1 chapter 10 8051 serial port programming in assembly and c
TRANSCRIPT
1
Chapter 10 8051 Serial Port Programming in
Assembly and C
2
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。
3
Sections
10.1 Basics of serial communication
10.2 8051 connection to RS232
10.3 8051 serial port programming in Assembly
10.4 Programming the second serial port
10.5 Serial port programming in C
4
Section 10.1Basics of Serial Communication
5
Figure 4-1. 8051 Pin DiagramPDIP/Cerdip
123
4567891011121314151617181920
403938
3736353433323130292827262524232221
P1.0P1.1P1.2
P1.3P1.4P1.5P1.6P1.7RST
(RXD)P3.0(TXD)P3.1
(T0)P3.4(T1)P3.5
XTAL2XTAL1
GND
(INT0)P3.2
(INT1)P3.3
(RD)P3.7(WR)P3.6
VccP0.0(AD0)P0.1(AD1)
P0.2(AD2)P0.3(AD3)P0.4(AD4)P0.5(AD5)P0.6(AD6)P0.7(AD7)
EA/VPPALE/PROG
PSENP2.7(A15)P2.6(A14)P2.5(A13)P2.4(A12)P2.3(A11)P2.2(A10)P2.1(A9)P2.0(A8)
8051(8031)
6
Inside Architecture of 8051
CPU
On-chip RAM
On-chip ROM for program code
4 I/O Ports
Timer 0
Serial Port
Figure 1-2. Inside the 8051 Microcontroller Block Diagram
OSC
Interrupt Control
External interrupts
Timer 1
Timer/Counter
Bus Control
TxD RxDP0 P1 P2 P3
Address/Data
Counter Inputs
7
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.
PC 8051
COM 1 port
RS232
MAX232 UART
8
Simplex vs. Duplex Transmission
• Simplex transmission: the data can sent in one direction.– Example: the computer only sends data to the printer.
• Duplex transmission: the data can be transmitted and received.
Transmitter Receiver
Transmitter
ReceiverReceiver
Transmitter
Device 1 Device 2
Device 2Device 1
9
Half vs. Full Duplex
• Half duplex: if the data is transmitted one way at a time.
• Full duplex: if the data can go both ways at the same time.– Two wire conductors for the data lines.
Transmitter
Receiver
Receiver
Transmitter
Transmitter
Receiver
Receiver
Transmitter
10
Figure 10-2. Simplex, Half-, and Full-Duplex Transfers
Half Duplex
Full Duplex
Transmitter
Receiver
Transmitter
Receiver
Receiver
Transmitter
Receiver
Transmitter
Transmitter ReceiverSimplex
11
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).
12
Figure 10-1. Serial versus Parallel Data Transfer (1/2)
Sender Receiver Sender Receiver
Serial Transfer Parallel Transfer
D0
D7
13
Figure 10-1. Serial versus Parallel Data Transfer (2/2)
Sender Receiver Sender Receiver
Serial Transfer Parallel Transfer
D0-D7D0
Other control lines
Other control lines
14
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.
11101000001011
‘A’
register8-bit character
register8 1
parallel-in serial-out
serial-in parallel-out
15
Asynchronous vs. Synchronous
• Serial communication uses two methods:– In synchronous communication, data is sent in blocks
of bytes.
– In asynchronous communication, data is sent without continuity.
bytebyte 10101111
preamble
10101010
senderreceiver
byte
senderreceiver start bit stop bit
bytebyte
time
byte byte
16
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
receiver-transmitter)
– UART (universal asynchronous receiver-transmitter)
• The 8051 chip has a built-in UART.
17
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!
18
Framing (1/3)
• Each character is placed in between start and stop bits. This is called framing.– Figure 10-3. Framing ASCII “A” (41H)
stopbit
startbit mark0 0 0 0 0 01 1
D7 D0goes out last goes out first
Time (D0 first)
mark
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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.
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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.
21
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
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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.
23
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
markstopbit
startbit mark00 10 01 10 00 1111 11
Time (D0 first)
8-bit character
stopbit
24
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
25
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.
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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.
TxD
RxD
DTE
RxD
TxD
DCE
GNDPC Com1 modem
GND
TxD
RxD
DTE
RxD
TxD
DCE
GNDPC Com1modemGND
Telephone Line
DTE viewDCE view
27
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.
PC 8051
COM 1 port
RS232
MAX232 UART
DTE view DTE view
28
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
TxD
RxD
DTE
TxD
RxD
DTE
ground8051-based board
8051-based board
TxD
RxD
DTE
TxD
RxD
DTE
groundPC Com1
PC Com1
29
RS232 Voltage Level
• The input and output voltage of RS232 is not of the TTL (Transistor-Transistor-Logic) compatible.– 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
-25V
3V
25V
logic 1
undefined
RS 232 Voltage
30
MAX232
• MAX232 IC chips are commonly referred to as line drivers.
PC 8051
COM 1 port
RS232
MAX232 UART
RS232 voltage level
TTL voltage level
31
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.
32
Figure 10-4. RS232 Connector DB-25
14
1
25
13
33
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
34
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
35
Figure 10-5. DB-9 9-Pin Connector
6
1
9
5
36
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)
37
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
38
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)
39
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.
40
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.
41
Section 10.28051 Connection to RS232
42
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.
43
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
P3.1TxD
DB-9
P3.0RxD
11 11
10 12
14
13
2
3
5
44
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 F. The most widely used value for these capacitors is 22F.
• 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
45
Figure 10-7 (a): Inside MAX232
MAX232
161
34
5
2
6
14
13
7
8
11
12
10
9
15
Vcc
C1
C2
+
+
+
+
C3
C4
TTL side RS232 side
T1IN
R1OUT
T2IN
R2OUT
T1OUT
R1IN
T2OUT
R2IN
22 F
46
Figure 10-7: (b) MAX232’s Connection to the 8051 (Null Modem)
8051
MAX232
P3.1TxD
DB-9
P3.0RxD
11 11
10 12
14
13
2
3
5
RS232-compatibleTTL-compatible
47
Figure 10-8: (a) Inside MAX233
MAX233
7
15
10
11
16
5
4
18
19
2
3
1
20
6
Vcc
1217
14
9TTL side RS232 side
T1IN
R1OUT
T2IN
R2OUT
T1OUT
R1IN
T2OUT
R2IN
13
48
Figure 10-8: (b) MAX233’s Connection to the 8051 (Null Modem)
8051
MAX233
P3.1TxD
DB-9
P3.0RxD
11 2
10 3
5
4
2
3
5
RS232-compatibleTTL-compatible
49
Section 10.38051 Serial Port Programming in Assembly
50
PC Baud Rates
• 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.
110 bps
150
300
600
1200
2400
4800
9600 (default)
19200
Note: Baud rates supported by
486/Pentium IBM PC BIOS.
51
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
XTAL oscillator
÷ 12÷ 32
By UART
Machine cycle frequency
28800 Hz
To timer 1 To set the Baud rate
921.6 kHz
11.0592 MHz
Timer 1
52
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
53
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.
54
Table 10–4 Timer 1 TH1 Register Values for Various Baud Rates
55
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.
56
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
57
SCON Register
• Serial control register: SCONSM0, 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)
SM0 SM1 SM2 REN TB8 RB8 TI RI(LSB)
(MSB)
* SCON is bit-addressable.
58
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 SM1 Mode Operating Mode Baud Rate
0 0 0 Shift register Fosc./12
0 1 1 8-bit UART Variable by timer1
1 0 2 9-bit UART Fosc./64 or Fosc./32
1 1 3 9-bit UART Variable
59
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
60
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.
61
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.
62
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 4 5TxD RxD
data0 data0 address=51
8 bitsTB8(the 9th bit)
time
63
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.
64
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.
65
Figure 10-9. SCON Serial Port Control Register (Bit Addressable)
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.
SM0 SM1 SM2 REN TB8 RB8 TI RI
66
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.
SBUF TxDbit by bit8-bit char
UARTTI
Ashiftzero detect add start and
stop bits
8-bit
67
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.
68
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
69
Programming the 8051 to Transfer data Serially (2/2)
5. Clear TI flag.– CLR TI
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.
TI=0 transfer data
raise when sending the stop bit
70
Example 10-2
Write a program for the 8051 to transfer letter “A” serially at 4800
baud, continuously.
Solution:
MOV TMOD,#20H ;timer 1, mode 2
MOV TH1,#-6 ;4800 baud rate
MOV SCON,#50H ;8-bit,1 stop, REN enabled
SETB TR1 ;start timer 1
AGAIN: MOV SBUF,#”A” ;letter “A” to be transferred
HERE: JNB TI,HERE ;wait for the last bit
CLR TI ;clear TI for next char
SJMP AGAIN ;keep sending A
71
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
72
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
73
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.
shift regiterRxDbit by bit 8-bit
UART RI
SBUFloadshift
A
REN=1
remove/check the start and stop bits
8-bit
74
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.
75
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
76
Programming the 8051 to Receive data Serially (2/2)
5. Clear RI flag.– CLR RI
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.
RI=0 receive data
raise when getting the stop bit
77
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 TMOD,#20H ;timer1, mode 2 (auto reload)
MOV TH1,#-6 ;4800 baud
MOV SCON,#50H ;8-bit, 1 stop, REN enabled
SETB TR1 ;start timer 1
HERE: JNB RI,HERE ;wait for char to come in
MOV A,SBUF ;save incoming byte in A
MOV P1,A ;send to port 1
CLR RI ;get ready to receive next byte
SJMP HERE ;keep getting data
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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.
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Example 10-5 (2/4)
Solution:
ORG 0
MOV P2,#0FFH ;make P2 an input port
MOV TMOD,#20H
MOV TH1,#0FAH ;4800 baud rate
MOV SCON,#50H
SETB TR1 ;start timer 1
MOV DPTR,#MYDATA ;load pointer for message
TxD
RxD
P1
P2
LED
SW
To PC
COM port
8051
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Example 10-5 (3/4)
MOV DPTR,#MYDATA ;load pointer for message
H_1: 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
B_1: 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
(a)
(c)
(b)
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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
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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.
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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.
SMOD -- -- -- GF1 GF2 PD IDL(LSB)
(MSB)
* PCON is not bit-addressable. See Appendix H. p410
84
SMOD Flag of the PCON Register
• Power control register: PCONMOV 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
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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.
XTALoscillator
÷ 12÷ 16
÷ 32
Machine cycle freq.
921.6 kHz
57600 Hz
28800 Hz
To timer 1 to set baud rate
11.0592 MHz SMOD = 1
SMOD = 0
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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
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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)
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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.
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Example 10-6 (2/2)
MOV A,PCON
SETB ACC.7
MOV PCON,A ;SMOD=1, double baud rate
MOV TMOD,#20H ;Timer 1, mode 2, auto reload
MOV TH1,#-3 ;19200 baud rate
MOV SCON,#50H ;8-bit data,1 stop bit, RI enabled
SETB TR1 ;start Timer 1
MOV A,#”B” ;transfer letter B
A_1:CLR TI ;make sure TI=0
MOV SBUF,A ;transfer it
H_1:JNB TI H_1 ;check TI
SJMP A_1 ;do again
90
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.
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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.
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Table 10–5 Baud Rate Comparison for SMOD = 0 and SMOD = 1
93
Section 10.4Programming the Second Serial Port
94
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
95
Table 10–6 SFR Byte Addresses for DS89C4x0 Serial Ports
as same as the original 8051 serial port
new SCON1,SBUF1
shared
Compare with p556 Table A-2
96
Table 10–7 SFR Addresses for the
DS89C4x0 (420, 430, etc.)
97
Figure 10–12 SCON0 and SCON1 Bit Addresses (TI and RI bits must be noted)
98
Figure 10–10 DS89C4x0 Pin Diagram.
Note: Notice P1.2 and P1.3 pins are used by Rx and Tx lines of the 2nd serial port
port 1
port 0
99
Figure 10–11 Connection with MAX232
(a) Inside MAX232 (b) MAX232’s Connection to the DS89C4x0
100
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
101
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 EQU 0C1H ;defined by yourself SCON1 EQU 0C0H TI1 BIT 0C1H RI1 BIT 0C0H
ORG 0H
MAIN: MOV TMOD,#20H
MOV TH1,#-6
MOV SCON1,#50H
SETB TR1
102
Example 10-11 (2/2)
ANAIN:
MOV A,#’A’ ;send ‘A’ continuously
ACALL SENDCON2
SJMP AGAIN
SENDCOM2: ;send by using port 1
MOV SBUF1,A
JNB TI1,HERE
CLR TI1
RET
END
103
Section 10.5Serial Port Programming in C
104
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.
105
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; } }
106
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
107
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