Transcript Document

Chapter 10
8051 Serial Port Programming in
Assembly and C
1
Objective
• 8051 提供序列傳輸(serial communication)的功
能，允許 programmer 每次一個 bit 一個 bit 的
傳送或接收外界的資料。
• 由於 IBM PC 經常與 8051 通訊，因此我們特別
強調 8051 利用 RS232 與 PC 的 COM port 溝通。
• 所以要瞭解序列傳輸的基本特性，RS232 的規
格，轉換電壓準位的MAX232 IC。
• 我們也要瞭解 8051 內部相關 registers 的功能，
學習撰寫 8051 序列傳輸的程式。
2
Sections
10.1
10.2
10.3
10.4
10.5
Basics of serial communication
8051 connection to RS232
8051 serial port programming in Assembly
Programming the second serial port
Serial port programming in C
3
Section 10.1
Basics of Serial Communication
4
Figure 4-1. 8051 Pin Diagram
PDIP/Cerdip
P1.0
P1.1
P1.2
P1.3
P1.4
P1.5
P1.6
P1.7
RST
(RXD)P3.0
(TXD)P3.1
(INT0)P3.2
(INT1)P3.3
(T0)P3.4
(T1)P3.5
(WR)P3.6
(RD)P3.7
XTAL2
XTAL1
GND
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
8051
(8031)
40
39
38
37
36
35
34
33
32
31
30
29
28
27
26
25
24
23
22
21
Vcc
P0.0(AD0)
P0.1(AD1)
P0.2(AD2)
P0.3(AD3)
P0.4(AD4)
P0.5(AD5)
P0.6(AD6)
P0.7(AD7)
EA/VPP
ALE/PROG
PSEN
P2.7(A15)
P2.6(A14)
P2.5(A13)
P2.4(A12)
P2.3(A11)
P2.2(A10)
P2.1(A9)
P2.0(A8)
5
Inside Architecture of 8051
External interrupts
Interrupt
Control
On-chip
ROM for
program
code
Timer/Counter
On-chip
RAM
Timer 1
Timer 0
Counter
Inputs
CPU
OSC
Bus
Control
4 I/O Ports
P0 P1 P2 P3
Serial
Port
TxD RxD
Address/Data
Figure 1-2. Inside the 8051 Microcontroller Block Diagram
6
8051 and PC
• The 8051 module connects to PC by using RS232.
• RS232 is a protocol which supports full-duplex,
synchronous/asynchronous, serial communication.
• We discuss these terms in following sections.
RS232
PC
COM 1 port
8051
MAX232
UART
7
Simplex vs. Duplex Transmission
• Simplex transmission: the data can sent in one
direction.
– Example: the computer only sends data to the printer.
Transmitter
Device 1
Receiver
Device 2
• Duplex transmission: the data can be transmitted
and received.
Transmitter
Receiver
Device 1
Transmitter
Receiver
Device 2
8
Half vs. Full Duplex
• Half duplex: if the data is transmitted one way at a
time.
Transmitter
Receiver
Receiver
Transmitter
• Full duplex: if the data can go both ways at the
same time.
– Two wire conductors for the data lines.
Transmitter
Receiver
Receiver
Transmitter
9
Figure 10-2. Simplex, Half-, and FullDuplex Transfers
Simplex
Transmitter
Receiver
Half Duplex
Transmitter
Receiver
Receiver
Transmitter
Transmitter
Receiver
Receiver
Transmitter
Full Duplex
10
Parallel vs. Serial
• Computers transfer data in two ways:
– Parallel
•
•
•
•
Data is sent a byte or more at a time (fast)
Only very short distance between two systems
The 8-bit data path is expensive
Example: printer, hard disks
– Serial
•
•
•
•
The data is sent one bit at a time (slow) with simple wire
Relative long distance (rarely distortion)
cheap
For long-distance data transfers using communication lines
such as a telephone, it requires a modem to modulate (0/1 to
analog) and demoulate (analog to 0/1).
11
Figure 10-1. Serial versus Parallel Data
Transfer (1/2)
Serial Transfer
Sender
Receiver
Parallel Transfer
Sender
D0
Receiver
D7
12
Figure 10-1. Serial versus Parallel Data
Transfer (2/2)
Serial Transfer
Parallel Transfer
D0
Sender
D0-D7
Receiver
Sender
Receiver
Other control lines
Other control lines
13
Serial Communication
• How to transfer data?
– Sender:
• The byte of data must be converted to serial bits using a
parallel-in-serial-out shift register.
• The bit is transmitted over a single data line.
– Receiver
• The receiver must be a serial-in-parallel-out shift register to
receive the serial data and pack them into a byte.
8
register
parallel-in
serial-out
11101000001011
1
‘A’
register
serial-in
parallel-out
8-bit
character
14
Asynchronous vs. Synchronous
• Serial communication uses two methods:
– In synchronous communication, data is sent in blocks
of bytes.
time
10101010 10101111
byte byte byte byte
preamble
receiver
sender
– In asynchronous communication, data is sent without
continuity.
byte
receiver
byte
byte
start bit
stop bit
sender
15
UART & USART
• It is possible to write software to use both methods,
but the programs can be tedious and long.
• Special IC chips are made for serial
communication:
– USART (universal synchronous-asynchronous receivertransmitter)
– UART (universal asynchronous receiver-transmitter)
• The 8051 chip has a built-in UART.
16
8051 Serial Communication
• The 8051 has serial communication capability
built into it.
– Full-duplex
– Asynchronous mode only.
• How to detect that a character is sent via the line
in the asynchronous mode?
– Answer: Data framing!
17
Framing (1/3)
• Each character is placed in between start and stop
bits. This is called framing.
– Figure 10-3. Framing ASCII “A” (41H)
Time (D0 first)
mark
stop
bit
0
goes out last D7
1
0
0
0
0
0
1 start mark
bit
D0 goes out first
18
Framing (2/3)
•
•
•
•
The start bit is 0 (low) and always one bit.
The LSB is sent out first.
The stop bits is 1 (high).
The stop bit can be one (if 8 bits used in ASCII) or
two bits (if 7 bits used in ASCII).
– It is programmed for data that is 7 or 8 bits.
• When there is no transfer, the signal is 1 (high),
which is referred to as mark.
19
Framing (3/3)
• We have a total of 10 bits for each character:
– 8-bits for the ASCII code
– 2-bits for the start and stop bits
– 20% overhead
• In some systems in order to maintain data integrity,
the parity bit is included in the data frame.
– In an odd-parity bit system the total number of bits,
including the parity bit, is odd.
– UART chips allow programming of the parity bit for
odd-, even-, and no-parity options.
20
Data Transfer Rate (1/2)
• How fast is the data transferred?
• Three methods to describe the speed:
– Baud rate is defined as the number of signal changes
per second.
• The rate of data transfer is stated in Hz (used in modem).
– Data rate is defined as the number of bits transferred
per second.
• Each signal has several voltage levels.
• The rate of data transfer is stated in bps (bits per second).
– Effective data rate is defined as the number of actual
data bits transferred per second.
• Redundant bits must be removed
21
Data Transfer Rate (2/2)
• The data transfer rate depends on communication
ports incorporated into that system.
– Ex: 100-9600 bps in the early IBM PC/XT
– Ex: 56K bps in Pentium-based PC
– The baud rate is generally limited to 100kHz.
22
Example of Data Transfer Rate (1/2)
• Assume that data is sent in the following
asynchronous mode:
– 2400 baud rate
– each signal has 4 voltage levels (-5V, -3V, 3V, 5V)
– one start bit, 8-bit data, 2 stop bits
Time (D0 first)
stop stop
mark
bit bit
00
11
10
01
10
00
8-bit character
11
11
start
mark
bit
23
Example of Data Transfer Rate (2/2)
• 2400 baud = 2400 signals per second =2400 Hz
• 4 voltage level
– Log24=2, 2 bits is sent in every signal change
– Data rate = 2 * 2400 Hz = 4800 bps
• Effective ratio = 8 / (1+8+2) =8/11
• Effective data rate = data rate * effective ratio =
4800 * 8 /11=3490.9 bps
24
RS232 Standard
• RS232 (Recommended Standard 232) is an interfacing
standard which is set by the Electronics Industries
Association (EIA) in 1960.
– RS232 is the most widely used serial I/O interfacing standard.
– RS232A (1963), RS232B (1965) and RS232C (1969), now is
RS232E
– RS-232 is a standard for connecting between a DTE (Data
Terminal Equipment) and a DCE (Data Circuit-terminating
Equipment).
• Define the voltage level, pin functionality, baud rate,
signal meaning, communication distance.
25
DTE and DCE
• DTE (Data Terminal Equipment)
– DTE refers to PC, 8051, or other equipments.
• DCE (Data Communication Equipment)
– DCE refers to communication equipment, such as
modems, that are responsible for transferring the data.
DTE view DCE view
DTE
TxD
RxD
RxD
TxD
PC GND
Com1
DCE
DCE
GND
modem
Telephone
Line
DTE
RxD
TxD
TxD
RxD
GND
modem
GND
PC Com1
26
IBM PC/compatible COM ports
• IBM PC has 2 COM ports.
– Both COM ports have RS232-type connectors.
– For mouse, modem
• We can connect the 8051 serial port to the COM
port of a PC for serial communication experiments.
DTE view
RS232
DTE view
PC
COM 1 port
8051
MAX232
UART
27
Null Modem Connection
• The simplest connection between a PC and
microcontroller requires a minimum of three pins,
TxD, RxD, and GND.
– Figure 10-6 shows null modem connection
DTE
DTE
DTE
DTE
TxD
TxD
TxD
TxD
RxD
RxD
RxD
RxD
8051-based
board
ground
8051-based
board
PC
Com1
ground
PC
Com1
28
RS232 Voltage Level
• The input and output voltage of
RS232 is not of the TTL (TransistorTransistor-Logic) compatible.
RS 232 Voltage
25V
– RS232 is older than TTL.
• We must use voltage converter (also
referred to as line driver) such as
MAX232 to convert the TTL logic
levels to the RS232 voltage level,
and vice versa.
– MAX232, TSC232, ICL232
logic 0
3V
-3V
undefined
logic 1
-25V
29
MAX232
• MAX232 IC chips are commonly referred to as
line drivers.
RS232
PC
COM 1 port
8051
MAX232
RS232 voltage
level
UART
TTL voltage
level
30
RS232 pins
• Figure 10-4 shows the RS232 connector DB-25.
• Table 10-1 shows the pins and their labels for the
RS232 cable.
– DB-25P : plug connector (male)
– DB-25S: socket connector (female)
• Figure 10-5 shows DB9 connector and Table 10-2
shows the signals.
– IBM version for PC.
• All the RS 232 pin function definitions of Tables
10-1 and 10-2 are from the DTE point of view.
31
Figure 10-4. RS232 Connector DB-25
1
14
13
25
32
Table 10-1: RS232 Pins (DB-25) for DTE (1/2)
Pin
Description
1
Protective ground
2
Transmitted data (TxD)
3
Received data (RxD)
4
Request to send (RTS)
5
Clear to send (CTS)
6
Data set ready (DSR)
7
Signal ground (GND)
8
Data carrier detect (DCD)
9/10
Reserved for data testing
11
Unassigned
12
Secondary data carrier detect
13
Secondary clear to send
33
Table 10-1: RS232 Pins (DB-25) for DTE (2/2)
Pin
Description
14
Secondary transmitted data
15
Transmit signal element timing
16
Secondary received data
17
Receive signal element timing
18
Unassigned
19
Secondary request to sent
20
Data terminal ready (DTR)
21
Signal quality detector
22
Ring indicator (RI)
23
Data signal rate select
24
Transmit signal element timing
25
Unassigned
34
Figure 10-5. DB-9 9-Pin Connector
1
6
5
9
35
Table 10-2: IBM PC DB-9 Signals for DTE
Pin
Description
1
Data carrier detect (DCD)
2
Received data (RxD)
3
Transmitted data (TxD)
4
Data terminal ready (DTR)
5
Signal ground (GND)
6
Data set ready (DSR)
7
Request to send (RTS)
8
Clear to send (CTS)
9
Ring indicator (RI)
36
DB9 - DB25 conversion
DB9
DB25
Function
1
8
Data carrier detect
2
3
Receive data
3
2
Transmit data
4
20
Data terminal ready
5
7
Signal ground
6
6
Data set ready
7
4
Request to send
8
5
Clear to send
9
22
Ring indicator
37
RS232 Handshaking Signals
• Many of the pins of the RS232 connector are used
for handshaking signals.
–
–
–
–
–
–
DTR (data terminal ready)
DSR (data set ready)
RTS (request to send)
CTS (clear to send)
DCD (carrier detect, or data carrier detect)
RI (ring indicator)
38
Communication Flow
• While signals DTR and DSR are used by the PC
and modem, respectively, to indicate that they are
alive and well.
• RTS and CTS control the flow of data.
• When the PC wants to send data, it asserts RTS.
• If the modem is ready (has room) to accept the data,
it sends back CTS.
• If, for lack of room, the modem does not activate
CTS, and PC will de-assert DTR and try again.
39
RS422 & RS485
• By using RS232, the limit distance between two
PCs is about 15m.
• It works well even the distance=30m.
• If you want to transfer data with long distance (ex:
300m), you can use RS422 or RS485.
40
Section 10.2
8051 Connection to RS232
41
TxD and RxD pins in the 8051
• Many of the pins of the RS232 connector are used
for handshaking signals. However, they are not
supported by the 8051 UART chips.
• In 8051, the data is received from or transmitted to
– RxD: received data (Pin 10, P3.0)
– TxD: transmitted data (Pin 11, P3.1)
• TxD and RxD of the 8051 are TTL compatible.
• The 8051 requires a line driver to make them
RS232 compatible.
– One such line driver is the MAX232 chip.
42
MAX232 (1/2)
• MAX232 chip converts from RS232 voltage levels
to TTL voltage levels, and vice versa.
– MAX232 uses a +5V power source which is the same
as the source voltage for the 8051.
8051
MAX232
P3.1
TxD
11
P3.0 10
RxD
11
12
14
2 5
13
3
DB-9
43
MAX232 (2/2)
• MAX232 has two sets of line drivers.
– Figure 10.7 shows the inside of MAX232.
– MAX232 requires four capacitors ranging from 1 to 22
mF. The most widely used value for these capacitors is
22mF.
• MAX233 performs the same job as the MAX232
but eliminates the need for capacitors.
– Note that MAX233 and MAX232 are not pin
compatible.
– Figure 10.8 (a) shows the inside of MAX233
– Figure 10.8 (b) shows the connection to the 8051
44
Figure 10-7 (a): Inside MAX232
Vcc
22 mF
+
C1
+
C2
11
12
10
9
TTL side
1
3
4
16
2
MAX232
6
5
T1IN
R1OUT
T1OUT
R1IN
T2IN
R2OUT
T2OUT
R2IN
15
C3
+
C4
+
14
13
7
8
RS232 side
45
Figure 10-7: (b) MAX232’s Connection to
the 8051 (Null Modem)
8051
MAX232
P3.1 11
TxD
11
P3.0 10
RxD
12
14
2 5
13
3
DB-9
TTL-compatible
RS232-compatible
46
Figure 10-8: (a) Inside MAX233
Vcc
13
14
12
17
2
3
1
20
TTL side
7
MAX233
11
15
16
10
T1IN
R1OUT
T1OUT
R1IN
T2IN
R2OUT
T2OUT
R2IN
6
9
5
4
18
19
RS232 side
47
Figure 10-8: (b) MAX233’s Connection to
the 8051 (Null Modem)
8051
MAX233
P3.1 11
TxD
2
P3.0 10
3
5
2 5
4
3
RxD
DB-9
TTL-compatible
RS232-compatible
48
Section 10.3
8051 Serial Port Programming in
Assembly
49
PC Baud Rates
110 bps
• PC supports several
baud rates.
– You can use Netterm,
terminal.exe, stty, ptty
to send/receive data.
• Hyperterminal
supports baud rates
much higher than the
ones list in the Table.
150
300
600
1200
2400
4800
9600 (default)
19200
Note: Baud rates supported by
486/Pentium IBM PC BIOS.
50
Baud Rates in the 8051 (1/2)
• The 8051 transfers and receives data serially at
many different baud rates by using UART.
• UART divides the machine cycle frequency by 32
and sends it to Timer 1 to set the baud rate.
• Signal change for each roll over of timer 1
11.0592 MHz
XTAL
oscillator
÷ 12
Machine cycle
frequency
921.6 kHz
÷ 32
28800 Hz
By UART To timer 1
To set the
Baud rate
Timer 1
51
Baud Rates in the 8051 (2/2)
• Timer 1, mode 2 (8-bit, auto-reload)
• Define TH1 to set the baud rate.
–
–
–
–
XTAL = 11.0592 MHz
The system frequency = 11.0592 MHz / 12 = 921.6 kHz
Timer 1 has 921.6 kHz/ 32 = 28,800 Hz as source.
TH1=FDH means that UART sends a bit every 3 timer
source.
– Baud rate = 28,800/3= 9,600 Hz
52
Example 10-1
With XTAL = 11.0592 MHz, find the TH1 value needed to have the
following baud rates. (a) 9600 (b) 2400 (c) 1200
Solution:
With XTAL = 11.0592 MHz, we have:
The frequency of system clock = 11.0592 MHz / 12 = 921.6 kHz
The frequency sent to timer 1 = 921.6 kHz/ 32 = 28,800 Hz
(a) 28,800 / 3 = 9600 where -3 = FD (hex) is loaded into TH1
(b) 28,800 / 12 = 2400 where -12 = F4 (hex) is loaded into TH1
(c) 28,800 / 24 = 1200 where -24 = E8 (hex) is loaded into TH1
Notice that dividing 1/12th of the crystal frequency by 32 is the
default value upon activation of the 8051 RESET pin.
53
Table 10–4 Timer 1 TH1 Register Values
for Various Baud Rates
54
Registers Used in Serial Transfer Circuit
•
•
•
•
SBUF (Serial data buffer)
SCON (Serial control register)
PCON (Power control register)
You can see Appendix H (pages 609-611) for
details.
• PC also has several registers to control COM1,
COM2.
55
SBUF Register
• Serial data register: SBUF
– An 8-bit register to store the serial data
– For a byte of data to be transferred via the TxD line, it
must be placed in the SBUF.
MOV SBUF,A
;send data from A
– SBUF holds the byte of data when it is received by the
8051’s RxD line.
MOV A,SBUF
;receive and copy to A
– Not bit-addressable
56
SCON Register
• Serial control register: SCON
SM0, SM1 Serial port mode specifier
REN
(Receive enable) set/cleared by software to
enable/disable reception.
TI
Transmit interrupt flag.
RI
Receive interrupt flag.
SM2 = TB8 = RB8 =0 (not widely used)
(MSB)
(LSB)
SM0 SM1 SM2 REN TB8 RB8
TI
RI
* SCON is bit-addressable.
57
SM0, SM1
• SM0 and SM1 determine the framing of data.
– SCON.6 (SM1) a1d SCON.7 (SM0)
– Only mode 1 is compatible with COM port of IBM PC.
– See Appendix A.3. and Appendix H
SM0
0
0
1
1
SM1
0
1
0
1
Mode
0
1
2
3
Operating Mode Baud Rate
Shift register Fosc./12
8-bit UART Variable by timer1
9-bit UART Fosc./64 or Fosc./32
9-bit UART
Variable
58
REN (Receive Enable)
• SCON.4
• Set/cleared by software to enable/disable reception.
– REN=1
• It enable the 8051 to receive data on the RxD pin of the 8051.
• If we want the 8051 to both transfer and receive data, REN
must be set to 1.
• SETB SCON.4
– REN=0
• The receiver is disabled.
• The 8051 can not receive data.
• CLR SCON.4
59
TI (Transmit Interrupt Flag)
• SCON.1
• When the 8051 finishes the transfer of the 8-bit
character, it raises the TI flag.
• TI is raised by hardware at the beginning of the
stop bit in mode 1.
• Must be cleared by software.
60
RI (Receive Interrupt)
• SCON.0
• Receive interrupt flag
• When the 8051 receives data serially via RxD, it
gets rid of the start and stop bits and place the byte
in the SBUF register.
• Set by hardware halfway through the stop bit time
in mode 1.
• Must be cleared by software.
61
SM2
• SCON.5
• SM2 enables the multiprocessor communication
for mode 2 & 3.
– SM2=0 : to receive data  two processors talks
– SM2=1 : to receive address  multiprocessor talks
1
2
3
TxD
0
4
5
RxD
data
TB8(the 9th bit) 8 bits
0
data
1 address=5
time
62
TB8 (Transfer Bit 8)
• SCON.3
• For multiprocessor communication
• The 9th bit will be transmitted in mode 2 & 3.
– TB8 = 1: address
– TB8 = 0: data
• Set/Cleared by software.
63
RB8 (Receive Bit 8)
• SCON.2
• For multiprocessor communication
• In serial mode 1, RB8 gets a copy of the stop bit
when an 8-bit data is received.
• In serial mode 2 & 3, RB8 gets a copy of the 9th
bit. In fact, it is the TB8 in transmitter.
64
Figure 10-9. SCON Serial Port Control
Register (Bit Addressable)
SM0 SM1 SM2 REN TB8 RB8
TI
RI
SM0
SCON.7
Serial port mode specifier
SM1
SCON.6
Serial port mode specifier
SM2
SCON.5
Used for multiprocessor communication. (Make it 0)
REN SCON.4
Set/cleared by software to enable/disable reception.
TB8
SCON.3
Not widely used.
RB8
SCON.2
Not widely used.
TI
SCON.1
Transmit interrupt flag. Set by hardware at the beginning of
the stop bit in mode 1. Must be cleared by software.
RI
SCON.0
Receive interrupt flag. Set by hardware halfway through the
stop bit time in mode 1. Must be cleared by software.
Note: Make SM2, TB8, and RB8 = 0.
65
Transfer Data with the TI flag (1/2)
•
The following sequence is the steps that the 8051
goes through in transmitting a character via TxD:
1. The byte character to be transmitted is written into the
SBUF register.
2. It transfers the start bit.
3. The 8-bit character is transferred one bit at a time.
4. The stop bit is transferred.
8-bit char
bit by bit
8-bit
TxD
SBUF
A
zero detect
TI
shift add start and
UART stop bits
66
Transfer Data with the TI flag (2/2)
•
Sequence continuous:
5. During the transfer of the stop bit, the 8051 raises the
TI flag, indicating that the last character was
transmitted and it is ready to transfer the next character.
6. By monitoring the TI flag, we know whether or not the
8051 is ready to transfer another byte.
– We will not overloading the SBUF register.
– If we write another byte into the SBUF before TI is raised, the
untransmitted portion of the previous byte will be lost.
– We can use interrupt to transfer data in Chapter 11.
7. After SBUF is loaded with a new byte, the TI flag bit
must be cleared by the programmer.
67
Programming the 8051 to Transfer data
Serially (1/2)
1. Use the timer 1 in mode 2
– MOV TMOD,#20H
2. Set the value TH1 to chose baud rate. Look at the
Table 10-4.
– MOV TH1,#0FDH
;Baud rate = 9600Hz
3. Set SCON register in mode 1.
– MOV SCON,#50H
4. Start the timer.
– SETB TR1
68
Programming the 8051 to Transfer data
Serially (2/2)
raise when sending
the stop bit
5. Clear TI flag.
– CLR TI
TI=0
transfer data
6. The character byte to be transferred serially is
written into the SBUF register.
– MOV SBUF,#’A’
7. Keep monitoring the Transmit Interrupt (TI) to
see if it is raised.
– HERE:
JNB TI, HERE
8. To transfer the next character, go to Step 5.
69
Example 10-2
Write a program for the 8051 to transfer letter “A” serially at 4800
baud, continuously.
Solution:
MOV
MOV
MOV
SETB
AGAIN: MOV
HERE: JNB
CLR
SJMP
TMOD,#20H
TH1,#-6
SCON,#50H
TR1
SBUF,#”A”
TI,HERE
TI
AGAIN
;timer 1, mode 2
;4800 baud rate
;8-bit,1 stop, REN enabled
;start timer 1
;letter “A” to be transferred
;wait for the last bit
;clear TI for next char
;keep sending A
70
Example 10-3 (1/2)
Write a program to transfer the message “YES” serially at 9600
baud, 8-bit data, 1 stop bit. Do this continuously.
Solution:
MOV TMOD,#20H ;timer 1, mode 2
MOV TH1,#-3
;9600 baud
MOV SCON,#50H
SETB TR1
AGAIN:MOV A,#”Y”
;transfer “Y”
ACALL TRANS
MOV A,#”E”
;transfer “E”
ACALL TRANS
MOV A,#”S”
;transfer “S”
ACALL TRANS
SJMP AGAIN
;keep doing it
71
Example 10-3 (2/2)
;serial data transfer subroutine
TRANS:MOV SBUF,A
;load SBUF
HERE: JNB TI,HERE
;wait for last bit to transfer
CLR TI
;get ready for next byte
RET
72
Receive Data with the RI flag (1/2)
•
The following sequence is the steps that the 8051
goes through in receiving a character via RxD:
1. 8051 receives the start bit indicating that the next bit
is the first bit of the character to be received.
2. The 8-bit character is received one bit at a time.
When the last bit is received, a byte is formed and
placed in SBUF.
bit by bit
RxD
remove/check the
start and stop bits
shift regiter
8-bit
SBUF
shift
REN=1
UART
RI
8-bit
load
A
73
Receive Data with the TI flag (2/2)
•
Sequence continuous:
3. The stop bit is received. During receiving the stop bit,
the 8051 make RI=1, indicating that an entire character
was been received and must be picked up before it
gets overwritten by an incoming character.
4. By monitoring the RI flag, we know whether or not the
8051 has received a character byte.
– If we fail to copy SBUF into a safe place, we risk the loss of
the received byte.
– We can use interrupt to transfer data in Chapter 11.
5. After SBUF is copied into a safe place, the RI flag bit
must be cleared by the programmer.
74
Programming the 8051 to Receive data
Serially (1/2)
1. Use the timer 1 in mode 2
– MOV TMOD,#20H
2. Set the value TH1 to chose baud rate. Look at the
Table 10-4.
– MOV TH1,#0FDH
;Baud rate = 9600Hz
3. Set SCON register in mode 1.
– MOV SCON,#50H
4. Start the timer.
– SETB TR1
75
Programming the 8051 to Receive data
Serially (2/2)
raise when getting
the stop bit
5. Clear RI flag.
– CLR RI
RI=0
receive data
6. Keep monitoring the Receive Interrupt (RI) to
see if it is raised.
– HERE:
JNB RI, HERE
7. When RI is raised, SBUF has the whole byte.
Move the content of SBUF to a safe place.
– MOV A,SBUF
8. To receive the next character, go to Step 5.
76
Example 10-4
Program the 8051 to receive bytes of data serially, and put them in
P1. Set the baud rate at 4800, 8-bit data, and 1 stop bit.
Solution:
MOV
MOV
MOV
SETB
HERE: JNB
MOV
MOV
CLR
SJMP
TMOD,#20H
TH1,#-6
SCON,#50H
TR1
RI,HERE
A,SBUF
P1,A
RI
HERE
;timer1, mode 2 (auto reload)
;4800 baud
;8-bit, 1 stop, REN enabled
;start timer 1
;wait for char to come in
;save incoming byte in A
;send to port 1
;get ready to receive next byte
;keep getting data
77
Example 10-5 (1/4)
Assume that the 8051 serial port is connected to the COM port
of the IBM PC, and on the PC we are using the terminal.exe
program to send and receive data serially.
P1 and P2 of the 8051 are connected to LEDs and switches,
respectively.
Write an 8051 program to
(a) send to the PC the message ”We Are Ready”,
(b)
receive any data sent by the PC and put it on LEDs
connected to P1, and
(c)
get data on switches connected to P2 and send it to
the PC serially.
The program should perform part (a) once, but parts (b) and
(c) continuously.
Use the 4800 baud rate.
78
Example 10-5 (2/4)
8051
To
PC
COM
port
TxD
RxD
P1
LED
P2
SW
Solution:
ORG
MOV
MOV
MOV
MOV
SETB
MOV
0
P2,#0FFH
TMOD,#20H
TH1,#0FAH
SCON,#50H
TR1
DPTR,#MYDATA
;make P2 an input port
;4800 baud rate
;start timer 1
;load pointer for message
79
Example 10-5 (3/4)
H_1:
(a)
(c)
(b)
B_1:
MOV DPTR,#MYDATA ;load pointer for message
CLR A
MOVC A,@A+DPTR
;get the character
JZ B_1
;if last character get out
ACALL SEND
INC DPTR
SJMP H_1
;next character
MOV A,P2
;read data on P2
ACALL SEND
;transfer it serially
ACALL RECV
;get the serial data
MOV P1,A
;display it on LEDs
SJMP B_1
;stay in loop indefinitely
80
Example 10-5 (4/4)
;-----------serial data transfer. ACC has the data
SEND: MOV SBUF,A
;load the data
H_2:
JNB TI,H_2
;stay here until last bit gone
CLR TI
;get ready for next char
RET
;-------------- Receive data serially in ACC
RECV: JNB RI,RECV
;wait here for char
MOV A,SBUF
;save it in ACC
CLR RI
;get ready for next char
RET
;---------------The message to send
MYDATA:DB
“We Are Ready”,0
END
81
Doubling the Baud Rate in the 8051
•
There are two ways to increase the baud rate of
data transfer in the 8051:
1. To use a higher frequency crystal.
•
•
It is not feasible in many situations since the system crystal is
fixed.
Many new crystal may not be compatible with the IBM PC
serial COM ports baud rate.
2. To change a bit in the PCON register.
– This is a software way by setting SMOD=1.
82
PCON Register
SMOD
Double baud rate. If Timer 1 is used to generate
baud and SMOD=1, the baud rate is doubled
when the Serial Port is used in modes 1,2,3.
GF1,GF0 General purpose flag bit.
PD
Power down bit. Setting this bit activates “Power
Down” operation in the 80C51BH. (precedence)
IDL
Idle Mode bit. Setting this bit activates “Idle Mode”
operation in the 80C51BH.
(MSB)
SMOD
(LSB)
--
--
--
GF1 GF2
PD
* PCON is not bit-addressable. See Appendix H. p410
IDL
83
SMOD Flag of the PCON Register
• Power control register: PCON
MOV A, PCON
SETB ACC.7
MOV PCON,A
–
–
–
–
–
;we don’t want to modify other bits
An 8-bit register
Not bit-addressable
SCOM=0: default
SCOM=1: double the baud rate
See Table 10-5
84
Table 10-5: Baud Rate Comparison for
SMOD = 0 and SMOD =1
TH1 (Decimal)
(Hex)
SMOD = 0
SMOD = 1
-3
FD
9,600
19,200
-6
FA
4,800
9,600
-12
F4
2,400
4,800
-24
E8
1,200
2,400
Note: XTAL = 11.0592 MHz.
11.0592 MHz
XTAL
oscillator
÷ 12
SMOD = 1
To
Machine
57600 Hz timer 1
÷
16
cycle freq.
to set
28800 Hz baud
921.6 kHz
÷ 32
rate
SMOD = 0
85
Baud Rates for SMOD=0
• When SMOD=0, the 8051 divides 1/12 of the
crystal frequency by 32, and uses that frequency
for timer 1 to set the baud rate.
–
–
–
–
XTAL = 11.0592 MHz
The system frequency = 11.0592 MHz / 12 = 921.6 kHz
Timer 1 has 921.6 kHz/32 = 28,800 Hz as source.
TH1=256 - Crystal frequency/(12*32*Baud rate)
• Default on reset
– See Appendix A, page 363
86
Baud Rates for SMOD=1
• When SMOD=1, the 8051 divides 1/12 of the
crystal frequency by 16, and uses that frequency
for timer 1 to set the baud rate.
–
–
–
–
XTAL = 11.0592 MHz
The system frequency = 11.0592 MHz / 12 = 921.6 kHz
Timer 1 has 921.6 kHz/16 = 57,600 Hz as source.
TH1=256 - Crystal frequency/(12*16*Baud rate)
87
Example 10-6 (1/2)
Assuming that XTAL = 11.0592 MHz for the following program,
state (a) what this program does, (b) compute the frequency used by
timer 1 to set the baud rate, and (c) find the baud rate of the data
transfer.
Solution:
(a) This program transfers ASCII letter B (01000010 binary)
continuously.
(b) and (c) With XTAL = 11.0592 MHz and SMOD = 1
11.0592 / 12 = 921.6 kHz machine cycle frequency.
921.6 /16 = 57,600 Hz frequency used by timer 1 to set the baud rate.
57,600 / 3 = 19,200, the baud rate.
88
Example 10-6 (2/2)
MOV
SETB
MOV
MOV
MOV
MOV
SETB
MOV
A_1:CLR
MOV
H_1:JNB
SJMP
A,PCON
ACC.7
PCON,A
TMOD,#20H
TH1,#-3
SCON,#50H
TR1
A,#”B”
TI
SBUF,A
TI H_1
A_1
;SMOD=1, double baud rate
;Timer 1, mode 2, auto reload
;19200 baud rate
;8-bit data,1 stop bit, RI enabled
;start Timer 1
;transfer letter B
;make sure TI=0
;transfer it
;check TI
;do again
89
Example 10-7 x
Find the TH1 value (in both decimal and hex) to set the baud
rate to each of the following: (a) 9600 Hz (b) 4800 Hz if
SMOD =1
Assume that XTAL = 11.0592 MHz.
Solution:
With XTAL = 11.0592 and SMOD = 1,
11.0592 / 12 = 921.6 kHz machine cycle frequency.
921.6 / 16 = 57,600 Hz frequency used by the timer 1
(a) 57,600 / 9600 = 6  TH1 = -6 or TH1 = FAH.
(b) 57,600 / 4800 = 12  TH1 = -12 or TH1 = F4H.
90
Example 10-8 x
Find the baud rate if TH1 = -2, SMOD = 1, and XTAL = 11.0592
MHz. Is this baud rate supported by IBM/compatible PCs?
Solution:
With XTAL = 11.0592 and SMOD = 1, we have timer 1
frequency
= 57,600 Hz. The baud rate is 57,600 / 2 = 28,800.
This baud rate is not supported by the BIOS of the PC; however,
the PC can be programmed to do data transfer at such a speed.
The software of many modems can do this.
Also, Hyperterminal in Windows 95 (and higher) supports this
and other baud rates.
91
Table 10–5 Baud Rate Comparison for
SMOD = 0 and SMOD = 1
92
Section 10.4
Programming the Second Serial Port
93
Second Serial Port
• Many of the new generations of the 8051
microcontrollers come with two serial ports.
– DS89C4x0, DS89C320
– But not all versions of 8051/52 microcontroller.
• There two ports are called as Port 0 and Port 1.
–
–
–
–
–
Port 0 is as same as the original 8051.
Port 1 uses new pins and new registers.
See Table 10-6 and Figure 10-12 for registers
See Figure 10-10 for 89C4x0
See Figure 10-11 for connection with MAX232
94
Table 10–6 SFR Byte Addresses for DS89C4x0
Serial Ports
shared
as same as the
original 8051
serial port
Compare with p556 Table A-2
new
SCON1,SBUF1
95
Table 10–7 SFR
Addresses for the
DS89C4x0
(420, 430, etc.)
96
Figure 10–12 SCON0 and SCON1 Bit Addresses
(TI and RI bits must be noted)
97
Figure 10–10 DS89C4x0 Pin Diagram.
port 1
port 0
Note: Notice P1.2 and P1.3
pins are used by Rx and Tx
lines of the 2nd serial port
98
Figure 10–11 Connection with MAX232
(a) Inside MAX232
(b) MAX232’s Connection
to the DS89C4x0
99
Assembly Language for Second Port
• The older 8051 assemblers do not support the
second serial port, we need to define them by their
SFR address.
– Example 10-11.
• Also, many C compiler do not support the second
serial port, we have to declare the byte addresses
of the new SFR register by using the SFR
keyword.
– Example 10-20
100
Example 10-11 (1/2)
Write a program for the second serial port of the DS89C4x0 to
continuously transfer the letter “A” serially at 4800 baud. User
8-bit data and 1 stop bit. User Timer 1.
Solution:
SBUF1
SCON1
TI1
RI1
ORG
MAIN: MOV
MOV
MOV
SETB
EQU 0C1H
;defined by yourself
EQU 0C0H
BIT 0C1H
BIT 0C0H
0H
TMOD,#20H
TH1,#-6
SCON1,#50H
TR1
101
Example 10-11 (2/2)
ANAIN:
MOV
A,#’A’
ACALL SENDCON2
SJMP AGAIN
SENDCOM2:
MOV SBUF1,A
JNB TI1,HERE
CLR TI1
RET
END
;send ‘A’ continuously
;send by using port 1
102
Section 10.5
Serial Port Programming in C
103
Transmitting and Receiving Data in C
• Using C to program the serial port.
– Example 10-15 does the same work as Example 10-11.
• Please read other examples by yourself.
104
Example 10-15
Write an 8051 C program to transfer the letter “A” serially at
4800 baud continuously. Use 8-bit data and 1 stop bit.
Solution :
#include <reg51.h>
void main(void) {
TMOD=0x20;
SCON=0x50;
TH1=0xFA; //4800 baud rate
TR1=1;
while (1) {
SBUF=‘A’;
while (TI==0);
TI=0;
} }
105
You are able to (1/2)
• Contract and compare serial versus parallel
communication
• List the advantages of serial communication over parallel
• Explain serial communication protocol
• Contrast synchronous versus asynchronous
communication
• Contrast half- versus full-duplex transmission
• Explain the process of data framing
• Describe data transfer rate and bps rate
106
You are able to (2/2)
•
•
•
•
•
•
•
•
Define the RS232 standard
Explain the use of the MAX232 and MAX232 chips
Interface the 8051 with an RS232 connector
Discuss the baud rate of the 8051
Describe serial communication features of the 8051
Program the 8051 for serial data communication
Program the 8051 serial port in Assembly and C
Program the second serial port of DS89C4x0 in
Assembly and C
107