Download 8051 Programming using Assembly

A
Registers
B
R0
DPTR
DPH
DPL
R1
R2
PC
PC
R3
R4
Some 8051 16-bit Register
R5
R6
R7
Some 8-bit Registers of
the 8051
A: Accumulator
B: Used specially in MUL/DIV
R0-R7: GPRs
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2
8051 Programming using
Assembly
The MOV Instruction – Addressing
Modes
MOV dest,source
MOV
MOV
MOV
MOV
A,#72H
A, #’r’
R4,#62H
B,0F9H
MOV
MOV
MOV
DPTR,#7634H
DPL,#34H
DPH,#76H
MOV
P1,A
; dest = source
;A=72H
;A=‘r’ OR 72H
;R4=62H
;B=the content of F9’th byte of RAM
;mov A to port 1
Note 1:
MOV
A,#72H
After instruction “MOV
≠
MOV
A,72H
A,72H ” the content of 72’th byte of RAM will replace in Accumulator.
8086
MOV
MOV
MOV
MOV
8051
AL,72H
AL,’r’
BX,72H
AL,[BX]
MOV
MOV
A,#72H
A,#’r’
MOV
A,72H
MOV
A,3
Note 2:
MOV
A,R3
≡
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Arithmetic Instructions
ADD A, Source
;A=A+SOURCE
ADD
A,#6
;A=A+6
ADD
A,R6
;A=A+R6
ADD
A,6
;A=A+[6] or A=A+R6
ADD
A,0F3H
;A=A+[0F3H]
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Set and Clear Instructions
SETB
CLR
SETB
SETB
SETB
SETB
SETB
bit
bit
C
P0.0
P3.7
ACC.2
05
; bit=1
; bit=0
; CY=1
;bit 0 from port 0 =1
;bit 7 from port 3 =1
;bit 2 from ACCUMULATOR =1
;set high D5 of RAM loc. 20h
Note:
CLR instruction is as same as SETB
i.e:
CLR
C
;CY=0
But following instruction is only for CLR:
CLR
A
;A=0
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SUBB
A,source ;A=A-source-CY
SETB C
SUBB A,R5
ADC
SETB C
ADC
;CY=1
;A=A-R5-1
A,source ;A=A+source+CY
;CY=1
A,R5
;A=A+R5+1
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DEC
INC
byte
byte
;byte=byte-1
;byte=byte+1
INC
DEC
DEC
R7
A
40H
; [40]=[40]-1
CPL
A
;1’s complement
Example:
L01:
MOV
CPL
MOV
ACALL
SJMP
A,#55H ;A=01010101 B
A
P1,A
DELAY
L01
 CALL
NOP & RET & RETI
All are like 8086 instructions.
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Logic Instructions
ANL byte/bit
ORL byte/bit
XRL byte
EXAMPLE:
MOV R5,#89H
ANL R5,#08H
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Rotate Instructions
• RR A Accumulator rotate right
• RL A Accumulator Rotate left
• RRC A Accumulator Rotate right through
the carry.
• RLC A Accumulator Rotate left through
the carry.
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Structure of Assembly language
and Running an 8051 program
EDITOR
PROGRAM
ORG
MOV
MOV
MOV
ADD
ADD
HERE: SJMP
END
0H
R5,#25H
R7,#34H
Myfile.lst
A,#0
A,R5
A,#12H
HERE
Myfile.asm
ASSEMBLER
PROGRAM
Other obj file
Myfile.obj
LINKER
PROGRAM
Myfile.abs
OH
PROGRAM
Myfile.hex
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11
Memory mapping in 8051
• ROM memory map in 8051 family
4k
0000H
8k
32k
0000H
0000H
0FFFH
DS5000-32
8751
AT89C51
1FFFH
8752
AT89C52
7FFFH
from Atmel Corporation
from Dallas Semiconductor
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• RAM memory space allocation in the 8051
7FH
Scratch pad RAM
30H
2FH
Bit-Addressable RAM
20H
1FH
Register Bank 3
18H
17H
Register Bank 2
10H
0FH
08H
(Stack) Register Bank 1
07H
Register Bank 0
00H
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13
8051 Flag bits and the PSW register
• PSW Register
CY
AC
F0
RS1
RS0
Carry flag
Auxiliary carry flag
Available to the user for general purpose
Register Bank selector bit 1
Register Bank selector bit 0
Overflow flag
User define bit
Parity flag Set/Reset odd/even parity
RS1
RS0
Register Bank
OV
--
P
PSW.7
PSW.6
PSW.5
PSW.4
PSW.3
PSW.2
PSW.1
PSW.0
CY
AC
-RS1
RS0
OV
-P
Address
0
0
0
00H-07H
0
1
1
08H-0FH
1
0
2
10H-17H
1
1
3
18H-1FH
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Instructions that Affect Flag Bits:
Note: X can be 0 or 1
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Example:
MOV
A,#88H
ADD
A,#93H
88
+93
---11B
CY=1
AC=0
10001000
+10010011
-------------00011011
P=0
9C
+64
---100
CY=1
AC=1
Example:
MOV
A,#38H
ADD
A,#2FH
38
+2F
---67
CY=0
AC=1
Example:
MOV
A,#9CH
ADD
A,#64H
10011100
+01100100
-------------00000000
P=0
00111000
+00101111
-------------01100111
P=1
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Addressing Modes
•
•
•
•
•
Immediate
Register
Direct
Register Indirect
Indexed
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Immediate Addressing Mode
MOV
MOV
MOV
MOV
MOV
A,#65H
A,#’A’
R6,#65H
DPTR,#2343H
P1,#65H
Example :
Num
…
MOV
MOV
…
ORG
data1:
EQU
30
R0,Num
DPTR,#data1
100H
db
“Example”
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Example
• Write the decimal value 4 on the SSD in
the following figure. Switch the decimal
point off.
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Register Addressing Mode
MOV
ADD
MOV
Rn, A
A, Rn
DPL, R6
MOV
MOV
DPTR, A
Rm, Rn
;n=0,..,7
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Direct Addressing Mode
Although the entire of 128 bytes of RAM can be accessed using direct
addressing mode, it is most often used to access RAM loc. 30 – 7FH.
MOV
MOV
MOV
MOV
R0, 40H
56H, A
A, 4
6, 2
; ≡ MOV A, R4
; copy R2 to R6
; MOV R6,R2 is invalid !
SFR register and their address
MOV
MOV
MOV
0E0H, #66H
0F0H, R2
80H,A
; ≡ MOV A,#66H
; ≡ MOV B, R2
; ≡ MOV P1,A
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Register Indirect Addressing Mode
•
In this mode, register is used as a pointer to the data.
MOV
A,@Ri
MOV
@R1,B
; move content of RAM loc.Where address is held by Ri into A
( i=0 or 1 )
In other word, the content of register R0 or R1 is sources or target in MOV, ADD and SUBB
insructions.
Example:
Write a program to copy a block of 10 bytes from RAM location sterting at 37h to RAM
location starting at 59h.
Solution:
MOV R0,37h
MOV R1,59h
MOV R2,10
L1: MOV A,@R0
MOV @R1,A
INC R0
INC R1
DJNZ R2,L1
; source pointer
; dest pointer
; counter
 jump
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Indexed Addressing Mode And On-Chip
ROM Access
• This mode is widely used in accessing data elements
of look-up table entries located in the program (code)
space ROM at the 8051
MOVC
A,@A+DPTR
A= content of address A +DPTR from ROM
Note:
Because the data elements are stored in the program
(code ) space ROM of the 8051, it uses the instruction
MOVC instead of MOV. The “C” means code.
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•
Example:
Assuming that ROM space starting at 250h contains “Hello.”, write a program to transfer the
bytes into RAM locations starting at 40h.
Solution:
ORG
0
MOV
DPTR,#MYDATA
MOV
R0,#40H
L1:
CLR
A
MOVC
A,@A+DPTR
JZ
L2
MOV
@R0,A
INC
DPTR
INC
R0
SJMP
L1
L2:
SJMP
L2
;------------------------------------ORG
250H
MYDATA:
DB
“Hello”,0
END
Notice the NULL character ,0, as end of string and how we use the JZ instruction to
detect that.
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• Example:
Write a program to get the x value from P1 and send x2 to P2, continuously .
Solution:
ORG
0
;code segment
MOV DPTR, #TAB1 ;moving data segment to data pointer
MOV A,#0FFH
;configuring P1 as input port
MOV P1,A
L01:
MOV A,P1
;reading value from P1
MOVC A,@A+DPTR
MOV
P2,A
SJMP
L01
;---------------------------------------------------ORG
300H
;data segment
TAB1: DB
0,1,4,9,16,25,36,49,64,81
END
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External Memory Addressing
• MOVX A, @R1 ; A
memory)
• MOVX A, @DPTR
• MOVX @DPTR, A
[R1] (in external
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16-bit, BCD and Signed
Arithmetic in 8051
Exercise:

Write a program to add n 16-bit number. Get n
from port 1. And sent Sum to SSD
a) in hex
b) in decimal

Write a program to subtract P1 from P0 and send
result to LCD
(Assume that “ACAL DISP” display A to SSD )
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MUL & DIV
• MUL
MOV
MOV
MUL
AB
A,#25H
B,#65H
AB
• MUL
MOV
MOV
MUL
AB
A,#25
B,#10
AB
;B|A = A*B
;25H*65H=0E99
;B=0EH, A=99H
;A = A/B, B = A mod B
;A=2, B=5
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Stack in the 8051
• The register used to access
the stack is called SP (stack
pointer) register.
7FH
Scratch pad RAM
30H
• The stack pointer in the
8051 is only 8 bits wide,
which means that it can take
value 00 to FFH. When
8051 powered up, the SP
register contains value 07.
2FH
Bit-Addressable RAM
20H
1FH
18H
17H
10H
0FH
08H
07H
00H
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Register Bank 3
Register Bank 2
(Stack) Register Bank 1
Register Bank 0
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Example:
MOV
MOV
MOV
PUSH
PUSH
PUSH
R6,#25H
R1,#12H
R4,#0F3H
6
1
4
0BH
0BH
0BH
0BH
0AH
0AH
0AH
0AH
F3
09H
09H
09H
12
09H
12
08H
08H
08H
25
08H
25
Start SP=07H
25
SP=08H
SP=09H
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SP=10H
30
Example (cont.)
POP
POP
POP
4
1
6
0BH
0BH
0BH
0BH
0AH
0AH
09H
0AH
F3
0AH
09H
12
09H
12
09H
08H
25
08H
25
08H
SP=10H
SP=09H
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25
SP=08H
08H
Start SP=07H
31
How to use the stack
• You can use the stack as temporary storage for
variables when calling functions
RLC A ;you can only rotate A
Call function
DIV AB ; A has the wrong value!!!!!
…
function: MOV A, #5 ;values are for example
sake
MOV B, #10
MUL AB ;you can only multiply on A
RET
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Example (correct)
RLC A ;you can only rotate A
PUSH A ;saving A and B on the stack before
PUSH B ;calling function
Call function
POP B ;restoring B
POP A ;and A (POP in reverse order)
DIV AB ; A has the wrong value!!!!!
…
function: MOV A, #5 ;values are for example sake
MOV B, #10
MUL AB ;you can only multiply on A
RET
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Saving PSW
• The Program Status Word registers contains flags that are often
important for correct program flow
• You can push PSW on the stack before calling a function
ADD A, R0
PUSH PSW
PUSH A ;saving A and R0 on the stack before
PUSH R0 ;calling function
Call function
POP R0
;restoring R0
POP A
;and A (POP in reverse order)
POP PSW
JC loop
;If this means the carry from the
;function then don’t push PSW
…
function: MOV A, #5 ;values are for example sake
ADD A, R2 ;the flags are set according to ADD result
RET
34
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LOOP and JUMP Instructions
 DJNZ:
Write a program to clear ACC, then
add 3 to the accumulator ten times
Solution:
MOV
MOV
AGAIN: ADD
DJNZ
MOV
A,#0;
R2,#10
A,#03
R2,AGAING ;repeat until R2=0 (10 times)
R5,A
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• Other conditional jumps :
JZ
Jump if A=0
JNZ
Jump if A/=0
DJNZ
Decrement and jump if A/=0
CJNE A,byte
Jump if A/=byte
CJNE reg,#data
Jump if byte/=#data
JC
Jump if CY=1
JNC
Jump if CY=0
JB
Jump if bit=1
JNB
Jump if bit=0
JBC
Jump if bit=1 and clear bit
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SJMP and LJMP:
LJMP(long jump)
LJMP is an unconditional jump. It is a 3-byte instruction in
which the first byte is the opcode, and the second and third
bytes represent the 16-bit address of the target location. The
20byte target address allows a jump to any memory location
from 0000 to FFFFH.
SJMP(short jump)
In this 2-byte instruction. The first byte is the opcode and the
second byte is the relative address of the target location. The
relative address range of 00-FFH is divided into forward and
backward jumps, that is , within -128 to +127 bytes of memory
relative to the address of the current PC.
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CJNE , JNC
Exercise:
Write a program that compare R0,R1.
If R0>R1 then send 1 to port 2,
else if R0<R1 then send 0FFh to port 2,
else send 0 to port 2.
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CALL Instructions
Another control transfer instruction is the CALL
instruction, which is used to call a subroutine.
• LCALL(long call)
In this 3-byte instruction, the first byte is the opcode
an the second and third bytes are used for the address
of target subroutine. Therefore, LCALL can be used
to call subroutines located anywhere within the 64K
byte address space of the 8051.
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• ACALL (absolute call)
ACALL is 2-byte instruction in contrast to LCALL,
which is 13 bytes. Since ACALL is a 2-byte instruction,
the target address of the subroutine must be within 2K
bytes address because only 11 bits of the 2 bytes are used
for the address. There is no difference between ACALL
and LCALL in terms of saving the program counter on
the stack or the function of the RET instruction. The only
difference is that the target address for LCALL can be
anywhere within the 64K byte address space of the 8051
while the target address of ACALL must be within a 2Kbyte range.
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Example
A
ORG
B
R5
R7
Address
Data
0H
VAL1 EQU 05H
MOV R5,#25H
LOOP: MOV R7,#VAL1
MOV
A,#0
ADD
A,R5
ADD
A,#12H
RRC A
DJNZ A, LOOP
SETB ACC.3
CLR A
CJNE A, #0, LOOP
HERE: SJMP HERE
END
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I/O Port Programming
Port 1（pins 1-8）

• Port 1 is denoted by P1.
– P1.0 ~ P1.7
• We use P1 as examples to show the operations on ports.
– P1 as an output port (i.e., write CPU data to the external pin)
– P1 as an input port (i.e., read pin data into CPU bus)
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A Pin of Port 1
Read latch
TB2
Vcc
Load(L1)
Internal CPU
bus
D
Write to latch
Clk
P1.X
pin
Q
P1.X
Q
M1
TB1
P0.x
Read pin
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8051 IC
43
Hardware Structure of I/O Pin
• Each pin of I/O ports
– Internal CPU bus：communicate with CPU
– A D latch store the value of this pin
• D latch is controlled by “Write to latch”
– Write to latch＝1：write data into the D latch
– 2 Tri-state buffer：
• TB1: controlled by “Read pin”
– Read pin＝1：really read the data present at the pin
• TB2: controlled by “Read latch”
– Read latch＝1：read value from internal latch
– A transistor M1 gate
• Gate=0: open
• Gate=1: close
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Tri-state Buffer
Output
Input
Tri-state control
(active high)
L
L
H
H
H
H
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Low
Highimpedance
(open-circuit)
45
Writing “1” to Output Pin P1.X
Read latch
Vcc
TB2
Load(L1) 2. output pin is
Vcc
1. write a 1 to the pin
Internal CPU
bus
D
Write to latch
Clk
1
Q
P1.X
pin
P1.X
Q
0
M1
output 1
TB1
Read pin
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8051 IC
46
Writing “0” to Output Pin P1.X
Read latch
Vcc
TB2
Load(L1) 2. output pin is
ground
1. write a 0 to the pin
Internal CPU
bus
D
Write to latch
Clk
0
Q
P1.X
pin
P1.X
Q
1
M1
output 0
TB1
Read pin
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8051 IC
47
Port 1 as Output（Write to a Port）
• Send data to Port 1：
BACK:
MOV
A,#55H
MOV
P1,A
ACALL
DELAY
CPL A
SJMP BACK
– Let P1 toggle.
– You can write to P1 directly.
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Reading Input v.s. Port Latch
• When reading ports, there are two possibilities：
– Read the status of the input pin. （from external pin value）
• MOV A, PX
• JNB P2.1, TARGET ; jump if P2.1 is not set
• JB
P2.1, TARGET ; jump if P2.1 is set
• Figures C-11, C-12
– Read the internal latch of the output port.
• ANL P1, A
; P1 ← P1 AND A
• ORL P1, A
; P1 ← P1 OR A
• INC P1
; increase P1
• Figure C-17
• Table C-6 Read-Modify-Write Instruction (or Table 8-5)
• See Section 8.3
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Reading “High” at Input Pin
Read latch
1.
TB2
write a 1 to the pin MOV
P1,#0FFH
Internal CPU bus
2. MOV A,P1
Vcc
external pin=High
Load(L1)
D
1
Q
1
P1.X pin
P1.X
Write to latch
Clk
0
Q
M1
TB1
Read pin
3. Read pin=1 Read latch=0
Write to latch=1
8051 IC
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Reading “Low” at Input Pin
Read latch
1.
Vcc
2. MOV A,P1
TB2
write a 1 to the pin
Load(L1)
external pin=Low
MOV P1,#0FFH
Internal CPU bus
D
1
Q
0
P1.X pin
P1.X
Write to latch
Clk
Q
0
M1
TB1
Read pin
3. Read pin=1 Read latch=0
Write to latch=1
8051 IC
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Port 1 as Input（Read from Port）
• In order to make P1 an input, the port must be programmed by writing 1 to
all the bit.
BACK:
MOV
MOV
MOV
MOV
SJMP
A,#0FFH
P1,A
A,P1
P2,A
BACK
;A=11111111B
;make P1 an input port
;get data from P0
;send data to P2
– To be an input port, P0, P1, P2 and P3 have similar methods.
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Instructions For Reading an Input Port
• Following are instructions for reading external pins of ports:
Mnemonics
Examples
Description
MOV A,PX
MOV A,P2
Bring into A the data at P2
pins
JNB PX.Y,..
JNB P2.1,TARGET
Jump if pin P2.1 is low
JB PX.Y,..
JB P1.3,TARGET
Jump if pin P1.3 is high
MOV C,PX.Y
MOV C,P2.4
Copy status of pin P2.4 to
CY
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Reading Latch
• Exclusive-or the Port 1：
MOV P1,#55H ;P1=01010101
ORL P1,#0F0H ;P1=11110101
1. The read latch activates TB2 and bring the data from the Q latch into
CPU.
• Read P1.0=0
2. CPU performs an operation.
• This data is ORed with bit 1 of register A. Get 1.
3. The latch is modified.
• D latch of P1.0 has value 1.
4. The result is written to the external pin.
• External pin (pin 1: P1.0) has value 1.
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Reading the Latch
1. Read pin=0 Read latch=1 Write to
latch=0 (Assume P1.X=0 initially)
Read latch
Vcc
TB2
Load(L1)
2. CPU compute P1.X OR 1
0
Internal CPU bus
D
1
Write to latch
3. write result to latch Read
pin=0
Read latch=0
Write to latch=1
0
Q
P1.X
Clk
1
4. P1.X=1
P1.X pin
0
M1
Q
TB1
Read pin
8051 IC
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Read-modify-write Feature
• Read-modify-write Instructions
– Table C-6
• This features combines 3 actions in a single instruction：
1. CPU reads the latch of the port
2. CPU perform the operation
3. Modifying the latch
4. Writing to the pin
– Note that 8 pins of P1 work independently.
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Port 1 as Input（Read from latch）
• Exclusive-or the Port 1：
MOV P1,#55H ;P1=01010101
AGAIN: XOR P1,#0FFH ;complement
ACALL DELAY
SJMP AGAIN
– Note that the XOR of 55H and FFH gives AAH.
– XOR of AAH and FFH gives 55H.
– The instruction read the data in the latch (not from the pin).
– The instruction result will put into the latch and the pin.
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Read-Modify-Write Instructions
Mnemonics
Example
ANL
ANL P1,A
ORL
ORL P1,A
XRL
XRL P1,A
JBC PX.Y, TARGET
JBC P1.1, TARGET
CPL
CPL P1.2
INC
INC
DEC
DEC P1
DJNZ PX, TARGET
DJNZ P1,TARGET
MOV PX.Y,C
MOV P1.2,C
CLR PX.Y
CLR P1.3
SETB PX.Y
SETB P1.4
P1
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You are able to answer this Questions:
• How to write the data to a pin？
• How to read the data from the pin？
– Read the value present at the external pin.
• Why we need to set the pin first？
– Read the value come from the latch（not from the external
pin）.
• Why the instruction is called read-modify write?
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Other Pins
• P1, P2, and P3 have internal pull-up resisters.
– P1, P2, and P3 are not open drain.
• P0 has no internal pull-up resistors and does not connects to
Vcc inside the 8051.
– P0 is open drain.
– Compare the figures of P1.X and P0.X. 
• However, for a programmer, it is the same to program P0, P1,
P2 and P3.
• All the ports upon RESET are configured as output.
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A Pin of Port 0
Read latch
TB2
Internal CPU
bus
D
Write to latch
Clk
P0.X
pin
Q
P1.X
Q
M1
TB1
P1.x
Read pin
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61
Port 0（pins 32-39）
• P0 is an open drain.
– Open drain is a term used for MOS chips in the same way
that open collector is used for TTL chips. 
• When P0 is used for simple data I/O we must connect it to
external pull-up resistors.
– Each pin of P0 must be connected externally to a 10K ohm
pull-up resistor.
– With external pull-up resistors connected upon reset, port 0
is configured as an output port.
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Port 0 with Pull-Up Resistors
Vcc
10 K
Port
P0.0
DS5000 P0.1
P0.2
8751
P0.3
P0.4
8951
P0.5
P0.6
P0.7
0
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Dual Role of Port 0
• When connecting an 8051/8031 to an external memory, the 8051
uses ports to send addresses and read instructions.
– 8031 is capable of accessing 64K bytes of external memory.
– 16-bit address：P0 provides both address A0-A7, P2 provides
address A8-A15.
– Also, P0 provides data lines D0-D7.
• When P0 is used for address/data multiplexing, it is connected to the
74LS373 to latch the address.
– There is no need for external pull-up resistors as shown in
Chapter 14.
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64
74LS373
PSEN
ALE
P0.0
74LS373
G
D
P0.7
OE
OC
A0
A7
D0
D7
EA
P2.0
A8
P2.7
A15
8051
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ROM
65
Reading ROM (1/2)
P0.0
2. 74373 latches the
address and send to
OE
ROM
OC
G 74LS373
A0
P0.7
A7
PSEN
ALE
1. Send address to
ROM
D
Address
D0
D7
EA
P2.0
A8
P2.7
A12
8051
ROM
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Reading ROM (2/2)
PSEN
ALE
P0.0
P0.7
2. 74373 latches the
address and send to
ROM
74LS373
G
D
Address
OE
OC
A0
A7
D0
D7
EA
3. ROM send the
instruction back
P2.0
A8
P2.7
A12
8051
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ROM
67
ALE Pin
• The ALE pin is used for de-multiplexing the
address and data by connecting to the G pin of
the 74LS373 latch.
– When ALE=0, P0 provides data D0-D7.
– When ALE=1, P0 provides address A0-A7.
– The reason is to allow P0 to multiplex address and
data.
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Port 2（pins 21-28）
• Port 2 does not need any pull-up resistors since
it already has pull-up resistors internally.
• In an 8031-based system, P2 are used to
provide address A8-A15.
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Port 3（pins 10-17）
• Port 3 does not need any pull-up resistors since it already
has pull-up resistors internally.
• Although port 3 is configured as an output port upon reset,
this is not the way it is most commonly used.
• Port 3 has the additional function of providing signals.
– Serial communications signal：RxD, TxD（Chapter 10）
– External interrupt：/INT0, /INT1（Chapter 11）
– Timer/counter：T0, T1（Chapter 9）
– External memory accesses in 8031-based system：/WR,
/RD（Chapter 14）
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Port 3 Alternate Functions
P3 Bit
Function
Pin
P3.0
P3.1
P3.2
P3.3
P3.4
P3.5
P3.6
P3.7
RxD
TxD
INT0
INT1
T0
T1
WR
RD
10
11
12
13
14
15
16
17
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
71
Generating Delays
• You can generate short delays using a register and
incrementing or decrementing its value
• Example:
mov r1, #0ah
loop: djnz r1, loop
• How much delay is that?
–
–
–
–
–
–
Djnz is a 2-byte instruction it takes two machine cycles
One machine cycle is 1/12 of the system clock period
For a 12 MHz system clock that is:
Machine cycle = 12/12 = 1 MHz
Machine period = 1/(1 MHz) = 10^(-6) s = 1 μs
Loop time = 10*2*1 μs = 20 μs
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Generating longer delays
• Each register is 8 bits long, so it can
increment 256 times before overflowing
• For larger delays, or when interrupts are
required 8051 uses two timers
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74
TMOD Register:
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75
TCON Register:
•
•
•
•
•
•
•
•
TF1: Timer 1 overflow flag.
TR1: Timer 1 run control bit.
TF0: Timer 0 overflag.
TR0: Timer 0 run control bit.
IE1: External interrupt 1 edge flag.
IT1: External interrupt 1 type flag.
IE0: External interrupt 0 edge flag.
IT0: External interrupt 0 type flag.
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Timer Mode Register
• Bit 7: Gate bit; when set, timer only runs while \INT high.
(T0)
• Bit 6: Counter/timer select bit; when set timer is an event
counter when cleared timer is an interval timer (T0)
• Bit 5: Mode bit 1 (T0)
• Bit 4: Mode bit 0 (T0)
• Bit 3: Gate bit; when set, timer only runs while \INT high.
(T1)
• Bit 2: Counter/timer select bit; when set timer is an event
counter when cleared timer is an interval timer (T1)
• Bit 1: Mode bit 1 (T1)
• Bit 0: Mode bit 0 (T1)
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Timer Modes
• M1-M0: 00 (Mode 0) – 13-bit mode (not
commonly used)
• M1-M0: 01 (Mode 1) - 16-bit timer mode
• M1-M0: 10 (Mode 2) - 8-bit auto-reload
mode
• M1-M0: 11 (Mode 3) – Split timer mode
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Timer Control Register (TCON)
• Bit 7 (TF1) 8FH : Timer 1 overflow flag; set by hardware
upon overflow, cleared by software
• Bit 6 (TR1) 8EH: Timer 1 run-control bit; manipulated by
software - setting starts timer 1, resetting stops timer 1
• Bit 5 (TF0) 8DH: Timer 0 overflow flag; set by hardware
upon overflow, cleared by software.
• Bit 4 (TR0) 8CH: Timer 0 run-control bit; manipulated by
software - setting starts timer 0, resetting stops timer 0
• Bit 3 (IE1) 8BH: External 1 Interrupt flag bit
• Bit 2 (IT1) 8AH:
• Bit 1 (IE0) 89H: External 0 Interrupt flag bit
• Bit 0 (IT0) 88H:
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Initializing and stopping timers
• MOV TMOD, #16H ;initialization
• SETB TR0
;starting timers
SETB TR1
• CLR TR0
; stop timer 0
CLR TR1
; stop timer 1
• MOV R7, TH0
; reading timers
• MOV R6, TL0
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80
Reading timers on the fly
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81
Generating delays using the timers
• To generate a 50 ms (or 50,000 us) delay we start the
timer counting from 15,536. Then, 50,000 steps later it
will overflow. Since each step is 1 us (the timer's clock is
1/12 the system frequency) the delay is 50,000 us. 0
MOV TMOD, #10H; set up timer 1 as 16-bit interval timer
CLR TR1 ; stop timer 1 (in case it was started in some
other subroutine)
MOV TH1, #3CH
MOV TL1, #0B0H ; load 15,536 (3CB0H) into timer 1
SETB TR1 ; start timer 1
JNB TF1, $; repeat this line while timer 1 overflow flag is
not set
CLR TF1; timer 1 overflow flag is set by hardware on
transition from FFFFH - the flag must be reset by
software
CLR TR1 ; stop timer 1
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Generating long delays
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83
Using timers to measure execution
time
• Timers are often used to measure the
execution time of a program
ORG 0H
MOV TMOD, #16H ;initialization
SETB TR0
;starting timer 0
…
;main
…
;program
CLR TR0
; stop timer 0
MOV R7, TH0 ; reading timer 0
MOV R6, TL0
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Interrupt :
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85
Interrupt Enable Register :
• EA : Global enable/disable.
•
---
•
•
•
•
•
•
ET2 :Enable Timer 2 interrupt.
ES :Enable Serial port interrupt.
ET1 :Enable Timer 1 interrupt.
EX1 :Enable External 1 interrupt.
ET0 : Enable Timer 0 interrupt.
EX0 : Enable External 0 interrupt.
: Undefined.
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86
Interrupt handling
• 8051 Interrupt Vector Table
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87
Interrupt Service Routines
• ORG 0
JMP main
• ORG 0003H ; external interrupt 0 vector
….
; interrupt handler code for external interrupt 0
RETI
ORG 0013H ; external interrupt 1 vector
….
;interrupt handler code for external interrupt 1
RETI
ORG 0030H ; main program
main:SETB IT0 ; set external interrupt 0 as edge activated
SETB IT1 ; set external interrupt 1 as edge activated
SETB EX0 ; enable external interrupt 0
SETB EX1 ; enable external interrupt 1
SETB EA ; global interrupt enable
…
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Examples
• Write a 8051 assembly program that
matches 8 switches with 8 LEDs
• Write a 8051 assembly program that uses
a two-digit SSD to display the temperature
as input from an ADC. Assume that the 05V range corresponds to 0-50 °C. The
ADC uses RD, WR and INT pins.
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8051 Programming Using C
Programming microcontrollers
using high-level languages
• Most programs can be written exclusively
using high-level code like ANSI C
• Extensions
– To achieve low-level (Assembly) efficiency,
extensions to high-level languages are
required
• Restrictions
– Depending on the compiler, some restrictions
to the high-level language may apply
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Keil C keywords
•
data/idata:
Description: The variable will be stored in internal data memory of controller.
example:
unsigned char data x;
//or
unsigned char idata y;
•
•
•
bdata:
Description: The variable will be stored in bit addressable memory of
controller.
example:
unsigned char bdata x;
//each bit of the variable x can be accessed as follows
x ^ 1 = 1; //1st bit of variable x is set
x ^ 0 = 0; //0th bit of variable x is cleared
xdata:
Description: The variable will be stored in external RAM memory of
controller.
example:
unsigned char xdata x;
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92
Keil C keywords
•
code:
Description: This keyword is used to store a constant variable in code and not data memory.
example:
unsigned char code str="this is a constant string";
•
_at_:
Description: This keyword is used to store a variable on a defined location in ram.
•
•
•
example:
CODE:
unsigned char idata x _at_ 0x30;
// variable x will be stored at location 0x30
// in internal data memory
sbit:
Description: This keyword is used to define a special bit from SFR (special function register)
memory.
example:
sbit Port0_0 = 0x80;
// Special bit with name Port0_0 is defined at address 0x80
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Keil C keywords
•
sfr:
Description: sfr is used to define an 8-bit special function register from sfr memory.
example:
sfr Port1 = 0x90;
// Special function register with name Port1 defined at addrress 0x90
•
sfr16:
Description: This keyword is used to define a two sequential 8-bit registers in SFR memory.
example:
sfr16 DPTR = 0x82;
// 16-bit special function register starting at 0x82
// DPL at 0x82, DPH at 0x83
•
using:
Description: This keyword is used to define register bank for a function. User can specify register bank 0
to 3.
example:
void function () using 2{
// code
}
// Funtion named "function" uses register bank 2 while executing its code
•
Interrupt:
Description: defines interrupt service routine
void External_Int0() interrupt 0{
//code
}
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Pointers
• //Generic Pointer
char * idata ptr;
//character pointer stored in data memory
int * xdata ptr1;
//Integer pointer stored in external data memory
//Memory Specific pointer
char idata * xdata ptr2;
//Pointer to character stored in Internal Data memory
//and pointer is going to be stored in External data
memory
int xdata * data ptr3;
//Pointer to character stored in External Data memory
//and pointer is going to be stored in data memory
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Writing hardware-specific code
• #include <REGx51.h> //header file for 89C51
void main(){
//main function starts
unsigned int i;
//Initializing Port1 pin1
P1_1 = 0; //Make Pin1 o/p
while(1){
//Infinite loop main application
//comes here
for(i=0;i<1000;i++)
; //delay loop
P1_1 = ~P1_1;
//complement Port1.1
//this will blink LED connected on Port1.1
}
}
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C and Assembly together
• extern unsigned long add(unsigned long, unsigned long);
void main(){
unsigned long a;
a = add(10,30); //calling Assembly function
while(1);
}
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C and Assembly together
•
name asm_test
?PR?_add?asm_test segment code
?DT?_add?asm_test segment data
;let other function use this data space for passing variables
public ?_add?BYTE
;make function public or accessible to everyone
public _add
;define the data segment for function add
rseg ?DT?_add?asm_test
?_add?BYTE:
parm1: DS 4 ;First Parameter
parm2: ds 4 ;Second Parameter
;either you can use parm1 for reading passed value as shown below
;or directly use registers used to pass the value.
rseg ?PR?_add?asm_test
_add:
;reading first argument
mov parm1+3,r7
mov parm1+2,r6
mov parm1+1,r5
mov parm1,r4
;param2 is stored in fixed location given by param2
;now adding two variables
mov a,parm2+3
add a,parm1+3
;after addition of LSB, move it to r7(LSB return register for Long)
mov r7,a
mov a,parm2+2
addc a,parm1+2
;store second LSB
mov r6,a
mov a,parm2+1
addc a,parm1+1
;store second MSB
mov r5,a
mov a,parm2
addc a,parm1
mov r4,a
ret
end
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The infinite loop
• A loop with no termination condition or one
that will never be met may be unwanted in
computer systems, but common in
embedded systems.
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99
Example 1
• Generate a 5V peek-to-peek 200μs period
square waveform on the DAC output
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100
Example 2
• Generate a 5V peek-to-peek 200μs period
sawtooth waveform on the DAC output
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101
Example 3
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
Generate a 5V peek-to-peek 2ms period sine waveform on the DAC output
code unsigned char Sine[180] = { /* Sine values */
127,131,136,140,145,149,153,158,162,166,170,175, 179,183,187,191,194,198,202,205,209,212,215,218,
221,224,227,230,232,235,237,239,241,243,245,246, 248,249,250,251,252,253,253,254,254,254,254,254,
253,253,252,251,250,249,248,246,245,243,241,239, 237,235,232,230,227,224,221,218,215,212,209,205,
202,198,194,191,187,183,179,175,170,166,162,158, 153,149,145,140,136,131,127,123,118,114,109,105,
101, 96, 92, 88, 84, 79, 75, 71, 67, 64, 60, 56, 52, 49, 45, 42, 39, 36, 33, 30, 27, 24, 22, 19, 17, 15, 13, 11,
9, 8, 6, 5, 4, 3, 2, 1, 1, 0, 0, 0, 0, 0, 1, 1, 2, 3, 4, 5, 6, 8, 9, 11, 13, 15, 17, 19, 22, 24, 27, 30, 33, 36, 39, 42,
45, 49, 52, 56, 60, 63, 67, 71, 75, 79, 84, 88, 92, 96, 101,105,109,114,118
};
/************************************************************
* START of the PROGRAM *
************************************************************ /
void main (void) {
unsigned char i;
/************************************************************
* Enable the D/A Converter *
************************************************************ /
ENDAC0 = 1; /* Enable DAC0 */
/************************************************************
* Create the waveforms on DAC0 *
************************************************************ /
while(1){ /* Run for ever */
for(i = 0; i < 179; i++)
DAC0 = Sine[i];
} * while(1) */
} /* main() */
}
102
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Mixed C/Assembly code (μVision Version 2.06)
• Parameter passing in registers
• Examples:
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103
Function return values
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Example
extern unsigned char add2_func(unsigned char, unsigned char);
void main(){
unsigned char a;
a = add2_func(10,30);
//a will have 40 after execution
while(1);
}
;assembly file “add2.asm”
NAME
_Add2_func
?PR?add2_func?Add2
SEGMENT CODE
PUBLIC add2_func
RSEG
?PR?Add2_func?Add2
add2_func:
mov a, r7
;first parameter passed to r7
add a, r5
;second parameter passed to r5
mov r7, a
;return parameter must be in r7
RET
END
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Calling C from Assembly
NAME
A_FUNC
?PR?a_func?A_FUNC
SEGMENT CODE
EXTRN
CODE (c_func)
PUBLIC
a_func
RSEG ?PR?a_func?A_FUNC
a_func:
USING
0
LCALL
c_func
RET
END
void c_func (void)
{
}
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Data Converters
• Analog to Digital Converters
(ADC)
– Convert an analog quantity
(voltage, current) into a
digital code
• Digital to Analog Converters
(DAC)
– Convert a digital code into
an analog quantity (voltage,
current)
Video (Analog - Digital)
Amplifier
Filters
Modulator
Analog
Preamplifier
Digital
A/D
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Image
enhancement
and coding
108
Temperature Recording by a Digital System
Temperature
(ºC)
Temperature
(ºC)
Sampling &
quantization
Time
Time
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109
Need for Data Converters
Digital processing and storage of physical quantities (sound, temperature, pressure
etc) exploits the advantages of digital electronics
– Better and cheaper technology compared to the analog
– More reliable in terms of storage, transfer and processing
• Not affected by noise
– Processing using programs (software)
• Easy to change or upgrade the system
– (e.g. Media Player 7  Media Player 8 ή Real Player)
• Integration of different functions
– (π.χ. Mobile = phone + watch + camera + games + email +
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Signals (Analog - Digital)
u(V
16)
1111
1110
14
1100
12
10
101
0
1001
8
100
0
0110
6
4
Analog Signal
• can take infinity values
• can change at any time
0101
0100
2
1
2
3
4
5
6
7
8
9
ADC
Digital Signal
• can take one of
2 values (0 or 1)
• can change only
at distinct times
Reconstruction of
an analog signal
from a digital one
(Can take only
predefined
values)
u(V)
t (S)
16
1111
1110
14
1100
12
1010
D0
D1
0
0
1
0
0
1
1
0
0
1
0
1
1
1
0
0
0
0
1001
10
1000
8
DAC6
011
0
0100
D2
1
0
1
1
0
1
1
1
0
D3
0
1
0
0
1
1
1
1
1
0101
4
2
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2
3
4
5
6
7
8
9 t (S)
111
QUANTIZATION ERROR
• The difference between the true and quantized value
of the analog signal
• Inevitable occurrence due to the finite resolution of
the ADC
• The magnitude of the quantization error at each
sampling instant is between zero and half of one LSB.
• Quantization error is modeled as noise (quantization
noise) u(V)
Analog signal value at
sampling time: 4.9 V
16
Quantized Analog signal
value: 5.0 V
14
12
Quantization error:
5.0 - 4.9 = 0.1 V
10
8
6
4
2
1
2
3
4
5
6
7
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8
9
t (S)
112
SAMPLING FREQUENCY
(RATE)
• The frequency at which digital values are sampled from the analog
input of an ADC
• A low sampling rate (undersampling) may be insufficient to
represent the analog signal in digital form
• A high sampling rate (oversampling) requires high bitrate and
therefore storage space and processing time
• A signal can be reproduced from digital samples if the sampling rate
is higher than twice the highest frequency component of the signal
(Nyquist-Shannon theorem)
• Examples of sampling rates
– Telephone: 4 KHz (only adequate for speech, ess sounds like
eff)
– Audio CD: 44.1 KHz
– Recording studio: 88.2 KHz
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Digital to Analog Converters
• The analog signal at the output
of a D/A converter is linearly
proportional to the binary code
at the input of the converter.
– If the binary code at the input
is 0001 and the output
voltage is 5mV, then
– If the binary code at the input
becomes 1001, the output
45mV
voltage will become ......
• If a D/A converter has 4 digital
inputs then the analog signal at
the output can have one out of
16 values.
……
• If a D/A converter has N digital
inputs then the analog signal at
the output can have one out of
2Ν values.
…….
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D3
D2
D1
D0
Vout
(mV)
0
0
0
0
0
0
0
0
1
5
0
0
1
0
10
0
0
1
1
15
0
1
0
0
20
0
1
0
1
25
0
1
1
0
30
0
1
1
1
35
1
0
0
0
40
1
0
0
1
45
1
0
1
0
50
1
0
1
1
55
1
1
0
0
60
1
1
0
1
65
1
1
1
0
70
1
1
1
1
75
114
Characteristics of Data
Converters
1.
2.
3.
4.
Number of digital lines
– The number bits at the input of a D/A (or output of an A/D) converter.
– Typical values: 8-bit, 10-bit, 12-bit and 16-bit
– Can be parallel or serial
Microprocessor Compatibility
– Microprocessor compatible converters can be connected directly on the microprocessor bus
as standard I/O devices
– They must have signals like CS, RD, and WR
•
Activating the WR signal on an A/D converter starts the conversion process.
Polarity
– Polar: the analog signals can have only positive values
– Bipolar: the analog signals can have either a positive or a negative value
Full-scale output
– The maximum analog signal (voltage or current)
– Corresponds to a binary code with all bits set to 1 (for polar converters)
– Set externally by adjusting a variable resistor that sets the Reference Voltage (or current)
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5.
6.
Characteristics of Data
Converters (Cont…)
Resolution
– The analog voltage (or current) that corresponds to a change of 1LSB in the
binary code
– It is affected by the number of bits of the converter and the Full Scale voltage
(VFS)
– For example if the full-scale voltage of an 8-bit D/A converter is 2.55V the the
resolution is:
VFS/(2N-1) = 2.55 /(28-1) 2.55/255 = 0.01 V/LSB = 10mV/LSB
Conversion Time
– The time from the moment that a “Start of Conversion” signal is applied to an A/D
converter until the corresponding digital value appears on the data lines of the
converter.
– For some types of A/D converters this time is predefined, while for others this
time can vary according to the value of the analog signal.
7. Settling Time
0.1Vo
– The time needed by the analog signal at
the output of a D/A converter to be within
10% of the nominal value.
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Vo
116
ADC RESPONSE TYPES
• Linear
– Most common
• Non-linear
– Used in telecommunications, since human
voice carries more energy in the low
frequencies than the high.
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ADC TYPES
• Direct Conversion
– Fast
– Low resolution
• Successive approximation
– Low-cost
– Slow
– Not constant conversion delay
• Sigma-delta
– High resolution,
– low-cost,
– high accuracy
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Interfacing with Data Converters
A19
A0
+5V
DAC
D7 V(+)
D6
D5
D4 Vout
D3
D2
D1
D0 Vref
A11
A10
A9
A8
A7
A6
A5
A4
CS
WR
Vout
10K
D7
D6
D5
D4
D3
D2
D1
D0
10K
8088 System
• Microprocessor compatible data converters are attached on
the microprocessor’s bus as standard I/O devices.
V(-)
IO/M'
WR
RD
S.K DHAR
119
Programming Example 1
Write a program to generate a positive ramp at
the output of an 8-bit D/A converter with a 2V
amplitude and a 1KHz frequency. Assume that
the full scale voltage of the D/A converter is
2.55V. The D/A converter is in P0 and the WR
signal is in P1.1
main()
{
do {
for (i=0;i<200;i++)
{
P1_0=1;
P0=i;
delayu(5);
2V
}
} while (1)
}
0V
f = 1KHZ
200 steps
of 10 mV each
==> 2V amplitude
200 steps of 5 us each
==> 1ms period or 1KHz frequency
S.K DHAR
120
D/A
Converters
example
Write a program to generate the waveform, shown below, at the output of
an 8-bit digital to analog converter. The period of the waveform should be
approximately 8 ms. Assume that a time delay function with a 1 μs
resolution is available. The full scale output of the converter is 5.12 V and
the address of the DAC is P0, while the WR signal is in P1.1.
V ( volts)
4
3
2
1
0
1
5
6
7
8
t ( msec )
Assuming that an 8-bit A/D converter is used to interface a temperature sensor
measuring temperature values in the temperature range 0 - 51.2 C
o
, specify: The resolution in of the system in C
o
o
The digital output word for a temperature of 32.5 C
The temperature corresponding to a digital output word of 01001110
S.K DHAR
121