~: Simple Programs in 8051 assembly language :~
Here some simple programs of 8051 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
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
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
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
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
Agin: Mov p0, #0ffh ; initialize port P0 as input port
Loop: Mova, 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 agin ; 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
