Here are the key steps for programming timer 0 in mode 2: 1. Select mode 2 for timer 0 by writing to the TMOD register: MOV TMOD,#02h 2. Write the initial count value to TH0: MOV TH0,#0FCh 3. Clear the timer flag TF0: CLR TF0 4. Start the timer by setting TR0: SETB TR0 5. The timer will now count down from the value in TH0 (0FCh) to 0 in TL0. 6. When TL0 underflows to 0, TF0 will be set. 7. Check TF0 to detect overflow:

1. Addressing Modes
Definition: The way a Microcontroller can get the
data is called as addressing mode
Types:
Direct byte addressing Mode
Register Addressing
Register Indirect
Immediate Mode
Register Specific
Register Indexed Mode

2. Addressing Modes
Direct Mode – specify data by its 8-bit address
Usually for RAM
Mov a, 70h ; copy contents of RAM at 70h to a
Mov R0,40h ; copy contents of RAM at 40h to a
Mov 56h,a ; put contents of a at 56h to a
Mov PSW,a ; put contents of a into PSW
MOV R2,#5 ;Put 5 in R2
MOV R2,5 ;Put content of RAM at 5 in R2

3. Addressing Modes
Register Addressing – either source or destination is one of
CPU register
MOV R0,A
MOV A,R7
ADD A,R4
ADD A,R7
MOV DPTR,#25F5H
MOV R5,DPL
MOV R,DPH
Note that MOV R4,R7 is incorrect

4. Addressing Modes
Register Indirect (@) – the address of the source or destination is specified in
registers
Uses registers R0 or R1 for 8-bit address:
mov @r0, #03
mov @r1, a
mov @r1, 50
Uses DPTR register for 16-bit addresses:
mov dptr, #1234 ; dptr  1234h
movx a, @dptr ; a  M[1234]
Note that 1234 is an address in external memory

5. Addressing Modes
Immediate Mode – Data is a part of an Instruction
mov A, #0 ;put 0 in the accumulator
;A = 00000000
mov R4, #11h ;put 11hex in the R4 register
;R4 = 00010001
mov B, #11 ;put 11 decimal in b register
;B = 00001011
mov DPTR,#7521h ;put 7521 hex in DPTR
;DPTR = 0111010100100001

6. Addressing Modes
Register Specific (@) – the source and destination registers
both are same.
Uses registers A or DPTR
SWAP A
SWAP DPTR

7. Addressing Modes
Register Indexed Mode – source or destination
address is the sum of the base address and the
accumulator(Index)
• Base address can be DPTR or PC
– mov dptr, #4000h
– mov a, #5
– movc a, @a + dptr ;a  M[4005]

8. 8051- INSTRUCTION SET
Data Copy or transfer instructions
Data processing instructions
Arithmetic instructions
Logical instructions
Program flow or Branching instructions
JUMP
CALL & RET
Single Bit operation Instructions

9. Data Transfer Instructions
• MOV dest, source dest  source
• Stack instructions
PUSH byte ;increment stack pointer, ;move
byte on stack
POP byte ;move from stack to byte,
;decrement stack pointer
• Exchange instructions
XCH a, byte ;exchange accumulator and byte
XCHD a, byte ;exchange low nibbles of
;accumulator and byte

10. MOV
MOV dest, source dest  source
mov a,#0ffh
mov p0,a
mov R0, a
mov a, R1
Movx a,@DPTR
MovC a,@DPTR

11. Stack
• Direct addressing mode must be used in Push and Pop
mov sp, #0x40 ; Initialize SP
Push 55 ; SP  SP+1,
; M[41]  M[55]
pop b ; b  M[55]
push acc
Push psw
Push b

12. Exchange Instructions
two way data transfer
XCH a, 30h ; a  M[30]
XCH a, R0 ; a  R0
XCH a, @R0 ; a  M[R0]
XCHD a, R0 ; exchange “digit”
R0[7..4] R0[3..0]a[7..4] a[3..0]
Only 4 bits exchanged

13. Data Processing Instructions
Arithmetic Instructions
Logic Instructions

14. Arithmetic Instructions
• Add
• Subtract
• Increment
• Decrement
• Multiply
• Divide

15. Arithmetic Instructions
Mnemonic Description
ADD A, byte add A to byte, put result in A
ADDC A, byte add with carry
SUBB A, byte subtract with borrow
INC A increment A
INC byte increment byte in memory
INC DPTR increment data pointer
DEC A decrement accumulator
DEC byte decrement byte
MUL AB multiply accumulator by b register
DIV AB divide accumulator by b register

16. ADD Examples
mov a, #3Fh
add a, #D3h
• What is the value of the C,
AC, OV flags after the
second instruction is
executed?
0011 1111
1101 0011
0001 0010
C = 1
AC = 1
OV = 0

17. Logic Instructions
 Bitwise logic operations
 (AND, OR, XOR, NOT)
 Clear
 Rotate
 Swap

18. Bitwise Logic
ANL  AND
ORL  OR
XRL  XOR
CPL  Complement
Examples:
00001111
10101100ANL
00001111
10101100ORL
00001111
10101100XRL
10101100CPL
00001100
10101111
10100011
01010011

19. CLR ( Set all bits to 0)
CLR A
CLR byte (direct mode)
CLR Ri (register mode)
CLR @Ri (register indirect mode)

20. Rotate
• Rotate instructions operate only on a
RL a
Mov a,#0xF0 ; a 11110000
RR a ; a 11100001
RR a
Mov a,#0xF0 ; a 11110000
RR a ; a 01111000

21. Rotate through Carry
RRC a
mov a, #0A9h; a  A9
add a, #14h ; a  BD (10111101), C0
rrc a ; a  01011110, C1
RLC a
mov a, #3ch ; a  3ch(00111100)
setb c ; c  1
rlc a ; a  01111001, C1
C
C

22. Swap
SWAP a
mov a, #72h ; a  72h
swap a ; a  27h

23. Program Flow Control
• Unconditional jumps
Program transfer is
permanent
• Conditional jumps
Call and return Program transfer is
temporary

24. Unconditional Jumps
• SJMP <rel addr 8-bit > ;
Short jump, relative address is 8-bit, so jump can
be up to 127 locations forward, or 128 locations
back.
• AJMP <address 11> ; Absolute jump to
anywhere within 2K block of program memory(07FF H)
• LJMP <address 16> ; Long
jump(64 k 0000-FFFF H)

25. Infinite Loops
Start: mov C, p3.7
mov p1.6, C
sjmp Start
Microcontroller application programs are almost always infinite loops!

26. Conditional jumps
Mnemonic Description
JZ <rel addr> Jump if a = 0
JNZ <rel addr> Jump if a != 0
JC <rel addr> Jump if C = 1
JNC <rel addr> Jump if C != 1
JB <bit>, <rel addr> Jump if bit = 1
JNB <bit>,<rel addr> Jump if bit != 1
JBC <bir>, <rel addr> Jump if bit =1, &clear
bit
CJNE A, direct, <rel addr> Compare A and memory,
jump if not equal

27. More Conditional Jumps
Mnemonic Description
CJNE A, #data <rel addr> Compare A and data, jump
if not equal
CJNE Rn, #data <rel addr> Compare Rn and data,
jump if not equal
CJNE @Rn, #data <rel addr> Compare Rn and memory,
jump if not equal
DJNZ Rn, <rel addr> Decrement Rn and then
jump if not zero
DJNZ direct, <rel addr> Decrement memory and
then jump if not zero

28. Call and Return
• Call is similar to a jump, but
– Call pushes PC on stack before branching
acall <address ll> ; stack  PC
; PC  address 11 bit
lcall <address 16> ; stack  PC
; PC  address 16 bit

29. Return
• Return is also similar to a jump, but
– Return instruction pops PC from stack to get
address to jump to
– ret ; PC  stack

30. Bit-Oriented Data Transfer
• transfers between individual bits.
• Carry flag (C) (bit 7 in the PSW) is used as a single-bit
accumulator
• RAM bits in addresses 20-2F are bit addressable
mov C, P0.0
mov C, 67h
mov C, 2ch.7

31. Bit-Oriented Data Transfer
• CLR Bit
Ex:- CLR C, CLR P1.2
• SETB Bit
Ex:- SETB C, SET P1.2, SETB PSW.3
• CPL Bit
CPL C
• Single bit Logical Operation Instructions
AND Operation: ANL C, PSW.2
OR Operation : ORL C , ACC.7

32. TIMER / COUNTER PROGRAMMING
Indicate which mode and which timer are selected for each of the
following.
(a) MOV TMOD,#01H (b) MOV TMOD,#20H
(c) MOV TMOD,#12H
Solution:
(a) TMOD = 00000001, mode 1 of timer 0 is selected.
(b) TMOD = 00100000, mode 2 of timer 1 is selected.
(c) TMOD = 00010010
mode 2 of timer 0, and mode 1 of timer 1 are selected.
timer 1 timer 0
GATE C/T M1 M0 GATE C/T M1 M0
Timer 1 Timer 0
(MSB) (LSB)

33. TIMER PROGRAMMING
Find the timer’s clock frequency and its period for various 8051-
based systems, with the following crystal frequencies.
(a) 12 MHz (b) 16 MHz (c) 11.0592 MHz
Solution:
(a) 1/12 × 12 MHz = 1 MHz and T = 1/1 MHz = 1
µs
(b) 1/12 × 16 MHz = 1.333 MHz and
T = 1/1.333 MHz = 0.75 µs
(c) 1/12 × 11.0592 MHz = 921.6 KHz;
T = 1/921.6 KHz = 1.085 µs
XTAL
oscillator ÷ 12

34. Mode 1 Programming
XTAL
oscillator ÷ 12
TR
TH TL TF
TF goes high when
FFFF 0
overflow
flag
C/T = 0
like MC for
89C51
Start timer

35. Steps of Mode 1
1. Chose mode 1 timer 0
– MOV TMOD,#01H
2. Set the original value to TH0 and TL0.(FCFF H)
– MOV TH0,#0FFH
– MOV TL0,#0FCH
3. You had better to clear the flag to monitor: TF0=0.
– CLR TF0
4. Start the timer.
– SETB TR0

36. Steps of Mode 1
5.The 8051 starts to count up by incrementing
the TH0-TL0.
FFFCH,FFFDH,FFFEH,FFFFH,0000H
FFFC FFFD FFFE FFFF 0000
TF0=0 TF0=0 TF0=0 TF0=0 TF0=1
TH0 TL0Start timer
Stop timer
TR0=1 TR0=0
Roll over

37. Steps of Mode 1
6. When TH0-TL0 rolls over from FFFFH to 0000, the 8051
set TF0=1.
7. Keep monitoring the timer flag (TF) to see if it is raised.
– AGAIN: JNB TF0, AGAIN
8. Clear TR0 to stop the process.
– CLR TR0
9. Clear the TF flag for the next round.
– CLR TF0

38. Initial Count Values
• The initial count value = FFFC.
• The number of counts = FFFFH-FFFCH+1 = 4
– we add one to 3 because of the extra clock needed
when it rolls over from FFFF to 0 and raises the TF
flag.
• The delay = 4
• If T=1.085 µs, then the delay = 4 X T
4.34 µs

39. Example
MOV TMOD,#01 ;Timer 0,mode 1(16-bit)
HERE: MOV TL0,#0F2H ;Timer value = FFF2H
MOV TH0,#0FFH
CPL P1.5
ACALL DELAY
SJMP HERE
DELAY:
SETB TR0 ;start the timer 0
AGAIN:JNB TF0,AGAIN
CLR TR0 ;stop timer 0
CLR TF0 ;clear timer 0 flag
RET

40. SOLUTION
In Example , calculate the amount of time delay in the DELAY
subroutine generated by the timer. Assume that XTAL = 11.0592
MHz.
Solution:
The timer works with the internal system clock.
frequency of internal system clock = 11.0592/12 = 921.6 KHz
machine cycle = 1 /921.6 KHz = 1.085 µs (microsecond)
The number of counts = FFFFH – FFF2H +1 = 14 (decimal).
The delay = number of counts × 1.085 µs = 14 × 1.085 µs = 15.19 µs
for half the clock.
For the entire period of a clock, it is T = 2 × 15.19 µs = 30.38 µs as
the time delay generated by the timer.

41. Largest Time Delay IN MODE 0
Solution:
TH0=TL0=0000 means that
the timer will count from 0000 to FFFF, and then roll over to raise
the TF0 flag.
As a result, it goes through a total of 65536 states. Therefore, we
have delay = (65536 – 0) × 1.085 µs = 71.1065 ms.

42. Find Timer COUNT Values
• Assume XTAL = 11.0592 MHz .
• How to find the inter values needed for the TH, TL to
produce exact delay of 20 ms?
Step-1 -Divide the desired time delay by 1.085 µs.
20ms ÷ 1.085 µs = 18433
Step2-Perform 65536 –n, where n is the decimal value we got in
Step 1.
65536-18433= 47103 = B7FFH
– Convert the result of Step 2 to hex, where B7FF is the initial hex
value to be loaded into the timer’s registers.
– Set TH=B7H, TL=FFH

43. Example 2
Assuming XTAL = 11.0592 MHz, write a program to generate a
square wave of 50 Hz frequency on pin P2.3.
Solution:
Look at the following steps.
(a) The period of the square wave = 1 / 50 Hz = 20 ms.
(b) The high or low portion of the square wave = 10 ms.
(c) 10 ms / 1.085 µs = 9216
65536 – 9216 = 56320 in decimal = DC00H in hex.
(d) TL1 = 00H and TH1 = DCH.

44. Mode 2 programming
XTAL
oscillator ÷ 12
TR
TL
TH
TF
TF goes high when FF
0
overflow
flag
C/T = 0
reload

45. Steps of Mode 2
1. Chose mode 2 timer 0
– MOV TMOD,#02H
2. Set the original value to TH0.
– MOV TH0,#FCH
3. Clear the flag to TF0=0.
– CLR TF0
4. Start the timer.
– SETB TR0
– Note that the instruction SETB TR0 dose not load TH0 to TL0. So TL0
still is 00H.

46. Steps of Mode 2 (2/3)
5. The 8051 starts to count up by incrementing
the TL0.
– TL0= ..., FCH,FDH,FEH,FFH,FCH
00 01 FC FD FE
TF0 = 0 TF0 = 0 TF0 = 0 TF0 = 0 TF0 = 0
TH0=FCH
TL0=00H
Start timer
TR0=1
FF
FC FD FE
TF0 = 0 TF0 =0
TF0 = 0 TF0 = 0 TF0 = 0
Clear TF0
TF0=0
FF
TL0=FCH
auto reload
TH0=FCH
roll over
auto reload: TL=-FCH immediately
TF0 = 1
roll over
...
00H
00H

47. Steps of Mode 2 (3/3)
6. When TL0 rolls over from FFH to 00, the 8051 set TF0=1. Also, TL0 is
reloaded automatically with the value kept by the TH0.
– TL0= FCH, FDH, FEH, FFH, FCH(Now TF0=1)
– The 8051 auto reload TL0=TH0=FCH.
– Go to Step 6 (i.e., TL0 is incrementing continuously).
•. Note that we must clear TF0 when TL0 rolls over. Thus, we can
monitor TF0 in next process.
•. Clear TR0 to stop the process.
00H

48. Example 9-14 (1/2)
Assuming that XTAL = 11.0592 MHz, find
(a) the frequency of the square wave generated on pin P1.0 in the following
program
(b) the smallest frequency achievable in this program, and the TH value to do
that.
MOV TMOD,#20H ;Timer 1,mode 2
MOV TH1,#5 ;not load TH1 again
SETB TR1 ;start (no stop TR1=0)
BACK:JNB TF1,BACK
CPL P1.0
CLR TF1 ;clear timer flag 1
SJMP BACK ;mode 2 is auto-reload

49. Example 9-14 (2/2)
Solution:
(a) First notice that target address of SJMP. In mode 2 we do not
need to reload TH since it is auto-reload.
Half period = (FFH – 05 +1) × 1.085 µs = 272.33 µs
Total period = 2 × 272.33 µs = 544.67 µs
Frequency = 1.83597 kHz.
(b) To get the smallest frequency, we need the largest period and that
is achieved when TH1 = 00.
Total period = 2 × 256 × 1.085 µs = 555.52 µs
Frequency = 1.8kHz.

50. Counter programming
• These timers can also be used as counters
counting events happening outside the 8051
by setting C/T=1.
• The counter counts up as pulses are fed from
– T0: timer 0 input (Pin 14, P3.4)
– T1: timer 1 input (Pin 15, P3.5)
T0
to
LCD
P3.4
P1
8051
a switch
TL0
TH0
Vcc
Counter 0

51. Counter (2/2)
• When the timer is used as a counter, it is a
pulse outside of the 8051 that increments
– TH0 & TL0 for counter 0.
– TH1 & TL1 for counter 1.
T1
to
LCD
P3.5
P1
8051
a switch
TL1
TH1
Vcc
Counter 1

52. Port 3 Pins Used For Timers 0 and 1
Pin Port Pin Function Description
14 P3.4 T0 Timer/Counter 0 external input
15 P3.5 T1 Timer/Counter 1 external input
GATE C/T=1 M1 M0 GATE C/T=1 M1 M0
Timer 1 Timer 0
(MSB) (LSB)

53. Counter Mode 1
• 16-bit counter (TH0 and TL0)
• TH0-TL0 is incremented when TR0 is set to 1 and an
external pulse (in T0) occurs.
• When the counter (TH0-TL0) reaches its maximum of
FFFFH, it rolls over to 0000, and TF0 is raised.
• Programmers should monitor TF0 continuously and stop
the counter 0.
• Programmers can set the initial value of TH0-TL0 and let
TF0 as an indicator to show a special condition. (ex: 100
people have come).

54. Counter Mode 2
• 8-bit counter.
– TL0 is incremented if TR0=1 and external pulse occurs.
• Auto-reloading
– TH0 is loaded into TL0 when TF0=1.
– It allows only values of 00 to FFH to be loaded into TH0.
– You need to clear TF0 after TL0 rolls over.
• See Figure 9.6, 9.7 for logic view
• See Examples 9-18, 9-19

55. Example
Assuming that clock pulses are fed into pin T1, write a program for
counter 1 in mode 2 to count the pulses and display the state of the
TL1 count on P2.
Solution:
We use timer 1 as an event counter where it counts up as clock pulses are fed into
pin3.5.
P2 is connected to 8 LEDs and
input T1 to pulse.
T1
to
LEDs
P3.5
P2
8051

56. Example
MOV TMOD,#01100000B ;mode 2, counter 1
MOV TH1,#0
SETB P3.5 ;make T1 input port
AGAIN:SETB TR1 ;start
BACK: MOV A,TL1
MOV P2,A ;display in P2
JNB TF1,BACK ;overflow
CLR TR1 ;stop
CLR TF1 ;make TF=0
SJMP AGAIN ;keep doing it
Notice in the above program the role of the instruction “SETB
P3.5”. Since ports are set up for output when the 8051 is powered
up , we must make P3.5 an input port by making it high.

57. 8051 Serial Communication Programming
TxD and RxD pins
• In 8051, the data is received from or transmitted
to
– RxD: received data (Pin 10, P3.0)
– TxD: transmitted data (Pin 11, P3.1)

58. Data Transfer Rate
• How fast is the data transferred?
• Methods to describe the speed:
– Baud rate is defined as the number of signal changes
per second.
– The rate of data transfer is stated in Hz.
– Date rate is defined as the number of bits transferred
per second.
– Each signal has several voltage levels.
– The rate of data transfer is stated in bps (bits per second).

59. Standard Baud Rates
110 bps
150
300
600
1200
2400
4800
9600 (default)
19200
Various Baud Rates

60. 60
Baud Rate Comparison for SMOD = 0 and
SMOD =1
XTAL
oscillato
r
÷ 12
÷ 16
÷ 32
Machine
cycle freq.
921.6 kHz
57600 Hz
28800 Hz
To timer
1 to set
baud
rate
11.0592 MHz
SMOD = 1
SMOD = 0

61. Find Count Value in TH1 to generate Baud rate
With XTAL = 11.0592 MHz, find the TH1 value needed to have the
following baud rates. (a) 9600 (b) 2400 (c) 1200. Assume SMOD = 0
Solution:
With XTAL = 11.0592 MHz, we have:
The frequency of system clock = 11.0592 MHz / 12 = 921.6 kHz
The frequency sent to timer 1 = 921.6 kHz/ 32 = 28,800 Hz
(a) TH1 = FF – {Fosc /(12X32X9600)}
TH1 = 256-{11.0952X106 /(12X32X9600)}
TH1 = 256-3 253(Decimal) FD H

62. 62
Baud Rate Comparison for SMOD = 0 and
SMOD =1
TH1 (Decimal) (Hex) SMOD = 0 SMOD = 1
253 FD 9,600 19,200
250 FA 4,800 9,600
244 F4 2,400 4,800
232 E8 1,200 2,400
Note: XTAL = 11.0592 MHz.
XTAL
oscillato
r
÷ 12
÷ 16
÷ 32
Machine
cycle freq.
921.6 kHz
57600 Hz
28800 Hz
To timer
1 to set
baud
rate
11.0592 MHz
SMOD = 1
SMOD = 0

63. Transfer steps
• The following sequence is the steps that the 8051
goes through in transmitting a character via TxD:
1. The byte character to be transmitted is written into the
SBUF register.
2. It transfers the start bit.
3. The 8-bit character is transferred one bit at a time.
4. The stop bit is transferred.
SBUF
TxD
bit by bit
8-bit char
UARTTI

64. Transfer data Serially
Example 1
Write a program for the 8051 to transfer letter “A” serially at 4800
baud, continuously. Use the timer 1 in mode 2. Assume SMOD = 0
Solution:
MOV TMOD,#20H ;timer 1, mode 2
MOV TH1,#-6 ;4800 baud rate
MOV SCON,#50H ;8-bit,1 stop,REN enabled
SETB TR1 ;start timer 1
AGAIN: MOV SBUF,#”A” ;letter “A” to be transferred
HERE: JNB TI,HERE ;wait for the last bit
CLR TI ;clear TI for next char
SJMP AGAIN ;keep sending A

65. Example 2
Write a program to transfer the message “YES” serially at 9600
baud, 8-bit data, 1 stop bit. Do this continuously.
Solution:
MOV TMOD,#20H ;timer 1, mode 2
MOV TH1,#-3 ;9600 baud
MOV SCON,#50H
SETB TR1
AGAIN:MOV A,#”Y” ;transfer “Y”
ACALL TRANS
MOV A,#”E” ;transfer “E”
ACALL TRANS
MOV A,#”S” ;transfer “S”
ACALL TRANS
SJMP AGAIN ;keep doing it

66. Example 2
;serial data transfer subroutine
TRANS:MOV SBUF,A ;load SBUF
HERE: JNB TI,HERE ;wait for last bit to transfer
CLR TI ;get ready for next byte
RET

67. Receive Data Serially
• The following sequence is the steps that the
8051 goes through in receiving a character via
RxD:
1. 8051 receives the start bit indicating that the next
bit is the first bit of the character to be received.
2. The 8-bit character is received one bit at a time.
When the last bit is received, a byte is formed and
placed in SBUF.
SBUFRxD
bit by bit
8-bit character
UART
RI

68. Example 1
Program the 8051 to receive bytes of data serially, and put them in
P1. Set the baud rate at 4800, 8-bit data, and 1 stop bit.
Solution:
MOV TMOD,#20H ;timer1, mode 2 (auto reload)
MOV TH1,#-6 ;4800 baud
MOV SCON,#50H ;8-bit, 1 stop, REN enabled
SETB TR1 ;start timer 1
HERE: JNB RI,HERE ;wait for char to come in
MOV A,SBUF ;save incoming byte in A
MOV P1,A ;send to port 1
CLR RI ;get ready to receive next byte
SJMP HERE ;keep getting data

69. HOME WORK
MOV A,PCON
SETB ACC.7
MOV PCON,A ;SMOD=1, double baud rate
MOV TMOD,#20H ;Timer 1, mode 2,auto reload
MOV TH1,#-3 ;19200 baud rate
MOV SCON,#50H ;8-bit data,1 stop bit, RI enabled
SETB TR1 ;start Timer 1
MOV A,#”B” ;transfer letter B
A_1:CLR TI ;make sure TI=0
MOV SBUF,A ;transfer it
H_1:JNB TI H_1 ;check TI
SJMP A_1 ;do again

70. SOLUTION
Assuming that XTAL = 11.0592 MHz for the following program,
state (a) what this program does, (b) compute the frequency used by
timer 1 to set the baud rate, and (c) find the baud rate of the data
transfer.Assume SMOD=1.
Solution:
(a) This program transfers ASCII letter B (01000010 binary)
continuously.
(b) and (c) With XTAL = 11.0592 MHz and SMOD = 1
11.0592 / 12 = 921.6 kHz machine cycle frequency.
921.6 /16 = 57,600 Hz frequency used by timer 1 to set the baud rate.
57,600 / 3 = 19,200, the baud rate.

71. INTERRUPT PROGRAMMING
TIMER INTERRUPT PROGRAMMING
• TIMER Interrupt occurred when Timer Flag (TFO
or TF1) comes to set condition.
• Overflow occurred at Roll-over of TIMER from
FFFF to 0000 will set TFO or TF1
1
TF0
jumps to
0000
....
000
B....
....
ISR of timer0
Timer 0 Interrupt Vector: 000BH
1
TF1
jumps to
0000
....
....
001
B....
ISR of timer1
Timer 1 Interrupt Vector: 001BH

72. Example 1
Write a program that continuously gets 8-bits data from P0 and
sends it to P1 while simultaneously creating a square wave of 200
µs period on pin P2.1. Use timer 0 to create the square wave.
Assume that XTAL = 11.0592 MHz.
Solution:
We will use timer 0 in mode 2 (auto reload).
TH0 = 100 µs /1.085 µs = 92 for half clock.
We must avoid using memory space allocated to interrupt vector
table. Therefore, we place the main memory in 0030H
1
TF0
jumps to
0000
....
000
B....
....
0030
....
ISR of timer0
main program
interrupt
vector
table
100
µs
100
µs200 µs
P2.1

73. Example 1
ORG 0000H
LJMP MAIN ;by-pass interrupt vector table
ORG 000BH ;Timer 0 interrupt vector table
CPL P2.1 ;toggle P2.1 pin
RETI ;return from ISR
The main program for initialization
ORG 0030H ;after vector table space
MIAN: MOV TMOD,#02H ;Timer 0,mode 2(auto reload)
MOV P0,0FFH ;make P0 an input port
MOV TH0,#0A4H ;TH0=A4H for -92
MOV IE,#82H ;IE=100000010(bin) enable Timer 0
SETB TR0 ;Start Timer 0
BACK: MOV A,P0 ;get data from P0 and put it to P1
MOV P1,A ;loop unless interrupted by TF0
SJMP BACK
END

74. Example 2
Create a square wave that has a high portion of 1085 µs and a low
portion of 15 µs. Assume XTAL = 1.0592 MHz. Use timer 1.
Solution:
we need to use mode 1 of timer 1.
1085 µs /1.085 µs = ? Count value for ON time
15 µs /1.085 µs = ? Count value for OFF time
1
TF1
jumps to
0000
....
001
B....
....
0030
....
ISR of timer1
main program
interrupt
vector
table
1085
µs
15
µs
P2.1

75. Example 2
ORG 0000H ;by-pass interrupt vector table
LJMP MAIN
ORG 001BH ;timer 1 interrupt vector table
LJMP ISR_T1 ;jump to ISR
;The main program for initialization
ORG 0030H ;after vector table
MAIN: MOV TMOD,#10H ;timer 1, mode 1
MOV P0,#0FFH ;make P0 an input port
MOV TL1,#018H ;TL1=18 the low byte of -1000
MOV TH1,#0FCH ;TH1-FC the high byte of -1000
MOV IE,#88H ;IE=1001000 enable timer 1.
SETB TR1 ;start timer 1

76. Example 2
ISR_T1:CLR TR1 ;stop Timer 1
CLR P2.1 ;P2.1=0, start of low portion
MOV TL1,#18H ;load T1 low byte value (2 MC)
MOV TH1,#0FCH ;load T1 high byte value (2 MC)
SETB TR1 ;starts timer 1 (1 MC)
SETB P2.1 ;P2.1=1, back to high (1 MC)
RETI ;return to main
END

77. Example -3
Write a program to generate a square wave of 50 Hz frequency on pin
P1.2. Assume that XTAL=11.0592 MHz. Find the count value to be
loaded in TIMER 0.
Solution:
(a) The period of the square wave = 1 / 50 Hz = 20 ms.
(b) The half square wave = 10 ms = (10 ms /1.085 µs) = 9216
65536 – 9216 = 56320 in decimal = DC00H in hex.
ORG 0
LJMP MAIN
ORG 000BH
CPL P1.2
RETI
50% 50%
20ms
P1.2

78. Example -3
;--main program for initialization
ORG 30H
MAIN: MOV TMOD,#00000001B ;timer 0, mode 1
MOV TL0,#00
MOV TH0,#0DCH
MOV IE,#10000010B ;enable timer 0
interrupt
SETB TR0
HERE: SJMP HERE
END
P1.2
8051
TL0
TH0
50 MHz square wave

79. Programming External Hardware Interrupts
• The 8051 has two external hardware interrupts:
– EX0: INT0, Pin 12 (P3.2)
– EX1: INT1, Pin 13 (P3.3)
• They are enabled and disabled using the IE register.
– EX0 by IE.0
– EX1 by IE.1
TF1 TR1 TF0 TR0 IE1 IT1 IE0 IT0
Timer 1 Timer0 for Interrupt
(MSB) (LSB)

80. Figure 4-1. 8051 Pin Diagram
PDIP/Cerdip
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
40
39
38
37
36
35
34
33
32
31
30
29
28
27
26
25
24
23
22
21
P1.0
P1.1
P1.2
P1.3
P1.4
P1.5
P1.6
P1.7
RST
(RXD)P3.0
(TXD)P3.1
(T0)P3.4
(T1)P3.5
XTAL2
XTAL1
GND
(INT0)P3.2
(INT1)P3.3
(RD)P3.7
(WR)P3.6
Vcc
P0.0(AD0
)P0.1(AD1)
P0.2(AD2
)P0.3(AD3)
P0.4(AD4)
P0.5(AD5)
P0.6(AD6)
P0.7(AD7)
EA/VPP
ALE/PROG
PSEN
P2.7(A15)
P2.6(A14)
P2.5(A13)
P2.4(A12)
P2.3(A11)
P2.2(A10)
P2.1(A9)
P2.0(A8)
8051
(8031)
external
hardware
interrupt


81. Activation of INT0
IE0
(TCON.1)
INT0
(Pin
3.2)
Interrupt Service
Table with addr.
0003H
Edge-triggered
level-triggered
IT 0 =0
IT 0 =1
INT0
INT0
IE0=1
IE0=0

82. Example 1
Assume that the INT1 pin is connected to a switch that is normally
high. Whenever it goes low, it should turn on an LED.
The LED is connected to P1.3 and is normally off.
When it is turned on, it should stay on for a fraction of a second.
As long as the switch is pressed low, the LED should stay on.
Solution:
Pressing the switch will cause the
LED to be turned on.
If it is kept activated,
the LED stays on.
To
LED
8051
P1.3
INT1
Vcc

83. Example 1
ORG 0000H
LJMP MAIN
ORG 0013H
SETB P1.3
MOV R3,#255
BACK: DJNZ R3,BACK
CLR P1.3
RETI
main program for initialization
ORG 0030H
MAIN: MOV IE,#10000100B ;enable INT1
HERE: SJMP HERE ;stay here until get interrupt
END 

84. Example 2
Assuming that pin 3.3 (INT1) is connected to a pulse generator,
write a program in which the falling edge of the pulse will send a
high to P1.3 which is connected to an LED (or buzzer). In other
words, the LED is turned on and off at the same rate as the pulses
are applied to the INT1 pin.
This is an edge-triggered version of Example 11-5. But in this
example, to turn on the LED again, the INT1 pulse must be brought
back high and then forced low to create a falling edge to activate
the interrupt.
INT1
to
LED
P3.3
P1.3
8051
a switch
pulse
generator

85. Example 2
ORG 0000H
LJMP MAIN
ORG 0013H
SETB P1.3
MOV R3,#255
BACK: DJNZ R3,HERE ;keep the buzzer on for a while
CLR P1.3
RETI
MAIN program for initialization
ORG 0030H
MAIN: SETB TCON.2 ;make INT1 edge-trigger interrupt
MOV IE,#10000100B ;enable External INT 1
HERE: SJMP HERE
END

86. Programming the Serial Communication
Interrupt
Write a program in which the 8051 reads data from P1 and writes it
to P2 continuously while giving a copy of it to the serial COM port
to be transferred serially. Assume that XTAL=11.0592. Set the baud rate at 9600.
Solution: ORG 0000
LJMP MAIN
ORG 23H
LJMP SERIAL ;jump to serial interrupt ISR
ORG 0030H
MAIN: MOV P1,#OFFH ;make P1 an input port
MOV TMOD,#20H ;timer 1,mode 2 (auto reload)
MOV TH1,#OFDH ;9600 baud rate
MOV SCON,#50H ;8-bit, 1 stop, REN enabled
MOV IE,#10010000B ;enable serial interrupt
SETB TR1 ;start timer 1

87. Example 1
;------ stay in loop indefinitely ------
BACK: MOV A,P1
MOV SBUF,A ;A has a copy of data
MOV P2,A
SJMP BACK
;------ Serial communication ISR ------
ORG 100H
SERIAL: JB TI,TRANS ;jump if TI is high
RETI
TRANS: MOV SBUF,A ;transmit the copy of P1
CLR TI ;clear TI
RETI
END