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
full opcode with instruction
www.pmr.poli.usp.br/lpa/pmr2310/INT%208085.pdf
See the bottom of the related link, below, starting on page 26, for the 8086 OP Code Sheet.
program to find prime number in 8085 microprocessor
Refer to link below - (cpu-world.com/Arch/8085.html)
Sp[[Q/Write a 8085 microprocessor program to find A inverse and A transpose if A is a 3x3 matrix|Answer]]ell chec[[Q/Write a 8085 microprocessor program to find A inverse and A transpose if A is a 3x3 matrix&action=edit&section=new|Answer it!]]k your answe[[Q/Discuss:Write a 8085 microprocessor program to find A inverse and A transpose if A is a 3x3 matrix|Disc]][[help/answering questions|guidelin]]Spell check your answeresussionr[[help/signing in|full benefits]] Save C[[Q/Write a 8085 microprocessor program to find A inverse and A transpose if A is a 3x3 matrix|Write a 8085 microprocessor program to find A inverse and A transpose if A is a 3x3 ]][[Q/Write a 8085 microprocessor program to find A inverse and A transpose if A is a 3x3 matrix&action=edit&section=new|Answering 'Write a 8085 microprocessor program to find A inverse and A transpose if A is a 3x3 matrix?']]matrix?ancel[[Q/How many animals are in West Texas|How many animals are in West Texas?]][[Q/How do you increase the number of four wheelers vehicles for servicing in a Service workshop|How do you increase the number of four wheelers vehicles for servicing in a]][[Q/How do you increase the number of four wheelers vehicles for servicing in a Service workshop|How do you increase the number of four wheelers vehicles for servicing in a Service workshop?]] Service workshop?[[Q/How do you increase the number of four wheelers vehicles for servicing in a Service workshop|How do you increase the number of four wheelers vehicles for servicing in a Service workshop?]]More Q&A
You add the three numbers, then divide the result by 3.
The 8085 has a 16 bit address bus. As such, it can access 216, or 65,536 bytes. System design, of course, will place limits on that, as you need to share this space with code, data, and stack.
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.
Hi, The equation to find the number of instructions with n-bit opcode is 2^n. If your opcode is n=4, the it's 2^4 which is 16. So with a four bit opcode you can have 16 different instructions. ---- 11 levels
Arrange the any one of the order and store the memory in order vice (ie.Ascending for Descending) Then print the second data of that array it is simple way
Lxih, 2200 mov c,m inxh mov a,m inxh cmp m jc l1 mov a,m dcr c jnz l2 inhx mov m,a hlt
The 8085 has five interrupts, INTR, RST5.5, RST6.5, RST7.5, and TRAP. It also has eight software interrupts, RST0, RST1, ..., RST7. The INTR interrupt requires a hardware response that is an opcode. Usually, the opcode is either a CALL instruction, in which case the interrupt vector can go anywhere in memory, or it is an RST instruction, in which case the vector is based on a table in low memory. In the case of RST instructions, either directly or via INTR, or the RSTx.5 interrupts, you simply multiply the interrupt number by 8 to get the vector address. The following table presents the vector addresses for all possible interrupts... RST 0 - 00H RST 1 - 08H RST 2 - 10H RST 3 - 18H RST 4 - 20H TRAP - 24H RST 5 - 28H RST5.5 - 2CH RST 6 - 30H RST 6.5 - 34H RST 7 - 38H RST 7.5 - 3CH
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sample program in sum of the series using the formula for s=n/2[2a+{n-1}d] in 8085