solution:

Figure

Worksheet No. 2

No. 1 What is the Boolean equation? What is the truth table?

No. 2 What is the Boolean equation? What is the truth table?

No. 3 What is the Boolean equation? What is the truth table?

No. 4 Construct the truth table.

No. 5 Draw the logic circuit whose Boolean equation is

Use the 7404 and the 7432 with pin numbers

No. 6 Draw the logic circuit whose Boolean equation is

Use the 7404, 7432 and the 7411 with pin numbers.

Digital Teaching Aid (DED Philippinen, 86 p.) Circuit Analysis and Design - Lesson 3 Lesson Plan (introduction...) Introduction Boolean laws and theorems Sum of product method Design example Handout No. 2 Worksheet No. 3

Digital Teaching Aid (DED Philippinen, 86 p.)

Circuit Analysis and Design - Lesson 3

Lesson Plan

Titel: Circuit Analysis and Design

Objectives:

- Know how to apply the basic Boolean laws
- Able to create digital circuits with the help of the sum of products method

Introduction

Circuit Analysis and Design

Boolean laws and theorems

Basic laws

(see also Handout No. 2)

Commutative law:

A + B = B + A

A B = B A

Fig. 3-1: Commutative law

Associative law:

A + (B + C) = (A + B) + C

A (B + C) = (A B) C

Fig. 3-2: Associative law

Distributive law:

A (B + C) = AB + AC

Fig. 3-3: Distributive law

Boolean relations about OR operations

A + 0 = A

Proof:

when A is 0

0 + 0 = 0

when A is 1

1 + 0 = 1

A + A = A

Proof:

when A is 0

0 + 0 = 0

when A is 1

1 + 1 = 1

A + 1 = 1

Proof:

when A is 0

0 + 1 = 1

when A is 1

1 + 1 = 1

A + A = 1

If one input is high, the output is high no matter what the other input is.

Boolean relations about AND operations

A · 1 = A
A · A = A
A · 0 = 0

HO: Check the equations above in the same way as we did it before.

A · A = 0

If one input is low, the output is low no matter what the other input is.

Double inversion

De Morgan's theorems

see also Lesson 1

Duality theorem

1. Change each OR sign to an AND sign

2. Change ech AND sign to an OR sign

3. Complement any 0 or 1 appearing in the expression

TIP: Proof it with a truth table

Ex: Simplify the following Boolean equation

Solution: (see also Handout No. 2)

HO: Simplify the following Boolean equation

X = B + B
X = B

Sum of product method

Fig. 3-4: Example truth table with fundamental products

Design example

Sum of products equation

Given is the following truth table:

Fig. 3-5: Example truth table

We have to locate each output 1 in the truth table and write down the fundamental product.

The next step is to OR the fundamental products:

Now we can derive the corresponding logic circuit:

Fig. 3-6: Logic circuit

HO: What is the sum of product circuit for the given truth table?

Fig. 3-7: Truth table

Handout No. 2

Boolean Algebra Theorems

Worksheet No. 3

No. 1 Simplify the Boolean equation and discribe the logic circuit:

No. 2 Simplify the following Boolean expressions:

No. 3 A digital system has a 4-bit input from 0000 to 1111. Design a logic circuit that produces a high output whenever the equivalent decimal input is greater than 13.

No. 4 In a heating plant the burner X has to be switched on, when the circulating pump A is actuated and the temperature probe B for the warm water supply or the room temperature probe C respond.

a) Develop the truth table
b) Write down the sum of products equation
c) Draw the logic circuit
d) Use Boolean algebra to simplify the equation
e) Draw the corresponding logic circuit.

Digital Teaching Aid (DED Philippinen, 86 p.) Karnaugh Mapping - Lesson 4 Lesson Plan (introduction...) Introduction Karnaugh map Truth table to Karnaugh map Pairs, Quads, and Octets Overlapping and Rolling Worksheet No. 4

Digital Teaching Aid (DED Philippinen, 86 p.)

Karnaugh Mapping - Lesson 4

Lesson Plan

Titel: Karnaugh Mapping

Objectives:

- Able to derive Karnaugh maps from a given truth table
- Know how to simplify Karnaugh maps

Introduction

Karnaugh Mapping

Simplify a Boolean equation

Karnaugh map

As you see, Boolean algebra can be used to create simpler circuits. But if you not quite familiar with Boolean theorems it is difficult to find the best simplification. Therefore Karnaugh mapping is the better alternative for simplification.

Truth table to Karnaugh map

The starting point for digital circuit design is usually the truth table which gives us the following information:

Under which input condition occurs a certain output condition.

Two variable map

Ex:

Fig. 4-1: Two variable Karnaugh map

The truth table outputs are translated into the Karnaugh map. Each position in the map represents a certain fundamental product.

Three variable map

Ex:

Fig. 4-2: Three variable Karnaugh map

Four variable map

Many digital systems process 4 bit numbers. For this reason, logic circuits are often designed to handle 4 input variables.

Ex:

(see Fig. 4-3)

Fig. 4-3: Four variable Karnaugh map

Pairs, Quads, and Octets

Pairs

Fig. 4-4: Four variable simplification

As you see in Fig. 4-4, only one variable goes from uncomplement to complement. Whenever this happens, you can eliminate the variable that changes form.

Ex:

Fig. 4-5: Pairs

Whenever you see a pair first encircle it and then simplify to get the simplified Boolean expression:

Quad

Fig. 4-6: Quad

Quad: A group of 4 one's that are horizontally or vertically adjacent. End to end or in form of a square.

A quad eliminates two variables and their complements.

Encircle the quad and step through the different one's in the quad and determine which two variables go from complement to uncomplement (or vs), these are the variables that drop out.

Ex:

Fig. 4-7: Quad

The variables B and D can be eliminated. So we get the following equation:

X = A C

Octet

Fig. 4-8: Octet

An octet eliminates three variables and their complements.

Karnaugh Simplifications

Process:

1. Draw the Karnaugh map
2. Look for octets and encircle them.
3. Look for quads and encircle them.
4. Look for pairs and encircle them.
5. Simplify and write down the equation.

Ex:

Fig. 4-9: Karnaugh map

Overlapping and Rolling

Overlapping groups

Ex:

Fig. 4-10: Karnaugh map

Groups can overlap to get a simpler equation:

Rolling the map

Ex:

Fig. 4-11: Karnaugh map

Instead of encircling two pairs:

We can roll the map and encircle a quad:

HO: Simplify the following map.

Solution:

HO: Simplify the following map.

Solution:

Worksheet No. 4

No. 1 Simplify the following Karnaugh maps and write down the Boolean equation for every map.

No. 2 Translate each output (v, w, x, y, z) into a Karnaugh map, do the simplification and write down the Boolean equation.

Digital Teaching Aid (DED Philippinen, 86 p.) Karnaugh Mapping II - Lesson 5 Lesson Plan (introduction...) Introduction Product of sums method Don't care conditions Worksheet No. 5

Digital Teaching Aid (DED Philippinen, 86 p.)

Karnaugh Mapping II - Lesson 5

Lesson Plan

Titel: Karnaugh Mapping II

Objectives:

- Understand the product of sums method
- Able to use “Don't care conditions” for more effective simplifications

Introduction

Karnaugh Mapping II

Product of sums method

Instead of using the sum of products method we can also take the product of sums method which is based on the fundamental sums:

Fig. 5-1: Truth table with fundamental sums

Product of sums simplification

A equation gained through the sum of product method can be realized into a NAND-NAND logic circuit.

Fig. 5-2: NAND-NAND logic circuit

If we simplify the Karnaugh map in Fig. 5-2 we get the following equation:

Based on this equation we can draw the logic circuit like in Fig. 5-2. This solution is made with the help of the sum of product method. If we want to try the product of sums method we have to go through the following steps:

1. Complement the Karnaugh map and draw the complementary NAND-NAND circuit:

Fig. 5-3: Complementary NAND-NAND circuit

The simplified equation is:

2. Convert the complementary NAND-NAND circuit to a NOR-NOR circuit:

Fig. 5-4: NOR-NOR circuit

we have changed all NAND to NOR and complemented all signals.

3. Now we got a product of sums solution for X. The last step is to compare both circuits (NAND-NAND, NOR-NOR) to find out which circuit is the simpler one, meaning the cheaper one. Because we need less gates. In our case the sum of product solution (NOR-NOR circuit) is the better solution.

Summary of process

1. Convert the truth table into a Karnaugh map

- Write the sum of products equation
- Draw the NAND-NAND circuit

2. Complement the Karnaugh map

- Draw the complementary NAND-NAND circuit

3. Convert the complementary NAND-NAND circuit to a NOR-NOR circuit.

- Change all NAND to NOR
- Complement all signals

4. Compare NAND-NAND circuit with NOR-NOR circuit.

Don't care conditions

In some digital systems, certain input conditions never occur during normal operations; therefore, the corresponding output never appears.

Þ It is indicated in the truth table by an X.

Fig. 5-5: Don't care conditions

Don't care conditions are like wild cards, you can let them stand for what ever you like:

Fig. 5-6: Karnaugh map and simplified equation

HO: Simplify the following Boolean function:

That has the don't care conditions:

Solution:

First we have to translate the function into a truth table:

Now we have to convert it into a Karnaugh map and simplify it:

Figure

Worksheet No. 5

No. 1

a) Draw the Karnaugh map.

b) Encircle all octets, quads and pairs you can find.

c) What is the simplified Boolean equation for the Karnaugh map?

d) Draw the logic circuit.

e) Suppose the last six entries of the truth table are changed to don't cares. Using the Karnaugh map, show the simplified circuit.

f) What is the simplified NOR-NOR circuit?

Digital Teaching Aid (DED Philippinen, 86 p.) Coding - Lesson 6 Lesson Plan (introduction...) Introduction ASCII code Excess-3 code Gray code Encoder Decoder Code converter Worksheet No. 6

Digital Teaching Aid (DED Philippinen, 86 p.)

Coding - Lesson 6

Lesson Plan

Titel: Coding

Objectives:

- Get an idea about the purpose of codes
- Understand the principle of decoder, encoder, and code converter
- Able to design logic circuits for code conversion

Introduction

Coding

Ex: Coding ® Alphanumeric information in and out of a computer.

ASCII code

ASCII Code (American Standard Code for Information Interchange)

ASCII code is a 7 bit code whose format is

X6 X5 X4 X3 X2 X1 X0

i.e. the letter A is coded as:

1 0 0 0 0 0 1

Coding scheme:

Ex; The letter B via ASCII in binary:

1 0 0 0 0 1 0

Parity bit

During sending/transmitting data, 1 bit errors may occur. To catch these errors, a parity bit is usually transmitted along with the original bits:

X7 X6 X5 X4 X3 X2 X1 X0
|
parity bit

A parity checker at the receiving end can test for even or odd parity. I.e the transmitting station will set the parity bit always in that way to put the whole number in even parity. If the parity checker at the receiving end determines odd parity it will report an error message.

Now we have 8 bits (1 byte), this is the ideal length because most digital equipment is set up to handle bytes of data.

BCD Code (Binary Coded Decimal)

Each decimal place (0...9) is represented by a binary code.

The BCD code is a 4 bit code which is very common along digital systems.

Excess-3 code

Also an important 4 bit code

Ex: Convert a decimal number in excess-3 code:

Fig. 6-1: Convertion table, dec ® BCD® excess-3

Gray code

Each Gray code number differs from the preceding number by a single bit.

Fig. 6-2: Convertion table, dec ® Gray ® BCD

Encoder

An encoder converts an active input signal into a coded output signal.

Ex: Decimal to BCD encoder

(see Fig. 6-3 on the next page) If switch 9 is pressed:

A B C D = 1 0 0 1

The circuit in Fig. 6-3 is also available as TTL device 74147.

Fig. 6-3: Decimal to BCD encoder

Decoder

A decoder converts a coded input signal into an active output signal.

Ex; Develop a decoder circuit which converts BCD code into decimal.

Fig. 6-4: Truth table, BCD ® decimal

To optimize the circuit we use a Karnaugh map for simplification:

Fig. 6-5: Karnaugh map

For every number in the Karnaugh map we can write an equation:

Now we can draw the logic circuit:

Fig. 6-6: Decoder circuit, BCD ® decimal

Code converter

Code converter are logical networks which convert codes into another.

Fig. 6-7: TTL device, BCD ® excess-3 converter

Fig. 6-8: Convertion table, dec ® BCD® excess-3

Worksheet No. 6

No. 1 Develop a decoder circuit for the conversion of the Gray code into the decimal system. Use a Karnaugh map to get a simplified circuit!

No. 2 Develop a code converter for the conversion of the excess 3 code into the 8421 code (BCD).

Complete the truth table!

Use Karnaugh maps to get a simplified circuit!

The lower six rows are redundant.

Digital Teaching Aid (DED Philippinen, 86 p.) Flip-Flops - Lesson 7 Lesson Plan (introduction...) Introduction RS Flip-Flops D - Flip-Flop (D-FF) JK Flip-Flop (JK-FF) Worksheet No. 7

Digital Teaching Aid (DED Philippinen, 86 p.)

Flip-Flops - Lesson 7

Lesson Plan

Titel: Flip-Flops

Objectives:

- Understand the Flip-Flop principle
- Know the three basic Flip-Flops (RS, D, JK)
- Able to analyze timing diagrams

Introduction

Flip-Flops

Sometimes there is a need of digital devices or circuits whose output will remain unchanged, once set, even if there is a change in input.

RS Flip-Flops

A Flip-Flop is a bistable electronic circuit that has two stable states.

Þ Output is either 0 or 5V dc

The Flip-Flop can be regarded as a memory device. It can be used to store one binary digit at the output.

Fig. 7-1: RS Flip-Flop, logic circuit and device symbol

HO: What is the truth table for the circuit above?

Solution:

Fig. 7-2: Truth table, RS Flip-Flop

If both inputs (R, S) are high at once, the output can not be determined before; therefore, it is a forbidden state.

Ex: Create a RS Flip-Flop with NAND gates.

Fig. 7-3: RS Flip-Flop with NAND gates

Note: The inputs (R, S) are indicated with an overbar so they are inverted.

Fig. 7-4: Logic symbol, RS Flip-Flop with inverted inputs

Clocked RS Flip-Flop

Fig. 7-5: Clocked RS Flip-Flop

This Flip-Flop ca be enabled or disabled.

ENABLE ® low: R and S will have no effect on the output

ENABLE ® high: R and S inputs will be directly transmitted to the output

Fig. 7-6: Logic symbol, Clocked RS Flip-Flop

Timing diagram

A timing diagram is a drawing to determine the time dependent actions of logic devices.

Fig. 7-7: Timing diagram of a clocked RS Flip-Flop

Fig. 7-7 shows that the inputs (R, S) effect the output (Q) only when the clock signal (CLK) is high.

RS Flip-Flop application: Bounce free switch

D - Flip-Flop (D-FF)

The generation of two signals to drive a Flip-Flop is a disadvantage in many applications. This has led to the D-FF, a circuit that needs only a single data input.

Fig. 7-8: D Flip-Flop, logic circuit and truth table

CLK ® low: D can change without effect on the output

CLK ® high: Q is forced to equal the value of D

Ex: Create a 4 bit data memory with D-latches (D-FF).

Fig. 7-9: Data storage with D-FF

In Fig. 7-9, when the clock goes high, input data is loaded into the Flip-Flops and appears at the output.

Suppose the data input is:

D₃ D₂ D₁ D₀ = 1 0 1 0

When the clock goes high this 4 bit word is loaded into the D-latches, resulting in an output of:

Q₃ Q₂ Q₁ Q₀ = 1 0 1 0

Flip-Flop switching time

Fig. 7-10: Timing diagram, FF switching time

JK Flip-Flop (JK-FF)

Flip - Flops can be used to build counters, JK-FF are the ideal elements for that purpose.

Fig. 7-11: JK-FF, logic circuit

Ex: What is the truth table for the circuit above?

Fig. 7-12: Truth table, JK-FF

Fig. 7-13: JK-FF's, logic symbols

Preset (PR) and Clear (CLR) are input signals to get a definite start point.

JK Master-Slave FF (JK-MS-FF)

Fig. 7-14: JK-MS-FF, logic circuit

Regardless what the master does, the slave copies it. The slave copies the master on the negative clock edge. This circuit provides a way to avoid racing.

Fig. 7-15: JK-MS-FF, logic symbol

Available as TTL device: 74 LS 76

Worksheet No. 7

No. 1 A JK master slave FF has its inputs tied to + 5V, and a series of pulses are applied to its CLK input Describe the Q output.

Figure

No. 2 The signal drives a clocked RS - FF. If Q is low before point A in time: At what point does Q becomes a 1? What does Q reset to 0?

Figure

No. 3 Use the information in the preceding problem and draw the waveform Q.

No. 4 The signal drives a D-FF. What is the value of stored in the FF after the clock pulse is over?

Figure

No. 5 A normal JK-FF, J = K = 1. A 1 MHz is applied to the CLK input it has a propagation delay t_p of 50 ns. Draw the input squarewave and the out put expected at Q. Be sure to show the propagation delay time.

Digital Teaching Aid (DED Philippinen, 86 p.) Counter - Lesson 8 Lesson Plan (introduction...) Introduction Asynchronous counter Synchronous counter Seven segment indicator and decoder Worksheet No. 8

Digital Teaching Aid (DED Philippinen, 86 p.)

Counter - Lesson 8

Lesson Plan

Titel: Counter

Objectives:

- Understand the principle of asynchronous and synchronous counter
- Able to design asynchronous counter
- Know the seven segment principle

Introduction

Counter

Counters are beside the logic operations the basic elements of all digital controls.

Since the clock pulses occur at known intervals, the counter can be used as an instrument for measuring time and therefore period or frequency.

On principle there are two counting organisations:

- Asynchronous (serial, ripple) counter

Each Flip-Flop is triggered by the previous Flip-Flop, and thus the counter has a cumulative settling time. (Delay time, tp)

- Synchronous (parallel) counter

Here, every Flip-Flop is triggered by the clock (synchronous), and thus the settling time is simply equal to the delay time of a single Flip-Flop.

Asynchronous counter

Fig. 8-1: Asynchronous counter with JK-MS-FF in toggle mode

To understand how this counter works lets have a look at the timing diagram:

Fig. 8-2: Timing diagram, asynchronous mod 8 counter

The frequency of waveform C is one half that at B, but is only one-eighth the clock frequency.

The FF's are negative edge triggered, hence output signals change only at the falling side of the clock pulse.

Fig. 8-3: Truth table, asynchronous mod 8 counter

A three Flip-Flop counter is often referred to as a mod 8 (modulus 8) counter since it has 8 states.

The largest decimal number which is represented by a 3 Flip-Flop counter is:

Count Down mode

Switching the clock inputs of each Flip-Flop to the outputs causes the counting sequence to start at 111 down to 000.

HO: How many Flip-Flops are required to construct a mod-128 counter? A mod-32? What is the largest decimal number that can be stored in a mod-64 counter?

Solution:

* mod-128 must have 7 Flip-Flops (27 = 128)
* mod-32 must have 5 Flip-Flops
* mod-64, the largest decimal number is 63

Synchronous counter

Fig. 8-4: Synchronous counter, with JK-MS-FF in toggle mode

The negative clock transistion is the mechanism that toggles each FF. Therefore, whenever a Flip-Flop changes state, it toggles at exactly the same time as all the other Flip-Flops. In other words, all Flip-Flops change state in synchronism. This advantage needs an increase in hardware (additional gates), hence the asynchronous counter is the cheaper solution for time uncritical applications.

Seven segment indicator and decoder

Seven segment indicators

Seven segment indicators are made by seven LED's. LED's are light emitting diodes. (LED function: Free electrons recombine with holes near the junction. As the free electrons “fall” from a higher energy level to lower one, they give up energy in form of heat and light)

Fig. 8-5: seven LED's labeled a through g

Depending on which LED's are light up it is possible to indicate decimal numbers from 0 to 9.

Fig. 8-6: Seven segment indicator, common anode type

Two types of indicators are available:

- common anode
- common cathode

The diodes are controlled by ground (common anode) or by 5V (common cathode).

Fig. 8-7: Seven segment indicator, common cathode type

Seven segment decoder

Seven segment decocer are devices to drive seven segment indicators.

Ex: The 7448

(see Fig. 8-8 on the next page)

Fig. 8-8: Seven segment indicator and decoder

The 7448 is a decoder to drive common cathode indicators.

When the BCD input is set to:

A B C D = 0 1 0 0

The internal logic of the decoder forces the LED's b, c, f, and g to conduct.

Worksheet No. 8

No. 1 Draw the logic diagram, truth table, and the waveform for a three flip-flop serial counter that uses JK master slave flip-flops sensitive to positive clock transition.

No. 2 Determine the number of possible states in a counter composed of the following number of flip-flops: