mp and mc 2015

8/10/2019 mp and mc 2015

1/150

PSNCET MICROPROCESSOR ND MICROCONTROLLER L B M NU L

Page 1

PSN COLLEGE OF ENGINEERING AND TECHNOLOGY

Melathediyoor, Tirunelveli(Approved by AICTE and Recognized by UGC Section 2 (f)

An ISO 9001:2008 Certified Institution(Accredited by NBA and NAAC, Affiliated to Anna University of Technology,

Tirunelveli)

DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING

MICROPROCESSOR

AND

MICROCONTROLLER

Prepared by

T.Rajesh M .E.,(Ph.D) AP/ECE

K.Manikandan M .E., AP/ECE

LABORATORY

MANUAL

8/10/2019 mp and mc 2015

2/150


Page 2

SYLLABUS

MICROPROCESSOR AND MICROCONTROLLER LABORATORY

LIST OF EXPERIMENTS

Microprocessor (8085/8086):

1. Program for 8 bit Arithmetic operation using 8085/86

2. Program for 16 bit Arithmetic operation using 8085/86

3. Program for sorting and searching using 8085/86

4. Program for String manipulation operation using 8085/86

Microcontroller:

1. Programming using Arithmetic, Logical and bit manipulation instructions of 8051

2. Program for sorting and searching using 8051

3. Code conversion using 8051.

Interfacing:

1. Analog to Digital/ Digital to Analog

2. Interfacing and programming of stepper motor control

3. Communication between 8051 Microcontroller and PC

4. Interfacing and programming 8279, 8259, 8255 and 8253.

5. Traffic Light Controller.

TOTAL : 45 PERIODS

L T P C

0 0 3 2

8/10/2019 mp and mc 2015

3/150


Page 3

MICROPROCESSOR AND MICROCONTROLLER LABORATORY

Sl.No Cycle I Page No

1. Program for 8 bit Arithmetic operation using 8085

2. (a) Program for 16 bit Arithmetic operation using 8085

2. (b) Program for 16 bit Arithmetic operation using 8086

3. (a) Program for sorting and searching using 8085

3. (b) Program for sorting and searching using 8086

4. Program for String manipulation operation using 8086

5. Programming using Arithmetic, Logical and bit manipulation instructions of

8051

Cycle II

6. Program for sorting and searching using 8051

7. Code conversion using 8051.

8. Analog to Digital/ Digital to Analog

9. Interfacing and programming of stepper motor control

10. Communication between 8051 Microcontroller and PC

11. Interfacing and programming 8279, 8259, 8255 and 8253.

12. Traffic Light Controller.

8/10/2019 mp and mc 2015

4/150


Page 4

EXP.NO.

TITLE:

Date:_________STUDYOFINTEL8085MICROPROCESSORKIT

Aim: TostudytheworkingofIntel8085Microprocessortrainerkit.

Figure1.1 VimicrosMicro-85Microprocessorkit

1.1INTRODUCTION

This section briefs the hardware and software facilities available in both the trainers

Micro-85 EBl and Micro-85 EB2. Micro-85 EBl is a powerful Microprocessor Trainer

with basic features such as 24 TTL lines using 8255, Hardware Single Stepping and

SoftwareSingleSteppingofuserprograms.

In addition to the above features, Micro-85 EB2 has RS232C compatible serial port,

Bus Expansion for interfacing VBMB series of add-on cards and 24 TTL I/O lines.A

separate switch is provided for learning more about hardware. interrupts. There is

also provision to add multi output power supply for interfacing experiment boards.

Most of the control signals are terminated .at test points for easy analysis on CRO or

logicprobe.

8/10/2019 mp and mc 2015

5/150


Page 5

The differences in the specification of Micro-85 EBl and Micro-85EB2 are highlighted

in this manual. The users are therefore requested to go through the Hardware

specificationcarefully.

1.2 SPECIFICATIONS

1.2.1HARDWARESPECIFICATIONS

1)PROCESSOR,CLOCKFREQUENCY:Intel8085Aat6.144MHzclock..

2)MEMORY:

MonitorEPROM

EPROM-Expansion

SystemRAM

MonitorDataArea

UserRAMArea

RAM-Expansion

:0000-1FFF

:2000-3FFF&COOO-FFFF

:4000-5FFF

:4000-40FF(Reserved)

:4100-5FFF

:6000-BFFF

Note:TheRAMareafrom4000-40FFshouldnotbeaccessedbythe

usersinceit isusedasscratchpadbytheMonitorprogram.

3)INPUT/OUTPUT:

Parallel:48TTL:I/O linesusingtwonumberof8255

(only24:I/Oline.availableinmicro-85EB1).

Serial :OnenumberofRS232C compatibleSerial interface

using8251A-USART.

Timer :Threechannel16-bitProgrammableTimerusing8253.-Channel0 isusedasbaudrateclockgenerator for8251AUSART.

-Channel1 isusedfor insinglesteppinguserprograms.-Channel2 isusedforHardwareSingleSteppinguserprograms.

4)DISPLAY:- 6digit,0.3",7SegmentRedLEDDisplaywithfilter.- 4digits foraddressdisplayand2digitsfordatadisplay.

5)KEYBOARD:- 21Keyssoftkeyboard includingcommandkeysandhexa-decimalkeys.

6)AUDIOCASSETTEINTERFACEwithfilemanagement.

7)BATTERYBACKUP:-OnboardBatterybackupfacility isprovided fortheavailableRAM.

8/10/2019 mp and mc 2015

6/150


Page 6

8) HARDWARESINGLESTEP:Thisfacilityallowstheusertoexecuteprogramsatmachinecycle level

usingaseparateswitch.

9) SYSTEMPOWER CONSUMPTION:

Micro-85EB2

+5V@ 1Amp

+12V@ 200mA-12V@ l00mA

Micro-85EBl

+5V@ 500mA

10) POWER SUPPLYSPECIFICATIONS:[External]

Micro-85EB2

Input: 230VAC @ 50,Hz

Micro-85EB1

230VAC @ 50Hz

Output:+5V @ 1.5A +5V@ 600mA

+12V@ 150mA

-12V@ 150mA+30V@ 250mA(Unregulated)

11) PHYSICALCHARACTERISTICS:

Micro-85EB PCB :230mmx170mm[LxB] Weight :1Kg.

12) TESTPOINTS: TestpointsprovidedforMR*,MW*,INTA*,IO/M*,IOR*,IOW*,S0,S1,INTA.

Thisenablestheusertostudythehardwaretimingeasily.

8/10/2019 mp and mc 2015

7/150


Page 7

1.2.2 SOFTWARESPECIFICATIONS

Micro-S5 EB contains a high performance 8K bytes monitor program. It is designed

to respond to user input, RS232C serial communications, tape interface etc. The

following is a simple description of the key functions. Out of the 21 keys in the

keyboard 16 are hexadecimal, command and register keys and the remaining arestand-alonekeys.

KEY FUNCTIONSUMMARY

ThisRESkeyallowsyou toterminateanypresentactivityandtoreturnyourMicro-85EB to an initialized state.When pressed, the ..85 sign-

RES on message appears in the display for a few seconds and the monitorwilldisplaycommandprompt- inthe leftmostdigit.

Maskable interruptconnectedtoCPU'sRST7.5 interrupt.

INT

Decrementtheaddressbyoneanddisplay itcontents

(or)

DEC Displaythepreviousregistercontents.

Executeaparticularprogramafterselectingtheaddress

EXECthroughGOcommand.

Incrementaddressbyoneanddisplay itscontents

(or)

DEC Displaythenextregistercontent.

The16Hexadecimalkeyshaveeitheradualoratripleroletoplay.

i)ItfunctionsasaHexkeyentrywhenaaddressordataentry isrequired.

ii)ItfunctionsastheRegisterkeyentryduringRegistercommand.

iii)It functionsascommandkeywhenpresseddirectlyaftercommandprompt.

NOTE: TheHex-keyfunctionsummarybelow is intheorder:

i) Hexkey.ii) Commandkey

iii) Registerkey.

8/10/2019 mp and mc 2015

8/150


Page 8

KEY FUNCTIONSUMMARY

i. Hexkeyentry"0"

0E

SUB

DREG

2C

TW

ii.Thiskey isforsubstitutingmemorycontentsWhenNEXTkey is

pressed immediate1yafterthis ittakestheusertothestart

addressforenteringuserprograms,4100Hex(UserRAM).iii.Registerkey"E"

i. Hexkeyentry"1"

ii.Examinethe8085Aregistersandmodifythesame.

iii.Registerkey"D"

i. Hexkeyentry"2"

ii.Writesdatafrommemoryontoaudiotape.

iii.Registerkey"C"

3 i. Hexkeyentry3B

TRii. Retrievedata fromanaudiotapetomemory.

iii.RegisterkeyB

4i. Hexkeyentry4.

F ii. Blocksearch forabyte.BLOC

iii.RegisterkeyF.

5i. Hexkeyentry5.

A ii. FillablockofRAMmemorywithdesireddata.

FILLiii.RegisterkeyA.

6i. Hexkeyentry"6"

L ii. Transmit/Receivedatato/fromtheserialport.SER

TheTW/TR keysareusedforsending/receivingrespectively.

iii.RegisterkeyL.

7 i. Hexkeyentry"7"H

F2 ii. Registerkey"H"

8/10/2019 mp and mc 2015

9/150


Page 9

8I

GO

i. Hexkeyentry"8"

ii. Startrunningaparticularprogram

iii.Registerkey"I"

9i. Hexkeyentry"9"

PL

SNG

APH

F3

ii. Singlestepaprogram instructionbyinstruction.

iii.RegisterkeyPCL.

i Hexkeyentry"A"

ii. FunctionkeyF3F3[0]=Inputabytefromaport

P3[1]=Outputabytetoaport

iii.RegisterkeyPCH"

iv.UsedwithSNG keyforhardwaresinglestepping.

B i Hexkeyentry"B"

SLBC

ii. Checkaparticularblockforblank.

iii.Registerkey"SPL"

C i. Hexkeyentry"C"SH

MOVii. Moveblockofmemorytoanotherblock

iii. RegisterkeySPH

D i. Hexkeyentry"D"

CMPii. Comparetwomemoryblocks.

E i. Hexkeyentry"E"

INSii. Insertbytes intomemory(RAM).

Fi.Hexkeyentry"F"

DEL ii.Deletebytesfrommemory(RAM).

8/10/2019 mp and mc 2015

10/150


Page 10

EXP.NO.

TITLE:

Date:_________

STUDYOFINTEL8086MICROPROCESSOR

FEATURES:

1. 16-bitDatabus2. Computes16bit/32bitdata.3. 20-bitaddressbus.4. Morememoryaddressingcapability(2

20=1MB)

5. 16bitFlagregisterwith9Flags6. Canbeoperated inMinimummodeandMaximummode

7. Hastwostagepipelinedarchitecture8. No internalclockgeneration9. 40pinDIPIC -HMOStechnology10.Operateson+5Vsupplyvoltage11.Hasmorepowerful instructionset

Memory:

Program, data and stack memories occupy the same memory space. The total

addressable memory size is 1MB KB.As the most of the processor instructions use

16-bit pointers the processor can effectively address only 64 KB of memory. To

access memory outside of 64 KB the CPU uses special segment registers to specify

where the code, stack and data 64 KB segments are positioned within 1 MB of

memory(seethe"Registers"sectionbelow).

8/10/2019 mp and mc 2015

11/150


Page 11

16-bitpointersanddataarestoredas:

address

address+1

: low-orderbyte

:high-orderbyte

32-bitaddressesarestoredin"segment:offset"formatas:

address

address+1

address+2

address+3

: low-orderbyteofsegment

:high-orderbyteofsegment

: low-orderbyteofoffset

:high-orderbyteofoffset

PhysicalmemoryaddresspointedbySEGMENT:OFFSETpair iscalculatedas:

Physicaladdress=(*16)+

Programmemory - program can be located anywhere in memory. Jump and

call instructions can be used for shortjumps within currently selected 64 KB

code segment, as well as for farjumps anywhere within 1 MB of memory. All

conditionaljump instructions can be used tojump within approximately +127 -

-127 bytesfromcurrent instruction.

Data memory - the processor can access data in any one out of 4

availablesegments, which limits the size of accessible memory to 256 KB (if

all four segments point to different 64 KB blocks). Accessing data from the

Data, Code,Stack or Extra segments can be usually done by prefixing instructionswith the DS:,CS:, SS: or ES: (some registers and instructions by default may use

the ES or SSsegments insteadofDSsegment).

Word data can be located at odd or even byte boundaries. The processor uses two

memory accesses to read 16-bit word located at odd byte boundaries. Reading

worddatafromevenbyteboundariesrequiresonlyonememoryaccess.

Stackmemory can be placed anywhere in memory. The stack can be located at

odd memory addresses, but it is not recommended for performance reasons (see

"DataMemory"above).

Reservedlocations:

0000h - 03FFh are reserved for interrupt vectors. Each interrupt vector is a

32-bitpointer informatSEGMENT:OFFSET.

FFFF0h -FFFFFh-afterRESET theprocessoralwaysstartsprogram execution

attheFFFF0haddress.

8/10/2019 mp and mc 2015

12/150


Page 12

Interrupts:

Theprocessorhasthefollowing interrupts:

INTR is a maskable hardware interrupt. The interrupt can be enabled/disabled

using STI/CLI instructions or using more complicated method of updating the

FLAGS register with the help of thePOPF instruction. When an interrupt occurs, the

processor storesFLAGS register into stack, disables further interrupts, fetches from

the bus one byte representing interrupt type, andjumps to interrupt processing

routine address of which is stored in location 4 * . Interrupt

processingroutineshouldreturnwiththeIRET instruction.

NMI is a non-maskable interrupt. Interrupt is processed in the same way as the

INTR interrupt. Interrupt type of the NMI is 2, i.e. the address of the NMI

processing routine is stored in location 0008h. This interrupt has higher priority

thenthemaskable interrupt.

Softwareinterruptscanbecausedby:

INT instruction-breakpoint interrupt.This isatype3 interrupt.

INT instruction - any one interrupt from available 256

interrupts.

INTO instruction- interruptonoverflow

Single-step interrupt - generated if the TF flag is set. This is a type 1

interrupt.When the CPUprocesses this interrupt itclearsTF flagbeforecalling

the interruptprocessingroutine.

Processor exceptions: divide error (type 0), unused opcode (type 6) and

escapeopcode(type7).

Softwareinterruptprocessing isthesameasforthehardware interrupts.

I/Oports:

8086 can interface maximum of 65536 nos of 8-bit I/O ports. These ports can be

alsoaddressedas3276816-bitI/Oports.

8/10/2019 mp and mc 2015

13/150


Page 13

Registers:

Most of the registers contain data/instruction offsets within 64 KB memory

segment. There are four different 64 KB segments for instructions, stack, data and

extra data. To specify where in 1 MB of processor memory these 4 segments are

locatedtheprocessorusesfoursegmentregisters:

Code segment (CS) is a 16-bit register containing address of 64 KB segment with

processor instructions. The processor uses CS segment for all accesses to

instructions referenced by instruction pointer (IP) register. CS register cannot be

changed directly. The CS register is automatically updated during farjump, far call

andfarreturn instructions.

Stack segment (SS) is a 16-bit register containing address of 64KB segment with

program stack. By default, the processor assumes that all data referenced by the

stack pointer (SP) and base pointer (BP) registers is located in the stack segment.

SSregistercanbechangeddirectlyusingPOP instruction.

Data segment (DS) is a 16-bit register containing address of 64KB segment with

program data. By default, the processor assumes that all data referenced by

general registers (AX, BX, CX,DX)and index register (SI,DI) is located in thedata

segment.DSregistercanbechangeddirectlyusingPOPandLDS instructions.

Extra segment (ES) is a 16-bit register containing address of 64KB segment,

usually with program data. By default, the processor assumes that the DI register

references the ES segment in string manipulation instructions. ES register can be

changeddirectlyusingPOPandLESinstructions.

It is possible to change default segments used by general and index registers by

prefixing instructionswithaCS,SS,DSorESprefix.

All general registers of the 8086 microprocessor can be used for arithmetic and

logicoperations.Thegeneralregistersare:

Accumulator (AX) register consists of 2 8-bit registersAL andAH, which can be

combined together and used as a 16-bit registerAX.AL in this case contains the

low-order byte of the word, andAH contains the high-order byte.Accumulator can

beused forI/Ooperationsandstringmanipulation.

Base(BX) register consists of 2 8-bit registers BL and BH,which can becombined

together and used as a 16-bit register BX. BL in this case contains the low-order

byteof the word, and BH contains the high-order byte. BX register usually contains

adatapointerusedforbased,based indexedorregister indirectaddressing.

8/10/2019 mp and mc 2015

14/150


Page 14

Count (CX) register consists of 2 8-bit registers CL and CH, which can be

combined together and used as a 16-bit register CX. When combined, CL register

contains the low-order byte of the word, and CH contains the high-order byte.

Count register can be used as a counter in string manipulation and shift/rotate

instructions.

Data(DX) registerconsistsof2 8-bit registers DL and DH,whichcanbecombinedtogetherand usedasa16-bit registerDX.Whencombined,DL registercontainsthe

low-order byte of the word, and DH contains the high-order byte. Data register can

be used as a port number in I/O operations. In integer 32-bit multiply and divide

instruction the DX register contains high-order word of the initial or resulting

number.

Thefollowingregistersarebothgeneraland indexregisters:

StackPointer(SP) isa16-bitregisterpointingtoprogramstack.

Base Pointer (BP) is a 16-bit register pointing to data in stack segment.

BPregister isusuallyused forbased,based indexedorregister indirectaddressing.

Source Index (SI) is a 16-bit register. SI is used for indexed, based indexed

and register indirect addressing, as well as a source data address in string

manipulationinstructions.

Destination Index (DI) is a 16-bit register. DI is used for indexed, based

indexedand register indirect addressing, as well as a destination data address

in stringmanipulation instructions.

Otherregisters:

InstructionPointer(IP) isa16-bitregister.

FlagRegisterisa16-bitregistercontaining9nosofonebitflags:

Overflow Flag (OF) - set if the result is too large positive number, or is too

smallnegativenumbertofit intodestinationoperand.

Direction Flag (DF) - if set then string manipulation instructions will auto-

decrement index registers. If cleared then the index registers will be auto-

incremented.

Interrupt-enableFlag(IF)-settingthisbitenablesmaskable interrupts.

Single-stepFlag(TF)- ifsetthensingle-step interruptwilloccurafterthenext

instruction.

SignFlag(SF)-set ifthemostsignificantbitoftheresult isset.

ZeroFlag(ZF)-set iftheresult iszero.

8/10/2019 mp and mc 2015

15/150


Page 15

Auxiliary carryFlag (AF) - set if there was a carry from or borrow to bits 0-3

intheALregister.

ParityFlag(PF)-set ifparity(thenumberof"1"bits) inthe low-orderbyteof

theresult iseven.

Carry Flag (CF) - set if there was a carry from or borrow to the most

significantbitduringlastresultcalculation.

InstructionSet:

8086instructionsetconsistsofthe following instructions:

Datamoving instructions.

Arithmetic - add, subtract, increment, decrement, convert byte/word and

compare.

Logic-AND,OR,exclusiveOR,shift/rotateandtest.

Stringmanipulation- load,store,move,compareandscanforbyte/word.

Control transfer - conditional, unconditional, call subroutine and return from

subroutine.

Input/Output instructions.

Other-setting/clearing flagbits,stackoperations,software interrupts,etc.

Addressingmodes:

Implied-thedatavalue/dataaddress is implicitlyassociatedwiththe instruction.

Register-referencesthedata inaregisteror inaregisterpair.

Immediate-thedata isprovided inthe instruction.

Direct - the instruction operand specifies the memory address where data is

located.

Register indirect - instruction specifies a register containing an address, where

datais located.ThisaddressingmodeworkswithSI,DI,BXandBPregisters.

Based - 8-bit or 16-bit instruction operand is added to the contents of a baseregister(BXorBP),theresultingvalueisapointerto locationwheredataresides.

Indexed - 8-bit or 16-bit instruction operand is added to the contents of an index

register(SIorDI),theresultingvalue isapointerto locationwheredataresides.

8/10/2019 mp and mc 2015

16/150


Page 16

Based Indexed - the contents of a base register (BX or BP) is added to the

contents of an index register (SI or DI), the resulting value is a pointer to location

wheredataresides.

Based Indexedwith displacement - 8-bit or 16-bit instruction operand is added

to the contents of a base register (BX or BP) and index register (SI or DI), the

resultingvalue isapointerto locationwheredataresides.

Result:

ThefeaturesoftheIntel8085&8086microprocessorswerestudiedandoperations

ofthecorrespondingkitswereunderstood.

8/10/2019 mp and mc 2015

17/150


Page 17

Fig.Flowchartfor8-bitAdditionwithcarry

8/10/2019 mp and mc 2015

18/150


Page 18

EXP.NO. Date:_________

TITLE: 8-BITARITHMETICOPERATIONS (8-BITADDITION)

AIM: Towriteanassembly languageprogram fortheadditionoftwo8-bitnumbers.

APPARATUSREQUIRED:

S.No ItemDescription Qty

1 8085-microprocessorkit. 1

2 Powersupplyunit 1

THEORY:

The two operands (i.e., 8-bit numbers) are loaded into two registersA & B, using

immediate addressing mode instructions and then added usingADD instruction. The

result is stored in the desired memory location. The overflow in addition is checked

byverifying thestatusofCarry(Cy)flagandaccordinglyeither00or01 isstored

inthe locationnexttotheresult.

ALGORITHM:

1.Starttheprogram

2.InitializetheC registeras00H.

3.Movethedata1anddata2toAccumulatorandBregisterrespectively.

4.AddBregistertothecontentofaccumulator

5.Ifthere isnocarry,gotostep6,else incrementC register.

6.Storethecontentofaccumulator tothememorylocation4500H.

7.MovethecontentofC registertoaccumulator.

8.Storethecontentofaccumulator tothememorylocation4501H.

9.Stoptheprogram.

8/10/2019 mp and mc 2015

19/150


Page 19

PROGRAM:

AddressOpcode&Operand Label Mnemonics Comments

4100 0E,00 MVIC,00 Clearcregister

4102 3E,Data1 MVIA,Data1Movedata1toaccumulator

4104 06,Data2 MVIB,Data2 Movedata2toBRegister

4106 80 ADDB AddBRegtoaccumulator

4107 D2,0B,41 JNC GOJumponNocarrytolocationGO

410A 0C INR C IncrementC Register

410B 32,00,45 GO: STA4500 Storetheresult

410E 79 MOVA,C MovecarrytoAcc

410F 32,01,45 STA4501 Storethecarry

4112 76 HLT Stoptheprogram

MANUALCALCULATION:

8/10/2019 mp and mc 2015

20/150


Page 20

SAMPLEDATA:

Resultof8-bitadditionwithoutcarry:

INPUT OUTPUT

Address Data Address Data

4103 05H4500

(Result) 0BH

4105 06H4501

(Carry) 00H

Resultof8-bitadditionwithcarry:

INPUT OUTPUT


4103 51H

4500

(Result) 3CH

4105 EBH4501

(Carry) 01H

EXERCISE:

Executetheprogramwithyourowndataandobservetheresults.Checktheresult

withyourmanualcalculation.

INPUT OUTPUT


41034500

(Result)

41054501

(Carry)

RESULT:

Thustheassembly languageprogram for8-bitaddition isexecutedandtheresults

areverified.

8/10/2019 mp and mc 2015

21/150


Page 21

Fig.Flowchartfor8-bitsubtraction

8/10/2019 mp and mc 2015

22/150


Page 22


TITLE: 8-BITARITHMETICOPERATION(8-BITSUBTRACTION)

AIM:Towriteanassembly languageprogramforthesubtractionoftwo8-bit

numbers,using8085.

APPARATUSREQUIRED:



2 Powersupplyunit 1

THEORY:

Out of the twooperands forsubtraction, the firstoperand is loaded intoAccumulator

and the second operand is subtracted directly from memory, using register indirect

addressing mode instructions. The result is stored in desired memory location and

the borrow in subtraction is checked by verifying the status of Carry (Cy) flag and

accordinglyeither00or01 isstored inthelocationnexttoresult.

ALGORITHM:

1.Starttheprogram.

2.LoadtheHLpairwith16-bitaddressofdata location.

3.MovethecontentofmemoryaddressinHLtoaccumulator.

4.Incrementtheaddress inHLpair.

5.Subtractthecontentofmemory fromaccumulator.

6.JumponNo-carrytostep8.

7.IncrementtheC-register.

8.StorethecontentofaccumulatorandC-register.

9.Stoptheprogram.

8/10/2019 mp and mc 2015

23/150


Page 23

PROGRAM:


4100 21,50,41 LXIH,4150 LoaddatatoHLRegister

4103 7E MOVA,M MoveData1toAcc

4104 0E,00 MVIC,00 ClearC-register.

4106 23 INXH Incrementaddress.

4107 96 SUBM SubtractData2 fromAcc

4108 D2,0C,41 JNC NEXTJumponNo-carrytothe locationNEXT.

410B 0C INR C IncrementC register

410C 32,52,41 NEXT: STA4152 Storetheresult

410F 79 MOVA,C Movecarrytoacc

4110 32,53,41 STA4153 Storethecarry4113 76 HLT Stoptheprogram

MANUALCALCULATION:

8/10/2019 mp and mc 2015

24/150


Page 24

SAMPLEDATA:

Resultof8bitsubtractionwithoutcarry:

INPUT OUTPUT


4150 68H4152

(Result) 14H

4151 54H4153

(Borrow) 00H

Resultof8bitsubtraction withcarry:

INPUT OUTPUT


4150 57H

4152

(Result) F1H

4151 66H4153

(Borrow) 01H

EXERCISE:



INPUT OUTPUT


41504152

(Result)

41514153

(Borrow)

RESULT:

Thustheassembly languageprogram for8-bitsubtractionisexecutedandthe

resultsareverified.

8/10/2019 mp and mc 2015

25/150


Page 25

Fig.Flowchartfor8-bitMultiplication

8/10/2019 mp and mc 2015

26/150


Page 26


TITLE: 8-BITARITHMETICOPERATION(8-BITMULTIPLICATION)

AIM:Towriteanassembly languageprogramforthemultiplicationoftwo8-bitnumbers,using8085.

APPARATUSREQUIRED:



2 Powersupplyunit 1

THEORY:

Usingthe immediateaddressingmode instructions,thetwooperandstobemultiplied

are loaded into two registers say, B and C. By the method of repeated addition, the

multiplication operation is performed (i.e., first number is repeatedly added to

accumulator as per the second number Eg. 03 x 04=>Acc+ 03 (04 times) ). The

overflow in multiplication is checked every time after each addition, by verifying the

statusof Carry (Cy) flag andaccordinglya register,sayD is incremented.The result

inaccumulator isstored inthedesiredmemory locationandthecontentofDregister

isstored inthe locationnexttoresult.

ALGORITHM:

1. Starttheprogram.

2. CleartheAccumulatorandDRegister.

3. LoadtheData1toBregisterandData2toC register.

4. AddthecontentofBtoaccumulatoruntilC becomeszero.

5. IfNo-carry,gotostep6.

6. IncrementtheD-register.

7. DecrementtheC register.

8. IfC register isNon-zero,Jumptostep4.

9. Storethecontentofaccumulatorto5000H.

10.MovethecontentofD-registertoAccumulator.

11.Storethecontentofaccumulatorto5001H.

12.Stoptheprogram.

8/10/2019 mp and mc 2015

27/150


Page 27

PROGRAM:


4500 3E,00 MVIA,00 Movedatatoaccumulator

4502

06,Data1

MVIB,Data1

Movemultiplicandtob

register4504 0E,Data2 MVIC,Data2

Movethemultipliertocregister

4506 16,00 MVID,00 ClearD-regforcarry

4508 80 LOOP2: ADDB repetitiveaddition

4509 D2,0D,45 JNC LOOP1JumponnocarrytoanlocationLOOP1

450C 14 INR DIncrementtheD-register,ifcarryoccurs.

450D 0D LOOP1: DCR C DecrementC-register

450E C2,08,45 JNZLOOP2JumponnozerotoLocationLOOP2

4511 32,00,50 STA5000 Storeresultantproduct

4514 7A MOVA,DMovecarrytoaccumulator

4515 32,01,50 STA5001 Storethecarry.

4518 76 HLT Stoptheprogram.

MANUALCALCULATION

8/10/2019 mp and mc 2015

28/150


Page 28

SAMPLEDATA:

Resultof8bitmultiplicationwithoutcarry:

INPUT OUTPUT


4503 05H5000

(Result) 19H

4505 05H5001

(Carry) 00H

Resultof8bitmultiplicationwithcarry:

INPUT OUTPUT


4503 41H 5000(Result) 45H

4505 25H5001

(Carry) 01H

EXERCISE:



INPUT OUTPUT


45035000

(Result)

45055001

(Carry)

RESULT:Thustheassembly languageprogram for8-bitmultiplicationisexecutedandthe

resultsareverified.

8/10/2019 mp and mc 2015

29/150


Page 29

Fig. Flowchartfor8-bitDivision

8/10/2019 mp and mc 2015

30/150


Page 30


TITLE: 8-BITARITHMETICOPERATION (8-BITDIVISION)

AIM: Towriteanassembly languageprogram forthedivisionoftwo8bit

numbers,using8085.

APPARATUSREQUIRED:



2 Powersupplyunit 1

THEORY:

Using the immediate addressing mode instructions, the Divisor and Dividend are

loaded directly into Accumulator and B register. By the method of successive

subtraction, the division is carried out (i.e., Divisor is repeatedly subtracted from

Dividend,untilthedividendbecomessmallerthandivisor Eg.09/02=>0902(4

times) => Remainder=01 and Quotient=04). The C register is incremented every

time after each subtraction, to keep count of the quotient. The final content in

accumulator will be remainder of the division and it is stored in the desired memory

locationand thecontentofC registercontaining thequotient isstored inthe location

nexttoresult.

ALGORITHM:

1. Starttheprogram.

2. ClearC Register.

3. LoadtheDivisortoAccumulator.

4. LoadtheDividendtoB-register.

5. ComparetheB-registervaluewiththeaccumulator.

6. IfAccumulatorcontent issmallertoBreg.,thenJumptostep10.

7. SubtractB-registervaluewithaccumulator.

8. IncrementtheC-register.

9. Jumptostep4.

10.StorethecontentsofaccumulatorandC-registerintomemory.

11.Stoptheprogram.

8/10/2019 mp and mc 2015

31/150


Page 31

PROGRAM:


4500 3E,Divisor MVIA,Divisor MoveDivisortoAcc

4502 06,Dividend MVIB,Dividend MoveDividendtoBreg.

4504 0E,00 MVIC,00 ClearC register

4506 B8 LOOP2: CMPB CompareAccandBreg.

4507 DA,0F,45 JC LOOP1JumponCarrytolocationLOOP1

450A 90 SUBBRepetitivesubtractionfordivision

450B 0C INR C IncrementC register

450C C3,06,45 JMPLOOP2 Jumpto locationLOOP2

450F 32,00,50 LOOP1: STA5000 StoretheRemainder

4512 79 MOVA,C MoveC reg.toAcc

4513 32,01,50 STA5001 StoretheQuotient.


MANUALCALCULATION:

8/10/2019 mp and mc 2015

32/150


Page 32

SAMPLEDATA:

Resultof8-bitDivisionwithoutremainder:

INPUT OUTPUT


4501 06H 5000(R) 00H

4503 03H 5001(Q) 02H

Resultof8bitdivisionwithremainder:

INPUT OUTPUT


4501 25H 5000(R) 02H

4503 05H 5001(Q) 07H

EXERCISE:



INPUT OUTPUT


4501 5000(R)

4503 5001(Q)

RESULT:

Thustheassembly languageprogram for8-bitdivision isexecutedandtheresults

areverified.

8/10/2019 mp and mc 2015

33/150


Page 33

Fig. Flowchartfor16-bitAddition

8/10/2019 mp and mc 2015

34/150


Page 34


TITLE: 16-BITARITHMETICOPERATION (16-BITADDITION)

AIM:Towriteanassembly languageprogramfortheadditionoftwo16-bit

numbers,using8085.

APPARATUSREQUIRED:



2 Powersupplyunit 1

THEORY:

Using the direct addressing mode instructions, the two 16-bit numbers are loaded

into HL and DE register pairs from memory. Using the double addition instruction

(DAD rp),thecontentsofHLandDEareaddedtogetherand theresultsarestored in

theHL registerpairagain. TheB register is used tocheck theoverflowoftheabove

addition,by verifying thecarry flag.The result in theHL registerpair is stored inthe

desired memory location and the content of B register is stored in the location next

toresult.

ALGORITHM:

1. Starttheprogram.

2. CleartheB-register.

3. LoadtheData1toHLregisterpair.

4. ExchangethecontentofHLtotheDEregisterpair.

5. LoadtheData2toHLpairregister.

6. AddthecontentofHLandDEpair.

7. Ifnooverflow inaddition(nocarry),gotostep9.

8. IncrementthecontentofB-register.

9. StorethecontentofHLpairtotheaddress5504H.

10.StorethecontentofB-registertotheaddress5506H.

11.Stoptheprogram.

8/10/2019 mp and mc 2015

35/150


Page 35

PROGRAM:


4500 06,00 MVIB,00 ClearBregister

4502 2A,00,55 LHLD5500 LoadData1toHLregister

4505 EB XCHGShiftthedata inHLtoDEregister

4506 2A,02,55 LHLD5502 LoadtheData2toHLpair

4509 19 DADDAddthecontentoftheHLandDEpair

450A D2,0E,45 JNC LOOPJumponnocarrytoanaddress

450D 04 INR B IncrementtheBregister

450E 22,04,55 LOOP SHLD5504 Storetheresult

4511 78 MOVA,B MovecarrytoBregister.

4512 32,06,55 STA5506 Storethecarry.


MANUALCALCULATION:

8/10/2019 mp and mc 2015

36/150


Page 36

SAMPLEDATA:

Resultof16-bitAdditionwithoutcarry:

INPUT OUTPUT


5500,01 7167H 5504,05(Result) F7ECH

5502,03 8685H5506

(Carry) 00H

Resultof16-bitAdditionwithcarry:

INPUT OUTPUT


5500,01 FF03H5504,05

(Result) 06FEH

5502,03 EF03H5506(Carry) 01H

Exercise:



INPUT OUTPUT

Address Data Address Data5500,01

5504,05

(Result)

5502,035506

(Carry)

RESULT:

Thustheassembly languageprogram for16-bitaddition isexecutedandtheresults

areverified.

8/10/2019 mp and mc 2015

37/150


Page 37

Fig. Flowchartfor16-bitSubtraction

8/10/2019 mp and mc 2015

38/150


Page 38


TITLE: 16-BITARITHMETICOPERATION (16-BITSUBTRACTION)

AIM:Towriteanassembly languageprogramforthesubtractionoftwo16-bit

numbers,using8085.

APPARATUSREQUIRED:



2 Powersupplyunit 1

THEORY:

Using the direct addressing mode instructions, the two 16-bit numbers are loaded

into HL and DE register pairs. The lower bytes of the two numbers are subtracted

first using SUB instruction and the higher bytes of the same two numbers are

subtracted along with borrow using SBB instruction. The result in the accumulator

aftereachsubtractionisstoredinthetwosubsequentdesiredmemory locations.

ALGORITHM:

1. Starttheprogram.


3. ExchangethecontentofHLtotheDEregisterpair.


5. MovethecontentofLregistertoAccumulator.

6. SubtractthecontentofEregisterfrom theAccumulator.

7. Storetheresult inaccumulatortothememory location,5100H.

8. MovethecontentofHregistertoAccumulator.

9. SubtractthecontentofDregisterfromtheAccumulator,alongwithborrow.

10.Storetheresult inAccumulatortothememory location,5101H.

11.Stoptheprogram.

8/10/2019 mp and mc 2015

39/150


Page 39

PROGRAM:


4500 2A,00,55 LHLD5500LoadtheData2toHLregister.

4503 EB XCHG TransferthecontentofHLtoDEregisterpair.

4504 2A,02,55 LHLD5002LoadtheData2toHLpairregister.

4507 7D MOVA,LMovethecontentofLregistertoaccumulator

4508 93 SUBESubtractEregisterandaccumulator

4509 32,00,51 STA5100 Storetheresult

450C 7C MOVA,HMoveHregistertoaccumulator

450D 9A SBBDSubtractDregisterandaccumulator

450E 32,01,51 STA5101 Storetheresult.


MANUALCALCULATION:

8/10/2019 mp and mc 2015

40/150


Page 40

SAMPLEDATA:

Resultof16-bitSubtraction:

INPUT OUTPUT


5500,01(Data2) 6677H

5100(Result) 1111H

5502,03

(Data1) 7788H - -

EXERCISE:



INPUT OUTPUT


5500,01

(Data2)5100

(Result)5502,03

(Data1) - -

RESULT:

Thustheassembly languageprogram for16-bitsubtractionisexecutedandthe

resultsareverified.

8/10/2019 mp and mc 2015

41/150


Page 41

EXPT NO: 2(b) Program for 16 bit Arithmetic operation using 8086 DATE:

AIM:To perform basic 16 bit arithmetic operations in 8086 Microprocessor.

APPARATUS REQUIRED:

Sl.No. Apparatus required Quantity

1 8086 Microprocessor Kit 1

2 DC Power Supply 5V 1

ALGORITHM:

ADDITION:

1. Set SI register as pointer for data

2. Get the first data in AX register

3. Get the second data in BX register

4. Clear CL register

5. Get the sum in AX register

6. Store the sum in memory

7. Check for carry .If carry flag is set then go to next step,otherwise go to step-9

8. Increment CL register9. Store the carry in memory

10.Stop

SUBTRACTION:

1. Set SI register as pointer for data

2. Get the minuend in AX register

3. Get the subtrahend in BX register

4. Clear CL register to account for sign

5. Subtract the content of BX from AX ,the difference will be in AX

6. Check for carry .If carry flag is set then go to next step,otherwise go to step 9

7. Increment CL retgister by one

8. Take 2s Complement of the difference in AX register(For this complement Ax and add one.

9. Store the magnitude of difference in memory

10.Store the sign bit in memory

11.Stop

8/10/2019 mp and mc 2015

42/150


Page 42

MULTIPLICATION:

1. Load the address of data in SI register

2. Get the first data in AX register

3. Get the second data in BX register

4. Multiply the content of AX and BX .The product will be in AX and DX

5. Save the product (AX and DX) in memory6. Stop

DIVISION:

1. Load the address of data in SI register

2. Get the lower word of dividend in AX register

3. Get the upper word of dividend in DX register

4. Perform division to get quotient in AX and remainder in DX

5. Save the quotient (AX) and the remainder (DX) in memory

RESULT:

Thus the Basic 16 bit arithmetic operations have been performed successfully on 8086

microprocessor by Assembly Language Programming (ALP).

8/10/2019 mp and mc 2015

43/150


Page 43

Program:Add two 16-bit numbers with considering the carry.

Flow Chart

Store carry (CL register) in memory

Stop

START

Load the address of data in SI register

Get the 1st

data in AX register

Get the 2nd

data in BX register

Clear CL register

Get the sum (in AX register)

Is

=

Increment CL register

Store the sum in memor

YES

NO

8/10/2019 mp and mc 2015

44/150


Page 44

Program

Address Label Mnemonics Opcode Comments

8000 MOV SI,1100H BE 81 00 Set SI register as

pointer for data

8003 MOV AX,[SI] 8B 04 Get the first data

in AX register

8005 MOV BX,[SI+2] 8B 5C 02 Get the second

data in BX

register

8008 MOV CL,00H B1 00 Clear the CL

register for carry

800A MOV AX,BX 03 C3 Add the two

data,sum will be

in AX register

800C MOV [SI+4],AX 89 44 04 Store the sum in

memory

location(1104H)

800F JNC AHEAD 73 02 Check the status

of carry flag

8011 INC CL FE C1 If carry flag is

set,increment CL

b one

8013 AHEAD MOV [SI+6],CL 88 4C 06 Store carry in

memory location

(1106H)

8016 INT 3 CC

8/10/2019 mp and mc 2015

45/150


Page 45

Program:Subtract two 16-bit numbers along with considering the borrow.

Flow Chart

Stop

Increment CL register by one

Complement the content of AX

re ister

Add 01H to AX registerStore the sign bit

START


Get the minuend in AX register

Get the subtrahend in BX register

Clear CL register

Subtract BX from AX

Is

CF=1?

Store the difference in memory

YES

NO

8/10/2019 mp and mc 2015

46/150


Page 46

Program:


8000 MOV SI,1100H BE 81 00 Load the address

of data in SI

register

8003 MOV AX,[SI] 8B 04 Get the minuendin AX register

8005 MOV BX,[SI+2] 8B 5C 02 Get the

subtrahend in BX

register

8008 MOV CL,00H B1 00 Clear the CL

register to

account for sign

800A SUB AX,BX 2B C3 Get the

difference in AX

register

800C JNC STORE 73 07 Check the status

of carry flag

800E INC CL FE C1 If carry flag is set

,increment by

one

8010 NOT AX F7 D0 Take 2s

complement of

the difference

8012 ADD AX,0001H 05 00 01 Add 01H to AXregister

8015 STORE MOV[SI +4],AX 89 44 04 Store difference

in memory

location (1104H)

8018 MOV [SI+6],CL 88 4C 06 Store sign bit in

memory location

(1106H)

801B INT 3 CC

8/10/2019 mp and mc 2015

47/150


Page 47

Program:Multiply two 16-bit numbers.

Flow Chart

Program:


8000 MOV SI,1100H BE 81 00 Load the address

of data in SI

register

8003 MOV AX,[SI] 8B 04 Get the minuend

in AX register

8005 MOV BX,[SI+2] 8B 5C 02 Get the

subtrahend in BX

register

8008 MUL BX F7 E3

800A STORE MOV[SI +4],AX 89 44 04 Store differencein memory

location (1104H)

800D MOV [SI+6],DX 89 54 06 Store sign bit in

memory location

(1106H)

8010 INT 3 CC


Get the first data in AX register

Get the second data in BX re ister

Multi l the content of AX and BX

Save the lower word (AX) of product in memory

Save the upper word (DX) of product in memory

Start

Stop

8/10/2019 mp and mc 2015

48/150


Page 48

Program:Division two 16-bit numbers.

Flow Chart

Program


8000 MOV SI,1100H BE 81 00 Load the address of

data in SI register

8003 MOV AX,[SI] 8B 04 Get the minuend in

AX register

8005 MOV DX,[SI+2] 8B 54 02 Get the subtrahend in

BX register

8008 MOVBX ,[SI +4] 8B 5C 04

800B STORE DIV BX F7 F3 Store difference in

memory location(1104H)

800D MOV [SI+6],AX 89 44 06

8010 MOV [SI+8],DX 89 54 08 Store sign bit in

memory location

(1106H)

8013 INT 3 CC

Start

Get the lower word of dividend in AX register

Save the remainder (DX) in memory

Get the upper word of dividend in DX register

Divide AX and DX with BX

Save the quotient (Ax) in memory

Stop

8/10/2019 mp and mc 2015

49/150


Page 49

Result:

Addition

Sample Data Input Data: Data =F048H ----Output Data:Sum=009AH, Carry=01H

INPUT OUTPUT

Memory Address Content Memory Address Content

8100 48 8104 9A

8101 F0 8105 008102 52 8106 01

8103 10

Subtraction

Sample Data: Input Data:Minuend =840CH-----Output Data:Difference=2EBEH , Sign Bit =01H

Multiplication:

Sample Data: Input Data:Data 1=EF1AH----Output Data:BFC28A20H, Data 2 =CD50H

INPUT OUTPUT


8100 1A 8104 20

8101 EF 8105 8A

8102 50 8106 C2

8103 CD 8107 BF

Division:

Sample Data: Input Data:Divident=71C2580AH,Divisor=F6F2H ,Output Data:Quotient=75EEH

Remainder=290EH

INPUT OUTPUT

Memory Address Content Memory Address Content8100 0A 8106 EE

8101 58 8107 75

8102 C2

8103 71

8104 F2 8108 0E

8105 F6 8109 29

INPUT OUTPUT


8100 0C 1104 BE

8101 84 1105 2E8102 CA 1106 01

8103 B2

8/10/2019 mp and mc 2015

50/150


Page 50

Fig.3.1aFlowchartforSortinginAscendingorder

8/10/2019 mp and mc 2015

51/150


Page 51

EXP.NO.

TITLE: SORTINGANDSEARCHINGUSING8085-

Date:_________

SORTINGPROGRAM

AIM:Towritean assembly language program forarrangean arrayof 8-bit numbers

inascendinganddescendingorder,using8085.

APPARATUSREQUIRED:



2 Powersupplyunit 1

THEORY:

This program uses the register indirect addressing mode instructions, to access the

data during sorting. The Bubble-Sort technique is used to sort the numbers in either

ascending order or descending order. The numbers to be sorted is stored in the

memory as a array with the first location containing the count of the data in the

array. The sorted numbers are stored back again in the same source location of the

array.

(I)ALGORITHM:ASENDINGORDER:

1. Starttheprogram

2. LoadthedataaddresstoHLregisterpairandInitializeBregisterwith00

3. Movethearraysizecount intoC-register,thendecrementtheC-register

by1and incrementHLregisterpairby1.

4. LoadthefirstdatainthememorytoAccumulator.

5. ComparethesubsequentmemorywithAccumulator.

6. JumponCarry,whenM isgreaterAandgotoStep9.

7. MovethememoryM toDregister

8. DecrementtheaddressofHLpairandmovetheDregistercontenttoMand incrementthevaluebyHLby1.

9. Move01toBregisteranddecrementtheC-register.

10. IfC isNon-zero,Jumptocomparisonofnextdata.

11. IfBbecomeszeroafterdecrement,gotostep1.

11.Stoptheprogram.

8/10/2019 mp and mc 2015

52/150


Page 52

PROGRAM:


4100 06,00 AHEAD: MVIB,00 Clearthecounter

4102 21,50,41 LXIH,4150LoadDataaddrtoHLpair.

4105 4E MOVC,M MoveArraysizetoC-reg

4106 0D DCR C DecrementC reg

4107 23 INXH Incrementaddr inHL

4108 7E LOOP2: MOVA,M Movedata-I inHLtoA


410A BE CMPMComparedata-IIofMemory(HL)withA

410B DA,15,41 JC LOOP1 Jumponcytoadd

410E 56 MOVD,MMovethedata inHLtoD

reg410F 77 MOVM,A

MovecontentofacctoHL.

4110 2B DCXHDecrementaddress inHLregister.

4111 72 MOVM,DMovecontentofDtoMreg

411223 INXH

IncrementaddressofHLregister

4113 06,01 MVIB,01 MovedatatoBreg

4115 0D LOOP1: DCR C DecrementC reg

4116 C2,08,41 JNZLOOP2 Jumponnozero

4119 05 DCR B DecrementBreg

411A CA,00,41 JZAHEAD Jumponzerotoahead

411D 76 HLT Stoptheprogram

8/10/2019 mp and mc 2015

53/150


Page 53

SAMPLEDATA:

ResultoftheSortinginAscendingOrder

INPUT OUTPUT

Address Data Address Data4150

(ArraySize) 05H4150

(ArraySize) 05H

4151 78H 4151 25H

4152 A6H 4152 37H

4153 42H 4153 42H

4154 25H 4154 78H

4155 37H 4155 A6H

EXERCISE:



INPUT OUTPUT


4150

(ArraySize)4150

(ArraySize)

4151 4151

4152 4152

4153 4153

4154 4154

4155 4155

4156 4156

4157 4157

8/10/2019 mp and mc 2015

54/150


Page 54

Fig. FlowchartforSortinginDescendingorder

8/10/2019 mp and mc 2015

55/150


Page 55

(II)ALGORITHM:DESENDINGORDER:

1. Starttheprogram

2. LoadthedataaddresstoHLand initializeBregisterwith 00

3. MovetheArraysizecount intoC-registerthendecrementtheC-registerby

1and incrementHLby1.4. Loadthedata-I inthememory toA.

5. ComparethesubsequentmemorycontentwithA.

6. JumponNo-carry(whenM isgreaterA),gotoStep9.

7. MovethememoryM toDregister

8. DecrementthevalueofHLpairandmovetheDregistercontenttoMand

incrementthevaluebyHLby1.

9. MoveBregisteranddecrementtheC-register.

10. JumponNon-Zerotothenextcomparison.Ifzero,thendecrementBand

gotostep1.11.Stoptheprogram.

8/10/2019 mp and mc 2015

56/150


Page 56

PROGRAM:


4100 06,00 AHEAD: MVIB,00 Clearthecounter

4102 21,50,41 LXIH,4150 LoadDataaddrtoHLpair.

4105 4E MOVC,M MoveArraysizetoC-reg

4106 0D DCR C DecrementC reg


4108 7E LOOP2: MOVA,M Movedata-I inHLtoA


410A BE CMPMComparedata-IIofMemory(HL)withA

410B D2,15,41 JNC LOOP1 JumponNo-CarrytoLoop1

410E 56 MOVD,M Movethedata inMtoDreg410F 77 MOVM,A MovecontentofAtoM.

4110 2B DCXH Decrementaddress inHLreg.

4111 72 MOVM,D MovecontentofDreg.toM

411223 INXH IncrementAddr inHLreg

4113 06,01 MVIB,01 MovedatatoBreg

4115 0D LOOP1: DCR C DecrementC reg

4116 C2,08,41 JNZLOOP2 JumponNon-zerotoLOOP2

4119 05 DCR B DecrementBreg411A CA,00,41 JZAHEAD JumponZerotoAHEAD

411D 76 HLT Stoptheprogram

8/10/2019 mp and mc 2015

57/150


Page 57

SAMPLEDATA:

ResultoftheSortinginDescendingorder:

INPUT OUTPUT


4150

(ArraySize) 05H4150

(ArraySize) 05H

4151 78H 4151 A6H

4152 A6H 4152 78H

4153 42H 4153 42H

4154 25H 4154 37H

4155 37H 4155 25H

EXERCISE:



INPUT OUTPUT


4150

(ArraySize)4150

(ArraySize)

4151 4151

4152 41524153 4153

4154 4154

4155 4155

4156 4156

4157 4157

RESULT:

Thustheassembly languageprogram forsortingAscending&Descendingorder isexecutedandtheresultsareverified.

8/10/2019 mp and mc 2015

58/150


Page 58

Fig. FlowChartforSearching

8/10/2019 mp and mc 2015

59/150


Page 59

EXP.NO.


Date:__________

SEARCHINGPROGRAM

AIM:Towriteanassembly languageprogramtosearchthegivendata inanarray.

APPARATUSREQUIRED:



2 Powersupplyunit 1

THEORY:

This program uses the register indirect addressing mode instructions, to access the

data for searching a number from the array stored in the memory. The numbers to

be searched is loaded into theAccumulator and compared with the numbers of the

given array. Here again, the array has the data count stored in its first location.At

the end of searching, this program provides the information about the number of

times the given number is found in the array. The result is stored in the desired

memory location.

ALGORITHM:

1. Starttheprogram.

2. LoadthedataaddresstotheHLregisterpair.

3. Loadthedata tobesearched inAccumulator.

4. Loadthedatacountofarray intheC register.

5. InitializetheBregisterwith 00.

6. ComparethememorycontentaddressedbyHLpairwithAccumulator.

7. Ifnotequal(zero flag isset);gotostep9.

8. IncrementtheBregister.

9. IncrementtheHLregisterpair.

10.DecrementtheC registerby1andifnotzero,gotostep6.

11.StorethecontentofBregister inmemory.

12.Stoptheprogram.

8/10/2019 mp and mc 2015

60/150


Page 60

PROGRAM:


4100 3A,50,42 LDA4250HLoaddatatobesearchedintoAcc frommemory

4103 21,51,42 LXIH,4251H Setpointerfordataarray

4106 06,00 MVIB,00H ClearBregister

4108 4E MOVC,M LoaddatacounttoC reg.

4109 23 LOOP: INXH IncrementHLreg.pair

410A BE CMPM CompareAcc&memory

410B C2,0F,41 JNZNEXTIfNotequal,goto locationNEXT

410E 04 INR B IncrementBregister

410F 0D NEXT: DCR C DecrementC register

4110 C2,09,41 JNZLOOPIfnotzero,gotolocationLOOP

4113 78 MOVA,B Movethedata fromBtoAcc

4114 32,00,42 STA4200 Storetheresult

4117 76 HLT Stoptheprogram

8/10/2019 mp and mc 2015

61/150


Page 61

SAMPLEDATA:

ResultoftheSearching:

INPUT OUTPUT


4250(Datatobesearched) 75H

4200

(Result) 02H

4251(Arraysize) 06H

Theresultshowsthenumberoftimes,

thegivennumberthatwasfound in

thearray.

4252 23H

4253 75H

4254 C1H

4255 A7H

4256 75H

4257 12H

EXERCISE:



INPUT OUTPUT


4250(Datatobesearched)

4200

(Result)4251

(Arraysize)

Theresultshowsthenumberoftimes,

thegivennumberthatwasfound in

thearray.

4252

4253

4254

4255

4256

RESULT:

Thustheassembly languageprogram forsearchingthegivendata fromanarray is

executedandtheresult isverified

8/10/2019 mp and mc 2015

62/150


Page 62

Fig. FlowchartforSortinginAscendingorderon8086.

8/10/2019 mp and mc 2015

63/150


Page 63

EXP.NO.


Date:_________

SORTINGPROGRAM

AIM:Towriteanassembly languageprogramforthesorting inascendingand

Descendingorder,using8086.

APPARATUSREQUIRED:



2 Powersupplyunit 1

THEORY:

The Bubble-Sort technique is used to sort the numbers in either ascending order or

descendingorder.The numbers tobesorted isstored in thememoryas a arraywith

the first location containing the count of the data in the array. The sorted numbers

arestoredbackagain inthesamesource locationofthearray.

(I)AL

GORITHM

:AS

CEND

INGOR

DER:

1.Starttheprogram

2.MovedatatoCXregisterandmoveCXdatatoDIregister

3.Move1200to theBXregisterandmovethecontentof1200tomemoryofAX

4.ComparethecontentofAXwith1202

5.Jumponborrowtorep

6.Movethedata inmemorytoAXandmoveDatafromAXtoAI

7.AddDatatoBXregisterandgoto loop2

8.Performnooperationandmovedata&DItoCX

9.Gotothe loop10.Stoptheprogram

8/10/2019 mp and mc 2015

64/150


Page 64

PROGRAM:


1000 B9,07,00 MOVCX,07 MovedatatoCXreg

1003 89CF LOOP1: MOVDI,CX MoveCXdatatoDI

1005 BB,00,12 MOVBX,1200 Move1200totheBXreg

1008 8B,07 LOOP2: MOVAX,AI[BX]Movethecontentof1200tomemoryofAX

100A 3B,47,02CMP

AX,AI[BX+2]ComparethecontentofAXwith1202H

100D 72,05 JBREP Jumponborrowtorep

100F 87,47,02XCHG

AX,AI[BX+2]Movethedata inmemorytoAX

1012 89,07 MOVAI[BX],AX MovedatafromAXtoAI

101A 83,C3,02 REP: ADDBX,02 AdddatatoBXregister

1017 F2,EF LOOPLOOP2 Goto loop2

1019 90 NOP Nooperation

101A 89,F9 MOVCX,DI MovedataofDItoCX

101C E2,E5 LOOPLOOP1 Goto loop1

101E F4 HLT Stoptheprocess

8/10/2019 mp and mc 2015

65/150


Page 65

SAMPLEDATA:

ResultofSortinginAscendingorder:

INPUT OUTPUT


1100 00FFH 1100 00CCH

1102 0100H 1102 00FEH

1104 1101H 1104 00FFH

1106 00FEH 1106 0100H

1108 00CCH 1108 1101H

110A CDEFH 110A 1234H

110C ABCDH 110C ABCDH

110E 1234H 110E CDEFH

EXERCISE:



INPUT OUTPUT


1100 1100

1102 11021104 1104

1106 1106

1108 1108

110A 110A

110C 110C

110E 110E

8/10/2019 mp and mc 2015

66/150


Page 66

Fig. FlowchartforSortinginDescendingorderon8086.

8/10/2019 mp and mc 2015

67/150


Page 67

(II)ALGORITHM:DESCENDINGORDER

1.Starttheprogram.

2.MovethedatatoCXregisterandmoveCXcontenttoD1register.

3.MovetheaddresstoBXregisterandmovethecountof1200toAX.

4.ComparethecontentofAXwith1202.5.JumponnoborrowtoREP.

6.ExchangeAXandA1registerandmoveAXregistertoBX.

7.AddBXregisterto2andcontinuethe loopuptoCXiszero.

8.PerformnooperationandmoveD1registertoCXregisterandcontinue

the loopuptoCXiszero.

9.Stoptheprogram.

8/10/2019 mp and mc 2015

68/150


Page 68

PROGRAM:


1000 B9,07,00 MOVCX,07 MovedatatoCXreg

1003 89CF LOOP1: MOVDI,CX MoveCXdatatoDI

1005 BB,00,12 MOVBX,1200 Move1200totheBXreg

1008 8B,07 LOOP2: MOVAX,AI[BX]Movethecontentof1200tomemoryofAX

100A 3B,47,02CMPAX,AI[BX+2]

ComparethecontentofAXwith1202H

100D 72,05 JNBREP JumponNo-borrowtoREP

100F 87,47,02XCHG

AX,AI[BX+2]Movethedata inmemorytoAX

1012 89,07 MOVAI[BX],AX MovedatafromAXtoAI

101A 83,C3,02 REP: ADDBX,02 AdddatatoBXregister

1017 F2,EF LOOPLOOP2 GotoLOOP2

1019 90 NOP Nooperation

101A 89,F9 MOVCX,DI MovedataofDItoCX

101C E2,E5 LOOPLOOP1 GotoLOOP1

101E F4 HLT Stoptheprocess

8/10/2019 mp and mc 2015

69/150


Page 69

SAMPLEDATA:

ResultofSortinginDescendingorder:

INPUT OUTPUT


1100 00FFH 1100 CDEFH

1102 0100H 1102 ABCDH

1104 1101H 1104 1234H

1106 00FEH 1106 1101H

1108 00CCH 1108 0100H

110A CDEFH 110A 00FFH

110C ABCDH 110C 00FEH

110E 1234H 110E 00CCH

EXERCISE:



INPUT OUTPUT


1100 1100

1102 1102

1104 1104

1106 1106

1108 1108

110A 110A

110C 110C

110E 110E

RESULT:

Thustheassembly languageprograms forsorting-ascending&descendingorderwas

executedareverified.

8/10/2019 mp and mc 2015

70/150


Page 70

EXPT NO: 3(b) Program for sorting and searching using 8086 DATE:

AIMTo search a number (largest, smallest) and Sorting (Ascending and Descending) of an array

using 8086.

APPARATUS REQUIRED:

Sl.No. Apparatus required Quantity

1 8086 Microprocessor Kit 1

2 DC Power Supply 5V 1

ALGORITHM

SEARCHING

SMALLEST ELEMENT IN AN ARRAY

1. Load the starting address of the array in SI register

2. Load the address of the result in DI register

3. Load the number of bytes in the array in CL register

4. Increment the array pointer(SI register)

5. Get the first byte of the array in AL register.

6. Decrement the byte count(CL register)


8. Get next byte of the array in BL register.

9. Compare current smallest (AL) and next byte (BL) of the array.

10.Move BL to AL

11.Decrement the byte count (CL register)

12.Check zero flag.If zero flag is reset then go to step-7,otherwise go to next step.13.Save the smallest data in memory pointed by DI

14.Stop

LARGEST ELEMENT IN AN ARRAY

1. Load the starting address of the array in SI register

2. Load the address of the result in DI register

3. Load the number of bytes in the array in CL register


5. Get the first byte of the array in AL register.

6. Decrement the byte count(CL register)


8. Get next byte of the array in BL register.

9. Compare current largest (AL) and next byte (BL) of the array.

10.Move BL to AL

11.Decrement the byte count (CL register)

8/10/2019 mp and mc 2015

71/150


Page 71

12.Check zero flag.If zero flag is reset then go to step-7,otherwise go to next step.

13.Save the largest data in memory pointed by DI

14.Stop

ASCENDING ORDERIN AN ARRAY

1. Set SI register as pointer for array

2. Set CL register as count for N-1 repeatitions3. Initialize array pointer

4. Set CH as count for N-1 comparisions

5. Increment the array pointer

6. Get an element of array in AL register


8. Compare the next element of the array with AL.

9. Check carry flag.If carry flag is set then go to step-12,otherwise go to next step.

10.Exchange the content of memory pointed by SI and the content of previous memory location

for this,exchange AL and memory pointed by SI,and then exchange AL and memory pointed

by SI-1

11.Decrement the count for comparisions(CH register)

12.Check zero flag .If zero flag is reset then go to step-6,otherwise go to next step

13.Decrement the count for repeatitions (CL register)


15.Stop

DESCENDING ORDER IN AN ARRAY

1. Set SI register as pointer for array

2. Set CL register as count for N-1 repetitions3. Initialize array pointer

4. Set CH as count for N-1 comparisions


6. Get an element of array in AL register


8. Compare the next element of the array with AL.

9. Check carry flag.If carry flag is reset then go to step-12,otherwise go to next step.

10.Exchange the content of memory pointed by SI and the content of previous memory location

(for this,exchange AL and memory pointed by SI,and then exchange AL and memory

pointed by SI-111.Decrement the count for comparisions(CH register)


13.Decrement the count for repetitions (CL register)


15.Stop

8/10/2019 mp and mc 2015

72/150


Page 72

SMALLEST ELEMENT IN AN ARRAY

Flow Chart

Load the address of array in SI register

and address of given data in DI register

Load the address of result in DI register

Compare AL and DL register

Set CL as byte count

Increment array pointer(SI)

Start

Get the first byte of array in AL register

Decrement the b te count

Increment array pointer(SI)

Get the next byte of array in BL register

Is

CF=1?

Move BL to AL

Decrement byte count (CL)

Is

ZF=1?

Store AL in memory

Stop

YES

YES

NO

NO

8/10/2019 mp and mc 2015

73/150


Page 73

PROGRAM:


8000 START MOV SI,1100 BE 11 00 Set SI register as

pointer for array

8003 MOV DI,1200 BF 12 00 Set DI register as

pointer for result

8006 MOV CL,[SI] 8A 0C Set CL as count for

elements in the array

8008 AGAIN INC SI 46 Increment the address

pointer

8009 MOV AL,[SI] 8A 04 Set first data as

Smallest

800B DEC CL FE C9 Decrement the count

800D INC SI 46 Make SI to point tonext data in array

800E MOV BL,[SI] 8A 1C Get the next data in

BL register

8010 CMP AL,BL 3A C3 Compare current

smallest data in AL

with BL

8012 JC AHEAD 72 02 If carry is set then AL

is less than BL hence

proceed to AHEAD8014 MOV AL,BL 8A C3 If carry is not set then

make BL as current

smallest

8016 AHEAD DEC CL FE C9 Decrement the count

8018 JNZ AGAIN 75 F3 If count is not zero

repeat search

801A MOV [DI],AL 88 05 Store the smallest data

in memory

801C INT 3 F4

8/10/2019 mp and mc 2015

74/150


Page 74

LARGEST ELEMENT IN AN ARRAY

FLOW CHART

Set SI register as array pointer

Set DI register as array pointer

Compare AL and DL register

Set CL as byte count

Increment arra ointer SI

Get the first byte of array in AL register

Decrement the byte count

Increment arra ointer SI

Get the next b te of arra in BL re ister

Is

CF=0?

Move BL to AL

Decrement byte count (CL)

Is

ZF=1?

Store AL in memory

Stop

Start

Yes

NO

NO

Yes

8/10/2019 mp and mc 2015

75/150


Page 75

PROGRAM:



pointer for array

8003 MOV DI,1200 BF 12 00 Set DI register as

pointer for result

8006 MOV CL,[SI] 8A 0C Set CL as count for

elements in the array

8008 AGAIN INC SI 46 Increment the address

pointer

8009 MOV AL,[SI] 8A 04 Set first data as largest

800B DEC CL FE C9 Decrement the count

800D INC SI 46 Make SI to point tonext data in array

800E MOV BL,[SI] 8A 1C Get the next data in

BL register

8010 CMP AL,BL 3A C3 Compare current

largest data in AL

with BL

8012 JNC AHEAD 73 02 If carry is set then AL

is greater than BL

hence proceed toAHEAD

8014 MOV AL,BL 8A C3 If carry is not set then

make BL as current

largest

8016 AHEAD DEC CL FE C9 Decrement the count

8018 JNZ AGAIN 75 F3 If count is not zero

repeat search

801A MOV [DI],AL 88 05 Store the largest data

in memory

801C INT 3 F4

8/10/2019 mp and mc 2015

76/150


Page 76

ASCENDING ORDER IN AN ARRAY

FLOW CHART

Yes

No

Load the address of the array in SI register

Load the count in CL re ister and decrement b one

Com are next element of arra with AL


Load the count in CH re ister and decrement b one

Increment the array pointer (SI)

Get an element of array pointer(SI)

Increment the arra ointer SI

Is

CF=1?

Exchange AL and memory pointed by SI

Exchange AL and memory pointed by (SI-1)

Is

ZF=1?

Decrement CL count

Stop

Decrement CH count

Is

ZF=1?

Start

No

Yes

8/10/2019 mp and mc 2015

77/150


Page 77

PROGRAM:



pointer for array

8003 MOV CL,[SI] 8A 0C Set CL as count for N-1

repeatitions8005 DEC CL FE C9 Decrement the count

8007 REPEAT MOV SI,1100 BE 11 00 Intialize pointer

800A MOV CH,[SI] 8A 2C Set Ch as count for N-1

comparisions

800C DEC CH FE CD

800E INC SI 46 Increment the pointer

800F REPCOM MOV AL,[SI] 8A 04 Get an element of array

in AL register

8011 INC SI 46

8012 CMP AL,[SI] 3A 04 Compare with next

element of array inmemory

8014 JC AHEAD 72 05 If AL is less thanmemory ,then go to

AHEAD

8016 XCHG AL,[SI] 86 04 If AL is lesser than

memory then exchangethe content of memorypointed by SI and the

previous memory

location

8018 XCHG AL,[SI-1] 86 44 FF

801B DCE CH FE CD Decrement the count fo

comparisions

801D JNZ REPCOM 75 F0 Repeat comparision

until CH count is zero

801F AHEAD DEC CH FE C9 Decrement the count fo

repeatitions8021 JNZ REPEAT 75 E4 Repeat N-1

comparisions until CLcount is zero

8023 INT 3 F4

8/10/2019 mp and mc 2015

78/150


Page 78

DESCENDING ORDER IN AN ARRAY

FLOW CHARTLoad the address of the array in SI register

Load the count in CL register and decrement by one

Com are next element of arra with AL


Load the count in CH re ister and decrement b one

Increment the array pointer (SI)

Get an element of arra ointer SI

Increment the arra ointer SI

Is

CF=0?

Exchange AL and memory pointed by SI

Exchange AL and memory pointed by (SI-1)

Is

ZF=1?

Decrement CL count

Decrement CH count

Is

ZF=1?

Start

Sto

No

Yes

No

Yes

Yes

No

8/10/2019 mp and mc 2015

79/150


Page 79

PROGRAM:



pointer for array

8003 MOV CL,[SI] 8A 0C Set CL as count for N-1

repeatitions8005 DEC CL FE C9 Decrement the count

8007 REPEAT MOV SI,1100 BE 11 00 Intialize pointer

800A MOV CH,[SI] 8A 2C Set Ch as count for N-1

comparisions

800C DEC CH FE CD

800E INC SI 46 Increment the pointer

800F REPCOM MOV AL,[SI] 8A 04 Get an element of array

in AL register

8011 INC SI 46

8012 CMP AL,[SI] 3A 04 Compare with next

element of array inmemory

8014 JNC AHEAD 73 05 If AL is less thanmemory ,then go to

AHEAD

8016 XCHG AL,[SI] 86 04 If AL is lesser than

memory then exchangethe content of memorypointed by SI and the

previous memory

location

8018 XCHG AL,[SI-1] 86 44 FF

801B DCE CH FE CD Decrement the count fo

comparisions

801D JNZ REPCOM 75 F0 Repeat comparision

until CH count is zero

801F AHEAD DEC CL FE C9 Decrement the count fo

repeatitions8021 JNZ REPEAT 75 E4 Repeat N-1

comparisions until CLcount is zero

8023 INT 3 F4

8/10/2019 mp and mc 2015

80/150


Page 80

Result:

Smallest from an array

INPUT OUTPUT


8100 H:

8200 H:

8101 H:

8102 H:8103 H:

8104 H:

8105 H:

8106 H:

Largest from an array

INPUT OUTPUT


8100 H:

8200 H:

8101 H:

8102 H:

8103 H:8104 H:

8105 H:

8106 H:

Sort the array in ascending order

INPUT OUTPUT


1100 H:

1101 H:1101 H:

1102 H: 1102 H:

1103 H: 1103 H:

1104 H: 1104 H:

1105 H: 1105 H:

1106 H: 1106 H:

1107 H: 1107 H:

Sort the array in descending order

INPUT OUTPUT


1100 H:

1101 H:1101 H:1102 H: 1102 H:

1103 H: 1103 H:

1104 H: 1104 H:

1105 H: 1105 H:

1106 H: 1106 H:

1107 H: 1107 H:

8/10/2019 mp and mc 2015

81/150


Page 81

Fig.FlowchartforStringManipulation

8/10/2019 mp and mc 2015

82/150


Page 82

EXP.NO.

TITLE:

Date:_________

STRINGMANIPULATIONUSING8086

AIM: Towriteanassembly languageprogramtomoveabyteofstringoflengthFF

fromasourcetoadestination.

APPARATUSREQUIRED:



2 Powersupplyunit 1

THEORY:

A group of similar data stored in consecutive memory locations, representing a

variablecanbecelledas a String. Variousoperationscanbeperformedonthestring

data like, string copy, string compare, string store, string load, etc. The following

programhelpstocopyastringdata fromasourcelocationtoadestination location.

ALGORITHM:

1. Starttheprogram.2. SettheSItopointthesourcearrayandDIatdestination location.

3. MovethestringsizetoCXregister.

4. DirectionFlag isclearedsothatSI&DIwillautoincrementaftereach loop.

5. MovethebytesofthestringusingMOVSB instruction.

6. Stop

8/10/2019 mp and mc 2015

83/150


Page 83

PROGRAM:


1000 BE,00,11 MOVSI,SourceLoadoffsetaddressofSourcetoSIregister

1003 BF,00,12 MOVDI, DestinationLoadoffsetaddressof

destinationtoDIregister1006 B9,FF,00 MOVCX,00FFH

NumberofarrayelementsinCXregister

1009 FC CLD ClearDirectionFlag(D)

100A A4 NEXT: MOVSB Movestringbyte

100B E2,FD LOOPNEXTDecrementCXandCheck

forZero.Ifnotzero,goto locationNEXT

100D F4 HLT Stoptheprogram

8/10/2019 mp and mc 2015

84/150


Page 84

SAMPLEDATA:

ResultofStringmanipulation:

INPUT(Source) OUTPUT(Destination)


1100 xx 1200 xx

To xx To xx

11FF xx 12FF xx

AsCXRegisteris loadedwithstringsize FF

256bytesofdataatsource location

(starting from1100H)willbecopiedtodestination location

(at address1200H)

EXERCISE:

ResultofStringmanipulation:

INPUT(Source) OUTPUT(Destination)


1100 1200

1101 1201

1102 1202

1103 1203

1104 1204

1105 1205

1106 12061107 1207

1108 1208

Executetheprogramwithfollowingmodifications.

a) Move smaller sized string data, by changing the string size stored in CX

register.

b) Change the auto-increment of SI & DI registers toauto-decrement bysetting

the direction flag to 1.Also store the last address of the string in SI & DI

register insteadofstartingaddressofthestring.

RESULT:

Thusthestringwasmovedfromsourcetodestinationusingtheassemblylanguage

of8086.

8/10/2019 mp and mc 2015

85/150


Page 85

FlowChart:

PROGRAM(16-bitAddition):


4100 C3 CLR C Clearcarry

4101 74,Data1L MOVA,#DATA1 MoveData1LtoAcc

4103 24,Data2L ADDA,#DATA2 AddData2LwithAcc

4105 90,41,50MOVDPTR,#4150h

Movecontent in4500toDPTR.

4108 F0 MOVX@DPTR,A MovedatatoDPTR location

4109 A3 INC DPTR IncrementDPTR

410A 74,Data1H MOVA,#DATA1 MoveData1HtoAcc

410C 34,Data2HADDC A,#DATA2 AddAccwithData2H&Carry

410E F0 MOVX@DPTR,AMovefromAcctoDPTR location

410F 80,FE HERE: SJMPHERE End

8/10/2019 mp and mc 2015

86/150


Page 86

EXP.NO: Date:_________

TITLE: PROGRAMMINGARITHMETIC,LOGICALAND

BITMANIPULATIONUSING8051

AIM: To write an assembly language program for the arithmetic, logical and bit

manipulationusing8051.

APPARATUSREQUIRED:


1 8051MicroControllerkit. 1

2 PowerSupplyunit 1

(A)ARITHMETICOPERATION:

(I)16BITADDITION:

ALGORITHM:

1.Starttheprogram.

2.GettheLSBof1st

and2nd

operands.

3.AddtheLSBofthetwooperandsandstoreit inmemory.

4.GettheMSBof1st

and2ndoperands.

5.AddtheMSBandstoretheresult inmemory

6.Stoptheprogram.

OUTPUTOF16-BITADDITION:

INPUT OUTPUT


4102 4500

4104 4501410B - -

410D - -

8/10/2019 mp and mc 2015

87/150


Page 87

FlowChart:

PROGRAM(8-bitSubtraction):


4100 C3 CLR C Clearcarry

4101 74,Data1 MOVA,#DATA1 Movedata1toacc

4103 94,Data2 SUBBA,#DATA2 Adddata2withacc

4105 90,45,00MOVDPTR,

#4500hMove4500toDPTR.

4108 FO MOVX@DPTR,AMoveAccvaluetoDPTRlocation

4109 80,FE HERE: SJMPHERE End

8/10/2019 mp and mc 2015

88/150


Page 88

(II)8-BITSUBTRACTION:

ALGORITHM:

1. Starttheprogram.

2. Clearthecarryflagand loadthe firstoperand inaccumulator.

3. Getthe2nd

operandandsubtract it fromaccumulator.

4. Storetheresult inmemory.

5. Stoptheprogram.

OUTPUTOF8-BITSubtractionwithoutCarry:

INPUT OUTPUT


4102 4500

4104 - -

8/10/2019 mp and mc 2015

89/150


Page 89

FlowChart:

PROGRAM(8-bitMultiplication):

AddressOpcode&

OperandLabel Mnemonics Comments

4100 74,Data1 MOVA,#DATA1 MoveData1 toAcc

4102 75,F0,Data2 MOVB,#DATA2 MoveData2 toAcc

4105 A4 MULAB MultiplyAccandB

4106 90,45,00MOVDPTR,#4500h Move4500toDPTR.

4109 FO MOVX@DPTR,AMoveAccvaluetoDPTR

location

410A A3 INC DPTR INC DPTR

410B E5,F0 MOVA,BMoveBregistervalueto

Acc410D FO MOVX@DPTR,A

MoveAccvaluetoDPTRlocation

410E 80,FE HERE: SJMPHERE End

8/10/2019 mp and mc 2015

90/150


Page 90

(III)8-BITMULTIPLICATION:

ALGORITHM:

1. Starttheprogram.

2. Loadthe1st operand inAand2

ndoperand inB.

3. MultiplyAandBcontentsusingMUL instruction.4. Storetheresult inmemory.

5. Stoptheprogram.

OUTPUTOF8-BITMultiplication:

INPUT OUTPUT


4101 4500

4104 4501

8/10/2019 mp and mc 2015

91/150


Page 91

FlowChart:

PROGRAM(8-BitDivision):


4100 74,Data1 MOVA,#DATA1 MoveDividendtoAcc

4102 75,F0,Data2 MOVB,#DATA1 MoveDivisortoB

4105 84 DIVAB DivideAbyB

4106 90,45,00MOVDPTR,#4500h Move4500toDPTR.

4109 FO MOVX@DPTR,AMoveAccvaluetoDPTRlocation

410A A3 INC DPTR IncrementDPTR

410B E5,F0 MOVA,B MoveB-reg.valuetoAcc

410D FO MOVX@DPTR,A Storetheresult

410E 80,FE HERE: SJMPHERE End

8/10/2019 mp and mc 2015

92/150


Page 92

(IV)8BITDIVISION:

ALGORITHM:

1. Starttheprogram.

2. Get1st operand inAand2

ndinB.

3. DivideAbyBcontentsusingdivision instruction.4. Storetheresult inmemory.

5. Stoptheprogram.

OUTPUTOF8-BITDivision:

INPUT OUTPUT


41014500

(Quotient)

4104 4501(Remainder)

8/10/2019 mp and mc 2015

93/150


Page 93

FlowChart:

PROGRAM(OR):



4102 44,Data2 ORLA,#DATA2 ORAccandData2

4104 90,45,00MOVDPTR,#4500H Move4500toDPTR.

4107F

O MOVX@DP

TR,A Storetheresult4108 80,FE HERE SJMPHERE End

8/10/2019 mp and mc 2015

94/150


Page 94

(B)LOGICALOPERATION:

(I) OROPERATION:

ALGORITHM:

1. Starttheprogram.

2. Load1st

operand inAccumulator.

3. Get2nd

operandandperformOR betweenAccand2nd

Operand.

4. Storetheresult inmemory.

5. Stoptheprogram.

OUTPUTOFOROperation:

INPUT OUTPUT

Address Data Address Data4101 4500

4103 - -

8/10/2019 mp and mc 2015

95/150


Page 95

FlowChart:

PROGRAM(AND):



4102 54,Data2 ADLA,#DATA2 ANDAccandData2

4104 90,45,00MOVDPTR,#4500H Move4500toDPTR.

4107 FO MOVX@DPTR,A Storetheresult

4108 80,FE HERE: SJMPHERE End

8/10/2019 mp and mc 2015

96/150


Page 96

(II)ANDOPERATION:

ALGORITHM:

1. Starttheprogram.

2. Get1st operand inAccumulator.

3. Get2nd

operandandperformANDaccumulatorcontent&2nd

operand

4. Storetheresult inmemory.5. Stoptheprogram.

OUTPUTOFANDOperation:

INPUT OUTPUT


4101 4500

4103 - -

RESULT:

Thustheprograms involvingarithmetic, logicalandbitmanipulationusing8051are

executedand itsresultsareverified.

8/10/2019 mp and mc 2015

97/150


Page 97

Fig. BlockdiagramofADCconversion

8/10/2019 mp and mc 2015

98/150


Page 98

EXP.NO.

TITLE:

INTERFACING ADC AND DAC USING 8085 Date:_________

AIM: Towriteanassembly languageprogramto interfaceADC andDAC with8085microprocessorkit.

APPARATUSREQUIRED:



2 CRO 1

3 ADC &DAC interfaceboard 14 FlatribbonCable 1

THEORY:

In a real time applications, processing of input data, conversion of data from digital

to analog and vice versa, are indispensable. The following program initiates the

analog to digital conversion process, checks theEOC pin ofADC 0809 as to whether

theconversion isoverand then inputs thedata to theprocessor.Italso instructs the

processortostoretheconverteddigitaldata inthememory.

ADC:

HARDWAREDETAILS:

ADC 0809 isamonolithicCMOSdevice,withan8-bitanalog-to-digitalconverter,8

channelmultiplexerandmicroprocessorcompatiblecontrol logic.

SelectedAnalogChannel

AddressLinesintheMultiplexerofADC 0809AddrC AddrB AddrA

IN0 0 0 0IN1 0 0 1IN2 0 1 0IN3 0 1 1IN4 1 0 0IN5 1 0 1

IN6 1 1 0IN7 1 1 1

Aparticular inputchannel isselecte

mp and mc 2015

Documents