Simple Programs in 8051 Assembly Language

Here some simple assembly language programs for 8051 microcontroller are given to understand the operation of different instructions and to understand the logic behind particular program. First the statement of the program that describes what should be done is given. Then the solution is given which describes the logic how it will be done and last the code is given with necessary comments.

Statement 1: – exchange the content of FFh and FF00h

Solution: – here one is internal memory location and other is memory external location. so first the content of ext memory location FF00h is loaded in acc. then the content of int memory location FFh is saved first and then content of acc is transferred to FFh. now saved content of FFh is loaded in acc and then it is transferred to FF00h.

Mov dptr, #0FF00h ; take the address in dptr

Movx a, @dptr ; get the content of 0050h in a

Mov r0, 0FFh ; save the content of 50h in r0

Mov 0FFh, a ; move a to 50h

Mov a, r0 ; get content of 50h in a

Movx @dptr, a ; move it to 0050h

Duplication and Subtraction

Statement 2: – store the higher nibble of r7 in to both nibbles of r6

Solution: –first we shall get the upper nibble of r7 in r6. Then we swap nibbles of r7 and make OR operation with r6 so the upper and lower nibbles are duplicated

Mov a, r7 ; get the content in acc

Anl a, #0F0h ; mask lower bit

Mov r6, a ; send it to r6

Swap a ; xchange upper and lower nibbles of acc

Orl a, r6 ; OR operation

Mov r6, a ; finally load content in r6

Statement 3: – treat r6-r7 and r4-r5 as two 16 bit registers. Perform subtraction between them. Store the result in 20h (lower byte) and 21h (higher byte).

Solution: – first we shall clear the carry. Then subtract the lower bytes afterward then subtract higher bytes.

Clr c ; clear carry

Mov a, r4 ; get first lower byte

Subb a, r6 ; subtract it with other

Mov 20h, a ; store the result

Mov a, r5 ; get the first higher byte

Subb a, r7 ; subtract from other

Mov 21h, a ; store the higher byte

Division & Data Transfer

Statement 4: – divide the content of r0 by r1. Store the result in r2 (answer) and r3 (reminder). Then restore the original content of r0.

Solution:-after getting answer to restore original content we have to multiply answer with divider and then add reminder in that.

Mov a, r0 ; get the content of r0 and r1

Mov b, r1 ; in register A and B

Div ab ; divide A by B

Mov r2, a ; store result in r2

Mov r3, b ; and reminder in r3

Mov b, r1 ; again get content of r1 in B

Mul ab ; multiply it by answer

Add a, r3 ; add reminder in new answer

Mov r0, a ; finally restore the content of r0

Statement 5: – transfer the block of data from 20h to 30h to external location 1020h to 1030h.

Solution: – here we have to transfer 10 data bytes from internal to external RAM. So first, we need one counter. Then we need two pointers one for source second for destination.

Mov r7, #0Ah ; initialize counter by 10d

Mov r0, #20h ; get initial source location

Mov dptr, #1020h ; get initial destination location

Nxt: Mov a, @r0 ; get first content in acc

Movx @dptr, a ; move it to external location

Inc r0 ; increment source location

Inc dptr ; increase destination location

Djnz r7, nxt ; decrease r7. if zero then over otherwise move next

Various Comparison Programs

Statement 6: – find out how many equal bytes between two memory blocks 10h to 20h and 20h to 30h.

Solution: – here we shall compare each byte one by one from both blocks. Increase the count every time when equal bytes are found

Mov r7, #0Ah ; initialize counter by 10d

Mov r0, #10h ; get initial location of block1

Mov r1, #20h ; get initial location of block2

Mov r6, #00h ; equal byte counter. Starts from zero

Nxt: Mov a, @r0 ; get content of block 1 in acc

Mov b, a ; move it to B

Mov a, @r1 ; get content of block 2 in acc

Cjne a, b, nomatch ; compare both if equal

Inc r6 ; increment the counter

Nomatch: inc r0 ; otherwise go for second number

Inc r1

djnz r7, nxt ; decrease r7. if zero then over otherwise move next

Statement 7: – given block of 100h to 200h. Find out how many bytes from this block are greater then the number in r2 and less then number in r3. Store the count in r4.

Solution: – in this program, we shall take each byte one by one from given block. Now here two limits are given higher limit in r3 and lower limit in r2. So we check first higher limit and then lower limit if the byte is in between these limits then count will be incremented.

Mov dptr, #0100h ; get initial location

Mov r7, #0FFh ; counter

Mov r4, #00h ; number counter

Mov 20h, r2 ; get the upper and lower limits in

Mov 21h, r3 ; 20h and 21h

Nxt: Movx a, @dptr ; get the content in acc

Cjne a, 21h, lower ; check the upper limit first

Sjmp out ; if number is larger

Lower: jnc out ; jump out

Cjne a, 20h, limit ; check lower limit

Sjmp out ; if number is lower

Limit: jc out ; jump out

Inc r4 ; if number within limit increment count

Out: inc dptr ; get next location

Djnz r7, nxt ; repeat until block completes

Interrupt Counting, Subroutines & Scan n Mulitply

Statement 8:- the crystal frequency is given as 12 MHz. Make a subroutine that will generate delay of exact 1 ms. Use this delay to generate square wave of 50 Hz on pin P2.0

Solution: – 50 Hz means 20 ms. And because of square wave 10 ms ontime and 10 ms offtime. So for 10 ms we shall send 1 to port pin and for another 10 ms send 0 in continuous loop.

<_x0021_xml:namespace prefix=”st1″ ns=”urn:schemas-microsoft-com:office:smarttags”/>Loop: Setb p2.0 ; send 1 to port pin

Mov r6, #0Ah ; load 10d in r6

Acall delay ; call 1 ms delay ×10 = 10 ms

Clr p2.0 ; send 0 to port pin

Mov r6, #0Ah ; load 10d in r6

Acall delay ; call 1 ms delay ×10 = 10 ms

Sjmp loop ; continuous loop

Delay: ; load count 250d

Lp2: Mov r7, #0FAh

Lp1: Nop ; 1 cycle

Nop ; 1+1=2 cycles

Djnz r7, lp1 ; 1+1+2 = 4 cycles

Djnz r6, lp2 ; 4×250 = 1000 cycles = 1000 µs = 1 ms

ret

Statement 9:-count number of interrupts arriving on external interrupt pin INT1. Stop whencounter overflows and disable the interrupt. Give the indication on pinP0.0

Solution: –as we know whenever interrupt occurs the PC jumps to one particular location where it’s ISR is written. So we have to just write one ISR that will do the job

Movr2, #00h ; initialize the counter

Movie, #84h ; enable external interrupt 1

Here: Sjmp here ; continuous loop

Org 0013h ; interrupt 1location

Incr2 ; increment the count

Cjner2, #00h, out ; check whether it overflows

Movie, #00h ; if yes then disable interrupt

Clr p0.0 ; and give indication

Out : reti ; otherwise keep counting

Statement 10: –continuously scan port P0. If data is other then FFh write a subroutine that will multiply it with 10d and send it to port P1

Solution: –here we have to use polling method. We shall continuously pole port P0 if there is any data other then FFh. If there is data we shall call subroutine

Again: Mov p0, #0ffh ; initialize port P0 as input port

Loop: Mov a, p0 ; get the data in acc

Cjne a, #0FFh, dat ; compare it with FFh

Sjmp loop ; if same keep looping

Dat: acall multi; if different call subroutine

Sjmp again ; again start polling

Multi

Mov b,#10d ; load 10d in register B

Mul ab ; multiply it with received data

Mov p1, a ; send the result to P1

Ret ;return to main program