Philips 80c51 Family Programmer/'s Guide
Philips Semiconductors
80C51 family programmer’s guide
80C51 Family
and instruction set
PROGRAMMER’S GUIDE AND INSTRUCTION SET
register bank contains eight 1-byte registers 0 through 7. Reset
initializes the stack pointer to location 07H, and it is incremented
Memory Organization
once to start from location 08H, which is the first register (R0) of
the second register bank. Thus, in order to use more than one
Program Memory
register bank, the SP should be initialized to a different location
The 80C51 has separate address spaces for program and data
of the RAM where it is not used for data storage (i.e., the higher
memory. The Program memory can be up to 64k bytes long. The
part of the RAM).
lower 4k can reside on-chip. Figure 1 shows a map of the 80C51
program memory.
2. Bit Addressable Area: 16 bytes have been assigned for this
segment, 20H-2FH. Each one of the 128 bits of this segment can
The 80C51 can address up to 64k bytes of data memory to the chip.
be directly addressed (0-7FH). The bits can be referred to in two
The MOVX instruction is used to access the external data memory.
ways, both of which are acceptable by most assemblers. One
The 80C51 has 128 bytes of on-chip RAM, plus a number of Special
way is to refer to their address (i.e., 0-7FH). The other way is
Function Registers (SFRs). The lower 128 bytes of RAM can be
with reference to bytes 20H to 2FH. Thus, bits 0-7 can also be
accessed either by direct addressing (MOV data addr) or by indirect
referred to as bits 20.0-20.7, and bits 8-FH are the same as
addressing (MOV @Ri). Figure 2 shows the Data Memory
21.0-21.7, and so on. Each of the 16 bytes in this segment can
organization.
also be addressed as a byte.
3. Scratch Pad Area: 30H through 7FH are available to the user as
Direct and Indirect Address Area
data RAM. However, if the stack pointer has been initialized to
The 128 bytes of RAM which can be accessed by both direct and
this area, enough bytes should be left aside to prevent SP data
indirect addressing can be divided into three segments as listed
destruction.
below and shown in Figure 3.
1. Register Banks 0-3: Locations 0 through 1FH (32 bytes). The
Figure 2 shows the different segments of the on-chip RAM.
device after reset defaults to register bank 0. To use the other
register banks, the user must select them in software. Each
FFFF
FFFF
60k
BYTES
EXTERNAL
OR
64k
BYTES
EXTERNAL
1000
AND
0FFF
4k BYTES
INTERNAL
0000
0000
SU00567
Figure 1. 80C51 Program Memory
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and instruction set
0FFF
INTERNAL
FF
64k
BYTES
SFRs
EXTERNAL
DIRECT ADDRESSING
ONLY
80
AND
7F
DRIECT AND INDIRECT
ADDRESSING
00
0000
SU00568
Figure 2. 80C51 Data Memory
8 BYTES
78
7F
70
77
68
6F
60
67
58
5F
SCRATCH
PAD
50
57
AREA
48
4F
40
47
38
3F
30
37
28
... 7F
2F
BIT
ADDRESSABLE
20
0 ...
27
SEGMENT
18
3
1F
10
2
17
REGISTER
BANKS
08
1
0F
00
0
07
SU00569
Figure 3. 128 Bytes of RAM Direct and Indirect Addressable
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and instruction set
Table 1.
80C51 Special Function Registers
DIRECT
BIT ADDRESS, SYMBOL, OR ALTERNATIVE PORT FUNCTION
SYMBOL
DESCRIPTION
RESET VALUE
ADDRESS
MSB
LSB
ACC*
Accumulator
E0H
E7
E6
E5
E4
E3
E2
E1
E0
00H
B*
B register
F0H
F7
F6
F5
F4
F3
F2
F1
F0
00H
DPTR
Data pointer (2 by-
tes)
DPH
Data pointer high
83H
00H
DPL
Data pointer low
82H
00H
AF
AE
AD
AC
AB
AA
A9
A8
IE*
Interrupt enable
A8H
EA
–
–
ES
ET1
EX1
ET0
EX0
0x000000B
BF
BE
BD
BC
BB
BA
B9
B8
IP*
Interrupt priority
B8H
–
–
–
PS
PT1
PX1
PT0
PX0
xx000000B
87
86
85
84
83
82
81
80
P0*
Port 0
80H
AD7
AD6
AD5
AD4
AD3
AD2
AD1
AD0
FFH
97
96
95
94
93
92
91
90
P1*
Port 1
90H
–
–
–
–
–
–
T2EX
T2
FFH
A7
A6
A5
A4
A3
A2
A1
A0
P2*
Port 2
A0H
A15
A14
A13
A12
A11
A10
A9
A8
FFH
B7
B6
B5
B4
B3
B2
B1
B0
P3*
Port 3
B0H
RD
WR
T1
T0
INT1
INT0
TxD
Rxd
FFH
PCON1
Power control
87H
SMOD
–
–
–
GF1
GF0
PD
IDL
0xxxxxxxB
D7
D6
D5
D4
D3
D2
D1
D0
PSW*
Program status word
D0H
CY
AC
F0
RS1
RS0
OV
–
P
00H
SBUF
Serial data buffer
99H
xxxxxxxxB
9F
9E
9D
9C
9B
9A
99
98
SCON*
Serial controller
98H
SM0
SM1
SM2
REN
TB8
RB8
TI
RI
00H
SP
Stack pointer
81H
07H
8F
8E
8D
8C
8B
8A
89
88
TCON*
Timer control
88H
TF1
TR1
TF0
TR0
IE1
IT1
IE0
IT0
TH0
Timer high 0
8CH
00H
TH1
Timer high 1
8DH
00H
TL0
Timer low 0
8AH
00H
TL1
Timer low 1
8BH
00H
TMOD
Timer mode
89H
GATE
C/T
M1
M0
GATE
C/T
M1
M0
00H
NOTES:
*
Bit addressable
1. Bits GF1, GF0, PD, and IDL of the PCON register are not implemented on the NMOS 8051/8031.
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and instruction set
8 BYTES
F8
FF
F0
F7
B
E8
EF
E0
E7
ACC
D8
DF
D0
D7
PSW
C8
CF
C0
C7
B8
BF
IP
B0
B7
P3
A8
AF
IE
A0
A7
P2
98
9F
SCON
SBUF
90
P1
97
88
TCON
8F
TMOD
TL0
TL1
TH0
TH1
80
P0
SP
DPL
DPH
PCON
87
BIT ADDRESSABLE
SU00570
Figure 4. SFR Memory Map
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and instruction set
Those SFRs that have their bits assigned for various functions are listed in this section. A brief description of each bit is
provided for quick reference. For more detailed information refer to the Architecture Chapter of this book.
PSW: PROGRAM STATUS WORD. BIT ADDRESSABLE.
CY
AC
F0
RS1
RS0
OV
–
P
CY
PSW.7
Carry Flag.
AC
PSW.6
Auxiliary Carry Flag.
F0
PSW.5
Flag 0 available to the user for general purpose.
RS1
PSW.4
Register Bank selector bit 1 (SEE NOTE 1).
RS0
PSW.3
Register Bank selector bit 0 (SEE NOTE 1).
OV
PSW.2
Overflow Flag.
–
PSW.1
Usable as a general purpose flag.
P
PSW.0
Parity flag. Set/cleared by hardware each instruction cycle to indicate an odd/even number of ‘1’ bus in
the accumulator.
NOTE:
1. The value presented by RS0 and RS1 selects the corresponding register bank.
RS1
RS0
REGISTER BANK
ADDRESS
0
0
0
00H-07H
0
1
1
08H-0FH
1
0
2
10H-17H
1
1
3
18H-1FH
PCON: POWER CONTROL REGISTER. NOT BIT ADDRESSABLE.
SMOD
–
–
–
GF1
GF0
PD
IDL
SMOD
Double baud rate bit. If Timer 1 is used to generate baud rate and SMOD = 1, the baud rate is doubled when the Serial
Port is used in modes 1, 2, or 3.
–
Not implemented, reserved for future use.*
–
Not implemented reserved for future use.*
–
Not implemented reserved for future use.*
GF1
General purpose flag bit.
GF0
General purpose flag bit.
PD
Power Down Bit. Setting this bit activates Power Down operation in the 80C51. (Available only in CMOS.)
IDL
Idle mode bit. Setting this bit activates Idle Mode operation in the 80C51. (Available only in CMOS.)
If 1s are written to PD and IDL at the same time, PD takes precedence.
*
User software should not write 1s to reserved bits. These bits may be used in future 8051 products to invoke new features.
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and instruction set
INTERRUPTS:
To use any of the interrupts in the 80C51 Family, the following three steps must be taken.
1. Set the EA (enable all) bit in the IE register to 1.
2. Set the corresponding individual interrupt enable bit in the IE register to 1.
3. Begin the interrupt service routine at the corresponding Vector Address of that interrupt. See Table below.
INTERRUPT SOURCE
VECTOR ADDRESS
IE0
0003H
TF0
000BH
IE1
0013H
TF1
001BH
RI & TI
0023H
In addition, for external interrupts, pins INT0 and INT1 (P3.2 and P3.3) must be set to 1, and depending on whether the
interrupt is to be level or transition activated, bits IT0 or IT1 in the TCON register may need to be set to 1.
ITx = 0 level activated
ITx = 1 transition activated
IE: INTERRUPT ENABLE REGISTER. BIT ADDRESSABLE.
If the bit is 0, the corresponding interrupt is disabled. If the bit is 1, the corresponding interrupt is enabled.
EA
–
–
ES
ET1
EX1
ET0
EX0
EA
IE.7
Disables all interrupts. If EA = 0, no interrupt will be acknowledged. If EA = 1, each interrupt source is
individually enabled or disabled by setting or clearing its enable bit.
—
IE.6
Not implemented, reserved for future use.*
—
IE.5
Not implemented, reserved for future use.*
ES
IE.4
Enable or disable the serial port interrupt.
ET1
IE.3
Enable or disable the Timer 1 overflow interrupt.
EX1
IE.2
Enable or disable External Interrupt 1.
ET0
IE.1
Enable or disable the Timer 0 overflow interrupt.
EX0
IE.0
Enable or disable External Interrupt 0.
*
User software should not write 1s to reserved bits. These bits may be used in future 80C51 products to invoke new features.
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ASSIGNING HIGHER PRIORITY TO ONE OR MORE INTERRUPTS:
In order to assign higher priority to an interrupt the corresponding bit in the IP register must be set to 1.
Remember that while an interrupt service is in progress, it cannot be interrupted by a lower or same level interrupt.
PRIORITY WITHIN LEVEL:
Priority within level is only to resolve simultaneous requests of the same priority level.
From high to low, interrupt sources are listed below:
IE0
TF0
IE1
TF1
RI or TI
IP: INTERRUPT PRIORITY REGISTER. BIT ADDRESSABLE.
If the bit is 0, the corresponding interrupt has a lower priority and if the bit is 1 the corresponding interrupt has a higher priority.
–
–
–
PS
PT1
PX1
PT0
PX0
–
IP.7
Not implemented, reserved for future use.*
–
IP.6
Not implemented, reserved for future use.*
–
IP.5
Not implemented, reserved for future use.*
PS
IP.4
Defines the Serial Port interrupt priority level.
PT1
IP.3
Defines the Timer 1 interrupt priority level.
PX1
IP.2
Defines External Interrupt 1 priority level.
PT0
IP.1
Defines the Timer 0 interrupt priority level.
PX0
IP.0
Defines the External Interrupt 0 priority level.
*
User software should not write 1s to reserved bits. These bits may be used in future 80C51 products to invoke new features.
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and instruction set
TCON: TIMER/COUNTER CONTROL REGISTER. BIT ADDRESSABLE.
TF1
TR1
TF0
TR0
IE1
IT1
IE0
IT0
TF1
TCON.7
Timer 1 overflow flag. Set by hardware when the Timer/Counter 1 overflows. Cleared by hardware as
processor vectors to the interrupt service routine.
TR1
TCON.6
Timer 1 run control bit. Set/cleared by software to turn Timer/Counter 1 ON/OFF.
TF0
TCON.5
Timer 0 overflow flag. Set by hardware when the Timer/Counter 0 overflows. Cleared by hardware as
processor vectors to the service routine.
TR0
TCON.4
Timer 0 run control bit. Set/cleared by software to turn Timer/Counter 0 ON/OFF.
IE1
TCON.3
External Interrupt 1 edge flag. Set by hardware when External Interrupt edge is detected. Cleared by
hardware when interrupt is processed.
IT1
TCON.2
Interrupt 1 type control bit. Set/cleared by software to specify falling edge/low level triggered External
Interrupt.
IE0
TCON.1
External Interrupt 0 edge flag. Set by hardware when External Interrupt edge detected. Cleared by
hardware when interrupt is processed.
IT0
TCON.0
Interrupt 0 type control bit. Set/cleared by software to specify falling edge/low level triggered External
Interrupt.
TMOD: TIMER/COUNTER MODE CONTROL REGISTER. NOT BIT ADDRESSABLE.
GATE
C/T
M1
M0
GATE
C/T
M1
M0
Timer 1
Timer 0
GATE
When TRx (in TCON) is set and GATE = 1, TIMER/COUNTERx will run only while INTx pin is high (hardware control).
When GATE = 0, TIMER/COUNTERx will run only while TRx = 1 (software control).
C/T
Timer or Counter selector. Cleared for Timer operation (input from internal system clock). Set for Counter operation
(input from Tx input pin).
M1
Mode selector bit. (NOTE 1)
M0
Mode selector bit. (NOTE 1)
NOTE 1:
M1
M0
Operating Mode
0
0
0
13-bit Timer (8048 compatible)
0
1
1
16-bit Timer/Counter
1
0
2
8-bit Auto-Reload Timer/Counter
1
1
3
(Timer 0) TL0 is an 8-bit Timer/Counter controlled by the standart Timer 0
control bits. TH0 is an8-bit Timer and is controlled by Timer 1 control bits.
1
1
3
(Timer 1) Timer/Counter 1 stopped.
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TIMER SET-UP
Tables 2 through 5 give some values for TMOD which can be used to set up Timer 0 in different modes.
It is assumed that only one timer is being used at a time. If it is desired to run Timers 0 and 1 simultaneously, in any mode, the
value in TMOD for Timer 0 must be ORed with the value shown for Timer 1 (Tables 5 and 6).
For example, if it is desired to run Timer 0 in mode 1 GATE (external control), and Timer 1 in mode 2 COUNTER, then the value
that must be loaded into TMOD is 69H (09H from Table 2 ORed with 60H from Table 5).
Moreover, it is assumed that the user, at this point, is not ready to turn the timers on and will do that at a different point in the
program by setting bit TRx (in TCON) to 1.
TIMER/COUNTER 0
Table 2.
As a Timer:
TMOD
MODE
TIMER 0
INTERNAL
EXTERNAL
FUNCTION
CONTROL
CONTROL
(NOTE 1)
(NOTE 2)
0
13-bit Timer
00H
08H
1
16-bit Timer
01H
09H
2
8-bit Auto-Reload
02H
0AH
3
Two 8-bit Timers
03H
0BH
Table 3.
As a Counter:
TMOD
MODE
COUNTER 0
INTERNAL
EXTERNAL
FUNCTION
CONTROL
CONTROL
(NOTE 1)
(NOTE 2)
0
13-bit Timer
04H
0CH
1
16-bit Timer
05H
0DH
2
8-bit Auto-Reload
06H
0EH
3
One 8-bit Counter
07H
0FH
NOTES:
1. The timer is turned ON/OFF by setting/clearing bit TR0 in the software.
2. The Timer is turned ON/OFF by the 1-to-0 transition on INT0 (P3.2) when TR0 = 1 (hardware control).
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and instruction set
TIMER/COUNTER 1
Table 4.
As a Timer:
TMOD
MODE
TIMER 1
INTERNAL
EXTERNAL
FUNCTION
CONTROL
CONTROL
(NOTE 1)
(NOTE 2)
0
13-bit Timer
00H
80H
1
16-bit Timer
10H
90H
2
8-bit Auto-Reload
20H
A0H
3
Does not run
30H
B0H
Table 5.
As a Counter:
TMOD
MODE
COUNTER 1
INTERNAL
EXTERNAL
FUNCTION
CONTROL
CONTROL
(NOTE 1)
(NOTE 2)
0
13-bit Timer
40H
C0H
1
16-bit Timer
50H
D0H
2
8-bit Auto-Reload
60H
E0H
3
Not available
–
–
NOTES:
1. The timer is turned ON/OFF by setting/clearing bit TR1 in the software.
2. The Timer is turned ON/OFF by the 1-to-0 transition on INT1 (P3.2) when TR1 = 1 (hardware control).
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and instruction set
SCON: SERIAL PORT CONTROL REGISTER. BIT ADDRESSABLE.
SM0
SM1
SM2
REN
TB8
RB8
TI
RI
SM0
SCON.7
Serial Port mode specifier. (NOTE 1)
SM1
SCON.6
Serial Port mode specifier. (NOTE 1)
SM2
SCON.5
Enables the multiprocessor communication feature in modes 2 & 3. In mode 2 or 3, if SM2 is set to 1 then
RI will not be activated if the received 9th data bit (RB8) is 0. In mode 1, if SM2 = 1 then RI will not be
activated if a valid stop bit was not received. In mode 0, SM2 should be 0. (See Table 6.)
REN
SCON.4
Set/Cleared by software to Enable/Disable reception.
TB8
SCON.3
The 9th bit that will be transmitted in modes 2 & 3. Set/Cleared by software.
RB8
SCON.2
In modes 2 & 3, is the 9th data bit that was received. In mode 1, if SM2 = 0, RB8 is the stop bit that was
received. In mode 0, RB8 is not used.
TI
SCON.1
Transmit interrupt flag. Set by hardware at the end of the 8th bit time in mode 0, or at the beginning of the
stop bit in the other modes. Must be cleared by software.
RI
SCON.0
Receive interrupt flag. Set by hardware at the end of the 8th bit time in mode 0, or halfway through the
stop bit time in the other modes (except see SM2). Must be cleared by software.
NOTE 1:
SM0
SM1
Mode
Description
Baud Rate
0
0
0
Shift Register
FOSC./12
0
1
1
8-bit UART
Variable
1
0
2
9-bit UART
FOSC./64 or FOSC./32
1
1
3
9-bit UART
Variable
SERIAL PORT SET-UP:
Table 6.
MODE
SCON
SM2 VARIATION
0
10H
Single Processor
1
50H
Environment
2
90H
(SM2 = 0)
3
D0H
0
NA
Multiprocessor
1
70H
Environment
2
B0H
(SM2 = 1)
3
F0H
GENERATING BAUD RATES
Serial Port in Mode 0:
Mode 0 has a fixed baud rate which is 1/12 of the oscillator frequency. To run the serial port in this mode none of the
Timer/Counters need to be set up. Only the SCON register needs to be defined.
Baud Rate + Osc Freq
12
Serial Port in Mode 1:
Mode 1 has a variable baud rate. The baud rate is generated by Timer 1.
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USING TIMER/COUNTER 1 TO GENERATE BAUD RATES:
For this purpose, Timer 1 is used in mode 2 (Auto-Reload). Refer to Timer Setup section of this chapter.
Baud Rate +
K Osc Freq
32 12 [256 * (TH1)]
If SMOD = 0, then K = 1.
If SMOD = 1, then K = 2 (SMOD is in the PCON register).
Most of the time the user knows the baud rate and needs to know the reload value for TH1.
TH1 + 256 * K Osc Freq
384 baud rate
TH1 must be an integer value. Rounding off TH1 to the nearest integer may not produce the desired baud rate. In this case, the
user may have to choose another crystal frequency.
Since the PCON register is not bit addressable, one way to set the bit is logical ORing the PCON register (i.e., ORL
PCON,#80H). The address of PCON is 87H.
SERIAL PORT IN MODE 2:
The baud rate is fixed in this mode and is 1/32 or 1/64 of the oscillator frequency, depending on the value of the SMOD bit in
the PCON register.
In this mode none of the Timers are used and the clock comes from the internal phase 2 clock.
SMOD = 1, Baud Rate = 1/32 Osc Freq.
SMOD = 0, Baud Rate = 1/64 Osc Freq.
To set the SMOD bit: ORL PCON,#80H. The address of PCON is 87H.
SERIAL PORT IN MODE 3:
The baud rate in mode 3 is variable and sets up exactly the same as in mode 1.
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80C51 FAMILY INSTRUCTION SET
Table 7.
80C51 Instruction Set Summary
Interrupt Response Time: Refer to Hardware Description Chapter.
Instructions that Affect Flag Settings(1)
Instruction
Flag
Instruction
Flag
C
OV
AC
C
OV AC
ADD
X
X
X
CLR C
0
ADDC
X
X
X
CPL C
X
SUBB
X
X
X
ANL C,bit
X
MUL
0
X
ANL C,/bit
X
DIV
0
X
ORL C,bit
X
DA
X
ORL C,/bit
X
RRC
X
MOV C,bit
X
RLC
X
CJNE
X
SETB C
1
(1)Note that operations on SFR byte address 208 or bit addresses 209-215 (i.e., the PSW or bits in the PSW) will also affect flag settings.
Notes on instruction set and addressing modes:
Rn
Register R7-R0 of the currently selected Register Bank.
direct
8-bit internal data location’s address. This could be an Internal Data RAM location (0-127) or a SFR [i.e., I/O port,
control register, status register, etc. (128-255)].
@Ri
8-bit internal data RAM location (0-255) addressed indirectly through register R1 or R0.
#data
8-bit constant included in the instruction.
#data 16
16-bit constant included in the instruction
addr 16
16-bit destination address. Used by LCALL and LJMP. A branch can be anywhere within the 64k-byte Program
Memory address space.
addr 11
11-bit destination address. Used by ACALL and AJMP. The branch will be within the same 2k-byte page of
program memory as the first byte of the following instruction.
rel
Signed (two’s complement) 8-bit offset byte. Used by SJMP and all conditional jumps. Range is –128 to +127
bytes relative to first byte of the following instruction.
bit
Direct Addressed bit in Internal Data RAM or Special Function Register.
OSCILLATOR
MNEMONIC
DESCRIPTION
BYTE
PERIOD
ARITHMETIC OPERATIONS
ADD
A,Rn
Add register to Accumulator
1
12
ADD
A,direct
Add direct byte to Accumulator
2
12
ADD
A,@Ri
Add indirect RAM to Accumulator
1
12
ADD
A,#data
Add immediate data to Accumulator
2
12
ADDC
A,Rn
Add register to Accumulator with carry
1
12
ADDC
A,direct
Add direct byte to Accumulator with carry
2
12
ADDC
A,@Ri
Add indirect RAM to Accumulator with carry
1
12
ADDC
A,#data
Add immediate data to ACC with carry
2
12
SUBB
A,Rn
Subtract Register from ACC with borrow
1
12
SUBB
A,direct
Subtract direct byte from ACC with borrow
2
12
SUBB
A,@Ri
Subtract indirect RAM from ACC with borrow
1
12
SUBB
A,#data
Subtract immediate data from ACC with borrow
2
12
INC
A
Increment Accumulator
1
12
INC
Rn
Increment register
1
12
All mnemonics copyrighted © Intel Corporation 1980
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Table 7.
80C51 Instruction Set Summary (Continued)
OSCILLATOR
MNEMONIC
DESCRIPTION
BYTE
PERIOD
ARITHMETIC OPERATIONS (Continued)
INC
direct
Increment direct byte
2
12
INC
@Ri
Increment indirect RAM
1
12
DEC
A
Decrement Accumulator
1
12
DEC
Rn
Decrement Register
1
12
DEC
direct
Decrement direct byte
2
12
DEC
@Ri
Decrement indirect RAM
1
12
INC
DPTR
Increment Data Pointer
1
24
MUL
AB
Multiply A and B
1
48
DIV
AB
Divide A by B
1
48
DA
A
Decimal Adjust Accumulator
1
12
LOGICAL OPERATIONS
ANL
A,Rn
AND Register to Accumulator
1
12
ANL
A,direct
AND direct byte to Accumulator
2
12
ANL
A,@Ri
AND indirect RAM to Accumulator
1
12
ANL
A,#data
AND immediate data to Accumulator
2
12
ANL
direct,A
AND Accumulator to direct byte
2
12
ANL
direct,#data
AND immediate data to direct byte
3
24
ORL
A,Rn
OR register to Accumulator
1
12
ORL
A,direct
OR direct byte to Accumulator
2
12
ORL
A,@Ri
OR indirect RAM to Accumulator
1
12
ORL
A,#data
OR immediate data to Accumulator
2
12
ORL
direct,A
OR Accumulator to direct byte
2
12
ORL
direct,#data
OR immediate data to direct byte
3
24
XRL
A,Rn
Exclusive-OR register to Accumulator
1
12
XRL
A,direct
Exclusive-OR direct byte to Accumulator
2
12
XRL
A,@Ri
Exclusive-OR indirect RAM to Accumulator
1
12
XRL
A,#data
Exclusive-OR immediate data to Accumulator
2
12
XRL
direct,A
Exclusive-OR Accumulator to direct byte
2
12
XRL
direct,#data
Exclusive-OR immediate data to direct byte
3
24
CLR
A
Clear Accumulator
1
12
CPL
A
Complement Accumulator
1
12
RL
A
Rotate Accumulator left
1
12
RLC
A
Rotate Accumulator left through the carry
1
12
RR
A
Rotate Accumulator right
1
12
RRC
A
Rotate Accumulator right through the carry
1
12
SWAP
A
Swap nibbles within the Accumulator
1
12
DATA TRANSFER
MOV
A,Rn
Move register to Accumulator
1
12
MOV
A,direct
Move direct byte to Accumulator
2
12
MOV
A,@Ri
Move indirect RAM to Accumulator
1
12
All mnemonics copyrighted © Intel Corporation 1980
1997 Sept 18
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80C51 family programmer’s guide
80C51 Family
and instruction set
Table 7.
80C51 Instruction Set Summary (Continued)
OSCILLATOR
MNEMONIC
DESCRIPTION
BYTE
PERIOD
DATA TRANSFER (Continued)
MOV
A,#data
Move immediate data to Accumulator
2
12
MOV
Rn,A
Move Accumulator to register
1
12
MOV
Rn,direct
Move direct byte to register
2
24
MOV
RN,#data
Move immediate data to register
2
12
MOV
direct,A
Move Accumulator to direct byte
2
12
MOV
direct,Rn
Move register to direct byte
2
24
MOV
direct,direct
Move direct byte to direct
3
24
MOV
direct,@Ri
Move indirect RAM to direct byte
2
24
MOV
direct,#data
Move immediate data to direct byte
3
24
MOV
@Ri,A
Move Accumulator to indirect RAM
1
12
MOV
@Ri,direct
Move direct byte to indirect RAM
2
24
MOV
@Ri,#data
Move immediate data to indirect RAM
2
12
MOV
DPTR,#data16
Load Data Pointer with a 16-bit constant
3
24
MOVC
A,@A+DPTR
Move Code byte relative to DPTR to ACC
1
24
MOVC
A,@A+PC
Move Code byte relative to PC to ACC
1
24
MOVX
A,@Ri
Move external RAM (8-bit addr) to ACC
1
24
MOVX
A,@DPTR
Move external RAM (16-bit addr) to ACC
1
24
MOVX
A,@Ri,A
Move ACC to external RAM (8-bit addr)
1
24
MOVX
@DPTR,A
Move ACC to external RAM (16-bit addr)
1
24
PUSH
direct
Push direct byte onto stack
2
24
POP
direct
Pop direct byte from stack
2
24
XCH
A,Rn
Exchange register with Accumulator
1
12
XCH
A,direct
Exchange direct byte with Accumulator
2
12
XCH
A,@Ri
Exchange indirect RAM with Accumulator
1
12
XCHD
A,@Ri
Exchange low-order digit indirect RAM with ACC
1
12
BOOLEAN VARIABLE MANIPULATION
CLR
C
Clear carry
1
12
CLR
bit
Clear direct bit
2
12
SETB
C
Set carry
1
12
SETB
bit
Set direct bit
2
12
CPL
C
Complement carry
1
12
CPL
bit
Complement direct bit
2
12
ANL
C,bit
AND direct bit to carry
2
24
ANL
C,/bit
AND complement of direct bit to carry
2
24
ORL
C,bit
OR direct bit to carry
2
24
ORL
C,/bit
OR complement of direct bit to carry
2
24
MOV
C,bit
Move direct bit to carry
2
12
MOV
bit,C
Move carry to direct bit
2
24
JC
rel
Jump if carry is set
2
24
JNC
rel
Jump if carry not set
2
24
All mnemonics copyrighted © Intel Corporation 1980
1997 Sept 18
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80C51 family programmer’s guide
80C51 Family
and instruction set
Table 7.
80C51 Instruction Set Summary (Continued)
OSCILLATOR
MNEMONIC
DESCRIPTION
BYTE
PERIOD
BOOLEAN VARIABLE MANIPULATION (Continued)
JB
rel
Jump if direct bit is set
3
24
JNB
rel
Jump if direct bit is not set
3
24
JBC
bit,rel
Jump if direct bit is set and clear bit
3
24
PROGRAM BRANCHING
ACALL
addr11
Absolute subroutine call
2
24
LCALL
addr16
Long subroutine call
3
24
RET
Return from subroutine
1
24
RETI
Return from interrupt
1
24
AJMP
addr11
Absolute jump
2
24
LJMP
addr16
Long jump
3
24
SJMP
rel
Short jump (relative addr)
2
24
JMP
@A+DPTR
Jump indirect relative to the DPTR
1
24
JZ
rel
Jump if Accumulator is zero
2
24
JNZ
rel
Jump if Accumulator is not zero
2
24
CJNE
A,direct,rel
Compare direct byte to ACC and jump if not equal
3
24
CJNE
A,#data,rel
Compare immediate to ACC and jump if not equal
3
24
CJNE
RN,#data,rel
Compare immediate to register and jump if not
3
24
equal
CJNE
@Ri,#data,rel
Compare immediate to indirect and jump if not
3
24
equal
DJNZ
Rn,rel
Decrement register and jump if not zero
2
24
DJNZ
direct,rel
Decrement direct byte and jump if not zero
3
24
NOP
No operation
1
12
All mnemonics copyrighted © Intel Corporation 1980
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and instruction set
INSTRUCTION DEFINITIONS
ACALL addr11
Function:
Absolute Call
Description:
ACALL unconditionally calls a subroutine located at the indicated address. The instruction increments
the PC twice to obtain the address of the following instruction, then pushes the 16-bit result onto the
stack (low-order byte first) and increments the Stack Pointer twice. The destination address is obtained
by successively concatenating the five high-order bits of the incremented PC, opcode bits 7-5, and the
second byte of the instruction. The subroutine called must therefore start within the same 2k block of the
program memory as the first byte of the instruction following ACALL. No flags are affected.
Example:
Initially SP equals 07H. The label “SUBRTN” is at program memory location 0345 H. After executing the
instruction,
ACALL SUBRTN
at location 0123H, SP will contain 09H, internal RAM locations 08H and 09H will contain 25H and 01H,
respectively, and the PC will contain 0345H.
Bytes:
2
Cycles:
2
Encoding:
a10 a9 a8 1
0
0
0 1
a7 a6 a5 a4
a3 a2 a1 a0
Operation:
ACALL
(PC) ← (PC) + 2
(SP) ← (SP) + 1
(SP) ← (PC7-0)
(SP) ← (SP) + 1
(SP) ← (PC15-8)
(PC10-0) ← page address
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and instruction set
ADD A,<src-byte>
Function:
Add
Description:
ADD adds the byte variable indicated to the Accumulator, leaving the result in the Accumulator. The carry
and auxiliary-carry flags are set, respectively, if there is a carry-out from bit 7 or bit 3, and cleared
otherwise. When adding unsigned integers, the carry flag indicates an overflow occurred.
OV is set if there is a carry-out of bit 6 but not out of bit 7, or a carry-out of bit 7 but not bit 6; otherwise OV
is cleared. When adding signed integers, OV indicates a negative number produced as the sum of two
positive operands, or a positive sum from two negative operands.
Four source operand addressing modes are allowed: register, direct, register-indirect, or immediate.
Example:
The Accumulator holds 0C3H (11000011B) and register 0 holds 0AAH (10101010B). The instruction,
ADD A,R0
will leave 6DH (01101101B) in the Accumulator with the AC flag cleared and both the Carry flag and OV
set to 1.
ADD A,Rn
Bytes:
1
Cycles:
1
Encoding:
0
0
1 0
1
r
r
r
Operation:
ADD
(A) ← (A) + (Rn)
ADD A,direct
Bytes:
2
Cycles:
1
Encoding:
0
0
1 0
0
1
0 1
direct address
Operation:
ADD
(A) ← (A) + (direct)
ADD A,@Ri
Bytes:
1
Cycles:
1
Encoding:
0
0
1 0
0
1
1 i
Operation:
ADD
(A) ← (A) + ((Ri))
ADD A,#data
Bytes:
2
Cycles:
1
Encoding:
0
0
1 0
0
1
0 0
immediate data
Operation:
ADD
(A) ← (A) + #data
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and instruction set
ADDC A,<src-byte>
Function:
Add with Carry
Description:
ADDC simultaneously adds the byte variable indicated, the carry flag and the Accumulator contents,
leaving the result in the Accumulator. The carry and auxiliary-carry flags are set, respectively, if there is a
carry-out from bit 7 or bit 3, and cleared otherwise. When adding unsigned integers, the carry flag
indicates an overflow occurred.
OV is set if there is a carry-out of bit 6 but not out of bit 7, or a carry-out of bit 7 but not out of bit 6;
otherwise OV is cleared. When adding signed integers, OV indicates a negative number produced as the
sum of two positive operands, or a positive sum from two negative operands.
Four source operand addressing modes are allowed: register, direct, register-indirect, or immediate.
Example:
The Accumulator holds 0C3H (11000011B) and register 0 holds 0AAH (10101010B) with the carry flag set.
The instruction,
ADDC A,R0
will leave 6EH (01101110B) in the Accumulator with AC cleared and both the Carry flag and OV set to 1.
ADDC A,Rn
Bytes:
1
Cycles:
1
Encoding:
0
0
1 1
1
r
r
r
Operation:
ADDC
(A) ← (A) + (C) + (Rn)
ADDC A,direct
Bytes:
2
Cycles:
1
Encoding:
0
0
1 1
0
1
0 1
direct address
Operation:
ADDC
(A) ← (A) + (C) + (direct)
ADDC A,@Ri
Bytes:
1
Cycles:
1
Encoding:
0
0
1 1
0
1
1 i
Operation:
ADDC
(A) ← (A) + (C) + ((Ri))
ADDC A,#data
Bytes:
2
Cycles:
1
Encoding:
0
0
1 1
0
1
0 0
immediate data
Operation:
ADDC
(A) ← (A) + (C) + #data
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and instruction set
AJMP addr11
Function:
Absolute Jump
Description:
AJMP transfers program execution to the indicated address, which is formed at run-time by concatenating
the high-order five bits of the PC (after incrementing the PC twice), opcode bits 7-5, and the second byte
of the instruction. The destination must therefore be within the same 2k block of program memory as the
first byte of the instruction following AJMP.
Example:
The label “JMPADR” is at program memory location 0123H. The instruction,
AJMP JMPADR
is at location 0345H and will load the PC with 0123H.
Bytes:
2
Cycles:
2
Encoding:
a10 a9 a8 0
0
0
0 1
a7 a6 a5 a4
a3 a2 a1 a0
Operation:
AJMP
(PC) ← (PC) + 2
(PC10-0) ← page address
ANL <dest-byte>,<src-byte>
Function:
Logical-AND for byte variables
Description:
ANL performs the bitwise logical-AND operation between the variables indicated and stores the results in
the destination variable. No flags are affected.
The two operands allow six addressing mode combinations. When the destination is the Accumulator, the
source can use register, direct, register-indirect, or immediate addressing; when the destination is a direct
address, the source can be the Accumulator or immediate data.
Note: When this instruction is used to modify an output port, the value used as the original port data will
be read from the output data latch, not the input pins.
Example:
If the Accumulator holds 0C3H (11000011B) and register 0 holds 55H (01010101B) then the instruction,
ANL A,R0
will leave 41H (01000001B) in the Accumulator.
When the destination is a directly addressed byte, this instruction will clear combinations of bits in any
RAM location or hardware register. The mask byte determining the pattern of bits to be cleared would
either be a constant contained in the instruction or a value computed in the Accumulator at run-time. The
instruction,
ANL P1,#01110011B
will clear bits 7, 3, and 2 of output port 1.
1997 Sept 18
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and instruction set
ANL A,Rn
Bytes:
1
Cycles:
1
Encoding:
0
1
0 1
1
r
r
r
Operation:
ANL
(A) ← (A) ∧ (Rn)
ANL A,direct
Bytes:
2
Cycles:
1
Encoding:
0
1
0 1
0
1
0 1
direct address
Operation:
ANL
(A) ← (A) ∧ (direct)
ANL A,@Ri
Bytes:
1
Cycles:
1
Encoding:
0
1
0 1
0
1
1 i
Operation:
ANL
(A) ← (A) ∧ ((Ri))
ANL A,#data
Bytes:
2
Cycles:
1
Encoding:
0
1
0 1
0
1
0 0
immediate data
Operation:
ANL
(A) ← (A) ∧ #data
ANL direct,A
Bytes:
2
Cycles:
1
Encoding:
0
1
0 1
0
0
1 0
direct address
Operation:
ANL
(A) ←〈direct) ∧ (A)
ANL direct,#data
Bytes:
3
Cycles:
2
Encoding:
0
1
0 1
0
0
1 1
direct address
immediate data
Operation:
ANL
(direct) ←〈direct) ∧ #data
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and instruction set
ANL C,<src-bit>
Function:
Logical-AND for bit variables
Description:
If the Boolean value of the source bit is a logical 0 then clear the carry flag; otherwise leave the carry flag
in its current state. A slash (“/”) preceding the operand in the assembly language indicates that the logical
complement of the addressed bit is used as the source value, but the source bit itself is not affected. No
other flags are affected.
Only direct addressing is allowed for the source operand.
Example:
Set the carry flag if, and only if, P1.0 = 1, ACC.7 = 1, and OV = 0:
MOV
C,P1.0
;LOAD CARRY WITH INPUT PIN STATE
ANL
C,ACC.7;AND CARRY WITH ACCUM. BIT 7
ANL
C,/OV
;AND WITH INVERSE OF OVERFLOW FLAG
ANL C,bit
Bytes:
2
Cycles:
2
Encoding:
1
0
0 0
0
0
1 0
bit address
Operation:
ANL
(C) ← (C) ∧ (bit)
ANL C,/bit
Bytes:
2
Cycles:
2
Encoding:
1
0
1 1
0
0
0 0
bit address
Operation:
ANL
(C) ← (C) ∧ (bit)
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80C51 Family
and instruction set
CJNE <dest-byte>,<src-byte>,rel
Function:
Compare and Jump if Not Equal
Description:
CJNE compares the magnitudes of the first two operands, and branches if their values are not equal. The
branch destination is computed by adding the signed relative-displacement in the last instruction byte to
the PC, after incrementing the PC to the start of the next instruction. The carry flag is set if the unsigned
integer value of <dest-byte> is less than the unsigned integer value of <src-byte>; otherwise, the carry is
cleared. Neither operand is affected.
The first two operands allow four addressing mode combinations: the Accumulator may be compared with
any directly addressed byte or immediate data, and any indirect RAM location or working register can be
compared with an immediate constant.
Example:
The Accumulator contains 34H. Register 7 contains 56H. The first instruction in the sequence,
CJNE
R7,#60H,NOT_EQ
;
...
....
;
R7 = 60H.
NOT_EQ
JC
REQ_LOW
;
IF R7 < 60H.
;
...
....
;
R7 > 60H.
sets the carry flag and branches to the instruction at label NOT_EQ. By testing the carry flag, this
instruction determines whether R7 is greater or less than 60H.
If the data being presented to Port 1 is also 34H, then the instruction,
WAIT: CJNE A,P1,WAIT
clears the carry flag and continues with the next instruction in sequence, since the Accumulator does
equal the data read from P1. (If some other value was being input on P1, the program will loop at this
point until the P1 data changes to 34H.)
CJNE A,direct,rel
Bytes:
3
Cycles:
2
Encoding:
1
0
1 1
0
1
0 1
direct address
rel. address
Operation:
(PC) ← (PC) + 3
IF (A) < > (direct)
THEN
(PC) ← (PC) + relative offset
IF (A) < (direct)
THEN.
(C) ← 1
ELSE
(C) ← 0
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and instruction set
CJNE A,#data,rel
Bytes:
3
Cycles:
2
Encoding:
1
0
1 1
0
1
0 0
immediate data
rel. address
Operation:
(PC) ← (PC) + 3
IF (A) < > data
THEN
(PC) ← (PC) + relative offset
IF (A) < data
THEN
(C) ← 1
ELSE
(C) ← 0
CJNE Rn,#data,rel
Bytes:
3
Cycles:
2
Encoding:
1
0
1 1
1
r
r
r
immediate data
rel. address
Operation:
(PC) ← (PC) + 3
IF (Rn) < > data
THEN
(PC) ← (PC) + relative offset
IF (Rn) < data
THEN
(C) ← 1
ELSE
(C) ← 0
CJNE @Ri,#data,rel
Bytes:
3
Cycles:
2
Encoding:
1
0
1 1
0
1
1 i
immediate data
rel. address
Operation:
(PC) ← (PC) + 3
IF ((Ri)) < > data
THEN
(PC) ← (PC) + relative offset
IF ((Ri)) < data
THEN
(C) ← 1
ELSE
(C) ← 0
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and instruction set
CLR A
Function:
Clear Accumulator
Description:
The Accumulator is cleared (all bits reset to zero). No flags are affected.
Example:
The Accumulator contains 5CH (01011100B). The instruction,
CLR A
will leave the Accumulator set to 00H (00000000B).
Bytes:
1
Cycles:
1
Encoding:
1
1
1 0
0
1
0 0
Operation:
CLR
(A) ← 0
CLR bit
Function:
Clear bit
Description:
The indicated bit is cleared (reset to zero). No other flags are affected. CLR can operate on the carry flag
or any directly addressable bit.
Example:
Port 1 has previously been written with 5DH (01011101B). The instruction,
CLR P1.2
will leave the port set to 59H (01011001B).
CLR C
Bytes:
1
Cycles:
1
Encoding:
1
1
0 0
0
0
1 1
Operation:
CLR
(C) ← 0
CLR bit
Bytes:
2
Cycles:
1
Encoding:
1
1
0 0
0
0
1 0
bit address
Operation:
CLR
(bit) ← 0
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and instruction set
CPL A
Function:
Complement Accumulator
Description:
Each bit of the Accumulator is logically complemented (one’s complement). Bits which previously
contained a one are changed to a zero and vice-versa. No flags are affected.
Example:
The Accumulator contains 5CH (01011100B). The instruction,
CPL A
will leave the Accumulator set to 0A3H (10100011B).
Bytes:
1
Cycles:
1
Encoding:
1
1
1 1
0
1
0 0
Operation:
CPL
(A) ← (A)
CPL bit
Function:
Complement bit
Description:
The bit variable specified is complemented. A bit which had been a one is changed to zero and
vice-versa. No other flags are affected. CPL can operate on the carry or any directly addressable bit.
Note: When this instruction is used to modify an output pin, the value used as the original data will be read
from the output data latch, not the input pin.
Example:
Port 1 has previously been written with 5DH (01011101B). The instruction sequence,
CPL
P1.1
CPL
P1.2
will leave the port set to 5BH (01011011B).
CPL C
Bytes:
1
Cycles:
1
Encoding:
1
0
1 1
0
0
1 1
Operation:
CPL
(C) ← (C)
CPL bit
Bytes:
2
Cycles:
1
Encoding:
1
0
1 1
0
0
1 0
bit address
Operation:
CPL
(bit) ← (bit)
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and instruction set
DA A
Function:
Decimal-adjust Accumulator for Addition
Description:
DA A adjusts the eight-bit value in the Accumulator resulting from the earlier addition of two variable (each
in packed-BCD format), producing two four-bit digits. Any ADD or ADDC instruction may have been used
to perform the addition.
If Accumulator bits 3-0 are greater than nine (xxx1010-xxx1111), or if the AC flag is one, six is added to
the Accumulator, producing the proper BCD digit in the low-order nibble. This internal addition would set
the carry flag if a carry-out of the low-order four-bit field propagated through all high-order bits, but it would
not clear the carry flag otherwise.
If the carry flag is now set, or if the four high-order bits now exceed nine (1010xxx-111xxxx), these
high-order bits are incremented by six, producing the proper BCD digit in the high-order nibble. Again, this
would set the carry flag if there was a carry-out of the high-order bits, but wouldn’t clear the carry. The
carry flag thus indicates if the sum of the original two BCD variables is greater than 100, allowing multiple
precision decimal addition. OV is not affected.
All of this occurs during the one instruction cycle. Essentially, this instruction performs the decimal
conversion by adding 00H, 06H, 60H, or 66H to the Accumulator, depending on initial Accumulator and
PSW conditions.
Note: DA A cannot simply convert a hexadecimal number in the Accumulator to BCD notation, nor does
DA A apply to decimal subtraction.
Example:
The Accumulator holds the value 56H (01010110B) representing the packed BCD digits of the decimal
number 56. Register 3 contains the value 67H (01100111B) representing the packed BCD digits of the
decimal number 67. The carry flag is set.. The instruction sequence,
ADDC
A,R3
DA
A
will first perform a standard two’s-complement binary addition, resulting in the value 0BEH (10111110B) in
the Accumulator. The carry and auxiliary carry flags will be cleared.
The Decimal Adjust instruction will then alter the Accumulator to the value 24H (00100100B), indicating
the packed BCD digits of the decimal number 24, the low-order two digits of the decimal sum of 56, 67,
and the carry-in. The carry flag will be set by the Decimal Adjust instruction, indicating that a decimal
overflow occurred. The true sum 56, 67, and 1 is 124.
BCD variables can be incremented or decremented by adding 01H or 99H. If the Accumulator initially
holds 30H (representing the digits of 30 decimal), the the instruction sequence,
ADD
A,#99H
DA
A
will leave the carry set and 29H in the Accumulator, since 30 + 99 = 129. The low-order byte of the sum
can be interpreted to mean 30 – 1 = 29.
Bytes:
1
Cycles:
1
Encoding:
1
1
0 1
0
1
0 0
Operation:
DA
–contents of Accumulator are BCD
IF
[[(A3-0) > 9] ∨ [(AC) = 1]]
THEN(A3-0) ← (A3-0) + 6
AND
IF
[[(A7-4) > 9] ∨ [(C) = 1]]
THEN(A7-4) ← (A7-4) + 6
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80C51 Family
and instruction set
DEC byte
Function:
Decrement
Description:
The variable indicated is decremented by 1. An original value of 00H will underflow to 0FFH. No flags are
affected. Four operand addressing modes are allowed: accumulator, register, direct, or register-indirect.
Note: When this instruction is used to modify an output port, the value used as the original data will be
read from the output data latch, not the input pin.
Example:
Register 0 contains 7FH (01111111B). Internal RAM locations 7EH and 7FH contain 00H and 40H,
respectively. The instruction sequence,
DEC
@R0
DEC
R0
DEC
@R0
will leave register 0 set to 7EH and internal RAM locations 7EH and 7FH set to 0FFH and 3FH.
DEC A
Bytes:
1
Cycles:
1
Encoding:
0
0
0 1
0
1
0 0
Operation:
DEC
(A) ← (A) – 1
DEC Rn
Bytes:
1
Cycles:
1
Encoding:
0
0
0 1
1
r
r
r
Operation:
DEC
(Rn) ← (Rn) – 1
DEC direct
Bytes:
2
Cycles:
1
Encoding:
0
0
0 1
0
1
0 1
direct address
Operation:
DEC
(direct) ← (direct) – 1
DEC @Ri
Bytes:
1
Cycles:
1
Encoding:
0
0
0 1
0
1
1 i
Operation:
DEC
((Ri)) ← ((Ri)) – 1
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80C51 Family
and instruction set
DIV AB
Function:
Divide
Description:
DIV AB divides the unsigned eight-bit integer in the Accumulator by the unsigned eight-bit integer in
register B.
The Accumulator receives the integer part of the quotient; register B receives the integer remainder. The
carry and OV flags will be cleared.
Exception: if B had originally contained 00H, the values returned in the Accumulator and B-register will be
undefined and the overflow flag will be set. The carry flag is cleared in any case.
Example:
The Accumulator contains 251 (0FBH or 11111011B) and B contains 18 (12H or 00010010B). The
instruction,
DIV AB
will leave 13 in the Accumulator (0DH or 00001101B) and the value 17 (11H or 00010001B) in B, since
251 = (13 x 18) + 17. Carry and OV will both be cleared.
Bytes:
1
Cycles:
4
Encoding:
1
0
0 0
0
1
0 0
Operation:
DIV
(A)15-8 ← (A)/(B)
(B)7-0
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80C51 Family
and instruction set
DJNZ <byte>,<rel-addr>
Function:
Decrement and Jump if Not Zero
Description:
DJNZ decrements the location indicated by 1, and branches to the address indicated by the second
operand if the resulting value is not zero. An original value of 00H will underflow to 0FFH. No flags are
affected. The branch destination would be computed by adding the signed relative-displacement value in
the last instruction byte to the PC, after incrementing the PC to the first byte of the following instruction.
The location decremented may be a register or directly addressed byte.
Note: When this instruction is used to modify an output port, the value used as the original port data will
be read from the output data latch, not the input pins.
Example:
Internal RAM locations 40H, 50H, and 60H contain the values 01H, 70H, and 15H, respectively. The
instruction sequence,
DJNZ
40H,LABEL_1
DJNZ
50H,LABEL_2
DJNZ
60H,LABEL_3
will cause a jump to the instruction at LABEL_2 with the values 00h, 6FH, and 15H in the three RAM
locations. The first jump was not taken because the result was zero.
This instruction provides a simple was of executing a program loop a given number of times, or for adding
a moderate time delay (from 2 to 512 machine cycles) with a single instruction. The instruction sequence,
MOV
R2,#8
TOGGLE: CPL
P1.7
DJNZ
R2,TOGGLE
will toggle P1.7 eight times, causing four output pulses to appear at bit 7 of output Port 1. Each pulse will
last three machine cycles, two for DJNZ and one to alter the pin.
DJNZ Rn,rel
Bytes:
2
Cycles:
2
Encoding:
1
1
0 1
1
r
r
r
rel. address
Operation:
DJNZ
(PC) ← (PC) + 2
(Rn) ← (Rn) – 1
IF (Rn) > 0 or (Rn) < 0
THEN
(PC) ← (PC) + rel
DJNZ direct,rel
Bytes:
3
Cycles:
2
Encoding:
1
1
0 1
0
1
0 1
direct data
rel. address
Operation:
DJNZ
(PC) ← (PC) + 2
(direct) ← (direct) – 1
IF (direct) > 0 or (direct) < 0
THEN
(PC) ← (PC) + rel
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80C51 Family
and instruction set
INC <byte>
Function:
Increment
Description:
INC increments the indicated variable by 1. An original value of 0FFH will overflow to 00H. No flags are
affected. Three addressing modes are allowed: register, direct, or register-indirect.
Note: When this instruction is used to modify an output port, the value used as the original port data will
be read from the output data latch, not the input pins.
Example:
Register 0 contains 7EH (01111110B). Internal RAM locations 7EH and 7FH contain 0FFH and 40H,
respectively. The instruction sequence,
INC
@R0
INC
R0
INC
@R0
will leave register 0 set to 7FH and internal RAM locations 7EH and 7FH holding (respectively) 00H and
41H.
INC A
Bytes:
1
Cycles:
1
Encoding:
0
0
0 0
0
1
0 0
Operation:
INC
(A) ← (A) + 1
INC Rn
Bytes:
1
Cycles:
1
Encoding:
0
0
0 0
1
r
r
r
Operation:
INC
(Rn) ← (Rn) + 1
INC direct
Bytes:
2
Cycles:
1
Encoding:
0
0
0 0
0
1
0 1
direct address
Operation:
INC
(direct) ← (direct) + 1
INC @Ri
Bytes:
1
Cycles:
1
Encoding:
0
0
0 0
0
1
1 i
Operation:
INC
((Ri)) ← ((Ri)) + 1
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80C51 Family
and instruction set
INC DPTR
Function:
Increment Data Pointer
Description:
Increment the 16-bit data pointer by 1. A 16-bit increment (modulo 216) is performed; an overflow of the
low-order byte of the data pointer (DPL) from 0FFH to 00H will increment the high-order byte (DPH). No
flags are affected.
This is the only 16-bit register which can be incremented.
Example:
Registers DPH and DPL contain 12H and 0FEH, respectively. The instruction sequence,
INC
DPTR
INC
DPTR
INC
DPTR
will change DPH and DPL to 13H and 01H.
Bytes:
1
Cycles:
2
Encoding:
1
0
1 0
0
0
1 1
Operation:
INC
(DPTR) ← (DPTR) + 1
JB bit,rel
Function:
Jump if Bit set
Description:
If the indicated bit is a one, jump to the address indicated; otherwise proceed with the next instruction. The
branch destination is computed by adding the signed relative-displacement in the third instruction byte to
the PC, after incrementing the PC to the first byte of the next instruction. The bit tested is not modified. No
flags are affected.
Example:
The data present at input port 1 is 11001010B. The Accumulator holds 56 (01010110B). The instruction
sequence,
JB
P1.2,LABEL1
JB
ACC.2,LABEL2
will cause program execution to branch to the instruction at label LABEL2.
Bytes:
3
Cycles:
2
Encoding:
0
0
1 0
0
0
0 0
bit address
rel. address
Operation:
JB
(PC) ← (PC) + 3
IF (bit) = 1
THEN
(PC) ← (PC) + rel
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80C51 Family
and instruction set
JBC bit,rel
Function:
Jump if Bit is set and Clear bit
Description:
If the indicated bit is a one, branch to the address indicated; otherwise proceed with the next instruction.
The bit will not be cleared if it is already a zero. The branch destination is computed by adding the signed
relative-displacement in the third instruction byte to the PC, after incrementing the PC to the first byte of
the next instruction. No flags are affected.
Note: When this instruction is used to test an output pin, the value used as the original data will read from
the output data latch, not the input pin.
Example:
The Accumulator holds 56H (01010110B). The instruction sequence,
JBC
ACC.3,LABEL1
JBC
ACC.2,LABEL2
will cause program execution to continue at the instruction identified by the LABEL2, with the Accumulator
modified to 52H (01010010B).
Bytes:
3
Cycles:
2
Encoding:
0
0
0 1
0
0
0 0
bit address
rel. address
Operation:
JBC
(PC) ← (PC) + 3
IF (bit) = 1
THEN
(bit) ← 0
(PC) ← (PC) + rel
JC rel
Function:
Jump if Carry is set
Description:
If the carry flag is set, branch to the address indicated; otherwise proceed with the next instruction. The
branch destination is computed by adding the signed relative-displacement in the second instruction byte
to the PC, after incrementing the PC twice. No flags are affected.
Example:
The carry flag is cleared. The instruction sequence,
JC
LABEL1
CPL
C
JC
LABEL2
will set the carry and cause program execution to continue at the instruction identified by the label
LABEL2.
Bytes:
2
Cycles:
2
Encoding:
0
1
0 0
0
0
0 0
rel. address
Operation:
JC
(PC) ← (PC) + 2
IF (C) = 1
THEN
(PC) ← (PC) + rel
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80C51 Family
and instruction set
JMP @A+DPTR
Function:
Jump indirect
Description:
Add the eight-bit unsigned contents of the Accumulator with the sixteen-bit data pointer, and load the
resulting sum to the program counter. This will be the address for subsequent instruction fetches.
Sixteen-bit addition is performed (modulo 216): a carry-out from the low-order eight bits propagates
through the higher-order bits. Neither the Accumulator nor the Data Pointer is altered. No flags are
affected.
Example:
An even number from 0 to 6 is in the Accumulator. The following sequence of instructions will branch to
one of four AJMP instructions in a jump table starting at JMP_TBL:
MOV
DPTR,#JMP_TBL
JMP
@A+DPTR
JMP_TBL:
AJMP
LABEL0
AJMP
LABEL1
AJMP
LABEL2
AJMP
LABEL3
If the Accumulator equals 04H when starting this sequence, execution will jump to label LABEL2.
Remember that AJMP is a two-byte instruction, so the jump instructions start at every other address.
Bytes:
1
Cycles:
2
Encoding:
0
1
1 1
0
0
1 1
Operation:
JMP
(PC) ← (A) + (DPTR)
JNB bit,rel
Function:
Jump if Bit Not set
Description:
If the indicated bit is a zero, branch to the indicated address; otherwise proceed with the next instruction.
The branch destination is computed by adding the signed relative-displacement in the third instruction
byte to the PC, after incrementing the PC to the first byte of the next instruction. The bit tested is not
modified. No flags are affected.
Example:
The data present at input port 1 is 11001010B. The Accumulator holds 56H (01010110B). The instruction
sequence,
JNB
P1.3,LABEL1
JNB
ACC.3,LABEL2
will cause program execution to continue at the instruction at label LABEL2.
Bytes:
3
Cycles:
2
Encoding:
0
0
1 1
0
0
0 0
bit address
rel. address
Operation:
JNB
(PC) ← (PC) + 3
IF (bit) = 0
THEN
(PC) ← (PC) + rel
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and instruction set
JNC rel
Function:
Jump if Carry Not set
Description:
If the carry flag is a zero, branch to the address indicated; otherwise proceed with the next instruction. The
branch destination is computed by adding the signed relative-displacement in the second instruction byte
to the PC, after incrementing the PC twice to point to the next instruction. The carry flag is not modified.
Example:
The carry flag is set. The instruction sequence,
JNC
LABEL1
CPL
C
JNC
LABEL2
will clear the carry and cause program execution to continue at the instruction identified by the label
LABEL2.
Bytes:
2
Cycles:
2
Encoding:
0
1
0 1
0
0
0 0
rel. address
Operation:
JNC
(PC) ← (PC) + 2
IF (C) = 0
THEN
(PC) ← (PC) + rel
JNZ rel
Function:
Jump if Accumulator Not Zero
Description:
If any bit of the Accumulator is a one, branch to the indicated address; otherwise proceed with the next
instruction. The branch destination is computed by adding the signed relative-displacement in the second
instruction byte to the PC, after incrementing the PC twice. The Accumulator is not modified. No flags are
affected.
Example:
The Accumulator originally holds 00H. The instruction sequence,
JNZ
LABEL1
INC
A
JNZ
LABEL2
will set the Accumulator to 01H and continue at label LABEL2.
Bytes:
2
Cycles:
2
Encoding:
0
1
1 1
0
0
0 0
rel. address
Operation:
JNZ
(PC) ← (PC) + 2
IF A ≠ 0
THEN (PC) ← (PC) + rel
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and instruction set
JZ rel
Function:
Jump if Accumulator Zero
Description:
If all bits of the Accumulator are zero, branch to the indicated address; otherwise proceed with the next
instruction. The branch destination is computed by adding the signed relative-displacement in the second
instruction byte to the PC, after incrementing the PC twice. The Accumulator is not modified. No flags are
affected.
Example:
The Accumulator originally holds 01H. The instruction sequence,
JZ
LABEL1
DEC
A
JZ
LABEL2
will change the Accumulator to 00H and cause program execution to continue at the instruction identified
by the label LABEL2.
Bytes:
2
Cycles:
2
Encoding:
0
1
1 0
0
0
0 0
rel. address
Operation:
JZ
(PC) ← (PC) + 2
IF A = 0
THEN (PC) ← (PC) + rel
LCALL addr16
Function:
Long Call
Description:
LCALL calls a subroutine located at the indicated address. The instruction adds three to the program
counter to generate the address of the next instruction and then pushes the 16-bit result onto the stack
(low byte first), incrementing the Stack Pointer by two. The high-order and low-order bytes of the PC are
then loaded, respectively, with the second and third bytes of the LCALL instruction. Program execution
continues with the instruction at this address. The subroutine may therefore begin anywhere in the full
64k-byte program memory address space. No flags are affected.
Example:
Initially the Stack Pointer equals 07H. The label “SUBRTN” is assigned to program memory location
1234H. After executing the instruction,
LCALL SUBRTN
at location 0123H, the Stack Pointer will contain 09H, internal RAM locations 08H and 09H will contain
26H and 01H, and the PC will contain 1235H.
Bytes:
3
Cycles:
2
Encoding:
0
0
0 1
0
0
1 0
addr15-addr8
addr7-addr0
Operation:
LCALL
(PC) ← (PC) + 3
(SP) ← (SP) + 1
((SP)) ← (PC7-0)
(SP) ← (SP) + 1
((SP)) ← (PC15-8)
(PC) ← addr15-0
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and instruction set
LJMP addr16
(Implemented in 87C751 and 87C752 for in-circuit emulation only.)
Function:
Long Jump
Description:
LJMP causes an unconditional branch to the indicated address, by loading the high-order and low-order
bytes of the PC (respectively) with the second and third instruction bytes. The destination may therefore
be anywhere in the full 64k program memory address space. No flags are affected.
Example:
The label “JMPADR” is assigned to the instruction at program memory location 1234H. The instruction,
LJMP JMPADR
at location 0123H will load the program counter with 1234H.
Bytes:
3
Cycles:
2
Encoding:
0
0
0 0
0
0
1 0
addr15-addr8
addr7-addr0
Operation:
LJMP
(PC) ← addr15-0
MOV <dest-byte>,<src-byte>
Function:
Move byte variable
Description:
The byte variable indicated by the second operand is copied into the location specified by the first
operand. The source byte is not affected. No other register or flag is affected.
This is by far the most flexible operation. Fifteen combinations of source and destination addressing
modes are allowed.
Example:
Internal RAM location 30H holds 40H. The value of RAM location 40H is 10H. The data present at input
port 1 is 11001010B (0CAH). The instruction sequence,
MOV
R0,#30H
;R0 < = 30H
MOV
A,@R0
;A < = 40H
MOV
R1,A
;R1 < = 40H
MOV
B,@R1
;B < = 10H
MOV
@R1,P1
;RAM (40H) < = 0CAH
MOV
P2,P1
;P2 #0CAH
leaves the value 30H in register 0, 40H in both the Accumulator and register 1, 10H in register B, and
0CAH (11001010B) both in RAM location 40H and output on port 2.
MOV A,Rn
Bytes:
1
Cycles:
1
Encoding:
1
1
1 0
1
r
r
r
Operation:
MOV
(A) ← (Rn)
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80C51 Family
and instruction set
*MOV A,direct
Bytes:
2
Cycles:
1
Encoding:
1
1
1 0
0
1
0 1
direct address
Operation:
MOV
(A) ← (direct)
MOV A,@Ri
Bytes:
1
Cycles:
1
Encoding:
1
1
1 0
0
1
1 i
Operation:
MOV
(A) ← ((Ri))
MOV A,#data
Bytes:
2
Cycles:
1
Encoding:
0
1
1 1
0
1
0 0
immediate data
Operation:
MOV
(A) ← #data
MOV Rn,A
Bytes:
1
Cycles:
1
Encoding:
1
1
1 1
1
r
r
r
Operation:
MOV
(Rn) ← (A)
MOV Rn,direct
Bytes:
2
Cycles:
2
Encoding:
1
0
1 0
1
r
r
r
direct address
Operation:
MOV
(Rn) ← (direct)
MOV Rn,#data
Bytes:
2
Cycles:
1
Encoding:
0
1
1 1
1
r
r
r
immediate data
Operation:
MOV
(Rn) ← #data
*MOV A,ACC is not a valid instruction.
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and instruction set
MOV direct,A
Bytes:
2
Cycles:
1
Encoding:
1
1
1 1
0
1
0 1
direct address
Operation:
MOV
(direct) ← (A)
MOV direct,Rn
Bytes:
2
Cycles:
2
Encoding:
1
0
0 0
1
r
r
r
direct address
Operation:
MOV
(direct) ← (Rn)
MOV direct,direct
Bytes:
3
Cycles:
2
Encoding:
1
0
0 0
0
1
0 1
dir. addr. (src)
dir. addr. (dest)
Operation:
MOV
(direct) ← (direct)
MOV direct,@Ri
Bytes:
2
Cycles:
2
Encoding:
1
0
0 0
0
1
1 i
direct address
Operation:
MOV
(direct) ← ((Ri))
MOV direct,#data
Bytes:
3
Cycles:
2
Encoding:
0
1
1 1
0
1
0 1
direct address
immediate data
Operation:
MOV
(direct) ← #data
MOV @Ri,A
Bytes:
1
Cycles:
1
Encoding:
1
1
1 1
0
1
1 i
Operation:
MOV
((Ri)) ← (A)
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and instruction set
MOV @Ri,direct
Bytes:
2
Cycles:
2
Encoding:
1
0
1 0
0
1
1 i
direct address
Operation:
MOV
((Ri)) ← (direct)
MOV @Ri,#data
Bytes:
2
Cycles:
1
Encoding:
0
1
1 1
0
1
1 i
immediate data
Operation:
MOV
((Ri)) ← #data
MOV <dest-bit>,<src-bit>
Function:
Move bit data
Description:
The Boolean variable indicated by the second operand is copied into the location specified by the first
operand. One of the operands must be the carry flag; the other may be any directly addressable bit. No
other register or flag is affected.
Example:
The carry flag is originally set. The data present at input Port 3 is 11000101B. The data previously written
to output Port 1 is 35H (00110101B). The instruction sequence,
MOV
P1.3,C
MOV
C,P3.3
MOV
P1.2,C
will leave the carry cleared and change Port 1 to 39H (00111001B).
MOV C,bit
Bytes:
2
Cycles:
1
Encoding:
1
0
1 0
0
0
1 0
bit address
Operation:
MOV
(C) ← (bit)
MOV bit,C
Bytes:
2
Cycles:
2
Encoding:
1
0
0 1
0
0
1 0
bit address
Operation:
MOV
(bit) ← (C)
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and instruction set
MOV DPTR,#data16
Function:
Load Data Pointer with a 16-bit constant
Description:
The Data Pointer is loaded with the 16-bit constant indicated. The 16-bit constant is loaded into the
second and third bytes of the instruction. The second byte (DPH) is the high-order byte, while the third
byte (DPL) holds the low-order byte. No flags are affected.
This is the only instruction which moves 16 bits of data at once.
Example:
The instruction,
MOV DPTR,#1234H
will load the value 1234H into the Data Pointer: DPH will hold 12H and DPL will hold 34H.
Bytes:
3
Cycles:
2
Encoding:
1
0
0 1
0
0
0 0
immed. data15-8
immed. data7-0
Operation:
MOV
(DPTR) ← (#data15-0)
DPH DPL ← #data15-8 #data7-0
MOVC A,@A+<base-reg>
Function:
Move Code byte
Description:
The MOVC instructions load the Accumulator with a code byte, or constant from program memory. The
address of the byte fetched is the sum of the original unsigned eight-bit Accumulator contents and the
contents of a sixteen-bit base register, which may be either the Data Pointer or the PC. In the latter case,
the PC is incremented to the address of the following instruction before being added with the Accumulator;
otherwise the base register is not altered. Sixteen-bit addition is performed so a carry-out from the
low-order eight bits may propagate through higher-order bits. No flags are affected.
Example:
A value between 0 and 3 is in the Accumulator. The following instructions will translate the value in the
Accumulator to one of four values defined by the DB (define byte) directive:
REL_PC:
INC
A
MOVC
A,@A+PC
RET
DB
66H
DB
77H
DB
88H
DB
99H
If the subroutine is called with the Accumulator equal to 01H, it will return with 77H in the Accumulator.
The INC A before the MOVC instruction is needed to “get around” the RET instruction above the table. If
several bytes of code separated the MOVC from the table, the corresponding number would be added to
the Accumulator instead.
MOVC A,@A+DPTR
Bytes:
1
Cycles:
2
Encoding:
1
0
0 1
0
0
1 1
Operation:
MOVC
(A) ← ((A) + (DPTR))
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and instruction set
MOVC A,@A+PC
Bytes:
1
Cycles:
2
Encoding:
1
0
0 0
0
0
1 1
Operation:
MOVC
(PC) ← (PC) + 1
(A) ← ((A) + (PC))
MOVX <dest-byte>,<src-byte> (Not implemented in the 8XC752 or 8XC752)
Function:
Move External
Description:
The MOVX instructions transfer data between the Accumulator and a byte of external data memory, hence
the “X” appended to MOV. There are two types of instructions, differing in whether they provide an
eight-bit or sixteen-bit indirect address to the external data RAM.
In the first type, the contents of R0 or R1 in the current register bank provide an eight-bit address
multiplexed with data on P0. Eight bits are sufficient for external I/O expansion decoding or for a relatively
small RAM array. For somewhat larger arrays, port pins can be used to output higher-order address bits.
These pins would be controlled by an output instruction preceding the MOVX.
In the second type of MOVX instruction, The Data Pointer generates a sixteen-bit address. P2 outputs the
high-order eight address bits (the contents of DPH) while P0 multiplexes the low-order eight bits (DPL)
with data. The P2 Special Function Register retains its previous contents while the P2 output buffers are
emitting the contents of DPH. This form is faster and more efficient when accessing very large data arrays
(up to 64k bytes), since no additional instructions are needed to set up the output ports.
It is possible in some situations to mix the two MOVX types. A large RAM array with its high-order address
lines driven by P2 can be addressed via the Data Pointer, or with code to output high-order address bits to
P2 followed by a MOVX instruction using R0 or R1.
Example:
An external 256 byte RAM using multiplexed address/data lines is connected to the 8051 Port 0. Port 3
provides control lines for the external RAM. Ports 1 and 2 are used for normal I/O. Registers 0 and 1
contain 12H and 34H. Location 34H of the external RAM holds the value 56H. The instruction sequence,
MOVX
A,@R1
MOVX
@R0,A
copies the value 56H into both the Accumulator and external RAM location 12H.
MOVX A,@Ri
Bytes:
1
Cycles:
2
Encoding:
1
1
1 0
0
0
1 i
Operation:
MOVX
(A) ← ((Ri))
MOVX A,@DPTR
Bytes:
1
Cycles:
2
Encoding:
1
1
1 0
0
0
0 0
Operation:
MOVX
(A) ← ((DPTR))
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and instruction set
MOVX @Ri,A
Bytes:
1
Cycles:
2
Encoding:
1
1
1 1
0
0
1 i
Operation:
MOVX
((Ri)) ← (A)
MOVX @DPTR,A
Bytes:
1
Cycles:
2
Encoding:
1
1
1 1
0
0
0 0
Operation:
MOVX
((DPTR)) ← (A)
MUL AB
Function:
Multiply
Description:
MUL AB multiplies the unsigned eight-bit integers in the Accumulator and register B. The low-order byte
of the sixteen-bit product is left in the Accumulator, and the high-order byte in B. If the product is greater
than 255 (0FFH) the overflow flag is set; otherwise it is cleared. The carry flag is always cleared.
Example:
Originally the Accumulator holds the value 80 (50H). Register B holds the value 160 (0A0H).The
instruction,
MUL AB
will give the product 12,800 (3200H), so B is changed to 32H (00110010B) and the Accumulator is
cleared. The overflow flag is set, carry is cleared.
Bytes:
1
Cycles:
4
Encoding:
1
0
1 0
0
1
0 0
Operation:
MUL
(A)7-0 ← (A) x (B)
(B)15-8
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and instruction set
NOP
Function:
No Operation
Description:
Execution continues at the following instruction. Other than the PC, no registers or flags are affected.
Example:
It is desired to produce a low-going output pulse on bit 7 of Port 2 lasting exactly 5 cycles. A simple
SETB/CLR sequence would generate a one-cycle pulse, so four additional cycles must be inserted. This
may be done (assuming are enabled) with the instruction sequence,
CLR
P2.7
NOP
NOP
NOP
NOP
SETB
P2.7
Bytes:
1
Cycles:
1
Encoding:
0
0
0 0
0
0
0 0
Operation:
NOP
(PC) ← (PC) + 1
ORL <dest-byte>,<src-byte>
Function:
Logical-OR for byte variables
Description:
ORL performs the bitwise logical-OR operation between the indicated variables, storing the results in the
destination byte. No flags are affected.
The two operands allow six addressing mode combinations. When the destination is the Accumulator, the
source can use register, direct, register-indirect, or immediate addressing; when the destination is a direct
address, the source can be the Accumulator or immediate data.
Note: When this instruction is used to modify an output port, the value used as the original port data will
be read from the output data latch, not the input pins.
Example:
If the Accumulator holds 0C3H (11000011B) and R0 holds 55H (01010101B) then the instruction,
ORL A,R0
will leave the Accumulator holding the value 0D7H (11010111B).
When the destination is a directly addressed byte, the instruction can set combinations of bits in any RAM
location or hardware register. The pattern of bits to be set is determined by a mask byte, which may be
either a constant data value in the instruction or a variable computed in the Accumulator at run-time. The
instruction,
ORL P1,#00110010B
will set bits 5, 4, and 1 of output Port 1.
ORL A,Rn
Bytes:
1
Cycles:
1
Encoding:
0
1
0 0
1
r
r
r
Operation:
ORL
(A) ← (A) ∨ (Rn)
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and instruction set
ORL A,direct
Bytes:
2
Cycles:
1
Encoding:
0
1
0 0
0
1
0 1
direct address
Operation:
ORL
(A) ← (A) ∨ (direct)
ORL A,@Ri
Bytes:
1
Cycles:
1
Encoding:
0
1
0 0
0
1
1 i
Operation:
ORL
(A) ← (A) ∨ ((Ri))
ORL A,#data
Bytes:
2
Cycles:
1
Encoding:
0
1
0 0
0
1
0 0
immediate data
Operation:
ORL
(A) ← (A) ∨ #data
ORL direct,A
Bytes:
2
Cycles:
1
Encoding:
0
1
0 0
0
0
1 0
direct address
Operation:
ORL
(direct) ← (direct) ∨ (A)
ORL direct,#data
Bytes:
3
Cycles:
2
Encoding:
0
1
0 0
0
0
1 1
direct address
immediate data
Operation:
ORL
(direct) ← (direct) ∨ #data
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and instruction set
ORL C,<src-bit>
Function:
Logical-OR for bit variables
Description:
Set the carry flag if the Boolean value is a logical 1; leave the carry in its current state otherwise. A slash
(“/”) preceding the operand in the assembly language indicates that the logical complement of the
addressed bit is used as the source value, but the source bit itself is not affected. No other flags are
affected.
Example:
Set the carry flag if and only if P1.0 = 1, ACC.7 = 1, or OV = 0:
ORL
C,P1.0
;LOAD CARRY WITH INPUT PIN P10
ORL
C,ACC.7
;OR CARRY WITH THE ACC. BIT 7
ORL
C,/OV
;OR CARRY WITH THE INVERSE OF OV.
ORL C,bit
Bytes:
2
Cycles:
2
Encoding:
0
1
1 1
0
0
1 0
bit address
Operation:
ORL
(C) ← (C) ∨ (bit)
ORL C,/bit
Bytes:
2
Cycles:
2
Encoding:
1
0
1 0
0
0
0 0
bit address
Operation:
ORL
(C) ← (C) ∨ (bit)
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and instruction set
POP direct
Function:
Pop from stack
Description:
The contents of the internal RAM location addressed by the Stack Pointer is read, and the Stack Pointer is
decremented by one. The value read is then transferred to the directly addressed byte indicated. No flags
are affected.
Example:
The Stack Pointer originally contains the value 32H, and internal RAM locations 30H through 32H contain
the values 20H, 23H, and 01H, respectively. The instruction sequence,
POP
DPH
POP
DPL
will leave the Stack Pointer equal to the value 30H and the Data Pointer set to 0123H. At this point the
instruction,
POP
SP
will leave the Stack Pointer set to 20H. Note that in this special case the Stack Pointer was decremented
to 2FH before being loaded with the value popped (20H).
Bytes:
2
Cycles:
2
Encoding:
1
1
0 1
0
0
0 0
direct address
Operation:
POP
(direct) ← ((SP))
(SP) ← (SP) – 1
PUSH direct
Function:
Push onto stack
Description:
The Stack Pointer is incremented by one. The contents of the indicated variable is then copied into the
internal RAM location addressed by the Stack Pointer. Otherwise no flags are affected.
Example:
On entering an interrupt routine the Stack Pointer contains 09H. The Data Pointer holds the value 0123H.
The instruction sequence,
PUSH
DPL
PUSH
DPH
will leave the Stack Pointer set to 0BH and store 23H and 01H in internal RAM locations 0AH and 0BH,
respectively.
Bytes:
2
Cycles:
2
Encoding:
1
1
0 0
0
0
0 0
direct address
Operation:
PUSH
(SP) ← (SP) + 1
((SP)) ← (direct)
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and instruction set
RET
Function:
Return from subroutine
Description:
RET pops the high- and low-order bytes of the PC successively from the stack, decrementing the Stack
Pointer by two. Program execution continues at the resulting address, generally the instruction
immediately following an ACALL or LCALL. No flags are affected.
Example:
The Stack Pointer originally contains the value 0BH. Internal RAM locations 0AH and 0BH contain the
values 23H and 01H, respectively. The instruction,
RET
will leave the Stack Pointer equal to the value 09H. Program execution will continue at location 0123H.
Bytes:
1
Cycles:
2
Encoding:
0
0
1 0
0
0
1 0
Operation:
RET
(PC15-8) ← ((SP))
(SP) ← (SP) – 1
(PC7-0) ← ((SP))
(SP) ← (SP) – 1
RETI
Function:
Return from interrupt
Description:
RETI pops the high- and low-order bytes of the PC successively from the stack, and restores the interrupt
logic to accept additional interrupts at the same priority level as the one just processed. The Stack Pointer
is left decremented by two. No other registers are affected; the PSW is not automatically restored to its
pre-interrupt status. Program execution continues at the resulting address, which is generally the
instruction immediately after the point at which the interrupt request was detected. If a lower- or
same-level interrupt has been pending when the RETI instruction is executed, that one instruction will be
executed before the pending interrupt is processed.
Example:
The Stack Pointer originally contains the value 0BH. An interrupt was detected during the instruction
ending at location 0122H. Internal RAM locations 0AH and 0BH contain the values 23H and 01H,
respectively. The instruction,
RETI
will leave the Stack Pointer equal to 09H and return program execution to location 0123H.
Bytes:
1
Cycles:
2
Encoding:
0
0
1 1
0
0
1 0
Operation:
RETI
(PC15-8) ← ((SP))
(SP) ← (SP) – 1
(PC7-0) ← ((SP))
(SP) ← (SP) – 1
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and instruction set
RL A
Function:
Rotate Accumulator Left
Description:
The eight bits in the Accumulator are rotated one bit to the left. Bit 7 is rotated into the bit 0 position. No
flags are affected.
Example:
The Accumulator holds the value 0C5H (11000101B). The instruction,
RL A
leaves the Accumulator holding the value 8BH (10001011B) with the carry unaffected.
Bytes:
1
Cycles:
1
Encoding:
0
0
1 0
0
0
1 1
Operation:
RL
(An+1) ← (An), n = 0 – 6
(A0) ← (A7)
RLC A
Function:
Rotate Accumulator Left through the Carry flag
Description:
The eight bits in the Accumulator and the carry flag are together rotated one bit to the left. Bit 7 moves into
the carry flag; the original state of the carry flag moves into the bit 0 position. No other flags are affected.
Example:
The Accumulator holds the value 0C5H (11000101B), and the carry is zero. The instruction,
RLC A
leaves the Accumulator holding the value 8AH (10001010B) with the carry set.
Bytes:
1
Cycles:
1
Encoding:
0
0
1 1
0
0
1 1
Operation:
RLC
(An+1) ← (An), n = 0 – 6
(A0) ← (C)
(C) ← (A7)
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and instruction set
RR A
Function:
Rotate Accumulator Right
Description:
The eight bits in the Accumulator are rotated one bit to the right. Bit 0 is rotated into the bit 7 position. No
flags are affected.
Example:
The Accumulator holds the value 0C5H (11000101B). The instruction,
RR A
leaves the Accumulator holding the value 0E2H (11100010B) with the carry unaffected.
Bytes:
1
Cycles:
1
Encoding:
0
0
0 0
0
0
1 1
Operation:
RR
(An) ← (An+1), n = 0 – 6
(A7) ← (A0)
RRC A
Function:
Rotate Accumulator Right through the Carry flag
Description:
The eight bits in the Accumulator and the carry flag are together rotated one bit to the right. Bit 0 moves
into the carry flag; the original state of the carry flag moves into the bit 7 position. No other flags are
affected.
Example:
The Accumulator holds the value 0C5H (11000101B), and the carry is zero. The instruction,
RRC A
leaves the Accumulator holding the value 62 (01100010B) with the carry set.
Bytes:
1
Cycles:
1
Encoding:
0
0
0 1
0
0
1 1
Operation:
RRC
(An) ← (An+1), n = 0 – 6
(A7) ← (C)
(C) ← (A0)
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and instruction set
SETB <bit>
Function:
Set Bit
Description:
SETB sets the indicated bit to one. SETB can operate on the carry flag or any directly addressable bit. No
other flags are affected.
Example:
The carry flag is cleared. Output Port 1 has been written with the value 34H (00110100B). The
instructions,
SETB
C
SETB
P1.0
will leave the carry flag set to 1 and change the data output on Port 1 to 35H (00110101B).
SETB C
Bytes:
1
Cycles:
1
Encoding:
1
1
0 1
0
0
1 1
Operation:
SETB
(C) ← 1
SETB bit
Bytes:
2
Cycles:
1
Encoding:
1
1
0 1
0
0
1 0
bit address
Operation:
SETB
(bit) ← 1
SJMP rel
Function:
Short Jump
Description:
Program control branches unconditionally to the address indicated. The branch destination is computed
by adding the signed displacement in the second instruction byte to the PC, after incrementing the PC
twice. Therefore, the range of destinations allowed is from 128 bytes preceding this instruction to 127
bytes following it.
Example:
The label “RELADR” is assigned to an instruction at program memory location 0123H. The instruction,
SJMP RELADR
will assemble into location 0100H. After the instruction is executed, the PC will contain the value 0123H.
(Note: Under the above conditions the instruction following SJMP will be at 102H. Therefore, the
displacement byte of the instruction will be the relative offset (0123H-0102H) = 21H. Put another way, an
SJMP with a displacement of 0FEH would be a one-instruction infinite loop.)
Bytes:
2
Cycles:
2
Encoding:
1
0
0 0
0
0
0 0
rel. address
Operation:
SJMP
(PC) ← (PC) + 2
(PC) ← (PC) + rel
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and instruction set
SUBB A, <src-byte>
Function:
Subtract with borrow
Description:
SUBB subtracts the indicated variable and the carry flag together from the Accumulator, leaving the result
in the Accumulator. SUBB sets the carry (borrow) flag if a borrow is needed for bit 7, and clears C
otherwise. (If C was set before executing a SUBB instruction, this indicates that a borrow was needed for
the previous step in a multiple precision subtraction, so the carry is subtracted from the Accumulator along
with the source operand.) AC is set if a borrow is needed for bit 3, and cleared otherwise. OV is set if a
borrow is needed into bit 6, but not into bit 7, or into bit 7, but not bit 6.
When subtracting signed integers OV indicates a negative number produced when a negative value is
subtracted from a positive value, or a positive result when a positive number is subtracted from a negative
number.
The source operand allows four addressing modes: register, direct, register-indirect, or immediate.
Example:
The Accumulator holds 0C9H (11001001B), register 2 holds 54H (01010100B), and the carry flag is set.
The instruction,
SUBB A,R2
will leave the value 74H (01110100B) in the Accumulator, with the carry flag and AC cleared but OV set.
Notice that 0C9H minus 54H is 75H The difference between this and the above result is due to the carry
(borrow) flag being set before the operation. If the state of the carry is not known before starting a single
or multiple-precision subtraction, it should be explicitly cleared by a CLR C instruction
SUBB A,Rn
Bytes:
1
Cycles:
1
Encoding:
1
0
0 1
1
r
r
r
Operation:
SUBB
(A) ← (A) – (C) – (Rn)
SUBB A,direct
Bytes:
2
Cycles:
1
Encoding:
1
0
0 1
0
1
0 1
direct address
Operation:
SUBB
(A) ← (A) – (C) – (direct)
SUBB A,@Ri
Bytes:
1
Cycles:
1
Encoding:
1
0
0 1
0
1
1 i
Operation:
SUBB
(A) ← (A) – (C) – (Ri)
SUBB A,#data
Bytes:
2
Cycles:
1
Encoding:
1
0
0 1
0
1
0 0
immediate data
Operation:
SUBB
(A) ← (A) – (C) – (#data)
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and instruction set
SWAP A
Function:
Swap nibbles within the Accumulator
Description:
SWAP A interchanges the low- and high-order nibbles (four-bit fields) of the Accumulator (bits 3-0 and bits
7-4). The operation can also be thought of as a four-bit rotate instruction. No flags are affected.
Example:
The Accumulator holds the value 0C5H (11000101B). The instruction,
SWAP A
leaves the Accumulator holding the value 5CH (01011100B).
Bytes:
1
Cycles:
1
Encoding:
1
1
0 0
0
1
0 0
Operation:
SWAP
(A3-0) (A
←
→
7-4)
XCH A,<byte>
Function:
Exchange Accumulator with byte variable
Description:
XCH loads the Accumulator with the contents of the indicated variable, at the same time writing the
original Accumulator contents to the indicated variable. The source/destination operand can use register,
direct, or register-indirect addressing.
Example:
R0 contains the address 20H. The Accumulator holds the value 3FH (00111111B). Internal RAM location
20H holds the value 75H (01110101B). The instruction,
XCH A,@R0
will leave the RAM location 20H holding the values 3FH (00111111B) and 75H (01110101B) in the
Accumulator.
XCH A,Rn
Bytes:
1
Cycles:
1
Encoding:
1
1
0 0
1
r
r
r
Operation:
XCH
(A) (R
←
→ n)
XCH A,direct
Bytes:
2
Cycles:
1
Encoding:
1
1
0 0
0
1
0 1
direct address
Operation:
XCH
(A) (direct)
←
→
XCH A,@Ri
Bytes:
1
Cycles:
1
Encoding:
1
1
0 0
0
1
1 i
Operation:
XCH
(A) ((R
←
→
i))
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and instruction set
XCHD A,@Ri
Function:
Exchange Digit
Description:
XCHD exchanges the low-order nibble of the Accumulator (bits 3-0), generally representing a
hexadecimal or BCD digit, with that of the internal RAM location indirectly addressed by the specified
register. The high-order nibbles (bits 7-4) of each register are not affected. No flags are affected.
Example:
R0 contains the address 20H. The Accumulator holds the value 36H (00110110B). Internal RAM location
20H holds the value 75H (01110101B). The instruction,
XCHD A,@R0
will leave RAM location 20H holding the value 76H (01110110B) and 35H (00110101B) in the Accumulator.
Bytes:
1
Cycles:
1
Encoding:
1
1
0 1
0
1
1 i
Operation:
XCHD
(A3-0) ((Ri
←
→
3-0))
XRL <dest-byte>,<src-byte>
Function:
Logical Exclusive-OR for byte variables
Description:
XRL performs the bitwise logical Exclusive-OR operation between the indicated variables, storing the
results in the destination. No flags are affected.
The two operands allow six addressing mode combinations. When the destination is the Accumulator, the
source can use register, direct, register-indirect, or immediate addressing; when the destination is a direct
address, the source can be the Accumulator or immediate data.
(Note: When this instruction is used to modify an output port, the value used as the original port data will
be read from the output data latch, not the input pins.)
Example:
If the Accumulator holds 0C3H (11000011B) and register 0 holds 0AAH (10101010B) then the instruction,
XRL A,R0
will leave the Accumulator holding the value 69H (01101001B).
When the destination is a directly addressed byte, this instruction can complement combinations of bits in
any RAM location or hardware register. The pattern of bits to be complemented is then determined by a
mask byte, either a constant contained in the instruction or a variable computed in the Accumulator at
run-time. The instruction,
XRL P1,#00110001B
will complement bits 5, 4, and 0 of output Port 1.
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and instruction set
XRL A,Rn
Bytes:
1
Cycles:
1
Encoding:
0
1
1 0
1
r
r
r
Operation:
XRL
(A) ← (A) ∨ (Rn)
XRL A,direct
Bytes:
2
Cycles:
1
Encoding:
0
1
1 0
0
1
0 1
direct address
Operation:
XRL
(A) ← (A) ∨ (direct)
XRL A,@Ri
Bytes:
1
Cycles:
1
Encoding:
0
1
1 0
0
1
1 i
Operation:
XRL
(A) ← (A) ∨ (Ri)
XRL A,#data
Bytes:
2
Cycles:
1
Encoding:
0
1
1 0
0
1
0 0
immediate data
Operation:
XRL
(A) ← (A) ∨ #data
XRL direct,A
Bytes:
2
Cycles:
1
Encoding:
0
1
1 0
0
0
1 0
direct address
Operation:
XRL
(direct) ← (direct) ∨ (A)
XRL direct,#data
Bytes:
3
Cycles:
2
Encoding:
0
1
1 0
0
0
1 1
direct address
immediate data
Operation:
XRL
(direct) ← (direct) ∨ #data
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Document Outline
- Table of Contents
- List of Figures
- 1. 80C51 Program Memory
- 2. 80C51 Data Memory
- 3. 128 Bytes of RAM Direct and Indirect Addressable
- 4. SFR Memory Map
- List of Tables
- 1. 80C51 Special Function Registers
- 2. As a Timer:
- 3. As a Counter:
- 4. As a Timer:
- 5. As a Counter:
- 6.
- 7. 80C51 Instruction Set Summary
- Programmer's Guide and Instruction Set
- PSW: Program Status Word. Bit Addressable.
- PCON: Power Control Register. Not Bit Addressable.
- Interrupts
- IE: Interrupt Enable Register. Bit Addressable
- Assigning Higher Priority to One or More Interrupts
- Priority Within Level
- IP: Interrupt Priority Register. Bit Addressable.
- TCON: Timer/Counter Control Register. Bit Addressable.
- TMOD: Timer/Counter Mode Control Register. Not Bit Addressable.
- Timer Set-Up
- Timer/Counter 0
- Timer/Counter 1
- SCON: Serial Port Control Register. Bit Addressable.
- Serial Port Set-Up:
- Generating Baud Rates
- Using Timer/Counter 1 to Generate Baud Rates:
- Serial Port in Mode 2:
- Serial Port in Mode 3:
- 80C51 Family Instruction Set
- Instruction Definitions
