chapter 1 introduction to microprocessor and microcontroller

CHAPTER 1 INTRODUCTION TO MICROPROCESSOR AND MICROCONTROLLER1.1 NUMBER SYSTEM‣ THREE number systems those are normally used in digital electronics:Base 10 (decimal)Base 2 (binary)Base 16 (hexadecimal)1.1.1 BASE 10 (DECIMAL NUMBER)‣ It is a number system which we use in calculation in our daily life.‣ It consists of 10 symbols (0,1,2,3, … , 9) to represent some counts.1.1.2 BASE 2 (BINARY NUMBER)‣ It consists of 2 symbols that is “0” and “1”.‣ Normally we call it as “BIT”.‣ The computer can only understand this number. So all information which need to beprocessed in computer, they must change into this form of number.‣ In digital electronics, bit “0” is equal to 0 volt (in between 0 – 0.8 volt) and bit “1” isequal to 5 volt (in between 2.0 – 5 volt) which can lit a LED.LOGIC 15.0 VV1UnusedLogic “1” 2.0V – 5.0VLOGIC 02.0 V0 0Logic “0” 0V – 0.8Vt1.1.3 BASE 16 (HEXADECIMAL)‣ This system is used to represent binary system in most of calculation cases.‣ It has simplified the usage of a series of binary number into hexadecimal number, suchas, we can represent binary number 1010010110110001 2 into A5B1 16 in hexadecimalnumber.Prepared by Tan KL Page 1-1

‣ It uses 16 symbols to represent a number as shown in the table below:Decimal Binary Hex0 0000 01 0001 12 0010 23 0011 34 0100 45 0101 56 0110 67 0111 78 1000 8Decimal Binary Hex9 1001 910 1010 A11 1011 B12 1100 C13 1101 D14 1110 E15 1111 F1.1.4 NUMBER CONVERTIONDecimal BinaryMethod 1: always divide with “2” and collect the remainderExample: Convert 25 10 to binary numberResult afterdivision (rounded) remainder25/2 = 12 112/2 = 6 0LSB (Least Significant Bit)6/2 = 3 03/2 = 1 11/2 = 0 1MSB (Most Significant Bit)Hence, 25 10 = 11001 2Method 2: by adding the weight in every bit2 4 2 3 2 2 2 1 2 0 weight16 8 4 2 11 1 0 0 125 10 = 16 + 8 + 1 = 11001 2Prepared by Tan KL Page 1-2

Binary Hexadecimal : collecting 4 bit in one group from the back (LSB)Example: Convert 100111110101 2 to Hex number.1001 1111 0101 2= 9 F 5= 9F5 16Hexadecimal Binary : convert every number of Hex to 4 bit binaryExample: Convert 29B 16 to binary number.2 9 B 16= 0010 1001 1011= 001010011011 2 (“00” in the front can be neglected 1010011011 2 )Decimal Hexadecimal : convert the number into binary first, and then intohexadecimalExample : Convert 699 10 to Hex2 9 2 8 2 7 2 6 2 5 2 4 2 3 2 2 2 1 2 0512 256 128 64 32 16 8 4 2 11 0 1 0 1 1 1 0 1 12 B B= 2BB 16 (have to add “00” in the front to make 4 bit in one group)1.1.5 ADDING 2 HEX NUMBORExample: 6 B 16 0110 1011 2+ A F 16 + 1010 1111 21 1 A 16 1 0001 1010 2Carry, C = 1Numbers of “1” is odd, Parity, P = 1Auxiliary Carry, AC = 1Prepared by Tan KL Page 1-3

1.2 BASIC LOGIC GATE1.2.1 AND GATE1.2.2 OR GATE1.2.3 NOT GATEPrepared by Tan KL Page 1-4

1.2.4 EX-OR GATE (Exclusive OR)1.2.5 NAND GATE1.2.6 NOR GATEPrepared by Tan KL Page 1-5

Assignment 1:a. Convert the following numbers to Hexadecimal number:i. 11001100 2 ii. 456 10iii. 001110101110 2 iv. 1011 10v. 1110001010 2 vi. 1101 10b. Convert the following numbers to binary number:i. 5F6A 16 ii. 127 10iii. 101101 16 iv. 255 10v. FFFF 16 vi. 256 10c. Solve the problems below and find out the condition of Carry (C), Auxiliary Carry (AC)and Parity (P) bit:i. 5AHex ADD with 36Hex ii. 9DHex ADD with ABHexiii. 8B 16 AND with 97 16 iv. A6 16 AND with C3 16v. A7 16 OR with C1 16 vi. 93 16 OR with C7 16vii. 4C 16 EX-OR with 8A 16 viii. 23 16 EX-OR with EC 16ix. Data 78 16 is OR-ed with data 5C 16 first and then is AND-ed with data A7 16x. Data FA 16 is AND-ed with data D8 16 first and then EX-OR-ed with data 67 16Prepared by Tan KL Page 1-6

1.3 SAIZ AND UNIT OF DATABit0 1orNibble0 1 0 0Byte 1 0 1 1 0 0 0 1`Word 1 1 0 0 0 0 1 11 0 0 0 1 0 0 0Bit : is the smallest data in digital electronics. It consists of “0” or “1”Byte : data consists of 8 bit or 2 nibble dataNibble : data consists of half byte or 4 bit dataWord : data consists of 2 byte or 4 nibble or 16 bit dataLong Word : data consists of 2 Word or 4 byte or 32 bit dataExample :5 F 9 A Hex 0101 1111 1001 1010 2 16 bit data or 4 nibble data or 2 byte data or 1 Word data or halfLong Word dataPrepared by Tan KL Page 1-7

1.3.1 TERMS FOR MEMORY CAPACITYKilobyte (K) 2 10 Byte (1024 byte)Megabyte (M) 2 20 byte (1,048,576 byte)Gigabyte (G) Gigabyte (G) : 2 30 byte (1,073,741,824 byte) Terabyte (T) : 2 40 byte (???)1.4 THREE COMMON TERMS IN DIGITAL ELECTRONICSMicrocomputerMicroprocessorMicrocontroller1.4.1 BASIC ELEMENTS OF MICROCOMPUTERA complete set of acomputerMemoryOutputControlInpuALUMicroprocessorMicrocomputerPrepared by Tan KL Page 1-8

A microcomputer system consists of the following components:Peripheral InputPeripheral OutputPeripheral StorageCentral Processing Unit1.4.1 Peripheral Input It is a tool to be used to prepare data or information which will be processed by CentralProcessing Unit. Example: Joystick, Keyboard, Sensor, Scanner1.4.2 Peripheral Output It is a tool to be used to display data or information which has been processed byCentral Processing Unit. Example: Monitor, Printer, Speaker1.4.3 Peripheral Memory It is a tool to be used to store data or information which will be processed or has beenprocessed by Central Processing Unit. Example: Hard disk, CD-ROM, RAM, ROM, Pen drive1.4.4 Central Processing Unit (CPU) It is the most important part in microcomputer system. It acts as a “brain” of amicrocomputer system. It is also called Microprocessor. In here the data are gathered, processed and sent out to display. It can perform operations such as arithmetic (addition, deduction, multiplication anddivision) and logic (AND, OR, NOT etc.) operation and make decision after operation. A general internal architecture of a CPU/Microprocessor is shown in the next page. Example: 80486, Pentium II, AMD AtronPrepared by Tan KL Page 1-9

1.4.5 BASIC CONNECTION OF A MICROCOMPUTER SYSTEMPrepared by Tan KL Page 1-10

1.5 MICROPROCESSOR (CPU) It is the most important part in a microcomputer system. It acts as a “brain” of amicrocomputer system. In here the data are gathered, processed and put into display through peripheraloutput. It can perform operations such as arithmetic (addition, deduction, multiplication anddivision) and logic (AND, OR, NOT etc.) operation and make decision after operation.1.5.1 FUNCTION OF COMPONENTS IN A MICROPROCESSORi. ALU (Arithmetic Logic Unit) It performs arithmetic task such as addition, deduction, multiplication anddivision and logic task such as AND, OR, NOT and etc.Prepared by Tan KL Page 1-11

ii.ACC (Accumulator) It is a register which can store data for temporary before sending to process inALU and also can store the processed data before sending to store in thememory or display through peripheral output. It is just like a “entrance” for a data to go in and come out from ALU.iii.Flag register/PSW (Program Status Word ) register This register contains information regarding the result of a process which hasbeen carried out by ALU. It tells the user whether the result has Carry (CY), Auxiliary Carry (AC), Odd orEven Parity (P) or Overflow (OV) or the other way round. It also consists information regarding registers from which bank will be workingon (will be discussed further in Chapter 2).iv.Program Counter (PC) It controls the sequence in which the instructions in a program are performed. Normally, it does this by counting in the sequence, that is 0, 1, 2, 3, … At any given time, the count indicates the location in memory from which thenext location of information/instruction is to be taken.v. Stack Pointer (SP) This register will store the contains of Program Counter (location of memory)for a short time when there is subroutine/sub-program occurs in the program. This register will be taken place when some particular instructions, such asPUSH, POP, CALL and RET are used in the program.vi.Data Register This register is a temporary storage location for data going to or coming fromthe data bus.Prepared by Tan KL Page 1-12

vii.Address Register This register is another temporary storage location. It holds the address of the memory location or I/O device that is used in theoperation presently being performed.viii.Instruction Decoder After an data is pulled from memory and placed in the data register, the data isdecoded by this circuit. The decoder checks the code and decides which operation is to be performed.ix.Controller-sequencer This unit will produce a variety of control signals to carry out the instruction. Since each instruction is different, a different combination of control signals isproduced for each instruction.x. Data Bus It is a group of wires that is used to channel data (which to be processed orafter processed) from peripherals (memory or I/O devices) to CPU or vice versa.xi.Address Bus It is a group of wires that is used to channel locations of data which will be orhas been processed.xii.Control Bus It is a group of wires that is used to send control signals (such as MEMW,MEMR, IOR, IOW, interrupt and DMA) from CPU to peripheral (memory or I/Odevices) or vice versa.Prepared by Tan KL Page 1-13

1.6 MICROCONTROLLER (MC) It is also called computer in a single chip. Inside MC we can find the important components which are needed in microcomputersystem, such as CPU, RAM, ROM and peripheral I/O.Prepared by Tan KL Page 1-14

1.6.1 THE DIFFERENCES BETWEEN MICROPROCESSOR AND MICROCONTROLLERMicroprocessorMicrocontroller1. CPU, ROM, RAM, timer and I/O devices areall in seperated unit.1. CPU, ROM, RAM, timer and I/O devicesare all in a chip (Embedded in a chip).2. System designer can determine the capasityfor RAM, ROM and I/O devices.2. Capacity of RAM,ROM and I/O deviceshave been determined during fabrication.3. Capacity of memory and I/O port can beadded (expansived) at any time.3. Capacity of memory and I/O port cannotbe added (expansived) at any time.4. Suitable for project which has expandedspace.4. Suitable for project which has limitedspace.5. For general-purposes usages4. For specific-purposes usages1.7 Evolution Of MicroprocessorNameDate Transistors MicronsClockspeedDatawidthMIPS8080 1974 6,000 6 2 MHz 8 bits 0.648088 1979 29,000 3 5 MHz16 bits8-bitbus0.3380286 1982 134,000 1.5 6 MHz 16 bits 180386 1985 275,000 1.5 16 MHz 32 bits 580486 1989 1,200,000 1 25 MHz 32 bits 20Pentium 1993 3,100,000 0.8 60 MHz32 bits64-bitbus100Prepared by Tan KL Page 1-15

Pentium II 1997 7,500,000 0.35233MHz32 bits64-bitbus~300Pentium III 1999 9,500,000 0.25450MHz32 bits64-bitbus~510Pentium 4 2000 42,000,000 0.18 1.5 GHz32 bits64-bitbus~1,700Pentium 4"Prescott"2004 125,000,000 0.09 3.6 GHz32 bits64-bitbus~7,000The date is the year that the processor was first introduced. Many processors are reintroducedat higher clock speeds for many years after the original release date.Transistors is the number of transistors on the chip. You can see that the number oftransistors on a single chip has risen steadily over the years.Microns is the width, in microns, of the smallest wire on the chip. For comparison, a humanhair is 100 microns thick. As the feature size on the chip goes down, the number oftransistors rises.Clock speed is the maximum rate that the chip can be clocked at. Clock speed will makemore sense in the next section. Data Width is the width of the ALU. An 8-bit ALU can add/subtract/multiply/etc. two 8-bit numbers, while a 32-bit ALU can manipulate 32-bit numbers. An 8-bit ALU would haveto execute four instructions to add two 32-bit numbers, while a 32-bit ALU can do it inone instruction. In many cases, the external data bus is the same width as the ALU, butnot always. The 8088 had a 16-bit ALU and an 8-bit bus, while the modern Pentiums fetchdata 64 bits at a time for their 32-bit ALUs.MIPS stands for "millions of instructions per second" and is a rough measure of theperformance of a CPU. Modern CPUs can do so many different things that MIPS ratingslose a lot of their meaning, but you can get a general sense of the relative power of theCPUs from this column.Prepared by Tan KL Page 1-16

Exercise:a. List out 4 important components of a microcomputer system and briefly describe theirfunctions.b. Describe briefly the functions for the components below:i. Address Busii.iii.Data BusControl Busc. Draw a block diagram of a microprocessor and briefly describe function of it’s component.d. State the differences between microprocessor and microcontroller.e. State the function for the components below:i. ALU iv. Stack Pointerii. Accumulator v. Program Status Word Registeriii. Program Counter vi. Controller Sequencerf. Different microprocessors have different data bus such as 8-bit, 16-bit and 32-bit. State2 advantages and 2 disadvantages of a smaller size of data bus as compared to a wider sizeof data bus of microprocessors.Prepared by Tan KL Page 1-17

1.7 EXECUTING A PROGRAM Before a program can be run, it must be placed in memory. There are 2 phases involving in executing a program, those are the fetch phase andthe executing phase.CARRYFLAGACCUMULATORALUControl Unit(ControllerSequencer)CLOCK andCONTROLLINESPROGRAMCOUNTER(PC)InstructionDecoderADDRESSREGISTER(AR)DATAREGISTER(DR)HEX00H01H02H03H04H05H06HADDRESS CODE00000000 0111010000000001 0001000000000010 0010010000000011 0000011100000100 1111010100000101 0011000000000110 11110100MEMORYExample:MOV A, #10HADD A, #07HMOV 30H, AMOV A, #10HBinary code whichrepresents a programProgramPrepared by Tan KL Page 1-18

To execute a program, the program needs to store in memory first, that is inProgram/Code Memory (ROM). This means that the program needs to be translated intothe machine language (binary code) and stored in the memory. The program will be executed step-by-step from the first address which is 00000000B(00H). This first address (00H) will be always referred by MP when there is RESET operationis carring out.The Fetching Phase** The phase which CPU fetches the instruction from main memory via the data bus intothe Instruction Register (IR) so that it can be decoded and executedi. When the Microprocessor (MP) is ON-ed, the contain of Program Counter (PC) isset to the first address, that is 00H automatically.ii.Then, this address is passed to the Address Register (AR = 00H).iii.The contain of PC is incremented by 1 automatically. So the contain of PC is 01Hnow.iv.The contain of AR (00H) is placed on the Address Bus so that the MP will go to theaddressed location to obtain the data in the addressed location, which is 00H.v. Then, the data which is in the location of 00H is passed through the Data Bus andsend to the Data Register (DR) which is inside the MP.vi.From the DR, the data is passed to the Instruction Decoder to interpret the data.vii.In this case, the instruction MOV A (with code = 01110100B) is interpreted, that is,moving/copying something into Accumulator (”A” stands for Accumulator).Prepared by Tan KL Page 1-19

viii.Following this, the Control Unit (Controller Sequencer) produces control signals toget the data which is in the following location of memory (01H) and place it intoAccumulator, as what has requested by the instruction.The Execute Phase In this phase, CPU will carry out the requirement on the instruction. In this case, theCPU will move/copy data into Accumulator.i. The contain of PC (01H) is sent to AR (AR = 01H).ii.The contain of PC, as usual is incremented by 1 and become 02H now.iii.The contain of AR is placed on the Address Bus. Then, the location of the memoryis identified, which is 02H.iv.The data in the location of 02H is placed on Data Bus and is sent to the DR in theMP.v. From DR, the data straight away sends to Accumulator as what has requestedbefore (MOV A).vi.So, now the data ”10H” is inside the Accumulator. The instruction ”MOV A, #10H”has been executed.vii.These program-executional steps will be the same for any other instruction.1.8 TERMS OF PROCESSING DATA IN MICROPROCESSORInstruction Cycle It is the process by which a computer retrieves a program instruction from its memory,determine what actions the instruction required, and carry out the actions. This cycle is repeated continuously by the CPU, from booting up to shutting down of acomputer.Prepared by Tan KL Page 1-20

Exercise:a. With the help of a suitable block diagram, explain briefly the steps to execute ”ADD A,#74H”.Prepared by Tan KL Page 1-21