8051 Manual Fin

Microcontrollers Lab Manual 06ESL47

KVG COLLEGE OF ENGINEERINGElectronics and Communications

Micro Controller

Lab Manual

Dept. Of E & C Engg KVGCE,Sullia


1. DATA TRANSFER INSTRUCTIONS1. Write A Program to move a block of data within the internal RAM Algorithm

1. Initialize registers to hold a. the count, the number of data bytes to be moved,b. the starting addresses of source block c. starting addresses of destination block.

2. Get data from source location into accumulator 3. Transfer accumulator content to the destination location.4. Decrement the count register5. Update data pointers.6. Check whether all the bytes are transferred or not. ie, if count is not

zero repeat step 2 through 6. PROGRAM CODE:

ORG 0H

START1: MOV R0,#40H ;R0 POINTED TO INTERNAL RAM 40H MOV R1,#30H ;R1 POINTING TO INTERNAL RAM 030H

MOV R2,#5 ;R2 LOADED WITH NO. OF ELEMENTS IN `;THE ARRAY

START: MOV A,@R0 ;DATA TRANSFER MOV @R1,A INC R0 INC R1

DJNZ R2,START ;DECREMENT R2,IF NOT EQUAL TO ;0,CONTINUE WITH DATA

;TRANSFER PROCESS. HERE: SJMP HERE END

RESULT:Before Execution: Fill 5 locations at I:0040h with data bytes. 5 locations at I:0030h are blank.

After Execution: 5 locations I:0040h are filled up with data. These are copied to 5 locations at I:0030h .



2) Write an assembly language program to transfer n =10 bytes of data from location 8035h to location 8041h within External RAM.(without overlap). Write the code at C:030hAlgorithm1. Initialize registers to hold

a. the count, the number of data bytes to be moved,b. the starting addresses of source block c. starting addresses of destination block.

2. Get data from source location into accumulator and transfer to the destination location.

3. Update data pointers.4. Decrement the count register, repeat step 2-3 if count is not zero.

PROGRAM CODE: ORG 0000H

SJMP 30HORG 30HMOV DPH,#80H ; source/Desti address Higher order ByteMOV R0,#35H ;//source address Low ByteMOV R1,#41H ;//destination address Low ByteMOV R3,#0AH ;//count

BACK: MOV DPL, R0MOVX A,@DPTRMOV DPL, R1MOVX @DPTR,AINC R0INC R1DJNZ R3, BACK

HERE: SJMP HEREEND

RESULT:Before Execution: 10 locations at X:8035h are to be filled up with data.

After Execution: 10 locations at X:8041h are filled up with data from 8035h.



3) Write an ALP to move a block of 6 data bytes ,stored in internal RAM starting at location 10h to locations starting at 13h.(Block move with overlapped memory addresses)

Algorithm:1. Initialize registers to hold the count of number of data bytes to be

moved, the source address and the shift(difference between destination addresses and source address).

2. Compute addresses of last locations of destination and source blocks. and set the pointers(using bank registers R0 & R1)

3. Get data from source location into accumulator 4. Transfer accumulator content to the destination location.5. Decrement the count, update data pointers(Decrement R0&R1).6. Check whether all the bytes are transferred or not. ie, if count is not

zero repeat step 3 through 6. Note: To transfer the Block to a memory with overlapped memory address the pointers are to be pointed to the end of the blocks Program Code:

ORG 0MOV R1,#10H ;SOURCE BLOCKMOV R2,#6H ;NUMBER OF ELEMENTS TO BE ;MOVED(BLOCK

SIZE)MOV R3,#3H ;SHIFTMOV A,R1ADD A,R2MOV R1,AADD A,R3MOV R0,A

LOOP:DEC R0DEC R1MOV A,@R1MOV @R0,ADJNZ R2,LOOPEND

Results:Before Execution: locations from I:0010h are filled up with data bytes.

After Execution: data bytes from I:0010h are moved to i:0013h



4. EXCHANGE BLOCKS OF DATA.Write an ALP to exchange 5 bytes of data stored in locations starting from 0027h with 5 bytes in locations starting from 0041h of external memory.Use Bank Registers to hold memory addresses.Algorithm

1. Initialize bank registers to hold the array size (count), the block_1 address & the block_2 addresses.

2. Get data from block_1 location into accumulator and save in a register temporarily (Say / Use R2).

3. Get data from the block_2 location into accumulator and store in block_1 location

4. Get the data from temporary register(R2) and store in block_1 locations.

5. Decrement the count register and repeat from step 2 to 4 till count is zero.

Program Code:ORG 00HMOV R0,#27H //source addressMOV R1,#41H //destination addressMOV R3,#05H //count

BACK: MOVX A,@R0MOV R2,AMOVX A,@R1MOVX @R0,AMOV A, R2MOVX @R1,AINC R0INC R1DJNZ R3, BACK

HERE: SJMP HEREEND

ii) Write the program using XCH instructionORG 0000HSJMP 30HORG 30HMOV R0,#27H //source addressMOV R1,#41H //destination addressMOV R3,#05H //count

BACK: MOVX A,@R0MOV R2,AMOVX A,@R1XCH A, R2



MOVX @R1,aXCH A, R2MOVX @R0,aINC R0INC R1DJNZ R3, BACK

HERE: SJMP HEREEND

RESULT:Before Execution: 5 locations at X:0027h & X:0041h are filled up with data.

After Execution: The data at X:0027h & X:0041h are exchanged.

LAB Assignments: 1) Write an ALP to move a block of 10 data bytes, stored in external RAM starting at location 8050h to internal RAM locations at 30h.2) Write an ALP to move a block of 10 data bytes, stored in internal RAM starting at location 10h to locations in external RAM starting at 30h.(i) Use DPTR ii) Bank registers to point to External RAM locations.3) Write an ALP to move a block of 10 data bytes, stored in external RAM starting at location 8050h to locations starting at 8055h.4) Write an ALP to Reverse a block of 10 data bytes, stored in internal RAM starting at location 10h 5) Write an ALP to exchange 10 data bytes, stored in internal RAM starting at location 20h with data bytes, stored in internal RAM locations starting at 35h.6) Write an ALP to exchange 10 data bytes, stored in internal RAM starting at location 20h with data bytes, stored in external RAM locations starting at 35h7) Write an ALP to exchange 10 data bytes, stored in internal RAM starting at location 20h with data bytes, stored in external RAM locations starting at 8035h8) Write an ALP to exchange 10 data bytes, stored in external RAM starting at location 8020h with data bytes, stored in external RAM locations starting at 8035h



5) ASSEMBLY LANGUAGE PROGRAM TO SORT NUMBERS.//BUBBLE SORT PROGRAMWrite an assembly language program to sort an array of n= 6 bytes of data in ascending order stored from location 8035h.(use bubble sort algorithm)

ORG 0000HSJMP 30HORG 30HMOV R0,#05 //count n-1 -ARRAY SIZE-n- Pass Counter

L1: MOV dptr, #9000h //array stored from address 9000hMOV A,R0 //initialize exchange counterMOV R1,A

L2: MOVX a, @dptr //GET NUMBER FROM ARRAYMOV B, A //& STORE IN BINC dptrMOVX a, @dptr //next number in the arrayCLR C //reset borrow flagMOV R2, A //STORE IN R2SUBB A, B //2nd - 1st no.—no compare instruction in 8051JC NOEXCHG // JNC - FOR ASCENDING ORDERMOV A,B //EXHANGE THE 2 NOES IN THE ARRAYMOVX @dptr,aDEC DPL //DEC dptr-INSTRUCTION NOT PTRESENTMOV a,R2MOVX @dptr,aINC DPTR

NOEXCHG: DJNZ R1,L2 //decrement compare counterDJNZ R0,L1 //decrement pass counter

here: SJMP hereEND



Algorithm1. Store the elements of the array from the address 9000h2. Initialize a pass counter with array size-1 count (for number of

passes).3. Load compare counter with pass counter contents & initialize DPTR

to point to the start address of the array (here 9000h).4. Store the current and the next array elements pointed by DPTR in

registers B and r2 respectively.5. Subtract the next element from the current element.6. If the carry flag is set (for ascending order) then exchange the 2

numbers in the array.7. Decrement the compare counter and repeat through step 4 until the

counter becomes 0.8. Decrement the pass counter and repeat through step 3 until the

counter becomes 0.RESULT : Before Execution:Unsorted Array at 9000h

After Execution: Sorted Array (Descending order) at 9000h

4) Write an assembly language program to find the largest element in a given string of n = 6 bytes at location 4000h. Store the largest element at location 4062h. ORG 0000H

SJMP 30HORG 30H

MOV R3,#6 //length of the array MOV DPTR,#4000H //starting address of the array MOVX A,@DPTR MOV r1,a

NEXTBYTE: INC DPTR MOVX A,@DPTR

CLR C //reset borrow flag MOV R2,A //next number in the array SUBB A,R1 //OTHER Num - PREVIOUS LARGEST

no. JC skip // JNC for smallest element MOV A,r2 //UPDATE larger number in r1 MOV R1,A

skip:DJNZ R3,NEXTBYTE MOV DPL, #62H //LOCATION OF THE RESULT-4062H



MOV A,R1 //LARGEST NUMBER MOVX @DPTR,A //STORE AT #4062H OVER: SJMP OVER

ENDAlgorithm

1. Store the elements of the array from the address 4000h2. Store the length of the array in r3 and set it as counter.3. DPTR is loaded with starting address of the array.4. Store the first number of the array in r1 (r1 is assigned to hold the

largest number). 5. Increment DPTR.6. Subtract the number pointed by DPTR from the contents of r1 (to

compare whether the next array element is larger than the one in r1).7. If the element pointed by DPTR is larger then load the larger number

into r1.8. Decrement the counter and repeat steps through 5 until the counter

becomes 0.9. Store the largest number in r1 in address 4062h

RESULT:Before Execution:

After Execution: Location 4062 has the largest element.

2. ARITHMETIC INSTRUCTIONSASSEMBLY LANGUAGE PROGRAM ILLUSTRATING ADDITION, SUBTRACTION, MULTIPLICATION AND DIVISION .5) Write an ALP to perform the following:If x=0-perform w + v; else if x=1-perform w-v; else if x=2-perform w*v; elseif x=3-perform w/v, where w & v are eight bit numbers.

ORG 0000HSJMP 30HORG 30HMOV R0, #40HMOVX A,@R0MOV R1, A //R1 HAS CONDITION XINC R0MOVX A,@R0MOV B, A //B HAS 1ST NUMBER-vINC R0MOVX A,@R0 //A HAS 2ND NUMBER-wCJNE R1,#00,CKSUB



ADD A,B //PERFORM ADDITIONMOV B,#00 //B HAS CARRYJNC SKIPMOV B,#01H

SKIP:SJMP LASTCKSUB: CJNE R1,#01,CKMUL

CLR C //RESET BORROW FLAGSUBB A,BMOV B,#00 //B INDICATES BORROWJNC SKIP1MOV B,#0FFH //FF INDICATES NEGATIVE NUMBER

SKIP1:SJMP LASTCKMUL: CJNE R1,#02,CKDIV

MUL AB //16 bit product in AB with A having lower byte

SJMP LASTCKDIV: CJNE R1,#03,OTHER

DIV AB //Quotient in A & remainder in BSJMP LAST

OTHER:MOV A,#00MOV B,#00

LAST: INC R0MOVX @R0,AINC R0MOV A,BMOVX @R0,A

HERE:SJMP HEREEND

Algorithm1. Store the condition x in r1.2. Load the first and second numbers to A and B registers respectively 3. Compare the contents of r1 and perform the operations add, sub, etc

accordingly.4. Store the result present in A and B registers to the appropriate

memory locations.

RESULT:Before Execution: ADD SUB

After Execution: ADD After Execution: SUB

Before Execution: MUL After Execution: MUL



ASSEMBLY PROGRAM ILLUSTRATING SQUARE AND CUBE OPERATIONS.//cube is an example of 16-bit arithmetic operation//depending on flag condition, square or cube is performed// Flag is a bit in the bit addressable RAM, say 1st bit of location 20h is used, then bit address is 016) An eight bit number X is stored in external memory location 9000h. Write an ALP to compute (i) the square of the number X if LSB of data RAM 20h (bit address 01H) is set(ii) the cube of the number X if LSB of data RAM 20h (bit address 01H) is reset.Store your result at locations 9001, 9002, 9003h.

ORG 0000HSJMP 30HORG 30HMOV DPTR,#9000HMOVX A,@DPTR //GET NUMBER-XMOV R0,A //STORE IN R0MOV B,AMUL AB //SQUARE IT-X^2CLR C //FOR STORING RESULTJB 01,LAST //IF BIT 01 IS SET THEN END, ELSE DO

CUBEPUSH B //STORE UPPER PART OF SQUAREMOV B,A //B-LOWER PART OF X^2MOV A,R0 //A-XMUL AB //X*LOWER X^2INC DPTRMOVX @DPTR,A //STORE PARTIAL RESULTMOV A,BMOV R2,A //UPPER PART OF X*LOWER X^2 IN R2POP B //GET BACK UPPER PART OF SQUAREMOV A,R0 //A-XMUL AB //X*UPPER X^2ADD A,R2 //ADD TO PARTIAL RESULT

LAST:INC DPTRMOVX @DPTR,AMOV A,BADDC A,#00 //ADD CARRY TO B(FOR SQUARE RESULT,

C=0)INC DPTRMOVX @DPTR,A

HERE:SJMP HEREEND

RESULT:CUBE OF 56H IS 9B498 WHICH IS STORED AS 98, B4, 09 (LOWER BYTE FIRST)



To get square make the D1 bit of data memory 20h high, say FF,02,06,etc. The bit address is 01. Similarly bit address 78h correspond to D0 bit 0f data ram location 2Fh.

Algorithm1. Store the eight bit number x in A, r0 & B registers.2. Multiply A and B registers to obtain the square (say SQH:SQL) of

the number x.3. Check if bit 01 is set. If set go to end (storing the result), else do the

cube operations.4. The high part of the square result (SQH) is stored on the stack.5. Multiply the low part of the square result (SQL) with x (partial cube

result).6. Store the low part of the above result at 9001h & the high part in R2.7. Retrieve the high part of the square result (SQH) stored on the stack

& multiply with x. 8. Add the low part of the above result (SQH*X) with R2 and store in

9002h.9. Add the high part (SQH*X) with the resulting carry and store in 9003.

3. PROGRAM ILLUSTRATING BIT MANIPULATIONS7) Two eight bit numbers NUM1 & NUM2 are stored in external memory locations 8000h & 80001h respectively. Write an ALP to compare the 2 nos.Reflect your result as: if NUMI<NUM2, SET LSB of data RAM 2F (bit address 78H)IF NUM1>NUM2, SET MSB OF 2F(7FH). if NUM1 = NUM2-Clear both LSB & MSB of bit addressable memory location 2Fh ORG 0000H

SJMP 30HORG 30HMOV DPTR,#8000HMOVX A,@DPTRMOV R0,AINC DPTRMOVX A,@DPTRCLR CSUBB A,R0JZ EQUALJNC BIGSETB 78H



SJMP END1BIG:SETB 7FH

SJMP END1EQUAL:CLR 77H

CLR 7FHEND1:SJMP END1

ENDAlgorithm:

1. Store the elements of the array from the address 4000h2. Move the first number in r0 and the second number in register A

respectively3. Clear carry flag and subtract the two numbers, if the carry flag is 0(if

the nos are equal), Clear both LSB & MSB of bit addressable memory location 2Fh

4. If the carry bit is set then Set MSB of 2F(7FH), else LSB of data RAM 2F (bit address 78H).

RESULT: 1) Before Execution: X:08000h = 45 & X:8001 = 35After Executuion: D:02FH =012) Before Execution: X:08000h = 25 & X:8001 = 35After Executuion: D:02FH =803) Before Execution: X:08000h = 45 & X:8001 = 45After Executuion: D:02FH =00

4. LOGICAL INSTRUCTIONS8) ASSEMBLY PROGRAM ILLUSTRATING LOGICAL INSTRUCTIONS (BYTE LEVEL)3 eight bit numbers X, NUM1 & NUM2 are stored in internal data RAM locations 20h, 21h & 22H respectively. Write an ALP to compute the following.IF X=0; THEN NUM1 (AND) NUM2, IF X=1; THEN NUM1 (OR) NUM2,IF X=2; THEN NUM1 (XOR) NUM2, ELSE RES =00, RES IS 23H LOCATION

ORG 0000HSJMP 30HORG 30HMOV A, 20h //donot use #, as data ram 20h is to be accessedMOV R1,A //X IN R1MOV A,21H //A -NUM1CJNE R1,#0,CKORANL A, 22HSJMP END1

CKOR:CJNE R1,#01,CKXORORL A, 22HSJMP END1

CKXOR:CJNE R1,#02,OTHERXRL A, 22HSJMP END1

OTHER: CLR AEND1: MOV 23H,A //STORE RESULT



HERE: SJMP HERE END

Algorithm:1. Point to the data RAM register 20h and store the condition x.2. Point to 21h and 22h and move the first number to A register.3. Compare the contents of r1 and perform the operations accordingly.4. The result will be stored in 23H register.

RESULT: 1)Before Execution: D:020H =00, 21=0f, 22 = 12After Execution D:023H = 022)Before Execution: D:020H =01, 21=0f, 22 = 12After Execution D:023H = 1F3)Before Execution: D:020H =02, 21=0f, 22 = 12After Execution D:023H = 1D4)Before Execution: D:020H =34, 21=0f, 22 = 12After Execution D:023H = 00The above program can also be written as shown below (using indirect addressing)

ORG 0000HSJMP 30HORG 30Hmov r0,#20hMOV A,@R0 //ON CHIP DATA RAM-DONOT USE

MOVXMOV R1,A //X IN R1INC R0MOV A,@R0 //A -NUM1INC R0 // R0 POINTS TO NUM2CJNE R1,#0,CKORANL A, @R0SJMP END1

CKOR:CJNE R1,#01,CKXORORL A, @R0SJMP END1

CKXOR:CJNE R1,#02,OTHERXRL A, @R0SJMP END1

OTHER: CLR AEND1:INC R0

MOV @R0,A //STORE RESULT HERE:SJMP HERE END

Boolean variable instructions are also called as bit level logical instructions9) 3 eight bit numbers X, NUM1 & NUM2 are stored in internal data RAM locations 20h, 21h & 22H respectively. Write an ALP to compute the following.IF X=0; THEN LSB OF NUM1 (AND) LSB OF NUM2, IF X=1; THEN MSB OF NUM1 (OR)MSB OF NUM2 ,IF X=2; THEN COMPLEMENT MSB OF NUM1



STORE THE BIT RESULT IN RES, WHERE RES IS MSB OF 23H LOCATION

ORG 00HSJMP 30h ORG 30hMOV R0,20H //R0-XCJNE R0,#0,CK1MOV C,08H //LSB OF NUM1 (21H) - BIT ADDRESS -

08ANL C,10H //LSB OF NUM2 (22H) - BIT ADDRESS -

10SJMP LAST

CK1:CJNE R0,#1,CK2MOV C,0FH //MSB OF NUM1 (21H) - BIT ADDRESS -

0FANL C,17H //MSB OF NUM2 (22H) - BIT ADDRESS -

17SJMP LAST

CK2:CJNE R0,#2,CK3CPL 0FHMOV C,0FH //MSB OF NUM1 (21H) - BIT ADDRESS -

0FSJMP LASTCK3:CLR CLAST:MOV 1FH,C //RES IS MSB OF 23H LOCATION -1FHHERE:SJMP HERE

ENDRESULT: 20h = 00 => AND OF LSBs=1 (hence 80 in 23h location)

20h = 01 => OR of MSBs = 0 (hence 00 in 23h location)

20h = 01 =>complement of MSB of 21h location. Hence 21h is changed to A1 and 23h location has 80hBefore Execution After Execution

Algorithm:1. Move the condition X (from 20h location) into R0 register.



2. If X=0; then move LSB bit of 21h to carry flag and ‘AND’ Carry flag with LSB bit of 22h. Goto step5

3. If X=1; then move MSB bit of 21h to carry flag and ‘OR’ Carry flag with MSB bit of 22h. Goto step5

4. If X=0; then complement MSB bit of 21h and move it to carry flag. Goto step5

5. Store Carry flag at MSB bit of 23h location.

5. COUNTERSASSEMBLY PROGRAM ILLUSTRATING HEX UP/DOWN COUNTERS.

//counter program - hex/binary counters10) Write an ALP to implement (display) an eight bit up/down binary (hex) counters on watch window.Note: to run this program, after selecting DEBUG session in the main menu use

View-> Watch& call Stack window, in the Watches select watch 1(or 2) and

press F2 and enter a (for accumulator A)ORG 0HSJMP 30HORG 0HMOV a,#00

BACK: ACALL DELAYINC a //dec a for binary down counterJNZ BACK

HERE:SJMP HERE

DELAY: MOV r1,#0FFHDECR1:MOV r2,#0FFHDECR: MOV r3,#OFFH

DJNZ r3,$DJNZ r2,DECR

DJNZ r1,DECR1RETEND

RESULT: Accumulator A is incremented in binary from 00, 01,02…09,0A, 0B,…,0F,10,11,…FF



Algorithm:1. Move 00 to A register2. Call the delay subroutine for 1 second, in delay program move FFH to

registers r1, r2 and r3, loop and decrement until 0.3. Increment A register(decremant for down counter)

ASSEMBLY PROGRAM ILLUSTRATING BCD UP/DOWN COUNTERS.//counter program – BCD up/down counters11) Write an ALP to implement (display) an eight bit up/down BCD counters on watch window.

ORG 0H SJMP 30H

ORG 30HMOV a,#00

BACK:ACALL DELAYADD a,#99H //ADD 01 for BCD up counterDA A //for bcd counterJNZ BACK

HERE:SJMP HEREDELAY:MOV r1,#0FFHDECR1:MOV r2,#0FFHDECR:MOV r3, #0FFH

DJNZ r3,$DJNZ r2, DECRDJNZ r1, DECR1RETEND

Algorithm:4. Move 00 to A register5. Call the delay subroutine for 1 second (in delay program move FFH

to registers r1, r2 and r3, loop and decrement until 0).6. Increment A register(add 99h for down counter)7. Decimal adjust accumulator for the BCD up/down counter.

RESULT: Accumulator A is incremented in BCD from 00, 01, 02…09, 10, 11,…99.

6. SERIAL DATA TRANSMISSION

Program illustrating serial ascii data transmission (data-yE)Note-to use result of this program, after selecting DEBUG session in the main menu useView-> serial window #1. On running & halting the program, the data is seen in the serial window.12) Conduct an experiment to configure 8051 microcontroller to transmit characters (yE) to a PC using the serial port and display on the serial window.

ORG 0HSJMP 30H



ORG 30HMOV TMOD,#20H //timer 1; mode 2MOV TH1,#-3 //-3=FD loaded into TH1 for 9600 baud,

11.0592MHz.MOV SCON,#50H //8-bit, 1 stop bit, REN enabledSETB TR1 //Start timer 1

AGAIN:MOV A,#’y’ //transfer “y”ACALL TRANSMOV a,#’E’ //transfer “E”ACALL TRANS

AGAIN1:SJMP AGAIN1TRANS: MOV SBUF,a //load SBUFHERE:JNB TI,HERE //Wait for last bit to transfer

CLR TI //get ready for next byteRETEND

RESULT: yE is printed on the serial window each time the program is executed.

Theory: In serial transmission as opposed to parallel transmission, one bit at a time is transmitted. In serial asynchronous transmission, the data consists of a Start bit (high), followed by 8 bits of data to be transmitted and finally the stop bit. The byte character to be transmitted is written into the SBUF register. It transmits the start bit. The 8-bit character is transferred one bit at a time. The stop bit is transferred. After the transmission, the TI flag = 1 indicating the completion of transmission. Hence in the subroutine wait until TI is set. Later clear the TI flag and continue with transmission of the next byte by writing into the SBUF register. (The program can also be written in interrupt mode). The speed of the serial transmission is set by the baud rate which is done with the help of timer 1. (Refer Ayala). Timer1 must be programmed in mode 2 (that is, 8-bit, auto reload).Baud rate Calculation: Crystal freq/ (12*32) = (11.0592MHz)/(12*32) = 28800.Serial communication circuitry divides the machine cycle frequency(11.0592MHz)/(12) by 32 before it is being used by the timer to set the baud rate.To get 9600, 28800/3 is obtained by loading timer1 with -3 (i.e., FF – 3 = FD) for further clock division. For 2400 baud rate, 28800/12 => -12 = F4 in TH1.Algorithm:

1. Initialize timer 1 to operate in mode 2 by loading TMOD register.2. load TH1 with -3 to obtain 9600 baud.3. Initialize the asynchronous serial communication transmission

(SCON) register.4. Start timer1 to generate the baud rate clock.5. Transmit the characters “y” & “E” by writing into the SBUF register

and waiting for the TI flag.

7) TIMER DELAY PROGRAMProgram illustrating timer delay



13) Generate a 1second delay continuously using the on chip timer in interrupt mode.

ORG 0H //Reset VectorSJMP 30HORG 0BH //TF0 vectorSJMP ISRORG 30HMOV a,#00MOV R0,#0MOV R1,#0MOV TMOD,#02H //00000010-Run timer0 in mode 2MOV TH0,#118 //Set up timer 0 to overflow in 0.05msec MOV IE,#82H //%10000010 – Enable timer0

interruptSETB TCON.4 //Start the timer0

HERE:SJMP HEREISR: CLR TCON.4 //Disable timer0

INC r1 //r1*r2 = 100*200 = 20000 * 0.05msec = 1secCJNE r1,#100,SKIPMOV r1,#00INC r0CJNE r0,#200,SKIPMOV r0,#00HINC a

SKIP: SETB TCON.4 //Enable TimerRETI //Return from interrupt subroutine

END

RESULT: Accumulator A is incremented in binary from 00, 01,02…09,0A, 0B, …, 0F, 10, 11, …FF every 1 second (for 33MHz clock setting & every 3 seconds for 11.0598MHz)

Algorithm:1. Set up timer0 in mode 2 operation2. Load TH1 with 118 to generate an interrupt every 0.05msec.3. Reset registers a, r1 & r0.4. Repeat step 4 continuously5. On interrupt; ISR at 000B loaction goes to step 66. disable timer07. Update r1 & r0 8. Check if 20000 interrupts (=1 sec) over. Yes –increment accumulator a.9. enable timer & return from ISR.



Timerdelay = 12*(257-delay)/frequency Timerdelay=0.05msecDelay=256-((timerdelay * frequency)/12) =256-(0.05*10 -3 * 33*106)/12 =256-137.5 =118.5 //loaded in TH0To get 1sec delay1/0.05msec = 200*100 in the ISR(assuming 33 MHZ crystal frequency. For 11 MHz, the calculations change).



8. CONVERSION PROGRAMS

14) Write an ALP to implement decimal to hex conversion

ORG 0000HSJMP 30h ORG 30hMOV DPTR,#40H //2-digit decimal number to be converted is given

in data memory 40hMOVX A, @DPTRANL A, #0F0H //obtain upper decimal digitSWAP A //bring to the units placeMOV B,#0AH //MULTIPLY tens digit with #0A-toget tens

in hexMUL ABMOV r1,a //temporarily store the converted tens

valueMOVX A,@DPTR //get the decimal number againANL A,#0FH //obtain the units digitADD A,R1 //add to the converted tens valueINC DPTR //increment data addressMOVX @DPTR,A //converted hexadecimal number in next

location HERE:SJMP HERE

END

RESULT: before execution- X:0040H = 45 (Decimal/BCD) After Execution: X:0041h = 2D (hex value)

Algorithm 1. Move the decimal data to be converted from external memory 40h to

accumulator.2. AND A reg with 0f0h and obtain the upper MSB of the decimal digit

and swap the LSB and MSB of accumulator to bring the same to units place.

3. Move 0ah to B register and multiply with A reg to convert to hex value, store the converted tens value in r1

4. Get the LSB of the decimal number and add to the converted tens value

5. point to the next memory location and store the result (hexadecimal).

15) Write an ALP to implement hex to decimal conversionORG 0000HSJMP 30h ORG 30hMOV DPTR,#9000HMOVX A,@DPTR //Get hex numberMOV B,#10DIV AB //divide by 10 (0AH)



INC DPTRXCH A,BMOVX @DPTR,A //Store the remainder (in B) In units placeXCH A,BMOV B,#10 //Divide the quotient in A by 10DIV ABINC DPTRXCH A,BMOVX @DPTR,A //Store the remainder (in B) In tens placeXCH A,BINC DPTRMOVX @DPTR,A //Store the quotient (in A) in hundreds place

HERE:SJMP HEREEnd

RESULT: 9000H – FF (HEX NUMBER)9001 to 9003 – unpacked BCD number (decimal)- 5,5,2 (i.e., 255 stored Lower digit first)

Algorithm 1. Move the hex data to be converted to accumulator.2. Move 10 to B register and divide with A reg to convert to ascii value3. Store the converted LSB value in r74. Repeat the step 2 to obtain the converted MSB value 5. Store the same in r6

16) Write an ALP to implement BCD to ASCII conversionORG 0000HSJMP 30h ORG 30hMOV R1,#50HMOV A,@R1 //get BCD data byte from RAM location 50hMOV R2,A //Store in R2ANL A,#0FH //Get the lower nibbleORL A,#30H //Add/or with 30h i.e., 0-9 converted to 30-39hINC R1MOV @R1,A //Store the lower digit's ASCII codeMOV A,R2 //Get back the numberSWAP A //Swap nibbles in AANL A,#0FH //Get the upper BCD digitORL A,#30H //Convert to ASCIIINC R1MOV @R1,A //Store the upper digit's ASCII code

here: sjmp hereEND



RESULT: The BCD code 28 at D:0050h is converted to 2 ASCII codes-38h 32h

Algorithm ://Converts the BCD byte in A into two ASCII characters.

1. Move the BCD data to be converted to accumulator.2. Get the lower nibble(BCD digit) & ADD (or ORL) with 30h3. Store the converted ASCII value 4. Get the higher nibble(tens BCD digit) & ADD (or ORL) with 30h5. Store the converted ASCII value

17) Write an ALP to implement hexadecimal to ASCII conversion//This program also illustrates conditional branching (JNC), call and return instructions.

ORG 0000HSJMP 30h ORG 30hMOV R1,#50HMOV A,@R1 //get hexadecimal data byte from RAM location 50hMOV R2,A //Store in R2ANL A,#0FH //Get the lower nibbleACALL ASCII //Convert to ASCIIINC R1MOV @R1,A //Store the lower digit's ASCII codeMOV A,R2 //Get back the numberSWAP A //Swap nibbles in AANL A,#0FH //Get the upper BCD digitACALL ASCIIINC R1MOV @R1,A //Store the upper digit's ASCII code

here: sjmp here

ASCII:MOV R4,A //Store aCLR CSUBB A,#0AH //Check if digit >=0AMOV A,R4JNC SKIPADD A,#07H //Add 07 if >09SKIP:ADD A,#30H //Else add only 30h for 0-9RETEND

RESULT: The BCD code 2C at D:0050h is converted to 2 ASCII codes-43h(for 0B) & 32h (for 02) Another Example-BA



Algorithm ://Converts the hexadecimal byte in A into two ASCII characters.

1. Move the hexadecimal data to be converted to accumulator.2. Get the lower nibble & call ASCII routine3. Store the converted ASCII value 4. Get the higher nibble & call ASCII routine5. Store the converted ASCII value

ASCII subroutine1. If digit greater than 09,(for A-F) add 07h & 30h2. Else (i.e., for 0-9) add only 30h3. return

18) Write an ALP to implement ASCII to hexadecimal conversionORG 0000HSJMP 30h ORG 30hMOV R1,#50HMOV A,@R1 //get ascii byte from RAM location 50hCLR CSUBB A,#41HMOV A,@R1JC SKIPCLR CSUBB A,#07HSKIP:CLR CSUBB A,#30HINC R1MOV @R1,A //Store the hex code

here: sjmp hereEND

RESULT: The ASCII code 45 at D:0050h is converted to hexadecimal -0E at 51h

Note: For this program the input data should be only in the range 30h-39h & 41h to 46h.



Algorithm ://Converts the ASCII characters into hexadecimal number.

1. Move the ASCII character to be converted to accumulator.2. If character is greater than 41h,(for A-F), then subtract 07h &

30h3. Else (i.e., for 0-9) subtract only 30h4. Store the converted hexadecimal number.

Hardware Interfacing

1.Waveform Generation using Dual DAC2.Stepper Motor interface.3.4X4 hexadecimal Keyboard interface4.DC motor interface5.Calculator using Keyboard and Seven segment display.6.Elevator control.7.Temperature sensor.

Features of Embedded C • C is a simple programming language and so very easy to code.• Embedded C has most features of C-language with more stress on certain bit

manipulative instructions.• This feature makes it easy to write program for μC and μP.• Keil is a versatile software with a cross compiler that will convert the C

program to assembly language and thus the program can be executed on the desired target (say 8051).

Some of the bit manipulative instructions used are Symbol Operation & Bitwise AND | Bitwise OR ~ Bitwise NOT >> Shift Right << Shift Left ^ P0.0

1.Dual Dac Interface to generate

a.Square waveform b.Triangular Waveform



c.Ramp waveform d.Sine waveform

1.aAlgorithm for Square wave generation

Let initial, amplitude of the square wave be 2.5v(7F) and frequency count 100.

Output the values 00h(0ff) and 7fh(on) Values through P0. If amplitude key is pressed then increase the voltage in steps of 0.15v(8). If the frequency key is pressed increment the count in steps of 50. If the

count exceeds 1000 reset it back to 100. Every time amplitude and frequency changes output the value thro P0 and

note the waveform on CRO.

Program for square wave

#include <REG51xD2.H>sbit Amp = P3^3; /* Port line to change amplitude */sbit Fre = P3^2; /* Port line to change frequency */void delay(unsigned int x) /* delay routine */{ for(;x>0;x--);}main(){ unsigned char on = 0x7f,off=0x00; unsigned int fre = 100;

while(1) { if(!Amp) /* if user choice is to change amplitude */ { while(!Amp); /* wait for key release */

on+=0x08; /* Increase the amplitude */ }


8 0

5 P0 1 P1

μC

DualDAC

CRO

Xout


if(!Fre) /* if user choice is to change frequency */ { if(fre > 1000) /* if frequency exceeds 1000 reset to default */ fre = 100; while(!Fre); /* wait for key release */ fre += 50; } /* Increase the frequency */ P0=on; /* write amplitude to port */ P1=on; delay(fre); P0 = off; /* clear port */ P1 = off; delay(fre); }}

1.bAlgorithm for Triangular wave generation

Output the initial value 00 through P0. Increment it in steps of 1 until a count value of FFh (5V) is reached. Every

time repeat step 1. Decrement it in steps of 1 until a zero value is reached and repeat step 1.

Program for triangular wave:

#include <REG51xD2.H>main(){ unsigned char i=0; P0 = 0x00; /* P0 as Output port */while(1) { for(i=0;i<0xff;i++){ /* Generate ON pulse */ P1 = i; P0 = i; } for(i=0xfe;i>0x00;i--) /* Generate OFF pulse */

{P0 = i; P1 = i;}

}}

1.c.Algorithm for Ramp wave generation

Output the initial value 00 through P0. Increment it in steps of 1 until a count value of FFh (5V) is reached.

Every time repeat step 1. Repeat step 1 & 2 continuously.

Program for Ramp waveform

#include <REG51xD2.H>



main (){ Unsigned char i=0; P0 = 0x00; /* P0 as Output port */while (1) { for (i=0;i<0xff;i++) /* Generate ON pulse */ { P1 = i; P0 = i; } }}

1d.Algorothm for Sine wave

Compute different step values (θ = 20o,15o…) of sine using the equation V= 2.5V +2.5Vsinθ. . Output the values thro P0.

More the steps smoother will be sine wave. E.g.: θ = 0o

V= 2.5V +2.5Vsinθ = 2.5VThe value sent to DAC is 25.6X5V= 128.

Program for sine wave

#include <REG51xD2.H> main(){ static int a[13]={128,192,238,255,238,192,128,64,17,0,17,64,128}; unsigned char i=0; P0 = 0x00; /* P0 as Output port */ while (1) { for(i=0;i<13;i++) /* Output different values */ { P0 = a[i]; } }}

2.Stepper Motor

• Stepper motor unlike DC motor rotates in steps.• Stepper motor has 4 coils which forms the stator and a central rotor.• Rotation depends on excitation of stator coils.

step coil A coil B coil C coil D 1 0 0 0 1 2 1 0 0 0 3 0 1 0 0 4 0 0 0 1Anyone of these values forms the initial value. To get 360o revolution 200 steps are required.



Step angle= 360o /200 = 1.8o. (difference between 2 teeth).

Algorithm for Stepper Motor

• Configure P0 as output.• Apply the initial excitation of 11 to motor coils through P0.• For clockwise motion -Rotate right once the excitation and repeat step 2.• For anticlockwise motion -Rotate left once the excitation and repeat step 2.

//Program for stepper motor interface

#include <REG51xD2.H>void delay (unsigned int x) /* Delay Routine */{ for(;x>0;x--); return;}Main ( ){ unsigned char Val, i; P0=0x00; Val = 0x11; for (i=0;i<4;i++)

{ P0 = Val; Val = Val<<1; /* Val= Val>>1; for clockwise direction*/ delay (500); }

}

3. 4X4 HEX Keyboard

Algorithm for Keyboard Interface

• Configure P1 as output port to scan the rows and P0 as input port to read the column values.


8051µC

P0FRC 26pin

Cable

PS

Stepper MotorInterface Card

PS

Stepper Motor


• First select the last row by grounding that row. Scan the columns of entire row if a key is pressed in that row then one of the column reads ‘0’.

• If now key is pressed in the selected row all 1’s is returned. So scan next row. Repeat the action until all rows are scanned.

//Program for 4X4 hex keypad.

#include < REG51xD2.H>#include <intrins.h>#include "lcd.h"unsigned char rows,columns,result,abhi;unsigned char temp = 0;void delay(){ unsigned int i; for(i = 0; i <= 20000; i ++);}void Display(){ if(result > 0x09) { result += 0x37;

WriteChar(result); }else { result += 0x30;

WriteChar(result); }} void KeyScan(){ again: columns = 0x77; rows = 0x04; result = 0x0c; next: P1 = columns; columns >>=1;


8051µC

P0FRC 26pin

Cable

PS

4X4 KeyboardInterface Card

PS

4X4 Hex keypad


if(CY) columns = columns |0x08 ; temp = P0; temp = (temp & 0x0f); if(temp != 0x0f) { rot: temp >>= 1;

if(!CY){ ClrLcd(); return;}else{ result += 1; goto rot; }

}

else { result -= 0x04; rows --; if(rows == 0) goto again; else { goto next; } }}void main(){ P0 = 0xff; P1 = 0x00; InitLcd(); WriteString ("KEY PRESSED="); while(1) { KeyScan();

WriteString ("KEY PRESSED=");Display(); }

}

4.DC Motor

Algorithm for DC motor interface



• Configure P0,P1 as output port and P3 as input port.• Let initially the motor rotate with half speed count 7fh.• If “INR” button is pressed reduce the count because the speed is inversely

proportional to count.• If “DEC” button is pressed increase the count.

Program for DC motor

#include <REG51xD2.H> sbit inr= P3^2; //speed increment switch sbit dcr= P3^3; //speed decrement switch main() { unsigned char i=0x80; P0 = 0x7f; /*Run the motor at half speed.*/while (1) { if (!inr) {while (!inr);if(i>10) i=i-10; //increase the DC motor speed } if(!dcr) { while(!dcr); if(i<0xf0) i=i+10; //decrease the DC motor speed } P0=i; } }

5.Calculator using Keyboard and 7-segment displayAlgorithm


8051µC

P0FRC 26pin

CableP3.2(inr)P3.3(dec)

PS

DC MotorInterface Card

PS

DC Motor


• Read the numbers n1 and n2 from keyboard and display them on seven segment.

• Read the operand from the keypad if key pressed is B (+), C(-),D(*),E(/) then respective operation is performed.

• Result is displayed on 2 digit seven segment display.• If any time the key pressed value returned as 10h then clear the LCD.

Program for calculator

#include <REG51xD2.H>void DispChar(unsigned char ch);void ClrLED();unsigned char getkey();unsigned char getnum();unsigned char getOp();sbit Clk = P3^4; /* Clock line for 7 segment display */sbit Dat = P0^0; /* Data line for 7 segment display */main(){ unsigned char tmp=0x0ff,n1=0,n2,Op,Res; unsigned char NumTab[10] = { 0x0c0,0x0f9,0x0a4,0xb0,0x99,0x92,0x82,0x0f8,0x80,0x90 }; unsigned char OpTab[4] = { 0x88,0x0Bf,0xc8,0x0a1}; bit Neg=0; ClrLED(); /* Clear 7 segment display */ while(1) { Neg = 0; /* Negative flag */

n1=getnum(); /* Get 1st number */

Op = getOp() - 0x0B; /* Get Opcode. 0x0b is keycode of '+'(see keyboard schematics)*/

n2=getnum(); /* Get 2nd number */


8051µC

P0FRC 26pin

Cable

PS

Keypad

PS

7 SegDisplay


while(getkey()!=0x13); /* wait for '=' key */ClrLED();

switch(Op) /* Perform corresponding operation */{ case 0: Res = n1 + n2; break; case 1: if(n2>n1) /* check for negativity */ {

Neg = 1; Res = n2 - n1; break; }

Res = n1 - n2; break; case 2: Res = n1 * n2; break;

case 3: Res = n1 / n2; break; }DispChar(NumTab[Res%10]); /* Display number */DispChar(NumTab[Res/10]);if(Neg) /* if negative result display '-' */ DispChar(0x0Bf);

}}void DispChar(unsigned char ch) /* Routine to display char on 7 segment */{ unsigned char i,tmp; P0=0x00;for(i=0;i<8;i++) /* for all bits */ { tmp = ch & 0x80;

if(tmp) /* write data depending on MSB */ Dat = 1; else Dat = 0;Clk = 0; /* Give Clk Pulse for synchronization */Clk = 1;ch = ch << 1; /* Get next bit */

}}void ClrLED(){ unsigned char i; for(i=0;i<4;i++) DispChar(0x0ff); /* 0xff for clear segment ( see 7 segment manual for more info) */}unsigned char getkey() { unsigned char i,j,indx,t;



P2 = 0x00; /* P2 as Output port */ P0 = 0x0ff; indx = 0x00; /* Index for storing the first value of scanline */ while(1) { for(i=1;i<=4;i<<=1) /* for 4 scanlines */ { P2 = 0x0f & ~i; /* write data to scanline */

t = P0; /* Read readlines connected to P0*/ t = ~t;

if(t>0) /* If key press is true */ { for(j=0;j<=7;j++) /* Check for 8 lines */ { t >>=1; if(t==0) /* if get pressed key*/ { return(indx+j); /* Return index of the key pressed */

} } } indx += 8; /* If no key pressed increment index */

} }}unsigned char getnum() /* Method for getting number */{ unsigned char tmp; while(1) { tmp = getkey(); if(tmp < 0x0a || tmp==0x10) /* if pressed key is number, return */ return(tmp); }}unsigned char getOp() /* Method for getting Operator */{ unsigned char tmp; while(1) { tmp = getkey(); if((tmp > 0x0a && tmp <0x0f)|| tmp==0x10) /* if pressed key is a Operator, return */ return(tmp); } }

6.Elevator

Algorithm for elevator interface

• Read the floor request through input port P1.



• If the current floor and requested floor are the same no change light up the corresponding LED through P0.

• If the requested floor greaterthan current moving up of the lift is indicated by glowing of LED’s from current floor to the requested.

• If the requested floor lesserthan current moving down of the lift is indicated by glowing of LED’s from current floor to the requested.

Program for Elevator

#include <REG51F.H>void delay(unsigned int);main(){ unsigned char Flr[9] = {0xff,0x00,0x03,0xff,0x06,0xff,0xff,0xff,0x09}; unsigned char FClr[9] = {0xff,0x0E0,0x0D3,0xff,0x0B6,0xff,0xff,0xff,0x79}; unsigned char ReqFlr,CurFlr = 0x01,i,j; P0 = 0x00; P0 = 0x0f0;while(1) { P1 = 0x0f; ReqFlr = P1 | 0x0f0; while(ReqFlr == 0x0ff)

ReqFlr = P1 | 0x0f0; /* Read Request Floor from P1 */ReqFlr = ~ReqFlr; if(CurFlr == ReqFlr) /* If Request floor is equal to Current

Floor */{

P0 = FClr[CurFlr]; /* Clear Floor Indicator */ continue; } /* Go up to read again */else if(CurFlr > ReqFlr) /* If Current floor is > request floor */{ i = Flr[CurFlr] - Flr[ReqFlr]; /* Get the no of floors to travel

*/ j = Flr[CurFlr]; for(;i>0;i--) /*Move the indicator down */ {


8051µC

P0FRC 26pin

Cable

PS

Elevator interface

PS


delay(25000); }} else /* If Current floor is < request floor */{ i = Flr[ReqFlr] - Flr[CurFlr]; /* Get the no of floors to travel */ j = Flr[CurFlr]; for(;i>0;i--) /* Move the indicator Up */ { P0 = 0x0f0 | j;

j++; delay(25000); } }

CurFlr = ReqFlr; /* Update Current floor */P0 = FClr[CurFlr]; /* Clear the indicator */

}}void delay(unsigned int x){ for(;x>0;x--);}

7.Temperature Sensor

The interface card has a DAC to convert the actual temperature to digital this is compared with reference temperarture. Realay also a part of interface card will turn on and off to indicate if the actual temperature is above or below reference.

Algorithm for Temperature sensor

1. Configure P0 and P1 as o/p, P3 as input port.2. Set up a counter with intial value 0xff send it to dac thro P0 after a delay check if comparator o/p has gone low. 3. If low compare with set value if actual greaterthan set turn on the relay else turn off.


8051µC

P0,P2,P3FRC 26pin

Cable

PS

Temp Sensor Interface

PS

HeatSource


Program for temperature sensor.

#include <REG51xD2.H>sbit Cmp_Out = P3^4; /*Input Bit for Comparator output*/sbit Rel_Con = P0^0; /*Relay controller Bit i.e Heater Power supply control*/

/*1- Supply OFF, 0-Supply ON*/#define Dac_Data P1 /*DAC input Data PORT i.e. P1*/ void delay() { int l; for (l=0;l<=0x8;l++); }main(){ unsigned char DacIp; void delay(void); Dac_Data =0x00; /*Move 00h to Dac input*/ P0=0x00; /*make P0 as output*/ while(1) { DacIp= 0xff; /*DAC input Data counter*/ do {

DacIp++; /*Increment the DAC input Data*/Dac_Data = DacIp; /*Move the DAC data to DAC*/delay();

}while(Cmp_Out); /* Check comparator output for low */ if(DacIp > 0x20) /*Compare with the set value i.e.0x20*/ Rel_Con = 1;

else Rel_Con = 0; /* Relay ON, Supply OFF */

}}


8051 Manual Fin

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