Intel 8086 and 8088

According to the (Michael R. Lindeburg, PE) 3rd edition FE review manual, "All of the bits in a word are passed in parallel to the CPU. There is one data line per data bit". Therefore it takes 16 parallel data lines.

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By paying with the CHIP, it means that you have put an amount of money on your bankcard and you can pay small amounts without having to use a code (this system is currently in use in the Netherland). By pay with your pin code (a 4 figure code), you can have the payment deducted directly from your bank account.

The web address of the Greyhound Bus Museum is: http://greyhoundbusmuseum.org

A Microprocessor provides a memory address (within the address space of that micro, which is related to the number of address pins of the micro. For instance 32 pins can generate 4GB of addresses). In a system with DRAM modules, each DRAm chip contributes with only one bit to compose the word. Each bit has its own address, each DRAM chip has its own address within a module and each module has its own address within the memory system. A DRAM chip is composed by N matrices, so there are "matrix address", line address and column address. There are so many addresses and control signals, and the microprocessor knows nothing about them. Thus, the system must contain a memory controller which translates the address provided by the processor (some times called virtual) into fisical memory address. These translations are made by the hardware (only).

Another history is the virtual memory, where there is a mapping between the address provided by the processor and the address where the data is stored (part in ram and part on disk). The address space is broken in pages, and a mapping unit (which can use a cache like memory - TLB) mantain information about which page is on the disk and witch is on the RAM. This system is a combination of hardware and operating system. The precise explanation can be found in any operating system book.

Sorry by the English.

Instruction pointer (IP) is used to hold the offset of the next instruction to be fetched for BIU available from Code Segment whose base address is held in CS segment base register..

Pentium I processors are CISC.

SS

0x00000000 to 0xFFFFFFFF in hexadecimal

0 to 4294967295 in binary

Bytes:32768

Bits:262144

The development of this chip led to the invention of the first personal computer. With this invention, the use of computers spread from large businesses and the military to small businesses and homes.

The processor stops and goes to the halt state. If an interrupt occurs, it responds and then continues execution.

addressing mode is used to form an instruction format.

internal configuration 8086

The need for a timing diagram for a microprocessor is, primarily, to allow you to properly design the interface logic that will support the microprocessor.

You need to know what lines are used to execute various data transfers, what are the timing of those lines with respect to each other, and how the microprocessor expects the external logic to behave.

You can also use the timing diagram to understand how the microprocessor functions and, particularly, to know how long each instruction will take.

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jmp

.code

main proc

mov ax,@data

mov ds,ax

lea dx,msg ;printing msg

mov ah,09h

int 21h

mov ax,x ;ax=x

mov bx,y ;bx=y

cmp ax,0 ;jump to l3 if ax is negtive

jb l3

cmp bx,0 ;jump to l6 if bx is negative

jb l6

cmp ax,bx ;if ax<bx,then jump to l1

jl l1

sub ax,bx ;else normal sub

mov diff,ax ;diff=result is stored

jmp l2

l1: ;iff (+)ax<(+)bx

neg bx ;bx=-bx

clc

add ax,bx

neg ax ;-ans=ans

mov diff,ax

mov dx,2dh ;print '-'

mov ah,02h

int 21h

jmp l2

l3: ;iff (-)ax

neg ax ;-ax=ax

cmp bx,0 ;jump to l4 if bx is negative

jb l4

clc

add ax,bx ;ax=(+)ax+(+)bx

mov ax,diff

mov dx,2dh ;print '-'

mov ah,02h

int 21h

jmp l2

l4: ;if (-)ax & (-)bx

neg bx ;-bx=bx

cmp ax,bx ;if ax>bx then jump to l5

jg l5

sub ax,bx ;else ax-bx

mov diff,ax

mov dx,2dh ;print '-'

mov ah,02h

int 21h

jmp l3

l5: ;if(-)ax>(-)bx

xchg ax,bx ;exchange ax and bx

sub ax,bx ;ax-bx

mov diff,ax ;ans is positive

jmp l2

l6: ;iff (-)bx

neg bx ;-bx=bx

add ax,bx ;ax-(-)bx

mov diff,ax ;ans will be positive

mov ah,4ch

int 21h

main endp

Yes, as long as one of them is the accumulator...

ADD BX

... adds BX to AX and leaves the result in AX.

three ,A,B$C.

2^14 memory locations. In general for n-bit address bus, its 2^n

when conditional jump instruction is executed it has 10 m/c cycles bt when nt executed it has 7 m/c cycles....while unconditional jump instruction has 10 m/c cycles...

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The control and high order address buses are unidirectional.

The 8085 microprocessor has an 8-bit flag register, even though it only has 5 flags, because all of the registers in the 8085 are 8-bit or 16-bit. In fact, the flags register is considered to be part of the accumulator, as one "register pair", for purposes of stack push/pop and interrupt call/return.

Be warned, however, that you should not attempt to change the state of bits stored on the stack and then pop'ing them into AF, because some of the three "unused" bits might be undocumented but actually used, or their meaning might change between different versions of the chip.