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Microprocessor Chapter 2 The 6800 microprocessor.

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Presentation on theme: "Microprocessor Chapter 2 The 6800 microprocessor."— Presentation transcript:

1 Microprocessor Chapter 2 The 6800 microprocessor

2 Address Bus Data Bus Control Bus Arithmetic and Logic Unit Control Unit Program Counter Controller File Storage Hard Disk Floppy Disk Tape CD Memory RAM ROM PROM Driver Input Devices Keyboard Mouse Scanner Serial Port Serial Devices (e.g. Modem) Parallel Port Parallel Devices (e.g. Printer) Output Devices Monitor Loudspeaker Driver A Typical Computer Architecture (follow the links for explanations of the terms) Clock CPU



5 Microprocessor Characteristics Number of address wires Number of data wires Speed Architecture Language (Assembly language)

6 Characteristics of the 6800 Microp. 16 address wires => it can address 2 16 memory locations It is an 8 bit microprocessor => 8 data bits Speed may be 4 MHZ Architecture??

7 6800 architecture Architecture: –Registers –Logic and Arithmetic unit –Control unit –Program counter –Stack pointer –…

8 Accumulator A Accumulator B Condition Code Index Register X=XH : XL Stack Pointer, SP Program Counter, PC 6800 registers Instruction register Control unit Data bus from and to memory Address wires from PC ALU Temp 8 8 8

9 PC: Program counter Program counter contain the address of the memory containing the next instruction to be executed. It is incremented automatically after each instruction In 6800, it is a 16 bits register

10 First instruction: ABA ABA: opcode=?? First, the instruction is fetched in the instruction register (IR is used to fixx the command for the control unit during processing) The control unit command the registers in the following way: –Command register A to put its contain on the internal bus –Command the register TEMP to take the data on the bus –Command the register B to put its contain on the bus –Command the ALU to add –Command A register to take the result

11 instruction LDA A #$ If this instruction is written in the memory address: $5000, the program counter contain $5000 when executed. Procedure: –PC=5000 so the data 86, read from this memory address, goes to the instruction register –The control unit analyse this data and command the registers: Command the PC to be incremented, the data $90 in the address $5001 can be read now. Data are on the internal bus Command the register A to take these Data


13 Example Add: $3D and $E C=1, Z=0, V=0, N=0 H=0

14 Example Add: $BD and $A C=1, Z=0, V=1, N=0 H=1

15 Some instructions Instruction = Opcode + Operand LDA A #$4E = 86 4E STA A $2EA5 B7 2E A5 OpcodeOperand OpcodeOperand

16 First program Write a program which can add the register A to $4E and then put the result in the memory location $2346 Find the machine code of this program

17 Program Write a program which can add the two numbers: $2E and $8A and put the result in the memory location 349A


19 Stack pointer SP Used to point to the memory location where all registers will be saved when an interrupt occurs or when a subroutine is called It is a 16 bit register, contain an address of a memory



22 Interrupt During the processing of a program, the microprocessor can stop its work and execute another emergency program. This is the technique of interrupt. 2 hardware pins can be found in the 6800 microprocessor: IRQ and NMI (both active low) – IRQ: interrupt request –NMI: Non maskable interrupt If IRQ occurs, the address of its corresponding program is written in : FFF8 and FFF9 If NMI occurs, the address of its corresponding program is written in : FFFC and FFFD

23 Interrupt IRQ may be stopped by the bit I in the status register –I=1 IRQ will not be executed (stopped) –I=0 IRQ will be executed NMI non maskable interrupt, it is executed whatever the bit I SWI is a software interrupt: Program address: FFFA and FFFB






29 Addressing modes Immediate mode LDA A #$89 A=$89 LDA B #$F3 B=$F3 Inherent mode ABA A+B A INCA A+1 A DECB B-1 B INS S+1 S

30 Addressing modes Extended mode (2 bytes operand) LDA A $15CD STA A $078C Direct mode (one byte operand) LDA $12 Equilvalent to LDA $0012

31 Adresing modes Indexed (using X register. LDA A $05,X STA A $08,X This register is used as a base address when we want to make the addressing for a series of successive memories

32 Addressing modes Relative mode (Branch instructions) BEQ: Branch if equal: if z=1 BNE : branch if not equal : if z=0 BME : branch if minus: if N=1 B PL: branch if positive : if N=0 BCS : branch if carry set : if C=1 BCC : branch if carry clear : if C=0

33 Program LDA A #$89 LDA B $6789 INC A BPL here STA A $1928 INCB LDX #$98F5 here STA A $09 LDX #$9000 RTS Write the machine code of this program

34 Instructions ADD SUB …

35 Instructions ROR ROL Stack pointer instructions –PUSH –PULL

36 Programs P1: write a program to add the memory contents: from A9F0 to A9FF and put the result in the memory location P2: write a program to transfer the memory contents: from FF to A000 - A0FF. Write a program to compute the integer part of: ( )/4

37 A program CLI LDA A #$26 LDX $2608 LDA B $02,X (IRQ arrives here) CMP A $25FE BNE here ABA RTS here TAB RTS Interrupt program :LDA A #$50 LDA B $2602 ABA RTI ContentAdresses 00260B 12260A D2605 1F C25FF 5A25FE :: 00FFFF 60FFFE 00FFFD 50FFFC 30FFFB 50FFFA 40FFF9 50FFF8 30FFF7 50FFF6 40FFF5 50FFF4 20FFF3 30FFF2 FFF1 ::

38 6800 Pins power and ground Address bus (A0-A15) –tri-state: controlled by BA or TSC –VMA (valid memory address) and dummy access Data bus (D0-D7) R/W – high impedance if BA=1 and TSC is asserted Reset, HALT, NMI, IRQ BA (bus available), BS (bus status): –normal (00), interrupt or reset acknowledge (01), sync acknowledge (10), and halt acknowledge(11) Clocks E and Q

39 6800 Pins (continued) Busy: –read and modify cycles, double-byte operation, indirect or vector fetch AVMA: advance VMA –MPU will use the bus in the following cycle LIC: last instruction cycle –instruction fetch if transitions from high to low TSC: three-state control –direct control when E is low –latched on the rising edge of E

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