Chapter 1. Microprocessor architecture ECE Dr. Mohamed Mahmoud.

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1 Chapter 1 Microprocessor architecture ECE 3130 Dr. Mohamed Mahmoud The slides are copyright protected. It is not permissible to use them without a permission from Dr Mahmoud

2 Outline 1.1 Number Systems 1.2 Computer hardware organization 1.3 Memory system operation 1.4 Processor 1.5 Program Execution 1.6 HCS12 Microcontroller

3 - The binary number system has two digits 0 and 1 - The hexadecimal number system uses 16 digits: 0, 1,.., 9, A, B, C,.., F - The decimal number system uses 10 digits: 0, 1,.., 9 - Computer hardware uses binary numbers to perform all operations. - Humans are used to decimal number system. - Hexadecimal numbers have shorter representations than the binary system. - Conversion between the different number systems is needed. 1-1

4 - A prefix is used to indicate the base of a number. Number like 10 can be binary, hexadecimal and decimal. We use a prefix to define it. - Convert % to Hexadecimal = $45 - Convert $4F to Binary $4F = Unsigned numbers are always positive - Signed number B Most significant bit (B) = 1 negative, otherwise positive 1-2

5 1- Unsigned number %1111 = = 15 %0111 = = 7 Unsigned N-bit number can have numbers from 0 to 2 N Signed number %1111 is a negative number. To convert to decimal, calculate the two s complement The two s complement = one s complement +1 = % =%0001 = 1 then %1111 = -1 %0111 is a positive number = = 7. - Unsigned 8-bit number can have numbers from 0 to 255 % to % Signed 8-bit number can have numbers from -128 to +127 to % to %

6 - Signed N-bit number can have numbers from -2 N-1 to 2 N Unsigned 3-bit number can represent numbers from 0 (= %000) to 7 (=111) - Signed 3-bit number can represent numbers from -4 (= %100) to 3 (= %011) Signed Computer does not know whether a bit sequence represents a signed or an unsigned number -> It is the programmer s responsibility! 1-4

7 Outline 1.1 Number Systems 1.2 Computer hardware organization 1.3 Memory system operation 1.4 Processor 1.5 Program Execution 1.6 HCS12 Microcontroller

8 - A Computer consists of hardware and software - Software is computer program. - The hardware consists of processor, input devices, output devices and memory. Computer Hardware Organization 1-5

9 1- Processor - Also called Microprocessor (MP) or Central Processing Unit (CPU). - Executes programs. - Performs: (1) all of the computational operations and (2) the coordination of the usage of resources of a computer - Ex. can a memory and an input device write data on a shared bus simultaneously? NO. Control signals are issued by processor to coordinate the devices operations. 2- Input devices - Used to enter the programs to be executed and the data to be processed into the computer. - Examples: keyboard, keypad, switches, sensors, scanners, bar code readers, etc. 1-6

10 3- Output devices - The results of the operations done by the processor should be displayed or printed on a media so that the user can see them. - Examples: seven segment displays, light emitting diode (LED), liquid crystal displays (LCD), a monitor, etc. 4- Memory - Programs to be executed and data to be processed are stored in memory so that the processor can readily access them. 5- Buses - Address bus: The set of conductor wires that carry address signals. - Data bus: The set of conductor wires that carry data signals - Control bus: The set of conductor wires that carry control signals 1-7

11 Outline 1.1 Number Systems 1.2 Computer hardware organization 1.3 Memory system operation 1.4 Processor 1.5 Program Execution 1.6 HCS12 Microcontroller

12 Address: location number $FF $00 $3F $AF Content: A byte stored in a location A block diagram for the memory - Each memory location has address and content. - The content is one byte (8 bits). - If a memory chip has 16-bit address, then the chip has 2 16 = 64 Kbytes because 16 bits can point at 2 16 locations. - Byte = 8 bits, Word = 2 bytes = 16 bits, Nibble = 4 bits - Kilobyte (KB) = 2 10 bytes =1024 bytes -Megabyte(MB)=K 2 Bytes = 2 20 bytes = 1,048,576 bytes 1-8

13 16 bits 8 bits 8 bits 2 16 locations - The memory data bits are connected to the data bus. - The memory address bits are connected to the address bus. - RD (Read) in the MP is connected to OE (output enable) in the memory and WR (write) is connected to WE (write enable) - RD and WR are two control signals issued by the CPU to inform the memory to read or write. 1-9

14 Read Operation - The processor places the address of the memory location that it intends to read on the address bus and asserts the RD signal. (RD = 1 and WR = 0) - After some time, the data stored in the location referred by the address will be copied from the memory to the data bus to be read by the MP. The content of the memory location does not change. Write Operation - The processor places the address of the location that it intends to write on the address bus. - The processor also places the data that it intends to write on the data bus and asserts the WE signal (RD = 0 and WR =1) - The memory chip stores the data on the data bus in the selected memory location after some time. The content of the memory location changes. 1-10

15 Outline 1.1 Number Systems 1.2 Computer hardware organization 1.3 Memory system operation 1.4 Processor 1.5 Program Execution 1.6 HCS12 Microcontroller

16 - A processor consists of arithmetic logic unit (ALU), control unit, andregisters. - The ALU performs the arithmetic and logic operations requested by the program. - A simple ALU that can perform only four operations (addition, subtraction, AND, OR) is shown in Figure

17 - All four operations are performed simultaneously by different circuits whereas the instruction opcode tells the multiplexer to select one of the four units outputs as the result - The adder is used to perform addition and subtraction operations - The ALU is more complicated if the processor designer wants to implement more operations directly in the hardware. Remember how multiplexer works: Selector Select Output Inputs I 0 I 1 I 2 MUX output I 0 I 1 I 2 I 3 11 I

18 The ALU operations:- 1- Opcode = 00 - The comparator output = 0, MUX2 output = CIN, MUX1 output = B, the adder inputs are n-bit A and n-bit B. - The result of the adder is A + B + CIN, CIN is the input carry. - MUX3 selects the output of the adder and the carry A + B+ CIN 1-13

19 2- Opcode = 01 - The comparator output = 1, MUX2 output = 1, MUX1 output = B (one complement of B), the adder inputs are n-bit A and n- bit B. - The result of the adder is A + B + 1 = A - B. Remember B + 1 = - B. B + 1 is the two s complement of B. - MUX3 selects the output of the adder and the carry (called borrow) A + B + 1 = A - B 1-14

20 3- Opcode = 10 MUX3 selects the output of the AND. It is the n bits resulted from performing logic AND operation for A and B. 4- Opcode = 11 MUX3 selects the output of the OR. It is the n bits resulted from performing logic OR operation for A and B. Remember:- X can be zero or one The ALU in Fig. 1.1 can be expanded to perform more instructions 1-15

21 Instruction Set The instruction set of a microprocessor is the fixed set of operations the computer can be programmed to perform on data. These instructions are executed on hardware, i.e., there is a hardware circuit that executes them. 1-16

22 2- Registers - A register is a storage area inside the CPU. - A register is used to hold data or address during the execution of an instruction. - Registers provide fast access to operands for program execution. - Unlike memory, registers are not used to store the program or large data. - Accessing memory is much slower than accessing registers 3- Control Unit - The control unit executes the instructions by telling the ALU what operation it has to do (opcode) and issuing control signals to read/write from memory. 1-17

23 Outline 1.1 Number Systems 1.2 Computer hardware organization 1.3 Memory system operation 1.4 Processor 1.5 Program Execution 1.6 HCS12 Microcontroller

24 Compiler vs assembler High-level programming language source code for (i=1, i<10, i++) { Y = Y+i; } Written by programmers Compiler Machine code Stored in memory - Executable by the CPU Low-level programming language source code ldaa $10 adda $20 staa $30 Written by programmers assembler Machine code Stored in memory - Executable by the CPU 1-18

25 Compiler vs assembler - Compiler is more complex than assembler - Each assembly language instruction is translated to one machine language instruction - But, one statement written in high level language may be translated into tens or hundreds of machine language instructions and this translation is not usually optimal - The CPU can execute ONLY machine codes. - Obviously, programming by the machine code is hard to understand, enter, debug, and maintain for human humans. That is why we use assembly language that use English words. 1-19

26 Executing a program 1- The assembler converts the assembly program to machine code and stores it in the memory, e.g., starting from location $ Some instructions take more than a byte. -A program counter (PC) is a register that is used to keep track of the address of the instruction to be executed next. $0000 ld 0x20, #0 ld 0x21, #20 ld ptr, #0x2000 loop: and A, #0x03 bnz next ld inc 0x20 next: dbnz 0x21,loop 1-20

27 2- Two phases to execute each instruction:- 2.1 Instruction fetch: - Read the instruction from the memory location pointed by PC and store it in the instruction register (IR) - May need more than one reading operation if the instruction is more than one byte. - Update PC to point at the address of the next instruction. - Both PC and IR are two registers in the CPU. 2.2 Instruction Execution - The control unit decodes the instruction and issues appropriate control signals to execute it. - It may need to read/write the memory. 1-21

28 - The clock signal is a square waveform. - Instruction execution is timed by the clock signal. - At each cycle, some operations are performed. - Executing one instruction may need more than one cycle. 1-22

29 Instruction ld $20,#0 (machine code ) Place 0 in data memory at 0x20 1- Fetch instruction: - The Processor places contents of PC ($0000) on the address bus to read the instruction to be executed. -It reads the first byte (opcode) = $75. PC is incremented to $ From the opcode, the control unit recognizes that it needs to read the following two bytes after the opcode. It reads $20 and $00 and holds them in IR. PC is incremented after each read operation to be 3 to point at the next executed instruction. 2- Execute the instruction: - Address bus $20 and Data bus $00 - WR =1 to write 0 into the memory location $

30 Initially PC = 0 PC = PC +1 after reading the byte pointed by 0 (or 75) 1-24

31 Instruction ld $21,#$14 (machine code ) - Similar to previous instruction. - PC = $0006 and write $14 to memory location $ Instruction ld ptr,#$2000 (machine code ) - Load $2000 into the ptr register 1- Fetch instruction: - The address bus PC value ($06), request to read. - Opcode = $90 and PC = $07. - From the opcode, we need to read the following two locations. Read $20 and $00 and PC = 0x Execute instruction: ptr

32 The machine code of the program Interesting video

33 Summary Functional blocks required to run a program: A place to store the program, operands, and computation results memory and registers A way to keep track of which instruction is to be executed next program counter (PC) A place to temporarily stage an instruction while it is being executed instruction register (IR) A way to perform arithmetic and logic operations arithmetic logic unit (ALU) A way to coordinate and sequence the functions of the machine to execute the instructions control unit 1-27

34 Instruction Execution Cycle One or more read cycles to fetch the instruction from memory. One or more read cycles to fetch the memory operand(s) (optional) Perform the operation specified by the instruction One or more write cycles to write back the result to either a register or a memory location (optional) 1-28

35 Memory is used to store: - Instructions: binary codes prescribing instructions - Data: to be manipulated - Stack: will be explained later Instruction, data and stack areas in memory must be separated. Any overlapping means, the program will crash It is the programmer s responsibility. 1-29

36 Outline 1.1 Number Systems 1.2 Computer hardware organization 1.3 Memory system operation 1.4 Processor 1.5 Program Execution 1.6 HCS12 Microcontroller

37 Microprocessor VS Microcontroller - A microprocessor (MP) needs external devices such as memory, I/O ports, etc to run a program. - A microcontroller (MCU) contains a microprocessor, memory timer, and I/O ports within a single chip. - Microcontrollers have been used in almost every product that requires a certain amount of intelligence. - Microcontrollers have been used in printers, modems, MP3 players, cars, home appliances such as washing machines, microwave ovens, etc. 1-30

38 Embedded Systems - An embedded system is a special purpose computer system designed to perform a dedicated function. - Examples of embedded systems: MP3 players, traffic lights, digital watches Runs firmware which is stored on read only memory chips Interesting applications Smart mirror: Smart spoon: Smart baby bottle:

39 The HCS12 Features - 16-bit CPU (the operations can be done on 16 bit data) - Maximum operating clock frequency is 25 MHz - 2KB to 14KB of on-chip SRAM (12KB in MC9S12DG256 ) - 4K bytes of EEPROM in MC9S12DG KB to 512KB on-chip flash memory (256 KB in MC9S12DG256 ) - Timer and Pulse counters - I/O pins to interface to I/O devices - Serial communication - A/D 1-32

40 HCS12 microcontroller 1-33

41 - Static random access memory (SRAM) is volatile. Flash memory: - Non-volatile memory does not lose the information when there is no power, but volatile memory does. - Electrically erasable and programmable many times by the MP. - The most widely used nonvolatile memory technology. - In embedded systems, the program is stored in a nonvolatile memory. Why? 1-34

42 HCS12 registers: - Registers are referred to by their names whereas memory locations are referred to by their addresses - Used to perform general purpose operations such as arithmetic, logic, and program flow control. - All the registers are 16 bits except the accumulators A and B which are 8 bits each. - A and B can be concatenated to form a single 16-bit register accumulator referred as the D accumulator - When A changes B is unaffected and vice versa, but when D changes both A and B change. 1-35

43 The HCS12 CPU Registers 7 A 0 7 B 0 15 D 0 8-bit accumulator A and B or 16-bit double accumulator D 15 X 0 Index register X 15 0 Y Index register Y 15 SP 0 Stack pointer 15 PC 0 Program counter S X H I N Z V C Figure 1.10 HCS12 CPU registers. Condition code register Carry Overflow Zero Negative I Interrupt mask Half-Carry (from bit 3) X Interrupt Mask Stop Disable 1-36

44 - Index registers X and Y: 16 bits. Used to hold addresses. They are also used in several arithmetic instructions. - Stack pointer (SP): 16 bits. A stack is a first-in-first-out data structure. SP points to the top byte of the stack as shown in figure. - Program counter (PC): 16 bits holds the address of the next instruction to be executed. - Condition code register (CCR): 8-bit register used to - Flags are automatically updated after executing each instruction to reflect the status of the result. Lower address Top of stack Push SP Pop - Used to make a condition in conditional branch. Higher address 1-37

45 - Each bit in the CCR register is independent and has a different indication. - Zero flag (Z): set when the result is zero - Negative flag (N): set whenever the result is negative, i.e., most significant bit of the result is 1. - Carry/borrow flag (C): set when addition/subtraction generates a carry/borrow. - Overflow flag (V): Set when: the addition of two positive numbers results in a negative number or the addition of two negative numbers results in a positive number. 1-38

46 Overflow Problem: Registers have fixed number of bits which means they can store a maximum number Overflow occurs when two numbers are added or subtracted and the correct result is outside the range that can a register hold the given result is not correct Addition: C = 1 there is an overflow if the numbers are unsigned. V = 1 there is an overflow if the numbers are signed. Overflow cannot occur when adding numbers of opposite sign why? Subtraction: A - B There is no unsigned overflow but there is signed overflow why? C = 1, when there is a borrow or B > A, C is called borrow flag V =1, when (-ve) - (+ve) = (+ve) this is equivalent to ( ve) + (-ve) = (+ve) (+ve) - (-ve) = (-ve) this is equivalent to (+ve) + (+ve) = (-ve) 1-39

47 C = 1, V = 0, Z = 1, N =0 Signed numbers: = 0, no overflow and the result is correct Unsigned numbers: = 0, incorrect, the correct result is 256, overflow because the max. unsigned number for 8 bit number is By using more bits (9 bits or more) instead of 8 bits, the result is correct, no overflow C = 0, V = 0, Z = 0, N = 1 Signed numbers: = -1, no overflow and the result is correct Unsigned numbers: = 255, no overflow and the result is correct 1-40

48 C = 0, V = 1, Z = 0, N = 1 Unsigned numbers: = 128, no overflow and the result is correct Signed numbers: = -128, there is overflow, the result is incorrect, the max. positive number in 8 bits is 127 that is less than the correct answer 128. If we use 9 bit addition, the result will be correct because 128 can be represented in 9 bits C = 0, V = 0, Z = 0, N = C = 1, V = 1, Z = 0, N = 0 Unsigned numbers: = 54 should be 310, overflow, 255 is the max. number for 8 bit number Signed numbers: (-118) = 54, should be -202, overflow, the max. negative number in 8 bits is -128 that is less than the correct answer 1-41

49 If we use 9 bit addition, the result will be correct. Unsigned numbers: C = 0 Signed numbers: V = 0 Subtraction: A B = A + the two s complement of B V = 0, C = 0, N = 0, Z =0 C is called borrow flag and it is set when we need to borrow from the most significant byte Unsigned numbers: = 30 correct Signed numbers: = 30 correct V = 0 because (+ve) (+ve) no overflow 1-42

50 V = 1, C = 1, N = 1, Z = There is one set of flags whether the numbers are signed or not, e.g. N flag will consider the numbers Unsigned numbers: signed even if they are not - There is a borrow, = What happened is (92+256) -138 = 210, where 256 is the borrow - If we do muti-byte subtraction, the result (210) is right and we should subtract one from the next byte. - If you want to get the absolute difference 46, subtract the small number from the bigger one or = 46 Signed numbers:- 92 (-118) = -46 should be 210, V = 1 means the numbers should be represented in more bits V =

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