The 8051 Microcontroller and Embedded Systems Using Assembly and C PDF

 The 8051 Microcontroller and Embedded Systems Using Assembly and C Second Edition Muhammad Ali Mazidi Janice Gillispie Mazidi Rolin D. McKinlay CONTENTS ˆ Introduction to Computing ˆ The 805...

 The 8051 Microcontroller and Embedded Systems Using Assembly and C Second Edition Muhammad Ali Mazidi Janice Gillispie Mazidi Rolin D. McKinlay CONTENTS ˆ Introduction to Computing ˆ The 8051 Microcontrollers ˆ 8051 Assembly Language Programming ˆ Branch Instructions ˆ I/O Port Programming ˆ 8051 Addressing Modes ˆ Arithmetic & Logic Instructions And Programs ˆ 8051 Programming in C ˆ 8051 Hardware Connection and Hex File ˆ 8051 Timer/Counter Programming in Assembly and C ˆ 8051 Serial Port Programming in Assembly and C ˆ Interrupts Programming in Assembly and C ˆ 8051 Interfacing to External Memory ˆ 8051 Real World Interfacing I: LCD,ADC AND SENSORS ˆ LCD and Keyboard Interfacing ˆ 8051 Interfacing with 8255  INTRODUCTION TO COMPUTING The 8051 Microcontroller and Embedded Systems: Using Assembly and C Mazidi, Mazidi and McKinlay Chung-Ping Young 楊中平 Home Automation, Networking, and Entertainment Lab Dept. of Computer Science and Information Engineering National Cheng Kung University, TAIWAN  ‰ Numbering and coding systems OUTLINES ‰ Digital primer ‰ Inside the computer Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 2  ‰ Human beings use base 10 (decimal) NUMBERING AND CODING arithmetic SYSTEMS ¾ There are 10 distinct symbols, 0, 1, 2, …, 9 Decimal and ‰ Computers use base 2 (binary) system Binary Number ¾ There are only 0 and 1 Systems ¾ These two binary digits are commonly referred to as bits Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 3  ‰ Divide the decimal number by 2 NUMBERING AND CODING repeatedly SYSTEMS ‰ Keep track of the remainders ‰ Continue this process until the quotient Converting becomes zero from Decimal ‰ Write the remainders in reverse order to Binary to obtain the binary number Ex. Convert 2510 to binary Quotient Remainder 25/2 = 12 1 LSB (least significant bit) 12/2 = 6 0 6/2 = 3 0 3/2 = 1 1 1/2 = 0 1 MSB (most significant bit) Therefore 2510 = 110012 Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 4  ‰ Know the weight of each bit in a binary NUMBERING number AND CODING ‰ Add them together to get its decimal SYSTEMS equivalent Converting Ex. Convert 110012 to decimal from Binary to Weight: 24 23 22 21 20 Decimal Digits: 1 1 0 0 1 Sum: 16 + 8+ 0+ 0+ 1 = 2510 ‰ Use the concept of weight to convert a decimal number to a binary directly Ex. Convert 3910 to binary 32 + 0 + 0 + 4 + 2 + 1 = 39 Therefore, 3910 = 1001112 Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 5  ‰ Base 16, the NUMBERING AND CODING hexadecimal system, Decimal Binary Hex SYSTEMS is used as a 0 0000 0 1 0001 1 convenient 2 0010 2 Hexadecimal representation of 3 0011 3 4 0100 4 System binary numbers 5 0101 5 ¾ ex. 6 0110 6 7 0111 7 It is much easier to 8 1000 8 represent a string of 0s 9 1001 9 and 1s such as 10 1010 A 100010010110 as its 11 1011 B hexadecimal equivalent of 12 1100 C 896H 13 1101 D 14 1110 E 15 1111 F Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 6  ‰ To represent a binary number as its NUMBERING equivalent hexadecimal number AND CODING ¾ Start from the right and group 4 bits at a SYSTEMS time, replacing each 4-bit binary number with its hex equivalent Converting between Binary Ex. Represent binary 100111110101 in hex and Hex 1001 1111 0101 = 9 F 5 ‰ To convert from hex to binary ¾ Each hex digit is replaced with its 4-bit binary equivalent Ex. Convert hex 29B to binary 2 9 B = 0010 1001 1011 Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 7  ‰ Convert to binary first and then NUMBERING AND CODING convert to hex SYSTEMS ‰ Convert directly from decimal to hex by repeated division, keeping track of Converting the remainders from Decimal to Hex Ex. Convert 4510 to hex 32 16 8 4 2 1 1 0 1 1 0 1 32 + 8 + 4 + 1 = 45 4510 = 0010 11012 = 2D16 Ex. Convert 62910 to hex 512 256 128 64 32 16 8 4 2 1 1 0 0 1 1 1 0 1 0 1 62910 = 512+64+32+16+4+1 = 0010 0111 01012 = 27516 Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 8  ‰ Convert from hex to binary and then to NUMBERING AND CODING decimal SYSTEMS ‰ Convert directly from hex to decimal by summing the weight of all digits Converting from Hex to Ex. 6B216 = 0110 1011 00102 1024 512 256 128 64 32 16 8 4 2 1 Decimal 1 1 0 1 0 1 1 0 0 1 0 1024 + 512 + 128 + 32 + 16 + 2 = 171410 Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 9  ‰ Adding the digits together from the NUMBERING AND CODING least significant digits SYSTEMS ¾ If the result is less than 16, write that digit as the sum for that position Addition of Hex ¾ If it is greater than 16, subtract 16 from it Numbers to get the digit and carry 1 to the next digit Ex. Perform hex addition: 23D9 + 94BE 23D9 LSD: 9 + 14 = 23 23 – 16 = 7 w/ carry + 94BE 1 + 13 + 11 = 25 25 – 16 = 9 w/ carry B897 1 + 3+4=8 MSD: 2 + 9=B Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 10  ‰ If the second digit is greater than the NUMBERING AND CODING first, borrow 16 from the preceding SYSTEMS digit Ex. Perform hex subtraction: 59F – 2B8 Subtraction of Hex Numbers 59F LSD: 15 – 8 = 7 – 2B8 9 + 16 – 11 = 14 = E16 2E7 5–1–2=2 Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 11  ‰ The ASCII (pronounced “ask-E”) code NUMBERING assigns binary patterns for AND CODING ¾ Numbers 0 to 9 SYSTEMS ¾ All the letters of English alphabet, uppercase and lowercase ASCII Code ¾ Many control codes and punctuation marks ‰ The ASCII system uses 7 bits to represent each code Hex Symbol Hex Symbol Selected ASCII codes 41 A 61 a 42 B 62 b 43 C 63 c 44 D 64 d......... … 59 Y 79 y 5A Z 7A z Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 12  ‰ Two voltage levels can be represented DIGITAL PRIMER as the two digits 0 and 1 ‰ Signals in digital electronics have two Binary Logic distinct voltage levels with built-in tolerances for variations in the voltage ‰ A valid digital signal should be within either of the two shaded areas 5 4 Logic 1 3 2 1 Logic 0 0 Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 13  ‰ AND gate DIGITAL PRIMER Logic Gates Computer Science Illuminated, Dale and Lewis ‰ OR gate Computer Science Illuminated, Dale and Lewis Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 14  ‰ Tri-state buffer DIGITAL PRIMER ‰ Inverter Logic Gates (cont’) Computer Science Illuminated, Dale and Lewis ‰ XOR gate Computer Science Illuminated, Dale and Lewis Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 15  ‰ NAND gate DIGITAL PRIMER Logic Gates (cont’) Computer Science Illuminated, Dale and Lewis ‰ NOR gate Computer Science Illuminated, Dale and Lewis Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 16  DIGITAL PRIMER Half adder Logic Design Using Gates Full adder Digital Design, Mano Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 17  DIGITAL PRIMER 4-bit adder Logic Design Using Gates (cont’) Digital Design, Mano Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 18  ‰ Decoders DIGITAL PRIMER ¾ Decoders are widely used for address decoding in computer design Logic Design Address Decoders Using Gates (cont’) Address decoder for 9 (10012) Address decoder for 5 (01012) The output will be 1 if and The output will be 1 if and only if the input is 10012 only if the input is 01012 Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 19  ‰ Flip-flops DIGITAL PRIMER ¾ Flip-flops are frequently used to store data Logic Design Using Gates (cont’) Digital Design, Mano Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 20  ‰ The unit of data size INSIDE THE COMPUTER ¾ Bit : a binary digit that can have the value 0 or 1 Important ¾ Byte : 8 bits Terminology ¾ Nibble : half of a bye, or 4 bits ¾ Word : two bytes, or 16 bits ‰ The terms used to describe amounts of memory in IBM PCs and compatibles ¾ Kilobyte (K): 210 bytes ¾ Megabyte (M) : 220 bytes, over 1 million ¾ Gigabyte (G) : 230 bytes, over 1 billion ¾ Terabyte (T) : 240 bytes, over 1 trillion Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 21  ‰ CPU (Central Processing Unit) INSIDE THE ¾ Execute information stored in memory COMPUTER ‰ I/O (Input/output) devices ¾ Provide a means of communicating with Internal CPU Organization of ‰ Memory Computers ¾ RAM (Random Access Memory) – temporary storage of programs that computer is running ƒ The data is lost when computer is off ¾ ROM (Read Only Memory) – contains programs and information essential to operation of the computer ƒ The information cannot be changed by use, and is not lost when power is off – It is called nonvolatile memory Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 22  INSIDE THE COMPUTER Internal Address bus Organization of Computers Memory Peripherals (cont’) CPU (monitor, (RAM, ROM) printer, etc.) Data bus Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 23  ‰ The CPU is connected to memory and INSIDE THE COMPUTER I/O through strips of wire called a bus ¾ Carries information from place to place Internal ƒ Address bus Organization of ƒ Data bus ƒ Control bus Computers (cont’) Address bus RAM ROM Printer Disk Monitor Keyboard CPU Data bus Read/ Write Control bus Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 24  ‰ Address bus INSIDE THE ¾ For a device (memory or I/O) to be COMPUTER recognized by the CPU, it must be assigned an address Internal ƒ The address assigned to a given device must Organization of be unique Computers ƒ The CPU puts the address on the address bus, (cont’) and the decoding circuitry finds the device ‰ Data bus ¾ The CPU either gets data from the device or sends data to it ‰ Control bus ¾ Provides read or write signals to the device to indicate if the CPU is asking for information or sending it information Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 25  ‰ The more data buses available, the INSIDE THE COMPUTER better the CPU ¾ Think of data buses as highway lanes More about ‰ More data buses mean a more Data Bus expensive CPU and computer ¾ The average size of data buses in CPUs varies between 8 and 64 ‰ Data buses are bidirectional ¾ To receive or send data ‰ The processing power of a computer is related to the size of its buses Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 26  ‰ The more address buses available, the INSIDE THE larger the number of devices that can COMPUTER be addressed More about ‰ The number of locations with which a Address Bus CPU can communicate is always equal to 2x, where x is the address lines, regardless of the size of the data bus ¾ ex. a CPU with 24 address lines and 16 data lines can provide a total of 224 or 16M bytes of addressable memory ¾ Each location can have a maximum of 1 byte of data, since all general-purpose CPUs are byte addressable ‰ The address bus is unidirectional Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 27  ‰ For the CPU to process information, INSIDE THE the data must be stored in RAM or COMPUTER ROM, which are referred to as primary memory CPU’s Relation ‰ ROM provides information that is fixed to RAM and and permanent ROM ¾ Tables or initialization program ‰ RAM stores information that is not permanent and can change with time ¾ Various versions of OS and application packages ¾ CPU gets information to be processed ƒ first form RAM (or ROM) ƒ if it is not there, then seeks it from a mass storage device, called secondary memory, and transfers the information to RAM Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 28  ‰ Registers INSIDE THE COMPUTER ¾ The CPU uses registers to store information temporarily Inside CPUs ƒ Values to be processed ƒ Address of value to be fetched from memory ¾ In general, the more and bigger the registers, the better the CPU ƒ Registers can be 8-, 16-, 32-, or 64-bit ƒ The disadvantage of more and bigger registers is the increased cost of such a CPU Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 29  Address Bus INSIDE THE COMPUTER Program Counter Inside CPUs (cont’) Instruction Register Control Bus Data Bus Flags ALU Instruction decoder, timing, and control Internal Register A buses Register B Register C Register D Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 30  ‰ ALU (arithmetic/logic unit) INSIDE THE ¾ Performs arithmetic functions such as add, COMPUTER subtract, multiply, and divide, and logic functions such as AND, OR, and NOT Inside CPUs (cont’) ‰ Program counter ¾ Points to the address of the next instruction to be executed ƒ As each instruction is executed, the program counter is incremented to point to the address of the next instruction to be executed ‰ Instruction decoder ¾ Interprets the instruction fetched into the CPU ƒ A CPU capable of understanding more instructions requires more transistors to design Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 31  Ex. A CPU has registers A, B, C, and D and it has an 8-bit INSIDE THE data bus and a 16-bit address bus. The CPU can access COMPUTER memory from addresses 0000 to FFFFH Assume that the code for the CPU to move a value to Internal register A is B0H and the code for adding a value to Working of register A is 04H Computers The action to be performed by the CPU is to put 21H into register A, and then add to register A values 42H and 12H... Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 32  Ex. (cont’) INSIDE THE Action Code Data COMPUTER Move value 21H into reg. A B0H 21H Add value 42H to reg. A 04H 42H Internal Add value 12H to reg. A 04H 12H Working of Mem. addr. Contents of memory address Computers 1400 (B0) code for moving a value to register A (cont’) 1401 (21) value to be moved 1402 (04) code for adding a value to register A 1403 (42) value to be added 1404 (04) code for adding a value to register A 1405 (12) value to be added 1406 (F4) code for halt... Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 33  Ex. (cont’) INSIDE THE The actions performed by CPU are as follows: COMPUTER 1. The program counter is set to the value 1400H, indicating the address of the first instruction code to Internal be executed Working of 2. Computers ¾ The CPU puts 1400H on address bus and sends it (cont’) out ƒ The memory circuitry finds the location ¾ The CPU activates the READ signal, indicating to memory that it wants the byte at location 1400H 以動畫表示 ƒ This causes the contents of memory location 1400H, which is B0, to be put on the data bus and brought into the CPU... Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 34  Ex. (cont’) INSIDE THE 3. COMPUTER ¾ The CPU decodes the instruction B0 ¾ The CPU commands its controller circuitry to bring Internal into register A of the CPU the byte in the next Working of memory location Computers ƒ The value 21H goes into register A (cont’) ¾ The program counter points to the address of the next instruction to be executed, which is 1402H ƒ Address 1402 is sent out on the address bus to fetch the next instruction... Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 35  Ex. (cont’) INSIDE THE 4. COMPUTER ¾ From memory location 1402H it fetches code 04H ¾ After decoding, the CPU knows that it must add to Internal the contents of register A the byte sitting at the Working of next address (1403) Computers ¾ After the CPU brings the value (42H), it provides (cont’) the contents of register A along with this value to the ALU to perform the addition ƒ It then takes the result of the addition from the ALU’s output and puts it in register A ƒ The program counter becomes 1404, the address of the next instruction... Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 36  Ex. (cont’) INSIDE THE 5. COMPUTER ¾ Address 1404H is put on the address bus and the code is fetched into the CPU, decoded, and Internal executed Working of ƒ This code is again adding a value to register A Computers ƒ The program counter is updated to 1406H (cont’) 6. ¾ The contents of address 1406 are fetched in and executed ¾ This HALT instruction tells the CPU to stop incrementing the program counter and asking for the next instruction Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 37  8051 MICROCONTROLLERS The 8051 Microcontroller and Embedded Systems: Using Assembly and C Mazidi, Mazidi and McKinlay Chung-Ping Young 楊中平 Home Automation, Networking, and Entertainment Lab Dept. of Computer Science and Information Engineering National Cheng Kung University, TAIWAN  ‰ Microcontrollers and embedded OUTLINES processors ‰ Overview of the 8051 family Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 2  ‰ General-purpose microprocessors MICRO- CONTROLLERS contains AND ¾ No RAM EMBEDDED ¾ No ROM PROCESSORS ¾ No I/O ports Microcontroller ‰ Microcontroller has vs. General- ¾ CPU (microprocessor) Purpose ¾ RAM Microprocessor ¾ ROM ¾ I/O ports ¾ Timer ¾ ADC and other peripherals Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 3  Data bus MICRO- General- purpose CONTROLLERS Micro- Processor I/O Serial AND RAM ROM Port Timer COM Port EMBEDDED PROCESSORS CPU Address bus Microcontroller vs. General- Purpose Microcontroller Microprocessor CPU RAM ROM (cont’) Serial I/O Timer COM Port Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 4  ‰ General-purpose microprocessors MICRO- ¾ Must add RAM, ROM, I/O ports, and CONTROLLERS timers externally to make them functional AND ¾ Make the system bulkier and much more EMBEDDED expensive PROCESSORS ¾ Have the advantage of versatility on the amount of RAM, ROM, and I/O ports Microcontroller ‰ Microcontroller vs. General- ¾ The fixed amount of on-chip ROM, RAM, Purpose and number of I/O ports makes them ideal Microprocessor for many applications in which cost and (cont’) space are critical ¾ In many applications, the space it takes, the power it consumes, and the price per unit are much more critical considerations than the computing power Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 5  ‰ An embedded product uses a MICRO- microprocessor (or microcontroller) to CONTROLLERS do one task and one task only AND ¾ There is only one application software that EMBEDDED is typically burned into ROM PROCESSORS ‰ A PC, in contrast with the embedded Microcontrollers system, can be used for any number of for Embedded applications Systems ¾ It has RAM memory and an operating system that loads a variety of applications into RAM and lets the CPU run them ¾ A PC contains or is connected to various embedded products ƒ Each one peripheral has a microcontroller inside it that performs only one task Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 6  ‰ Home MICRO- ¾ Appliances, intercom, telephones, security systems, CONTROLLERS garage door openers, answering machines, fax AND machines, home computers, TVs, cable TV tuner, VCR, camcorder, remote controls, video games, EMBEDDED cellular phones, musical instruments, sewing PROCESSORS machines, lighting control, paging, camera, pinball machines, toys, exercise equipment Microcontrollers ‰ Office for Embedded ¾ Telephones, computers, security systems, fax Systems machines, microwave, copier, laser printer, color printer, paging (cont’) ‰ Auto ¾ Trip computer, engine control, air bag, ABS, instrumentation, security system, transmission control, entertainment, climate control, cellular phone, keyless entry Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 7  ‰ Many manufactures of general-purpose MICRO- microprocessors have targeted their CONTROLLERS microprocessor for the high end of the AND embedded market EMBEDDED ¾ There are times that a microcontroller is PROCESSORS inadequate for the task x86 PC ‰ When a company targets a general- Embedded purpose microprocessor for the Applications embedded market, it optimizes the processor used for embedded systems ‰ Very often the terms embedded processor and microcontroller are used interchangeably Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 8  ‰ One of the most critical needs of an MICRO- embedded system is to decrease CONTROLLERS power consumption and space AND EMBEDDED ‰ In high-performance embedded PROCESSORS processors, the trend is to integrate more functions on the CPU chip and let x86 PC designer decide which features he/she Embedded wants to use Applications ‰ In many cases using x86 PCs for the (cont’) high-end embedded applications ¾ Saves money and shortens development time ƒ A vast library of software already written ƒ Windows is a widely used and well understood platform Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 9  ‰ 8-bit microcontrollers MICRO- CONTROLLERS ¾ Motorola’s 6811 AND ¾ Intel’s 8051 EMBEDDED ¾ Zilog’s Z8 PROCESSORS ¾ Microchip’s PIC Choosing a ‰ There are also 16-bit and 32-bit Microcontroller microcontrollers made by various chip makers Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 10  ‰ Meeting the computing needs of the MICRO- CONTROLLERS task at hand efficiently and cost AND effectively EMBEDDED ¾ Speed PROCESSORS ¾ Packaging ¾ Power consumption Criteria for ¾ The amount of RAM and ROM on chip Choosing a Microcontroller ¾ The number of I/O pins and the timer on chip ¾ How easy to upgrade to higher- performance or lower power-consumption versions ¾ Cost per unit Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 11  ‰ Availability of software development MICRO- tools, such as compilers, assemblers, CONTROLLERS and debuggers AND EMBEDDED ‰ Wide availability and reliable sources PROCESSORS of the microcontroller ¾ The 8051 family has the largest number of Criteria for diversified (multiple source) suppliers Choosing a ƒ Intel (original) ƒ Atmel Microcontroller (cont’) ƒ Philips/Signetics ƒ AMD ƒ Infineon (formerly Siemens) ƒ Matra ƒ Dallas Semiconductor/Maxim Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 12  ‰ Intel introduced 8051, referred as MCS- OVERVIEW OF 51, in 1981 8051 FAMILY ¾ The 8051 is an 8-bit processor ƒ The CPU can work on only 8 bits of data at a time 8051 Microcontroller ¾ The 8051 had ƒ 128 bytes of RAM ƒ 4K bytes of on-chip ROM ƒ Two timers ƒ One serial port ƒ Four I/O ports, each 8 bits wide ƒ 6 interrupt sources ‰ The 8051 became widely popular after allowing other manufactures to make and market any flavor of the 8051, but remaining code-compatible Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 13  External Interrupts OVERVIEW OF 8051 FAMILY Counter Inputs On-chip Interrupt ROM On-chip Etc. Control Timer 0 8051 for code RAM Timer 1 Microcontroller (cont’) CPU OSC Bus I/O Serial Control Ports Port P0 P1 P2 P3 TXD RXD Address/Data Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 14  ‰ The 8051 is a subset of the 8052 OVERVIEW OF 8051 FAMILY ‰ The 8031 is a ROM-less 8051 ¾ Add external ROM to it 8051 Family ¾ You lose two ports, and leave only 2 ports for I/O operations Feature 8051 8052 8031 ROM (on-chip program 4K 8K 0K space in bytes) RAM (bytes) 128 256 128 Timers 2 3 2 I/O pins 32 32 32 Serial port 1 1 1 Interrupt sources 6 8 6 Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 15  ‰ 8751 microcontroller OVERVIEW OF 8051 FAMILY ¾ UV-EPROM ƒ PROM burner Various 8051 ƒ UV-EPROM eraser takes 20 min to erase Microcontrollers ‰ AT89C51 from Atmel Corporation ¾ Flash (erase before write) ƒ ROM burner that supports flash ƒ A separate eraser is not needed ‰ DS89C4x0 from Dallas Semiconductor, now part of Maxim Corp. ¾ Flash ƒ Comes with on-chip loader, loading program to on-chip flash via PC COM port Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 16  ‰ DS5000 from Dallas Semiconductor OVERVIEW OF 8051 FAMILY ¾ NV-RAM (changed one byte at a time), RTC (real-time clock) Various 8051 ƒ Also comes with on-chip loader Microcontrollers ‰ OTP (one-time-programmable) version (cont’) of 8051 ‰ 8051 family from Philips ¾ ADC, DAC, extended I/O, and both OTP and flash Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 17  8051 ASSEMBLY LANGUAGE PROGRAMMING The 8051 Microcontroller and Embedded Systems: Using Assembly and C Mazidi, Mazidi and McKinlay Chung-Ping Young 楊中平 Home Automation, Networking, and Entertainment Lab Dept. of Computer Science and Information Engineering National Cheng Kung University, TAIWAN  ‰ Register are used to store information INSIDE THE 8051 temporarily, while the information could be Registers ¾ a byte of data to be processed, or ¾ an address pointing to the data to be fetched ‰ The vast majority of 8051 register are 8-bit registers ¾ There is only one data type, 8 bits Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 2  ‰ The 8 bits of a register are shown from INSIDE THE 8051 MSB D7 to the LSB D0 ¾ With an 8-bit data type, any data larger Registers than 8 bits must be broken into 8-bit (cont’) chunks before it is processed most least significant bit significant bit D7 D6 D5 D4 D3 D2 D1 D0 8 bit Registers Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 3  ‰ The most widely used registers INSIDE THE 8051 ¾ A (Accumulator) ƒ For all arithmetic and logic instructions Registers ¾ B, R0, R1, R2, R3, R4, R5, R6, R7 (cont’) ¾ DPTR (data pointer), and PC (program counter) A B R0 DPTR DPH DPL R1 R2 PC PC (Program counter) R3 R4 R5 R6 R7 Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 4  MOV destination, source ;copy source to dest. INSIDE THE ¾ The instruction tells the CPU to move (in reality, 8051 COPY) the source operand to the destination operand MOV “#” signifies that it is a value Instruction MOV A,#55H ;load value 55H into reg. A MOV R0,A ;copy contents of A into R0 ;(now A=R0=55H) MOV R1,A ;copy contents of A into R1 ;(now A=R0=R1=55H) MOV R2,A ;copy contents of A into R2 ;(now A=R0=R1=R2=55H) MOV R3,#95H ;load value 95H into R3 ;(now R3=95H) MOV A,R3 ;copy contents of R3 into A ;now A=R3=95H Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 5  ‰ Notes on programming INSIDE THE ¾ Value (proceeded with #) can be loaded 8051 directly to registers A, B, or R0 – R7 ƒ MOV A, #23H MOV ƒ MOV R5, #0F9H If it’s not preceded with #, Instruction Add a 0 to indicate that it means to load from a memory location (cont’) F is a hex number and not a letter ¾ If values 0 to F moved into an 8-bit register, the rest of the bits are assumed all zeros ƒ “MOV A, #5”, the result will be A=05; i.e., A = 00000101 in binary ¾ Moving a value that is too large into a register will cause an error ƒ MOV A, #7F2H ; ILLEGAL: 7F2H>8 bits (FFH) Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 6  ADD A, source ;ADD the source operand INSIDE THE ;to the accumulator 8051 ¾ The ADD instruction tells the CPU to add the source byte to register A and put the result in register A ADD ¾ Source operand can be either a register or immediate data, but the destination must always Instruction be register A ƒ “ADD R4, A” and “ADD R2, #12H” are invalid since A must be the destination of any arithmetic operation MOV A, #25H ;load 25H into A MOV R2, #34H ;load 34H into R2 ADD A, R2 ;add R2 to Accumulator There are always ;(A = A + R2) many ways to write the same program, MOV A, #25H ;load one operand depending on the ;into A (A=25H) registers used ADD A, #34H ;add the second ;operand 34H to A Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 7  ‰ In the early days of the computer, 8051 programmers coded in machine language, ASSEMBLY consisting of 0s and 1s PROGRAMMING ¾ Tedious, slow and prone to error ‰ Assembly languages, which provided Structure of mnemonics for the machine code instructions, Assembly plus other features, were developed Language ¾ An Assembly language program consist of a series of lines of Assembly language instructions ‰ Assembly language is referred to as a low- level language ¾ It deals directly with the internal structure of the CPU Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 8  ‰ Assembly language instruction includes 8051 ¾ a mnemonic (abbreviation easy to remember) ASSEMBLY ƒ the commands to the CPU, telling it what those PROGRAMMING to do with those items ¾ optionally followed by one or two operands Structure of ƒ the data items being manipulated Assembly ‰ A given Assembly language program is Language a series of statements, or lines ¾ Assembly language instructions ƒ Tell the CPU what to do ¾ Directives (or pseudo-instructions) ƒ Give directions to the assembler Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 9  ‰ An Assembly language instruction 8051 ASSEMBLY consists of four fields: PROGRAMMING [label:] Mnemonic [operands] [;comment] ORG 0H ;start(origin) at location Structure of 0 MOV R5, #25H ;load 25H into R5 Assembly MOV R7, #34H ;load 34H into R7 Directives do not Language MOV A, #0 ;load 0 into generate A any machine ADD A, R5 ;add contentscode ofandR5aretoused A ;now A = A + only R5 by the assembler Mnemonics ADD A, R7 ;add contents of R7 to A produce ;now A = A + R7 opcodes ADD A, #12H ;add to A value 12H ;now A = A + 12H HERE: SJMP HERE ;stay in this loop END ;endComments of asm may source file be at the end of a The label field allows line or on a line by themselves the program to refer to a The assembler ignores comments line of code by name Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 10  ‰ The step of Assembly language ASSEMBLING AND RUNNING program are outlines as follows: AN 8051 1) First we use an editor to type a program, PROGRAM many excellent editors or word processors are available that can be used to create and/or edit the program ƒ Notice that the editor must be able to produce an ASCII file ƒ For many assemblers, the file names follow the usual DOS conventions, but the source file has the extension “asm“ or “src”, depending on which assembly you are using Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 11  2) The “asm” source file containing the ASSEMBLING program code created in step 1 is fed to AND RUNNING an 8051 assembler AN 8051 ƒ The assembler converts the instructions into PROGRAM machine code (cont’) ƒ The assembler will produce an object file and a list file ƒ The extension for the object file is “obj” while the extension for the list file is “lst” 3) Assembler require a third step called linking ƒ The linker program takes one or more object code files and produce an absolute object file with the extension “abs” ƒ This abs file is used by 8051 trainers that have a monitor program Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 12  4) Next the “abs” file is fed into a program ASSEMBLING called “OH” (object to hex converter) AND RUNNING which creates a file with extension “hex” AN 8051 that is ready to burn into ROM PROGRAM ƒ This program comes with all 8051 assemblers (cont’) ƒ Recent Windows-based assemblers combine step 2 through 4 into one step Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 13  EDITOR PROGRAM ASSEMBLING myfile.asm AND RUNNING AN 8051 ASSEMBLER PROGRAM PROGRAM myfile.lst Other obj files Steps to Create myfile.obj a Program LINKER PROGRAM myfile.abs OH PROGRAM myfile.hex Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 14  ‰ The lst (list) file, which is optional, is ASSEMBLING AND RUNNING very useful to the programmer AN 8051 ¾ It lists all the opcodes and addresses as PROGRAM well as errors that the assembler detected ¾ The programmer uses the lst file to find lst File the syntax errors or debug 1 0000 ORG 0H ;start (origin) at 0 2 0000 7D25 MOV R5,#25H ;load 25H into R5 3 0002 7F34 MOV R7,#34H ;load 34H into R7 4 0004 7400 MOV A,#0 ;load 0 into A 5 0006 2D ADD A,R5 ;add contents of R5 to A ;now A = A + R5 6 0007 2F ADD A,R7 ;add contents of R7 to A ;now A = A + R7 7 0008 2412 ADD A,#12H ;add to A value 12H ;now A = A + 12H 8 000A 80EF HERE: SJMP HERE;stay in this loop 9 000C END ;end of asm source file address Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 15  ‰ The program counter points to the PROGRAM COUNTER AND address of the next instruction to be ROM SPACE executed ¾ As the CPU fetches the opcode from the Program program ROM, the program counter is Counter increasing to point to the next instruction ‰ The program counter is 16 bits wide ¾ This means that it can access program addresses 0000 to FFFFH, a total of 64K bytes of code Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 16  ‰ All 8051 members start at memory PROGRAM COUNTER AND address 0000 when they’re powered ROM SPACE up ¾ Program Counter has the value of 0000 Power up ¾ The first opcode is burned into ROM address 0000H, since this is where the 8051 looks for the first instruction when it is booted ¾ We achieve this by the ORG statement in the source program Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 17  ‰ Examine the list file and how the code PROGRAM is placed in ROM COUNTER AND 1 0000 ORG 0H ;start (origin) at 0 ROM SPACE 2 0000 7D25 MOV R5,#25H ;load 25H into R5 3 0002 7F34 MOV R7,#34H ;load 34H into R7 4 0004 7400 MOV A,#0 ;load 0 into A 5 0006 2D ADD A,R5 ;add contents of R5 to A Placing Code in ;now A = A + R5 6 0007 2F ADD A,R7 ;add contents of R7 to A ROM ;now A = A + R7 7 0008 2412 ADD A,#12H ;add to A value 12H ;now A = A + 12H 8 000A 80EF HERE: SJMP HERE ;stay in this loop 9 000C END ;end of asm source file ROM Address Machine Language Assembly Language 0000 7D25 MOV R5, #25H 0002 7F34 MOV R7, #34H 0004 7400 MOV A, #0 0006 2D ADD A, R5 0007 2F ADD A, R7 0008 2412 ADD A, #12H 000A 80EF HERE: SJMP HERE Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 18  ‰ After the program is burned into ROM, PROGRAM COUNTER AND the opcode and operand are placed in ROM SPACE ROM memory location starting at 0000 ROM contents Address Code Placing Code in 0000 7D ROM 0001 25 (cont’) 0002 7F 0003 34 0004 74 0005 00 0006 2D 0007 2F 0008 24 0009 12 000A 80 000B FE Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 19  ‰ A step-by-step description of the PROGRAM COUNTER AND action of the 8051 upon applying ROM SPACE power on it 1. When 8051 is powered up, the PC has Executing 0000 and starts to fetch the first opcode Program from location 0000 of program ROM ƒ Upon executing the opcode 7D, the CPU fetches the value 25 and places it in R5 ƒ Now one instruction is finished, and then the PC is incremented to point to 0002, containing opcode 7F 2. Upon executing the opcode 7F, the value 34H is moved into R7 ƒ The PC is incremented to 0004 Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 20  ‰ (cont’) PROGRAM COUNTER AND 3. The instruction at location 0004 is ROM SPACE executed and now PC = 0006 4. After the execution of the 1-byte Executing instruction at location 0006, PC = 0007 Program 5. Upon execution of this 1-byte instruction (cont’) at 0007, PC is incremented to 0008 ƒ This process goes on until all the instructions are fetched and executed ƒ The fact that program counter points at the next instruction to be executed explains some microprocessors call it the instruction pointer Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 21  ‰ No member of 8051 family can access PROGRAM COUNTER AND more than 64K bytes of opcode ROM SPACE ¾ The program counter is a 16-bit register ROM Memory Byte Byte Byte Map in 8051 0000 0000 0000 Family 0FFF 8751 AT89C51 3FFF DS89C420/30 7FFF DS5000-32 Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 22  ‰ 8051 microcontroller has only one data 8051 DATA TYPES AND type - 8 bits DIRECTIVES ¾ The size of each register is also 8 bits ¾ It is the job of the programmer to break Data Type down data larger than 8 bits (00 to FFH, or 0 to 255 in decimal) ¾ The data types can be positive or negative Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 23  ‰ The DB directive is the most widely 8051 DATA TYPES AND used data directive in the assembler DIRECTIVES ¾ It is used to define the 8-bit data ¾ When DB is used to define data, the Assembler numbers can be in decimal, binary, hex, The “D” after the decimal Directives ASCII formats number is optional, but using “B” (binary) and “H” ORG 500H (hexadecimal) for the others is required DATA1: DB 28 ;DECIMAL (1C in Hex) DATA2: DB 00110101B ;BINARY (35 in Hex) DATA3: DB 39H ;HEX The Assembler will ORG 510H Place ASCII in quotation marks convert the numbers DATA4: DB “2591” The;ASCII AssemblerNUMBERS will assign ASCII into hex code for the numbers or characters ORG 518H DATA6: DB “My name is Joe” Define ASCII strings larger ;ASCII CHARACTERS than two characters Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 24  ‰ ORG (origin) 8051 DATA ¾ The ORG directive is used to indicate the TYPES AND beginning of the address DIRECTIVES ¾ The number that comes after ORG can be either in hex and decimal Assembler ƒ If the number is not followed by H, it is decimal Directives and the assembler will convert it to hex (cont’) ‰ END ¾ This indicates to the assembler the end of the source (asm) file ¾ The END directive is the last line of an 8051 program ƒ Mean that in the code anything after the END directive is ignored by the assembler Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 25  ‰ EQU (equate) 8051 DATA TYPES AND ¾ This is used to define a constant without DIRECTIVES occupying a memory location ¾ The EQU directive does not set aside Assembler storage for a data item but associates a directives constant value with a data label (cont’) ƒ When the label appears in the program, its constant value will be substituted for the label Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 26  ‰ EQU (equate) (cont’) 8051 DATA TYPES AND ¾ Assume that there is a constant used in DIRECTIVES many different places in the program, and the programmer wants to change its value Assembler throughout ƒ By the use of EQU, one can change it once and directives the assembler will change all of its occurrences (cont’) Use EQU for the counter constant COUNT EQU 25....... MOV R3, #COUNT The constant is used to load the R3 register Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 27  ‰ The program status word (PSW) FLAG BITS AND PSW REGISTER register, also referred to as the flag register, is an 8 bit register Program Status ¾ Only 6 bits are used Word ƒ These four are CY (carry), AC (auxiliary carry), P (parity), and OV (overflow) – They are called conditional flags, meaning that they indicate some conditions that resulted after an instruction was executed ƒ The PSW3 and PSW4 are designed as RS0 and RS1, and are used to change the bank ¾ The two unused bits are user-definable Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 28  CY AC F0 RS1 RS0 OV -- P FLAG BITS AND A carry from D3 to D4 CY PSW.7 Carry flag. PSW REGISTER AC PSW.6 Auxiliary carry flag. Carry out from the d7 bit -- PSW.5 Available to the user for general purpose Program Status RS1 PSW.4 Register Bank selector bit 1. Word (cont’) RS0 PSW.3 Register Bank selector bit 0. OV PSW.2 Overflow flag. Reflect the number of 1s The result of -- PSW.1 User definable bit. in register A signed number P PSW.0 Parity flag. Set/cleared by hardware each operation is too instruction cycle to indicate an odd/even large, causing number of 1 bits in the accumulator. the high-order bit to overflow RS1 RS0 Register Bank Address into the sign bit 0 0 0 00H – 07H 0 1 1 08H – 0FH 1 0 2 10H – 17H 1 1 3 18H – 1FH Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 29  Instructions that affect flag bits FLAG BITS AND Instruction CY OV AC PSW REGISTER ADD X X X ADDC X X X ADD SUBB X X X MUL 0 X Instruction And DIV 0 X PSW DA X RPC X PLC X SETB C 1 CLR C 0 CPL C X ANL C, bit X ANL C, /bit X ORL C, bit X ORL C, /bit X MOV C, bit X CJNE X Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 30  ‰ The flag bits affected by the ADD FLAG BITS AND PSW REGISTER instruction are CY, P, AC, and OV Example 2-2 ADD Show the status of the CY, AC and P flag after the addition of 38H and 2FH in the following instructions. Instruction And PSW MOV A, #38H (cont’) ADD A, #2FH ;after the addition A=67H, CY=0 Solution: 38 00111000 + 2F 00101111 67 01100111 CY = 0 since there is no carry beyond the D7 bit AC = 1 since there is a carry from the D3 to the D4 bi P = 1 since the accumulator has an odd number of 1s (it has five 1s) Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 31  Example 2-3 FLAG BITS AND Show the status of the CY, AC and P flag after the addition of 9CH PSW REGISTER and 64H in the following instructions. MOV A, #9CH ADD ADD A, #64H ;after the addition A=00H, CY=1 Instruction And Solution: PSW (cont’) 9C 10011100 + 64 01100100 100 00000000 CY = 1 since there is a carry beyond the D7 bit AC = 1 since there is a carry from the D3 to the D4 bi P = 0 since the accumulator has an even number of 1s (it has zero 1s) Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 32  Example 2-4 FLAG BITS AND Show the status of the CY, AC and P flag after the addition of 88H PSW REGISTER and 93H in the following instructions. MOV A, #88H ADD ADD A, #93H ;after the addition A=1BH, CY=1 Instruction And PSW Solution: (cont’) 88 10001000 + 93 10010011 11B 00011011 CY = 1 since there is a carry beyond the D7 bit AC = 0 since there is no carry from the D3 to the D4 bi P = 0 since the accumulator has an even number of 1s (it has four 1s) Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 33  ‰ There are 128 bytes of RAM in the REGISTER 8051 BANKS AND ¾ Assigned addresses 00 to 7FH STACK ‰ The 128 bytes are divided into three RAM Memory different groups as follows: Space 1) A total of 32 bytes from locations 00 to Allocation 1F hex are set aside for register banks and the stack 2) A total of 16 bytes from locations 20H to 2FH are set aside for bit-addressable read/write memory 3) A total of 80 bytes from locations 30H to 7FH are used for read and write storage, called scratch pad Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 34  RAM Allocation in 8051 8051 7F REGISTER Scratch pad RAM BANKS AND 30 STACK 2F Bit-Addressable RAM RAM Memory 20 Space 1F Allocation Register Bank 3 18 (cont’) 17 Register Bank 2 10 0F Register Bank 1 (stack) 08 07 Register Bank 0 00 Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 35  ‰ These 32 bytes are divided into 4 8051 REGISTER banks of registers in which each bank BANKS AND has 8 registers, R0-R7 STACK ¾ RAM location from 0 to 7 are set aside for bank 0 of R0-R7 where R0 is RAM location Register Banks 0, R1 is RAM location 1, R2 is RAM location 2, and so on, until memory location 7 which belongs to R7 of bank 0 ¾ It is much easier to refer to these RAM locations with names such as R0, R1, and so on, than by their memory locations ‰ Register bank 0 is the default when 8051 is powered up Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 36  8051 Register banks and their RAM address REGISTER Bank 3 Bank 0 Bank 1 Bank 2 BANKS AND STACK 7 R7 F R7 17 R7 1F R7 6 R6 E R6 16 R6 1E R6 Register Banks 5 R5 D R5 15 R5 1D R5 (cont’) 4 R4 C R4 14 R4 1C R4 3 R3 B R3 13 R3 1B R3 2 R2 A R2 12 R2 1A R2 1 R1 9 R1 11 R1 19 R1 0 R0 8 R0 10 R0 18 R0 Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 37  ‰ We can switch to other banks by use 8051 REGISTER of the PSW register BANKS AND ¾ Bits D4 and D3 of the PSW are used to STACK select the desired register bank ¾ Use the bit-addressable instructions SETB Register Banks and CLR to access PSW.4 and PSW.3 (cont’) PSW bank selection RS1(PSW.4) RS0(PSW.3) Bank 0 0 0 Bank 1 0 1 Bank 2 1 0 Bank 3 1 1 Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 38  Example 2-5 8051 MOV R0, #99H ;load R0 with 99H REGISTER MOV R1, #85H ;load R1 with 85H BANKS AND STACK Example 2-6 Register Banks MOV 00, #99H ;RAM location 00H has 99H MOV 01, #85H ;RAM location 01H has 85H (cont’) Example 2-7 SETB PSW.4 ;select bank 2 MOV R0, #99H ;RAM location 10H has 99H MOV R1, #85H ;RAM location 11H has 85H Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 39  ‰ The stack is a section of RAM used by 8051 the CPU to store information REGISTER temporarily BANKS AND ¾ This information could be data or an STACK address Stack ‰ The register used to access the stack is called the SP (stack pointer) register ¾ The stack pointer in the 8051 is only 8 bit wide, which means that it can take value of 00 to FFH ¾ When the 8051 is powered up, the SP register contains value 07 ƒ RAM location 08 is the first location begin used for the stack by the 8051 Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 40  ‰ The storing of a CPU register in the 8051 stack is called a PUSH REGISTER BANKS AND ¾ SP is pointing to the last used location of STACK the stack ¾ As we push data onto the stack, the SP is Stack incremented by one (cont’) ƒ This is different from many microprocessors ‰ Loading the contents of the stack back into a CPU register is called a POP ¾ With every pop, the top byte of the stack is copied to the register specified by the instruction and the stack pointer is decremented once Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 41  Example 2-8 8051 Show the stack and stack pointer from the following. Assume the REGISTER default stack area. BANKS AND MOV R6, #25H STACK MOV R1, #12H MOV R4, #0F3H PUSH 6 Pushing onto PUSH 1 Stack PUSH 4 Solution: After PUSH 6 After PUSH 1 After PUSH 4 0B 0B 0B 0B 0A 0A 0A 0A F3 09 09 09 12 09 12 08 08 25 08 25 08 25 Start SP = 07 SP = 08 SP = 09 SP = 0A Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 42  Example 2-9 8051 Examining the stack, show the contents of the register and SP after REGISTER execution of the following instructions. All value are in hex. BANKS AND POP 3 ; POP stack into R3 STACK POP 5 ; POP stack into R5 POP 2 ; POP stack into R2 Popping From Solution: Stack After POP 3 After POP 5 After POP 2 0B 54 0B 0B 0B 0A F9 0A F9 0A 0A 09 76 09 76 09 76 09 08 6C 08 6C 08 6C 08 6C Start SP = 0B SP = 0A SP = 09 SP = 08 Because locations 20-2FH of RAM are reserved for bit-addressable memory, so we can change the SP to other RAM location by using the instruction “MOV SP, #XX” Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 43  ‰ The CPU also uses the stack to save 8051 REGISTER the address of the instruction just BANKS AND below the CALL instruction STACK ¾ This is how the CPU knows where to resume when it returns from the called CALL subroutine Instruction And Stack Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 44  ‰ The reason of incrementing SP after 8051 REGISTER push is BANKS AND ¾ Make sure that the stack is growing STACK toward RAM location 7FH, from lower to upper addresses Incrementing ¾ Ensure that the stack will not reach the Stack Pointer bottom of RAM and consequently run out of stack space ¾ If the stack pointer were decremented after push ƒ We would be using RAM locations 7, 6, 5, etc. which belong to R7 to R0 of bank 0, the default register bank Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 45  ‰ When 8051 is powered up, register 8051 REGISTER bank 1 and the stack are using the BANKS AND same memory space STACK ¾ We can reallocate another section of RAM to the stack Stack and Bank 1 Conflict Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 46  Example 2-10 8051 Examining the stack, show the contents of the register and SP after REGISTER execution of the following instructions. All value are in hex. BANKS AND MOV SP, #5FH ;make RAM location 60H STACK ;first stack location MOV R2, #25H MOV R1, #12H Stack And Bank MOV R4, #0F3H PUSH 2 1 Conflict PUSH 1 (cont’) PUSH 4 Solution: After PUSH 2 After PUSH 1 After PUSH 4 63 63 63 63 62 62 62 62 F3 61 61 61 12 61 12 60 60 25 60 25 60 25 Start SP = 5F SP = 60 SP = 61 SP = 62 Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 47  JUMP, LOOP AND CALL INSTRUCTIONS The 8051 Microcontroller and Embedded Systems: Using Assembly and C Mazidi, Mazidi and McKinlay Chung-Ping Young 楊中平 Home Automation, Networking, and Entertainment Lab Dept. of Computer Science and Information Engineering National Cheng Kung University, TAIWAN  ‰ Repeating a sequence of instructions a LOOP AND certain number of times is called a JUMP loop INSTRUCTIONS ¾ Loop action is performed by DJNZ reg, Label Looping ƒ The register is decremented ƒ If it is not zero, it jumps to the target address referred to by the label A loop can be repeated a ƒ Prior to the start of loop the register is loaded maximum of 255 times, if with the counter for the number of repetitions R2 is FFH ƒ Counter can be R0 – R7 or RAM location ;This program adds value 3 to the ACC ten times MOV A,#0 ;A=0, clear ACC MOV R2,#10 ;load counter R2=10 AGAIN: ADD A,#03 ;add 03 to ACC DJNZ R2,AGAIN ;repeat until R2=0,10 times MOV R5,A ;save A in R5 Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 2  ‰ If we want to repeat an action more LOOP AND JUMP times than 256, we use a loop inside a INSTRUCTIONS loop, which is called nested loop ¾ We use multiple registers to hold the Nested Loop count Write a program to (a) load the accumulator with the value 55H, and (b) complement the ACC 700 times MOV A,#55H ;A=55H MOV R3,#10 ;R3=10, outer loop count NEXT: MOV R2,#70 ;R2=70, inner loop count AGAIN: CPL A ;complement A register DJNZ R2,AGAIN ;repeat it 70 times DJNZ R3,NEXT Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 3  ‰ Jump only if a certain condition is met LOOP AND JZ label ;jump if A=0 JUMP MOV A,R0 ;A=R0 INSTRUCTIONS JZ OVER ;jump if A = 0 MOV A,R1 ;A=R1 JZ OVER ;jump if A = 0 Conditional... Jumps OVER: Can be used only for register A, not any other register Determine if R5 contains the value 0. If so, put 55H in it. MOV A,R5 ;copy R5 to A JNZ NEXT ;jump if A is not zero MOV R5,#55H NEXT:... Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 4  ‰ (cont’) LOOP AND JNC label ;jump if no carry, CY=0 JUMP ¾ If CY = 0, the CPU starts to fetch and execute INSTRUCTIONS instruction from the address of the label ¾ If CY = 1, it will not jump but will execute the next Conditional instruction below JNC Jumps Find the sum of the values 79H, F5H, E2H. Put the sum in registers (cont’) R0 (low byte) and R5 (high byte). MOV R5,#0 MOV A,#0 ;A=0 MOV R5,A ;clear R5 ADD A,#79H ;A=0+79H=79H ; JNC N_1 ;if CY=0, add next number ; INC R5 ;if CY=1, increment R5 N_1: ADD A,#0F5H ;A=79+F5=6E and CY=1 JNC N_2 ;jump if CY=0 INC R5 ;if CY=1,increment R5 (R5=1) N_2: ADD A,#0E2H ;A=6E+E2=50 and CY=1 JNC OVER ;jump if CY=0 INC R5 ;if CY=1, increment 5 OVER: MOV R0,A ;now R0=50H, and R5=02 Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 5  8051 conditional jump instructions LOOP AND Instructions Actions JUMP JZ Jump if A ＝ 0 INSTRUCTIONS JNZ Jump if A ≠ 0 DJNZ Decrement and Jump if A ≠ 0 Conditional CJNE A,byte Jump if A ≠ byte Jump if byte ≠ #data Jumps CJNE reg,#data (cont’) JC Jump if CY ＝ 1 JNC Jump if CY ＝ 0 JB Jump if bit ＝ 1 JNB Jump if bit ＝ 0 JBC Jump if bit ＝ 1 and clear bit ‰ All conditional jumps are short jumps ¾ The address of the target must within -128 to +127 bytes of the contents of PC Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 6  The unconditional jump is a jump in ‰ LOOP AND which control is transferred JUMP unconditionally to the target location INSTRUCTIONS LJMP (long jump) Unconditional ¾ 3-byte instruction Jumps ƒ First byte is the opcode ƒ Second and third bytes represent the 16-bit target address – Any memory location from 0000 to FFFFH SJMP (short jump) ¾ 2-byte instruction ƒ First byte is the opcode ƒ Second byte is the relative target address – 00 to FFH (forward +127 and backward -128 bytes from the current PC) Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 7  ‰ To calculate the target address of a LOOP AND JUMP short jump (SJMP, JNC, JZ, DJNZ, etc.) INSTRUCTIONS ¾ The second byte is added to the PC of the instruction immediately below the jump Calculating ‰ If the target address is more than -128 Short Jump to +127 bytes from the address below Address the short jump instruction ¾ The assembler will generate an error stating the jump is out of range Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 8  Line PC Opcode Mnemonic Operand LOOP AND 01 0000 ORG 0000 JUMP 02 0000 7800 MOV R0,#0 03 0002 7455 MOV A,#55H INSTRUCTIONS 04 0004 6003 JZ NEXT 05 0006 08 INC R0 Calculating 06 0007 04 + AGAIN: INC A 07 0008 04 INC A Short Jump 08 0009 2477 NEXT: ADD A,#77H Address 09 000B 5005 JNC OVER (cont’) 10 000D E4 CLR A 11 000E F8 MOV R0,A 12 000F F9 + MOV R1,A 13 0010 FA MOV R2,A 14 0011 FB MOV R3,A 15 0012 2B OVER: ADD A,R3 16 0013 50F2 JNC AGAIN 17 0015 80FE + HERE: SJMP HERE 18 0017 END Department of Computer Science and Information Engineering HANEL National Cheng Kung University, TAIWAN 9  ‰ Call instruction is used to call subroutine CALL ¾ Subroutines are often used to perform tasks INSTRUCTIONS that need to be performed frequently ¾ This makes a program more structured in addition to saving memory space LCALL (long call) ¾ 3-byte instruction ƒ First byte is the opcode ƒ Second and third bytes are used for address of target subroutine – Subroutine is located anywhere within 64K byte address space ACALL (absolute call) ¾ 2-byte instruction ƒ 11 bits are used for address within 2K-byte range Department of Computer Science and Information E