8051-mazidi-solution

1. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
Chapter 2
Section 2.1:-
1. 8 bit
2. 8 bit
3. 8 bit
4. PSW (Program Status Word) is of 16 bit.
5. Necessary (for literal value).
6. 28H and it is kept in accumulator.
7. (a),(d),(g) are illegal and for f only 0 is required before F5H
8. (c),(d) are illegal.
9. 44H and kept in Accumulator (A).
10.1EH and kept in Accumulator (A).
Section 2.4
21.0000H
22.Program counter will look for the location 0000H and if the program is not
starting for that address, it will consider that there is no program written so
program has to start from the location 0000H.
26.Lowest Memory is 0000H and the Highest memory is FFFFH.

2. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
Section 2.5:-
29.Solved below,
(Data) (Locations)
E 200
A 201
R 202
T 203
H 204
9 205
8 206
7 207
- 208
6 209
5 20A
Section 2.6:-
31.8 bit
32.And 33.
D7 D6D5
D4 D3 D2 D1 D0
CY AC F0RS1 RS0 OV -- P
34.We know that in 8051 registers are of 8 bits CY Flag is raised when the carry
is generated beyond past the max value that a register can store like FFH +
1.
35.AC is raised when a carry is generated from D3 to D4. Like
Mov a,#0FH
Add a,#1

3. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
36.CLR C ;CY=0
CPL C; CY
37.G
a. CY=1
b. CY=0
c. CY=0
38.ORG 0000H
MOV A,#55H
ADD A,#55H
ADD A,#55H
ADD A,#55H
ADD A,#55H
END
39.RS0 and RS1.
40.On Startup Stack Location is 07H.
42.24 Bytes.
43.Register Bank 0
44.Register Bank 0 from 00H to 07H.
Register Bank 1 from 08H to 0FH.
Register Bank 2 from 10H to 17H.
Register Bank 3 from 18H to 1FH.
45.(a) 04H (b) 00H (c) 07H (d) 05H

4. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
48.
INSTRUCTIONS STACK Pointer
before
execution.
Stack pointer
after
execution.
STACK
PUSH 0 07H 08H 66H
PUSH3 08H 09H 7FH
PUSH 7 09H 0AH 5DH
POP 3 0AH 09H
POP 7 09H 08H
POP 0 08H 07H
49.NO.
POP 7
POP 3
POP 0
REST OF THE PROGRAM WILL REMAIN THE SAME.
50.After execution of ( Mov SP,#70H ), Stack Pointer location has now become
70H instead of 07H.
INSTRUCTIONS STACK Pointer
before
execution.
Stack pointer
after execution.
STACK
Push 5 70H 71H 66H
Push 2 71H 72H 7FH
Push 7 72H 73H 5DH
Pop 7 73H 72H
Pop 2 72H 71H
Pop 5 71H 70H

5. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
Chapter 3:-
Section 3.1:-
12.MOV R1,#100
HERE1:MOV R2,#10
HERE:DJNZ R2,HERE
DJNZ R1,HERE1
13.MOV R1,#100
HERE2MOV R2,#100
HERE1:MOV R3,#10
HERE:DJNZ R3,HERE
DJNZ R2,HERE1
DJNZ R1,HERE2
14.Multiplication is taking place so,
200*100=20,000 (times).
15. -128 Bytes
16. 127 Bytes
Section 3.2:-
17. 3 byte
18. 2 byte
19. 2 bytes
20. 64Kb
21. 2 bytes
22. 1 byte

6. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
23.That’s because Stack works on the concept of LIFO so if the push is implied
2 times for instance, then pop must be used 2 times exactly.
Section 3.3:-
27. T = 1.2usec
F = 1/T = 833.333KHz
System frequency = 833.333KHz*12 = 10MHz
28. F = 18MHz
F =18MHZ/12 =1.5MHz
T = 1/F =1/1.5MHz
T = 0.666usec
29. F = 12MHz
F =12MHZ/12 =1MHz
T = 1/F =1/1.5MHz
T = 1usec
30. F = 25MHz
F =25MHZ/12 = 2.08MHz
T = 1/F =1/2.08MHz
T = 0.48usec
32. F = 11.0592MHz
F =11.0592MHZ/12 = 921.6KHz
T = 1/F = 1/921.6KHZ
T = 1.085usec

7. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
DELAY:MOV R3,#150 1 machine cycle
HERE:NOP 1 mc
NOP 1
NOP 1
DJNZ R3,HERE 2 mc
RET 2
The time delay of the HERE loop is [150(2+1+1+1)]*1.085usec=0.813msec
Now for the instruction outside the loop (mov and ret),
(2+1)*1.085Usec = 3.25usec
Now 0.813ms + 3.25usec = 0.8162msec
33. F = 16MHz
F =16MHZ/12 = 1.33333MHz
T = 1/F = 1/1.33333MHZ
T = 0.75usec
DELAY:MOV R3,#200 1
HERE:NOP 1
NOP 1
NOP 1
DJNZ R3,HERE 2
RET 2
The time delay of the HERE loop is [200(2+1+1+1)]*0.75usec=0.75msec
Now for the instruction outside the loop (mov and ret),
(2+1)*0.75usec = 2.25usec
Now 0.75ms + 2.25usec = 0.75225msec
34. F = 11.0592MHz
F =11.0592MHZ/12 = 921.6KHz
T = 1/F = 1/921.6KHZ
T = 1.085usec

8. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
DELAY:MOV R5,#100 1
BACK: MOV R2,#200 1
AGAIN:MOV R3,#250 1
HERE:NOP 1
NOP 1
DJNZ R3,HERE 2
DJNZ R3,AGAIN 2
DJNZ R3,BACK 2
RET 2
The time delay of the HERE loop is [250(2+1)]*1.085usec=1.085msec
The time delay of the AGAIN loop it repeats 200 times so ,
1.085msec*200 = 0.217 + (3*200*1.085usec) = 0.2176s
The time delay of the BACK loop, it is repeated 100 times so
0.2176*100=21.76sec
Time Delay = 21.76sec
35.Try it yourself it’s just like 34 with only 2 loops instead of 3 loope.
36. To 39.
For the problems from 36 to 39 everything remains the same except the
time delay that is changed due to the change of Microcontroller now you
must take Clock of DS89C420/30 Microcontroller i.e equal to 1 rather than
12 that was for 8051.
40.Yes it is 12 times faster because it only have 1 clock and on the other hand
8051 have 12 clocks so if we decrease the clocks our microcontroller
becomes faster.

9. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
Chapter 4:-
Section 4.1:-
1. 40
2. VCC
40
TH
PIN And GND9
TH
PIN
3. 32 Pins
4. 8 Pins and from 32 to 39.
5. 8 Pins and from 1 to 8.
6. 8 Pins and from 21 to 28.
7. 8 Pins and from 10 to 17.
8. Input
9. P0 (Port 0)
10.P1 (Port 1)
11.ORG 000H
MOV P1,#0FFH; MAKE IT AN INPUT PORT
MOV A,P1
MOV P2,A
MOV P0,A
MOV P3,A
END
12.ORG 000H
MOV P2,#0FFH; MAKE IT AN INPUT PORT
MOV A,P2
MOV P1,A
MOV P0,A
END

10. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
13.P3.0 AND P3.1
14.0000H is the address upon reset.
15.
(A) ORG 0000H
BACK:MOV A,#0AAH
MOV P1,A
MOV P2,A
CALL DELAY
MOV A,#55H
MOV P1,A
MOV P2,A
SJMP BACK
(B) ORG 0000H
MOV A,#0AAH
BACK:MOV P1,A
MOV P2,A
CALL DELAY
CPL A
MOV P1,A
MOV P2,A
SJMP BACK
Section 4.2:-
16.All ports are bit addressable.
17.The advantages for Bit addressable mode is that u con manipulate aa single
bit without disturbing and other bits of the port by using Setb and Clr.
18.Setb P1.X OrClr P1.X where X can vary from 0 to 7.

11. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
19.No , a whole port cannot be complemented at a time.
20.ORG 0000H
SETB P1.2
SETB P1.5
BACK:CALL DELAY
CPL P1.2
CPL P1.5
SJMP BACK
END
21.ORG 0000H
SETB P2.5
SETB P1.7
SETB P1.3
BACK:CALL DELAY
CPL P1.3
CPL P1.7
CPL P2.5
SJMP BACK
END
22.ORG 0000H
SETB P1.3
BACK:JB P1.3,HERE
SJMP BACK
HERE:MOV A,#55H

12. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
MOV P2,A
END
23.ORG 0000H
SETB P2.7
BACK:JNB P2.7,HERE
SJMP BACK
HERE:MOVA,#55H
MOV P0,A
CALL DELAY
MOC A,#0AAH
MOV P0,A
SJMP HERE
END
24.ORG 0000H
SETB P2.0
JNB P2.0,HERE
MOV A,#99H
MOV PI,A
SJMP BACK
HERE:MOV A,#66H
MOV P1,A
BACK:
END

13. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
25.ORG 0000H
SETB P1.5
AGAIN:JB P1.5,HERE
SJMP AGAIN
HERE:CLR PI.3
CALL DELAY
SETB P1.3
CALL DELAY
CLR P1.3
END
26.ORG 0000H
BACK:MOV C,P1.3 ;C IS FOR CARRY FLAG.
MOV P1,4,C
SJMP BACK
END
27. 5
TH
Bit
28.ORG 0000H
BACK:MOV C,P1.7
MOV P1.0,C
MOV C,P1.6
MOV P1.7,C
SJMP BACK

14. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
Chapter 5:-
Section 5.1 and 5,2:-
3. See on page 123 Figure 5-1, and on page 124 figure 5-2.
4. Register bank 1 , 2 , 3 share the space with stack because by default stack
starts from 07H and after increment data is stored in 08H on the other
hand Register bank 1 address starts from 08H as well see page 123 fig 5-1.
6. It copies the contents of the location 0F0H into the accumulator rather than
the value 0F0H.
7. Same as question nos 6.
8. ORG 0000H
MOV R0,#50H
MOV R1,#40H
MOV R3,#30H
PUSH 00H
PUSH 01H
PUSH 03H
POP 1DH
POP 1EH
POP 1FH
END
9. Registers R0 and R1.
10.ORG 0000H
MOV A,#0FFH
MOV R7,#32
MOV R0,#50H
NAME:MOV @R0,A

15. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
INC R0
CALL DELAY
DJNZ R7,NAME
11.ORG 0000H
MOV DPTR,#400H
MOV R7,#10
MOV R0,#30H
HERE:CLR A
MOVC A,@A+DPTR
CALL DELAY
INC DPTR
MOV @R0,A
INC R0
DJNZ R7,HERE
BACK:SJMP BACK
END
12.ORG 0000
MOV P1,#0FFH
MOV A,P1
MOV R0,A ; R0=x
MOV B,R0 ; B=x
MUL AB ; MULTIPLY A WITH B ANSWER STORE IN A=x*x
DA A
MOV R1,A ;R1=x*x Or x^2
MOV A,R0 ;A=x
MOV B,#2 ; B=2
MUL AB ; A=2*x
DA A
MOV R7,#5

16. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
ADD A,R1 ; A=x^2 + 2*x
DA A
ADD A,R7 ;A=x^2 + 2*x + 5
END
13.ORG 0000H
LJMP MAIN
ORG 20H
MYDATA: DB 06,09,02,05,07
ORG 300H
Main:MOV R0,#30H
MOV DPTR,#MYDATA
MOV R7,#5
HERE:CLR A
MOVC A,@A+DPTR
CALL DELAY
INC DPTR
PUSH 0E0H; Push the Accumulator into stack
DJNZ R7,HERE
POP 01
POP 02
POP 03
POP 04
POP 0E0H
ADD A,R1
ADD A R2
ADD A,R3
ADD A,R4; A Holds the added data.

17. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
MOV @R0,A
END
Section 5.3:-
14.INVALID
15.VALID
16.VALID
17.All ports are bit addressable.
18.See for answer on page124 figure 5-2.
19. (b),(c),(d),((f),(g),(h) are valid.
20.ORG 0000H
AGAIN:SETB P1.5
CALL DELAY
CALL DELAY
CALL DELAY
CLR P1.5
CALL DELAY
SJMP AGAIN
DELAY:MOV R1,#240
HERE:DHNZ R1,HERE
RET
END
21.ORG 0000H
AGAIN:SETB P2.7
CALL DELAY
CALL DELAY
CALL DELAY

18. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
CALL DELAY
CLR P2.7
CALL DELAY
SJMP AGAIN
DELAY:MOV R1,#240
HERE:DJNZ R1,HERE
RET
END
22.ORG 0000H
SETB P1.4
HERE:JNB P1.4,HERE
CMD:SETB P2,7
CALL DELAY
CLR P2.7
CALL DELAY
SJMP CMD
DELAY:MOV R1,#240
HERE:DJNZ R1,HERE
RET
END
23.ORG 0000H
SETB P2.1
HERE:JB P1.4,HERE
MOV P0,#55H
SJMP $
END

19. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
24.80H TO 87H
25.90H TO 97H
26.A0H TO A7H
27.B0H TO B7H
28.Not bit addressable register.
29.88H TO 8FH
30.E0H TO E7H
31.F0H TO F7H
32.D0H TO D7H
33.(a) P0 (b)87H (c)TCON (d)TCON (e)P1H (f)P2 (g)P2 (h)P3 (i)PSW (j)PSW (K)B
34.ORG 0000H
SETB RS1; FOR SELECTING REGISTER BANK 2
SETB RS0;FOR SELECTING REGISTER BANK 2
MOV R3,A
MOV R5,B
END
35.CLR 0D7H
37.See example 5-14.
38.To check the carry flag there are instructions namely JC and JNC.
39. And 40 see example 5-14.
41.CY
0D7H
P
0D0H
AC
0D6H
OV
0D2H
42.For this question see page 124 fig 5-2.
46.For this question see page 123 fir 5-1.
47.(a)20H (b)28H (c)18H (d)2DH (e)53H (f)15H (g)2CH (h)2AH (i)14H (j)37H
(k)7FH

20. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
50.MOV 04,C
51.MOV 16H,0D6H ; Auxiliary carry
52.MOV 12H,0D0H
53.ORG 0000H
JB ACC.0,HERE
SJMP AGAIN
HERE :JB ACC.1,HERE1
SJMP AGAIN
HERE1:MOV B,#4
DIV AB
AGAIN:
END
54.ORG 0000H
JB ACC.7,LCD_DISPLAY
SJMP NACK
LCD_DISPLAY:
NACK:
END
55.ORG 0000H
JB 0F7H,HERE
SJMP NACK
LCD_DISPLAY:
NACK:
END

21. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
56.
(A)ORG 0000H
MOV R0,#24H; PROGRAM IS DONE WITH THE HELP OF FIG 5-1.
MOV A,#0FFH
MOV @R0,A
MOV R0,#25H
MOV A,#0FFH
MOV @R0,A
END
(B)ORG 0000H
SETB 20H
SETB 21H
SETB 22H
SETB 23H
SETB 24H
SETB 25H
SETB 26H
SETB 27H
SETB 28H
SETB 29H
SETB 2AH
SETB 2BH
SETB 2CH
SETB 2DH
SETB 2EH
SETB 2FH
END

22. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
57.ORG 0000H
MOV B,#8
DIV AB
CJNE B,#00,HERE
SJMP FIN
HERE:MOV R0,A
FIN:
END
58.ORG 0000H
MOV R1,#8
BACK:MOV A,R2
RRC A; Rotate Right through carry means instead of 8 bit rotation it
JC HERE ;include carry flag as an MSB(most significant bit).
INC R0
HERE:DJNZ R1,BACK
END
Section 5.4:-
67.ORG 0000G
MOV A,#55H
MOV R0,#0C0H
MOV R7,#16
HERE:MOV @RO,A
INC R0
CALL DELAY
DJNZ R7,HERE
END

23. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
68.ORG 0000H
MOV R7,#16
MOV R1,#60H
MOV R0,#0D0H
HERE:MOV A,@R1
INC R1
MOV @R0,A
INC R0
CALL DELAY
DJNZ R7,HERE
END

24. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
Chapter 6:-
Section 6.1:-
1. (a) AC=1 (b)AC=1 (c)AC=1 (e)AC=1
Cy= 0 CY=0 CY=1 CY=1
2. Already been done in the lab, see your lab files.
3. ORG 0000H
MOV DPTR,#MYDATA
MOV R7,#9
MOV R2,#00H
MOV R3,#00H
BACK:CLR A
MOVC A,@A+DPTR
MOV R3,A
PUSH 03
INC DPTR
DJNZ R7,BACK
MOV R7,#9
CLR A
AGAIN:POP 00
ADD A,R0
JNC HERE1
INC R2
HERE1:DJNZ R7,AGAIN
MOV R3,A
SJMP $
ORG 250H

25. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
MYDATA: DB 53,94,56,92,74,65,43,23,83
END
4. Just use DA( Decimal Adjust instruction in question 3)
5. (a) AND (b)
ORG 0000H
MOV R0,#40H
MOV R7,#16
MOV A,#55H
HERE:MOV @R0,A
INC R0
DJNZ R7,HERE
MOV R7,#16
MOV R1,#60H
MOV R0,#40H
CLR A
BACK:ADD A,@R0
JNC HERE1
INC @R1
HERE1:INC R0
DJNZ R7,BACK
MOV R0,#61H
MOV @R0,A
SJMP $
END
9. ORG 0000H
MOV R4,#00H

26. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
MOV A,#48H
MOV R0,#9AH
ADD A,R0
MOV R7,A
MOV A,#0BCH
MOV R0,#7FH
ADDC A,R0 ;ADC is add with carry, if by adding A and R0
MOV R6,A ;Carry generates .What will it do , it will add both A and R0 With
MOV A,#34H;the Upside to it, it will also add the carry flag if it generates.
MOV R0,#89H
ADDC A,R0
MOV R5,A
JNC HERE
INC R4
HERE:MOV R0,#40H
MOV A,R4
MOV @R0,A
INC R0
MOV A,R5
MOV @R0,A
INC R0
MOV A,R6
MOV @R0,A
INC R0
MOV A,R7
MOV @R0,A
END
Here R5=BE,R6=3BH,R7=E2.The and is BE3BE2H.

27. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
12.ORG 0000H
mov a,#77
mov b,#34
mulab
end
13.ORG 0000H
mov a,#77
mov b,#3
divab
end
14.No, Only on A and B.
15.ORG 0000H
MOV DPTR,#MYDATA
MOV R0,#30H
CALL TRANSFER
CALL ADDITION
CALL AVERAGE
LJMP FIN
TRANSFER:
MOV R7,#9
HERE:
CLR A
MOVC A,@A+DPTR
MOV @R0,A
INC DPTR
INC R0
DJNZ R7,HERE
RET

28. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
ADDITION:
MOV R7,#9
MOV R0,#30H
CLR A
HERE1:
ADD A,@R0
INC R0
DJNZ R7,HERE1
RET
AVERAGE:
MOV B,#9
DIV AB
MOV R7,A
RET
ORG 250H
MYDATA: DB 3,9,6,9,7,6,4,2,8
FIN:
END
Section 6.3:-
23.(a) A=40h (b)A=F6H (c)A=86H
Rest do it yourself just use Keil write instruction and see the result in
project window.
24.Just as in Question no 23 write those instruction in Kiel and view the result
of accumulator in Project window.

29. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
27.There is no such instruction like CJE.
28.In this question you must monitor the status of the carry flag after the
execution of CJNE. Write a program below and check the status of the carry
flag in the PSW resister.
(a) ORG 0000H
BACK:MOV A,#25H ;Here the carry flag will go high.Always remember
CJNE A,#44H,over ; that carry will only go high when the value of source
SJMP BACK ; of CJNE instruction is greater than its destination.
OVER:
END ;
(b)
ORG 0000H
back:mov a,#0ffh ;Here carry flag will not go high as the value of the
cjne a,#6fh,over ;destination is greater than that of the source.
sjmp back
over:
end
Rest of the parts of question 28 now you can do them on your own.
30.(a)MOV A,#56H
SWAP A; What swap do is swap the upper and lower nibble now A becomes
; A=65H
RR A
RR A

30. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
(C)CLR C
MOV A,#4DH ; A=0100 1101B
SWAP A; A=D4 OR A= 1101 0100B
RRC A ;9 BIT ROTATION, A= 0 0110 1010B. You can see the zero before
;8-bit that is the carry bit that you included through RRC instructin.
RRC A ; A= 0 0011 0101b
RRC A ; A= 1 0001 1010b
32.ORG 0000H
MOV P1,#0FFH
MOV R7,#8
MOV A,P1
AGAIN:RRC A
JC HERE
INC R0
HERE:DJNZ R7,AGAIN
END
33.ORG 0000H
MOV R7,#8
MOV A,#68H
AGAIN:RRC A
JC HERE
INC R0
HERE:DJNZ R7,AGAIN
END
34.ORG 0000H
MOV R7,#8
MOV A,#68H

31. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
AGAIN:RLC A
JC HERE
INC R0
HERE:DJNZ R7,AGAIN
END
40.ORG 0000H
MOV R7,#9
MOV P1,#0FFH
AGAIN1:MOV A,P1
ANL A,#0FH
ORL A,#30H
MOV R1,A
MOV R4,#34H
HERE:CJNE A,#30H,HERE1
SJMP BACK
HERE1:CJNE A,#31H,HERE2
SJMP BACK
HERE2:CJNE A,#32H,HERE3
SJMP BACK
HERE3:CJNE A,#33H,HERE4
SJMP BACK
HERE4:CJNE A,#34H,HERE5
SJMP BACK
HERE5:CJNE A,#35H,HERE6
SJMP BACK
HERE6:CJNE A,#36H,HERE7
SJMP BACK
HERE7:CJNE A,#37H,HERE8
SJMP BACK

32. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
HERE8:CJNE A,#38H,HERE9
SJMP BACK
HERE9:CJNE A,#39H,HERE10
SJMP BACK
HERE10:ANL A,#0FH
ADD A,#37H
CJNE A,#41H,HERE11
SJMP BACK
HERE11:CJNE A,#42H,HERE12
SJMP BACK
HERE12:CJNE A,#43H,HERE13
SJMP BACK
HERE13:CJNE A,#44H,HERE14
SJMP BACK
HERE14:CJNE A,#45H,HERE15
SJMP BACK
HERE15:CJNE A,#46H,AGAIN1
BACK:
MOV P2,A
SJMP $
END

33. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
43.This program is same as the check-sum program, right next to 6-36
example. The only difference is here you have to find the checksum byte of
a whole sentence and in the program you had to find the check sum of HEX
values. I point out the difference that has to make for this program the rest
of the program will remain the same,
------------------------------------------------------
DATA_ADDR EQU 400H
COUNT EQU 31 ; Nos of characters in the whole sentence.
RAM_ADDR EQU 20H
ORG 0000H
CALL COPY_DATA
CALL CAL_CHKSUM
CALL TEST_CHKSUM
COPY_DATA: --- ;THERE SUBROUTINES ARE PRESENTIS THE BOOK Pg 172.
--------------------
--------------------
RET
CAL_CHKSUM:---
---------------------
---------------------
RET
TEST_CHKSUM:---
----------------------
----------------------
RET
ORG 400H
MYBYTE: DB ‘Hello, my fellow World citizens’
END

34. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
46.ORG 0000H
MOV R7,#9
MOV P1,#0FFH
AGAIN1:MOV A,P1
ANL A,#0FH
ORL A,#30H
MOV R1,A
MOV R4,#34H
HERE:CJNE A,#30H,HERE1
SJMP BACK
HERE1:CJNE A,#31H,HERE2
SJMP BACK
HERE2:CJNE A,#32H,HERE3
SJMP BACK
HERE3:CJNE A,#33H,HERE4
SJMP BACK
HERE4:CJNE A,#34H,HERE5
SJMP BACK
HERE5:CJNE A,#35H,HERE6
SJMP BACK
HERE6:CJNE A,#36H,HERE7
SJMP BACK
HERE7:CJNE A,#37H,HERE8
SJMP BACK
HERE8:CJNE A,#38H,HERE9
SJMP BACK
HERE9:CJNE A,#39H,AGAIN1
BACK:
MOV P2,A

35. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
SJMP $
END

36. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
Chapter 9:-
1. 2 timers
2. 16 bit, Timer 0 and Timer 1.
3. TH0 AND TL0.
4. TH1 AND TL1.
5. NO,These register are not bit addressable.
6. 8 bit
7. TMOD is used to initialize the Timer0 or Timer1 and also Mode of timer to
which we have to use.it also let us to select that weather we have to use
Timer or Counter.
8. No
9. Use the Figure 9-3 of TMOD register.
Gate C/T M1 M0 Gate C/T M1 M0
0 1 1 0 0 1 1 0
10.Just Divide the XTAL values with 12 for frequencies and for the time period
take the inverse of frequence.
11.(a)13 bit (b)16 bit (c)8 bit
12.(a)Mode 0
8192 in decimal you can find out by 2^(13)=8192, that is
because our Mode is of 13 bit and 2000H
(b)Mode 1
65536 in decimal 2^(16)=65536, it is 16 bit and HEX Value is
FFFFH.
(c)Mode 2
256 in decimal 2^(8)-256, it is 8 bit and HEX value is FFH.

37. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
16.TF0 flag is raised when value of TH and TL rolls over, which means after
FFFFH in TH and TL is incrimented, TF0 goes high.
XTAL=11.0592Mhz
TH=1CH and TL=12H so by combining both it becomes 1C12H A 16 BIT
VALUE.
Frequency = 11.0592Mhz/12 = 921.6Khz
Time = 1/F = 1/921.6Khz = 1.085usec
Now 1C12H is a HEX value, you must subtract it with the max vale of 16 bit
in decimal i.e FFFFH=65536 and for 1C12H=7186. Now you must subtract
65536-7186=58350
value is in decimal convert it in HEX to get
58350=E3EEH. Now Multiply 58350 with 1.085usec,
58350*1.085usec = 63.3msec. Which means it generate the time delay of
63.3msec approximately.TF flag will roll over when 1C12H will complete its
counts till FFFFH or Simply it count like this 1C12,1C13,1C14,1C15 and
onwards till it reach FFFFH and after FFFFH it raises the TF0 flag. You can
also test it like this, just add 1C12H+E3EE=10000H, you can see that
10000H, now TL0=00H, TH0=00H and TF0=1,this shows that the value is
rolled over.
17.TF0 flag is raised when value of TH and TL rolls over, which means after
FFFFH in TH and TL is incremented, TF0 goes high.
XTAL=16Mhz
TH=1CH and TL=12H so by combining both it becomes 1C12H A 16 BIT
VALUE.
Frequency = 16Mhz/12 = 1.33333Mhz
Time = 1/F = 1/1.33333Mhz = 0.75usec

38. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
Now 1C12H is a HEX value, you must subtract it with the max vale of 16 bit
in decimal i.e FFFFH=65536 and for 1C12H=7186. Now you must subtract
65536-7186=58350
value is in decimal convert it in HEX to get
58350=E3EEH. Now Multiply 58350 with 1.085usec,
58350*0.75usec = 47.3msec. Which means it generate the time delay of
47.3msec approximately
18.TF0 flag is raised when value of TH and TL rolls over, which means after
FFFFH in TH and TL is incrimented, TF0 goes high.
XTAL=11.0592Mhz
TH=0F2H and TL=10H so by combining both it becomes F210H a 16 BIT
VALUE.
Frequency = 11.0592Mhz/12 = 921.6Khz
Time = 1/F = 1/921.6Khz = 1.085usec
Now F210H is a HEX value, you must subtract it with the max vale of 16 bit
in decimal i.e FFFFH=65536 and for F210H=61968. Now you must subtract
65536-61968=3568 this value is in decimal convert it in HEX to get
3568=0DFH. Now Multiply 3568 with 1.085usec,
3568*1.085usec = 3.87msec. Which means it generate the time delay of
3.87msec approximately.TF flag will roll over when F210H will complete its
counts till FFFFH or Simply it count like this F210,F211,F212,F213 and
onwards till it reach FFFFH and after FFFFH it raises the TF0 flag. You can
also test it like this, just add F210H+0DF0H=10000H, you can see that
10000H, now TL0=00H, TH0=00H and TF0=1,this shows that the value is
rolled over.
19.It can be done by only changing the XTAL value and calculate rest.

39. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
20.XTAL = 11.0592Mhz, and need to find the delay 0f 2msec, Timer 1 is
programed in mode 1.
Frequency = 11.0592Mhz/12 = 921.6Khz
Time Period (Machine Cycle)= 1/T = 1/921.6Khz = 1.085usec
For the Time delay of 2msec = 2msec/1.085usec = 1843 in decimal.
Now subtract it from the max value that can be stored in TH and TL
combined, so 65536-1843 = 63693, now convert it in HEX to get
63693 = F8CDH, this is the value to be loaded in TH and TL, now to write the
program.
ORG 0000H
MOV TMOD,#10H
MOV TL1,#0CDH
MOV TH1,#0F8H
SETB TR1;To start the Timer TR0 must be set high.
JNB TF1,$
CLR TR1;To stop the timer.
CLR TF1
END
21.Frequency = 16Mhz/12 = 1.33333Mhz
Time Period (Machine Cycle)= 1/T = 1/1.33333Mhz = 0.75usec
Time Period for 2msec = 2msec/0.75usec = 2667 in decimal.
65536-2667 = 62869 for HEX F595H
ORG 0000H
MOV TMOD,#10H
MOV TL1,#95H
MOV TH1,#0F5H
SETB TR1;To start the Timer TR0 must be set high.

40. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
JNB TF0,$
CLR TR1
CLR TF1
END
22.Frequency = 11.0592Mhz/12 = 921.6Khz
Time Period (Machine Cycle)= 1/T = 1/921.6Khz = 1.085usec
Time Period for 2msec = 2.5msec/1.085usec =2304 in decimal.
65536-2304 = 63232 for HEX F700H
ORG 0000H
MOV TMOD,#01H ;For Timer 0.
MOV TL0,#00H
MOV TH0,#0F7H
SETB TR0 ;To start the Timer TR0 must be set high.
JNB TF0,$
CLR TR0
CLR TR0
END
23.Frequency = 11.0592Mhz/12 = 921.6Khz
Time Period (Machine Cycle)= 1/T = 1/921.6Khz = 1.085usec
Time Period for 2msec = 0.2msec/1.085usec = 184in decimal.
65536-184 = 65352 for HEX FF48H
ORG 0000H
MOV TMOD,#10H ;For Timer 1.
MOV TL1,#48H
MOV TH1,#0FFH
SETB TR1;To start the Timer TR0 must be set high.
JNB TF1,$

41. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
CLR TF1
CLR TR1
END
24.Frequency = 20Mhz/12 = 1.66666Mhz
Time Period (Machine Cycle)= 1/T = 1/1.66666Khz = 0.6usec
Time Period for 100msec = 100msec/0.6usec = 166667 in decimal.
Now we can see that 166667 is greater than 65536 that is the max value
that can be stored in TH and TL.
So now divide 166667 with any value that makes it less than 65536, here I
divide it by 5,
166667/5 = 33333 in decimal now this value is less than 166667, OK now
remember the value that you divide 166667 with. Now,
65536-33333 = 32203 for HEX 7DCBH
ORG 0000H
MOV TMOD,#10H ;For Timer 1.
MOV R0,#5
HERE:MOV TL1,#0CBH ;You must load TH and TL again as it is not an
MOV TH1,#7DH ;auto-reload mode.
SETB TR1 ;To start the Timer TR0 must be set high.
HERE1:JNZTF1,HERE1
CLR TF1
CLR TR1
DJNZ R0,HERE
END

42. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
25.Frequency = 11.0592Mhz/12 = 921.6Khz
Time Period (Machine Cycle)= 1/T = 1/921.6Khz = 1.085usec
Time Period for the lowest square wave you must put the maximum value
in TH and TL i.e FFFEH and it should be divided by 2 because of the square
wave to get 7FFFH.
ORG 0000H
SETB P1.2
MOV TMOD,#10H ;For Timer 1.
HERE:MOV TL1,#0FFH
MOV TH1,#7FH
SETB TR1;To start the Timer TR0 must be set high.
JNB TF1,$
CLR TF1
CLR TR1
CPL P1.2
SJMP HERE
END
26.Frequency = 11.0592Mhz/12 = 921.6Khz
Time Period (Machine Cycle)= 1/T = 1/921.6Khz = 1.085usec
Time Period for the Highest square wave you must put the minimum value
in TH and TL i.e 0000H because it only has to increment all the way from
0000h to FFFFH to roll over mean from 0000H, 0001H, 0002H, till 10000H
ORG 0000H
SETB P1.2
MOV TMOD,#10H ;For Timer 1.
HERE:MOV TL1,#00H
MOV TH1,#0FFH
SETB TR1;To start the Timer TR0 must be set high.

43. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
JNb TF1,$
CPL P1.2
CLR TR1
CLR TF1
SJMP HERE
END
27. AND 28. Are solved just like question 25 and 26, only change that occurs is
to change the values of crystal frequency to 16Mhz instead of 12Mhz, and
then find the remaining value and write a program.
29.Timer goes through states that are from F1H, F2H, F3H, F4H, F5H, F6H, F7H,
F8H, F9H, FAH, FBH, FCH, FDH, FEH, FFH and lastly 100 which makes the
Timer in mode 2 to roll over, total of 16 state.
30.Frequency = 11.0592Mhz/12 = 921.6Khz
Time Period (Machine Cycle)= 1/T = 1/921.6Khz = 1.085usec
For 1Khz frequency,
T = 1/F = 1/1Khz = 1msec
Time of wave must be divided by 2 because of high time and low time.
T = 1msec/2 = 0.5msec.
Now time for Delay = 0.5msec/1.085usec = 461 in decimal for HEX 1CDH.
65536-461 = 65075 in decimal for HEX FE33H
is the value to load in TH
and TL.
ORG 0000H
MOV TMOD,#10H ;For Timer 1.

44. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
HERE:MOV TL1,#33H
MOV TH1,#0FEH
SETB TR1 ;To start the Timer TR0 must be set high.
JNB TF1,$
CLR TR1
CLR TF1
SJMP HERE
END
31.Frequency = 11.0592Mhz/12 = 921.6Khz
Time Period (Machine Cycle)= 1/T = 1/921.6Khz = 1.085usec
For 3Khz frequency,
T = 1/F = 1/3Khz = 0.3333msec
Time of wave must be divided by 2 because of high time and low time.
T = 0.33333msec/2 = 0.166666msec.
Now time for Delay = 0.166667msec/1.085usec = 154 in decimal for HEX
09AH.
65536-154 = 65382 in decimal for HEX FF66H is the value to load in TH and
TL.
ORG 0000H
MOV TMOD,#01H ;For Timer 1.
HERE:MOV TL0,#66H
MOV TH0,#0FFH
SETB TR0;To start the Timer TR0 must be set high.
JNB TF0,$
CLR TR0

45. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
CLR TF0
SJMP HERE
END
32.Frequency = 20Mhz/12 = 1.66667Mhz
Time Period (Machine Cycle)= 1/T = 1/1.66667Mhz = 0.6usec
For 0.5Khz frequency,
T = 1/F = 1/0.5Khz = 2msec
Time of wave must be divided by 2 because of high time and low time.
T = 2msec/2 = 1msec.
Now time for Delay = 1msec/0.6usec = 1667 in decimal for HEX 683H.
65536-1667 = 63869 in decimal for HEX F97DH is the value to load in TH
and TL.
ORG 0000H
MOV TMOD,#01H ;For Timer 1.
HERE:MOV TL0,#7DH
MOV TH0,#0F9H
SETB TR0 ;To start the Timer TR0 must be set high.
JNB TF0,$
CLR TR0
CLR TF0
SJMP HERE
END

46. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
33.Frequency = 20Mhz/12 = 1.66667Mhz
Time Period (Machine Cycle)= 1/T = 1/1.66667Mhz = 0.6usec
For 10Khz frequency,
T = 1/F = 1/10Khz = 50usec
Time of wave must be divided by 2 because of high time and low time.
T = 50usec/2 = 25usec.
Now time for Delay = 25usec/0.6usec = 42 in decimal for HEX 02AH.
65536-42 = 65494in decimal for HEX FFD6H is the value to load in TH and
TL.
ORG 0000H
HERE:MOV TMOD,#01H ;For Timer 1.
HERE:MOV TL0,#D6H
MOV TH0,#0FFH
SETB TR0 ;To start the Timer TR0 must be set high.
JNB TF0,$
CLR TR0
CLR TF0
SJMP HERE
END

47. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
34.Frequency = 11.0592Mhz/12 = 921.6Khz
Time Period (Machine Cycle)= 1/T = 1/921.6Khz = 1.085usec
Time (T) = 1sec/1.085usec = 921659.
921659
this value is greater than 65536 so 921659/17 = 54215 in
decimal and for HEX 0D3C7H.
65536-54215 = 11321 in decimal for HEX 2C39H is the value to load in TH
and TL.
ORG 0000H
MOV R0,#17 ;Value that we divided 921659 by.
MOV TMOD,#01H ;For Timer 1.
HERE:MOV TL0,#39H
MOV TH0,#2CH
SETB TR0 ;To start the Timer TR0 must be set high.
JNB TF0,$
CLR TR0
CLR TF0
DJNZ R0,HERE
END

48. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
35.Frequency = 16Mhz/12 = 1.33333Mhz
Time Period (Machine Cycle)= 1/T = 1/1.33333Mhz = 0.75usec
Time (T) = 0.25sec/0.75usec = 333333.
333333 this value is greater than 65536, so 333333/33 = 10101 in decimal
and for HEX 2775H.
65536-10101 = 55435 in decimal for HEX D88BH is the value to load in TH
and TL.
ORG 0000H
MOV R0,#33 ;Value that we divided 333333 by.
MOV TMOD,#01H ;For Timer 1.
HERE:MOV TL0,#8BH
MOV TH0,#0D8H
SETB TR0 ;To start the Timer TR0 must be set high.
JNB TF0,$
CLR TR0;To stop the Timer.
CLR TF0
DJNZ R0,HERE
END

49. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
36.Frequency = 11.0592Mhz/12 = 921.6Khz
Time Period (Machine Cycle)= 1/T = 1/921.6Khz = 1.085usec
Lowest square wave we must take the largest value i.e 256 i.e 0FFH is HEX
for TH because this is mode 2 = 2*256*1.085usec = 0.555msec and for
Frequence =1/T = 1.8Khz.
ORG 0000H
MOV TMOD,#20H
SETB P1.3
MOV TH1,#0FFH
AGAIN:SETB TR1
JNB TF1,$
CLR TR1
CLR TF1
CPL P1.3
SJMP AGAIN
END
37.Frequency = 11.0592Mhz/12 = 921.6Khz
Time Period (Machine Cycle)= 1/T = 1/921.6Khz = 1.085usec
Highest square wave we must take the smallest value i.e 00H for TH
because this is mode and the Maximum Frequency is 0.45Mhz,
T = 2*1*1.085usec = 2.1usec, and for the Frequency = 1/T = 1/2.1usec =
460Khz.

50. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech
ORG 0000H
MOV TMOD,#20H
SETB P1.3
MOV TH1,#01
AGAIN:SETB TR1
JNB TF1,$
CLR TR1
CLR TF1
CPL P1.3
SJMP AGAIN
END
38.And 39. Are the same, the only difference is instead of XTAL = 11.0596Mhz
its 16Mhz.
Rest of the Questions Inshaa Allah (God Will’s) I will E-mail you after EID.

51. Microcontroller Solutions
From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech