Intel 8086 and 8088

There are many instructions in the 8085. To find the hex code for a particular instruction, you need to look at the data sheet. For more information, please see the Related Link below.

program counter holds the address of the next instruction.

b/c the 8085 microprocessor is the first 8 bit microprocessor which Intel is produces in 1877 and this is the first general purpose 8 bit microprocessor.

there was an 8 bit general purpose register before 8085 named as 8008 but this microprocessor is not functional 8 bit microprocessor

how many interrupts in 8051

Flags are used to indicate the status of various things, usually following some kind of operation. For instance, the ZERO flag indicates the result is zero, the CARRY flag indicates the operation results in a carry, the NEGATIVE flag indicates the result is negative, etc.

Flags can be used in loop control. For instance you can decrement a counter at the end of a loop, and jump conditionally to the top of the loop if the counter is non zero.

ACI n |CE|*****| 7|Add with Carry Immediate |A=A+n+CY |

|ADC r |8F|*****| 4|Add with Carry |A=A+r+CY(21X)|

|ADC M |8E|*****| 7|Add with Carry to Memory |A=A+[HL]+CY |

|ADD r |87|*****| 4|Add |A=A+r (20X)|

|ADD M |86|*****| 7|Add to Memory |A=A+[HL] |

|ADI n |C6|*****| 7|Add Immediate |A=A+n |

|ANA r |A7|****0| 4|AND Accumulator |A=A&r (24X)|

|ANA M |A6|****0| 7|AND Accumulator and Memory|A=A&[HL] |

|ANI n |E6|**0*0| 7|AND Immediate |A=A&n |

|CALL a |CD|-----|18|Call unconditional |-[SP]=PC,PC=a|

|CC a |DC|-----| 9|Call on Carry |If CY=1(18~s)|

|CM a |FC|-----| 9|Call on Minus |If S=1 (18~s)|

|CMA |2F|-----| 4|Complement Accumulator |A=~A |

|CMC |3F|----*| 4|Complement Carry |CY=~CY |

|CMP r |BF|*****| 4|Compare |A-r (27X)|

|CMP M |BF|*****| 7|Compare with Memory |A-[HL] |

|CNC a |D4|-----| 9|Call on No Carry |If CY=0(18~s)|

|CNZ a |C4|-----| 9|Call on No Zero |If Z=0 (18~s)|

|CP a |F4|-----| 9|Call on Plus |If S=0 (18~s)|

|CPE a |EC|-----| 9|Call on Parity Even |If P=1 (18~s)|

|CPI n |FE|*****| 7|Compare Immediate |A-n |

|CPO a |E4|-----| 9|Call on Parity Odd |If P=0 (18~s)|

|CZ a |CC|-----| 9|Call on Zero |If Z=1 (18~s)|

|DAA |27|*****| 4|Decimal Adjust Accumulator|A=BCD format |

|DAD B |09|----*|10|Double Add BC to HL |HL=HL+BC |

|DAD D |19|----*|10|Double Add DE to HL |HL=HL+DE |

|DAD H |29|----*|10|Double Add HL to HL |HL=HL+HL |

|DAD SP |39|----*|10|Double Add SP to HL |HL=HL+SP |

|DCR r |3D|****-| 4|Decrement |r=r-1 (0X5)|

|DCR M |35|****-|10|Decrement Memory |[HL]=[HL]-1 |

|DCX B |0B|-----| 6|Decrement BC |BC=BC-1 |

|DCX D |1B|-----| 6|Decrement DE |DE=DE-1 |

|DCX H |2B|-----| 6|Decrement HL |HL=HL-1 |

|DCX SP |3B|-----| 6|Decrement Stack Pointer |SP=SP-1 |

|DI |F3|-----| 4|Disable Interrupts | |

|EI |FB|-----| 4|Enable Interrupts | |

|HLT |76|-----| 5|Halt | |

|IN p |DB|-----|10|Input |A=[p] |

|INR r |3C|****-| 4|Increment |r=r+1 (0X4)|

|INR M |3C|****-|10|Increment Memory |[HL]=[HL]+1 |

|INX B |03|-----| 6|Increment BC |BC=BC+1 |

|INX D |13|-----| 6|Increment DE |DE=DE+1 |

|INX H |23|-----| 6|Increment HL |HL=HL+1 |

|INX SP |33|-----| 6|Increment Stack Pointer |SP=SP+1 |

|JMP a |C3|-----| 7|Jump unconditional |PC=a |

|JC a |DA|-----| 7|Jump on Carry |If CY=1(10~s)|

|JM a |FA|-----| 7|Jump on Minus |If S=1 (10~s)|

|JNC a |D2|-----| 7|Jump on No Carry |If CY=0(10~s)|

|JNZ a |C2|-----| 7|Jump on No Zero |If Z=0 (10~s)|

|JP a |F2|-----| 7|Jump on Plus |If S=0 (10~s)|

|JPE a |EA|-----| 7|Jump on Parity Even |If P=1 (10~s)|

|JPO a |E2|-----| 7|Jump on Parity Odd |If P=0 (10~s)|

|JZ a |CA|-----| 7|Jump on Zero |If Z=1 (10~s)|

|LDA a |3A|-----|13|Load Accumulator direct |A=[a] |

|LDAX B |0A|-----| 7|Load Accumulator indirect |A=[BC] |

|LDAX D |1A|-----| 7|Load Accumulator indirect |A=[DE] |

|LHLD a |2A|-----|16|Load HL Direct |HL=[a] |

|LXI B,nn |01|-----|10|Load Immediate BC |BC=nn |

|LXI D,nn |11|-----|10|Load Immediate DE |DE=nn |

|LXI H,nn |21|-----|10|Load Immediate HL |HL=nn |

|LXI SP,nn|31|-----|10|Load Immediate Stack Ptr |SP=nn |

|MOV r1,r2|7F|-----| 4|Move register to register |r1=r2 (1XX)|

|MOV M,r |77|-----| 7|Move register to Memory |[HL]=r (16X)|

|MOV r,M |7E|-----| 7|Move Memory to register |r=[HL] (1X6)|

|MVI r,n |3E|-----| 7|Move Immediate |r=n (0X6)|

|MVI M,n |36|-----|10|Move Immediate to Memory |[HL]=n |

|NOP |00|-----| 4|No Operation | |

|ORA r |B7|**0*0| 4|Inclusive OR Accumulator |A=Avr (26X)|

|ORA M |B6|**0*0| 7|Inclusive OR Accumulator |A=Av[HL] |

|ORI n |F6|**0*0| 7|Inclusive OR Immediate |A=Avn |

|OUT p |D3|-----|10|Output |[p]=A |

|PCHL |E9|-----| 6|Jump HL indirect |PC=[HL] |

|POP B |C1|-----|10|Pop BC |BC=[SP]+ |

|POP D |D1|-----|10|Pop DE |DE=[SP]+ |

|POP H |E1|-----|10|Pop HL |HL=[SP]+ |

|SUB r |97|*****| 4|Subtract |A=A-r (22X)|

|SUB M |96|*****| 7|Subtract Memory |A=A-[HL] |

|SUI n |D6|*****| 7|Subtract Immediate |A=A-n |

|XRA r |AF|**0*0| 4|Exclusive OR Accumulator |A=Axr (25X)|

|XRA M |AE|**0*0| 7|Exclusive OR Accumulator |A=Ax[HL] |

|XRI n |EE|**0*0| 7|Exclusive OR Immediate |A=Axn |

|XTHL |E3|-----|16|Exchange stack Top with HL|[SP]<->HL

|SBB r |9F|*****| 4|Subtract with Borrow |A=A-r-CY |

|SBB M |9E|*****| 7|Subtract with Borrow |A=A-[HL]-CY |

|SBI n |DE|*****| 7|Subtract with Borrow Immed|A=A-n-CY

To calculate the number of address lines required for a 64 kB memory, first convert 64 kB into bytes: 64 kB = 64 × 1024 bytes = 65,536 bytes. The number of address lines needed can be determined using the formula (2^n = \text{total number of addresses}), where (n) is the number of address lines. Since 65,536 is (2^{16}), you need 16 address lines to address a 64 kB memory.

difference between micro operation and microinstruction

8086 has two ground pins because to make the pins count 40(even number(40) so that it fits perfectly in ic cabin). that's all there no specfic operation for the both grounds .both do same operation

32 byte

first connect the 8255 to the 8086 microprocessor and configure it using the CWR then connect the key board to the 8255

it means small. it is a prefix

no

To speak to, to direct to the attention of something, to deal with something, or to mark with a destination, as in addressing a letter.

A self-addressed envelope is an envelope with your address on the front of it in the MAIL TO area. If the ad/directions include SASE...this means self-addressed stamped envelope. In this case, you would include postage so the company/organization/person can mail your request to you.

that depends on the microcontroller. check the datasheet.

I have a Sony KDL-52XBR4 connected to my Mac-Mini via a DVI to HDMI cable and it works fine. All I did was change the resolution to get it working. I'm pretty sure is in the 1280x800 range (I'm moving and the TV is in transit so I can't check at the moment). BUT, it does work and I use it all the time to watch iTunes & surf on Safari to watch Hulu. There is one quirk, the MacMini will invert the colors periodically, but I found a script someone else wrote that resets the display settings.

This may not be a complete answer to what you are looking for, but if I were to guess I would have to say that IBM's became so popular because of the fact that they are one of the most sturdiest computers ever made. IBM iswidely known for their standards in durability Even though the IBM PC's don't have all of the cosmetic wows of say an Apple they are usually more durable and therefor last longer.

Bankruptcy is a Federal matter, your State makes no differece.

Under the new bankruptcy law taking effect on October 17, 2005, Chapter 7 cannot be filed unless the debtor was discharged from the previous Chapter 7 or bankruptcy more than eight years ago. The debtor cannot file a Chapter 13 unless: (1) the debtor received a discharge under Chapter 7, 11 or 12 more than four years ago; or (2) the debtor received a discharge under Chapter 13 more than two years ago.

name "str2bin"

; convert string number to binary!

; this program written in 8086 assembly language to

; convert string value to binary form.

; this example is copied with major modifications

; from macro "scan_num" taken from c:\emu8086\inc\emu8086.inc

;

; the original "scan_num" not only converts the string number

; but also reads the string from the keyboard and supports

; backspace key, this example is a shorten version

; of original "scan_num" macro.

; here we assume that the string number is already given,

; and the string number does not contain non-digit chars

; and it cannot cause buffer overflow (number is in word range

; and/or has only 4 digits).

; negative values are allowed in this example.

; the original "scan_num" does not allow to enter non-digits

; and it also checks for buffer overflow.

; you can the original file with other macro definitions

; in c:\emu8086\inc\emu8086.inc

org 100h

jmp start

; text data:

msg1 db 0Dh,0Ah, " enter any number from -32768 to 65535 inclusive, or zero to stop: $"

msg2 db 0Dh,0Ah, " binary form: $"

; buffer for int 21h/0ah

; fist byte is buffer size,

; second byte is number of chars actually read (set by int 21h/0ah).

buffer db 7,?, 5 dup (0), 0, 0

; for result:

binary dw ?

start:

; print welcome message:

mov dx, offset msg1

The differences are given below:

80286

1. Low data bus width (16 bit)

2. Returning from protected mode to

real mode is hard and complicated.

3. Small RAM/Memory

80386

1. High data bus width (32 bit)

2. Easy for 80386 3. Big RAM / Memory (Real memory = 4GB and virtual memory= 64TB)

The 8086/8088 is a 16 bit processor running on a 16 bit (8086) or 8 bit (8088) bus with a 20 bit address bus. In order to obtain the extra 4 bits of addressibility, Intel designed segment registers that are effectively multiplied by four and then added to the 16 bit offset address generated by the instruction. This yields 64K segments of 64KB each, although they overlap each other at a distance of 16 bytes.

; program for moving a string from one block of memory

;to another block of memory

data segment

str1 db 'suriya',0

str2 db 20 dup(0)

data ends

code segment

assume cs:code,ds:data,es:data

start: mov ax,data

mov ds,ax

mov es,ax

mov cx,6

lea si,str1

lea di,str2

cld

rep movsb

int 3h

code ends

end start