Microprocessor and microcontroller MODULE-4.pptx

MODULE-4
Microcontroller 8051 Peripherals
BECE204L – MICROPROCESSORS AND MICROCONTROLLERS
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MODULE-4 BECE204L – MICROPROCESSORS AND MICROCONTROLLERS

MODULE-4
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Microcontroller 8051 Peripherals
I/O Ports, Timers-Counters, Serial Communication and
Interrupts.

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I/O PORTS
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I/O PORTS
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 8051 microcontrollers have 4 I/O ports each comprising 8 bits which can be configured as
inputs or outputs.
 Hence, total 32 input/output pins allow the microcontroller to be connected with the
peripheral devices.
 Except port 1 all other ports are used for dual purpose
 All ports are bidirectional and they are constructed with a D type output latch.
 Pin configuration, i.e. whether it is to be configured as an input (1) or an output (0),
depends on its logic state.
 All the ports upon RESET are configured as input, ready to be used as input ports.

I/O PORTS
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 Port 0
 Port 0 occupies a total of 8 pins (pins 32-39) and it can be used for input or output.
 To use the pins of port 0, each pin must be connected externally to a 10k ohm pull-up
resistor. This is due to the fact that P0 is an open drain, unlike P1, P2, and P3.
 Dual role of port-0: Port 0 is also designated as AD0-AD7, allowing it to be used for both
address and data. If ALE=1, then it is address or else data.
 Port 1
 Port 1 occupies a total of 8 pins (pins 1-8) and it can be used for input or output.
 In contrast to port 0, this port does not need any pull-up resistors since it already has pull-
up resistors internally.
 Unlike other ports, port-1 doesn’t used for dual purpose.

I/O PORTS
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 Port 2
 Port 2 occupies a total of 8 pins (pins 21- 28) and it can be used as input or output.
 Just like P1, P2 does not need any pull-up resistors since it already has pull-up resistors
internally.
 Port 2 is also designated as A8 –A15, indicating its dual function and Port 0 provides the
lower 8 bits via A0 –A7.
 Port 3
 Port 3 occupies a total of 8 pins, (pins 10-17) and it can be used as input or output.
 P3 does not need any pull-up resistors, the same as P1 and P2 did not.
 Port 3 has the additional function of providing some extremely important signals such as
interrupts.

I/O PORTS
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 I/O Pin’s Current Limitations:
 When configured as outputs (logic 0), single
port pins can receive a current of 10mA.
 If all 8 bits of a port are active, a total current
must be limited to 15mA (port P0: 26mA).
 If all ports (32 bits) are active, total maximum
current must be limited to 71mA.
 When these pins are configured as inputs
(logic 1), built-in pull-up resistors provide very
weak current, but strong enough to activate up
to 4 TTL inputs of LS series.

I/O PORTS
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PORT 0 BITS P0.7 P0.6 P0.5 P0.4 P0.3 P0.2 P0.1 P0.0
BIT ADDRESS 87H 86H 85H 84H 83H 82H 81H 80H
BIT ADDRESS 97H 96H 95H 94H 93H 92H 91H 90H
BIT ADDRESS A7H A6H A5H A4H A3H A2H A1H A0H
BIT ADDRESS B7H B6H B5H B4H B3H B2H B1H B0H
8051 I/O PORTS & SFR BIT AND BYTE ADDRESS LOCATIONS

I/O PORTS
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Write an 8051 assembly code to configure port-1 as output port and send out an alternating
value 55H and AAH continuously with a delay between each data out.
EXAMPLE-1
MOV P1, #00H
BACK: MOV A, #55H
MOV P1,A
ACALL DELAY
MOV A, #AAH
MOV P1,A
ACALL DELAY
SJMP BACK
DELAY: MOV R3, #250
HERE: NOP
NOP
NOP
NOP
DJNZ R3, HERE
RET

I/O PORTS
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Write an 8051 assembly code to generate a square wave of 50% duty cycle on bit 0 of port 1.
Solution:
The 50% duty cycle means that HIGH (ON state) and LOW (OFF state) of the pulse should have the same
length. So, we have to toggle P1.0 with a time delay in between each state.
EXAMPLE-2
CLR P1.0
BACK: SETB P1.0
ACALL DELAY
CLR P1.0
ACALL DELAY
SJMP BACK
DELAY: MOV R3, #250
HERE: NOP
NOP
NOP
NOP
DJNZ R3, HERE
Exercise: Write an 8051 assembly code to generate a square wave of 25% duty cycle on bit 0 of port 1.

I/O PORTS
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Assume that bit P2.3 is an input and represents the condition of an oven. If it goes high, it means
that the oven is hot. Monitor the bit continuously. Whenever it goes high, send a high-to-low pulse
to port P1.5 to turn on a buzzer. Write an 8051 assembly code for this condition.
EXAMPLE-3
SETB P2.3 ; Configure P2.3 as input port pin
HERE: JNB P2.3,HERE ;keep monitoring for high
SETB P1.5 ;set bit P1.5=1
CLR P1.5 ;make high-to-low
SJMP HERE ;keep repeating
Exercise: Write an 8051 program to perform the following:
(a) Keep monitoring the P1.2 bit until it becomes high
(b) When P1.2 becomes high, write value 45H to port 0
(c) Send a high-to-low (H-to-L) pulse to P2.

I/O PORTS
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A switch is connected to pin P1.7. Write an 8051 program to check the status of SW and perform the
following:
(a) If SW=0, send letter ‘N’ to P2
(b) If SW=1, send letter ‘Y’ to P2
EXAMPLE-4
SETB P1.7 ;make P1.7 an input
AGAIN: JB P1.2,OVER ;jump if P1.7=1
MOV P2,#’N’ ;SW=0, issue ‘N’ to P2
SJMP AGAIN ;keep monitoring
OVER: MOV P2,#’Y’ ;SW=1, issue ‘Y’ to P2
SJMP AGAIN ;keep monitoring
Exercise: A switch is connected to pin P1.0 and an LED to pin P2.7. Write an 8051program to get the
status of the switch and send it to the LED

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TIMERS / COUNTERS
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TIMERS / COUNTERS
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TIMERS
 8051 has two timers/counters (Timer 0 and Timer 1), both are 16 bits counter. They can
be used either as
 Timers to generate a time delay
 To generate a waveform with specific frequency
 To generate baud rate signal for serial communication
 Event counters to count events happening outside the microcontroller
 Register related to work with 8051 timers are:
1. TH & TL Timer/counter register– To hold the value for generating time delay
2. TMOD Register - to define mode of timer operation
3. TCON Register – To control the timer operation

TIMERS / COUNTERS
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TIMERS/COUNTER REGISTERS

TIMERS / COUNTERS
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 8051 has two timers/counters registers (Timer 0 and Timer 1), each 16-bits wide and
accessed as two separate registers of low byte and high byte
 The low byte register is called TL0/TL1 and
 The high byte register is called TH0/TH1

TIMERS / COUNTERS
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 Steps to calculate values to be loaded into the TL and TH registers
1. Divide the desired time delay by 1.085us (if operating frequency is 11.0592 MHz)
2. Perform 65536 – n, where “n” is the decimal value got in Step1 (if 16-bit counter mode is
selected)
3. Convert the result of Step2 to hex value and represent in four digits (ex: xxyy)
4. If “xxyy” is the hex value to be loaded into the timer’s register, then set TH = xx and TL = yy
 Example: 500us time delay
Step1: 500us/1.085us = 461pulses
Step2: P=65536-461=65075
Step3: 65074 converted by hexa decimal =FE33
Step4: TH1=0xFE; TL1=0x33;

TIMERS / COUNTERS
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TMOD REGISTER
 Both timers 0 and 1 use the same register, called TMOD (timer mode), to set the various
timer operation modes
 TMOD is a 8-bit register:
 The lower 4 bits are for Timer 0
 The upper 4 bits are for Timer 1
 Timers of 8051 do starting and stopping by either software or hardware control
 Software: First set GATE=0, then start and stop of the timer are controlled by the TR(timer
start) bits TR0 and TR1. The SETB starts it, and it is stopped by the CLR instruction.
 Hardware: The starting and stopping the timer by an external source is achieved by
making GATE=1 in the TMOD register.

TIMERS / COUNTERS
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TMOD REGISTER
GATE C/T M1 M0 GATE C/T M1 M0
TIMER-1 TIMER-0
Gating Control:
 If GATE=1 – Timer/Counter is
enable only while INTx pin is HIGH
and the TRx control pin is SET
 If GATE=0, the timer is enabled
whenever the TRx control bit is set
Counter/Timer Selction:
 C/T=1 – Act as a counter and receive
external input signal from P3.5 pin for
T1, (P3.4 for T0)
 C/T=0 – Act as a timer and receive
input signal from internal system clock
M1 M0 Operating Mode
0 0
13-bit Timer Mode (Mode-0): 8-bit Timer/Counter
THx with TLx as 5-bit prescaler
0 1
16-bit Timer Mode (Mode-1): 16-bit Timer/Counter
THx and TLx are cascaded, no prescaler
1 0
8-bit Auto-reload Mode (Mode-2): 8-bit auto
reload Timer/Counter; THx holds the value which is
to be reloaded when TLx overflows each time
1 1 Split-timer Mode (Mode-3)
*Where x represent 0 for Timer0 and 1 for Timer 1

TIMERS / COUNTERS
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TCON REGISTER
TF1 TR1 TF0 TR0 IE1 IT1 IE0 IT0
 TFx: Timerx Overflow Flag
 TFx =1 means Timerx overflow
occurred (i.e. Timerx goes to its
max and roll over back to zero).
not occurred.
 TRx: Timerx Run Control Bit
 TRx =1 means Timerx start.
 TRx =0 means Timerx stop.
 IEx: External Interruptx Edge Flag
 IEx=1 means External
interruptx occurred.
interruptx Processed.
 ITx: External Interruptx Trigger Type Select Bit
 ITx=1 means Interrupt occurs on falling
edge at INTx pin.
 ITx=0 means Interrupt occur on a low
level at the INTx pin.

TIMERS / COUNTERS
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TIMER 0 MODE 0 (13-BIT TIMER)
 This is one of the rarities being kept only for the purpose of compatibility with the
previous versions of microcontrollers.
 This mode configures timer 0 as a 13-bit timer which consists of all 8-bits of TH0 and the
lower 5-bits of TL0.
 How does it operate? Each coming pulse causes the lower register bits to change their
states. After receiving 32 pulses, this register is loaded and automatically cleared, while
the higher byte (TH0) is incremented by 1.
 This process is repeated until registers count up 8192 pulses. After that, both registers are
cleared and counting starts from 0

TIMERS / COUNTERS
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TIMERS / COUNTERS
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 It is a 16-bit timer; therefore it allows values from 0000 to FFFFH to be loaded into the
timer’s registers TL and TH.
 After TH and TL are loaded with a 16-bit initial value, the timer must be started.
 After the timer is started. It starts count up until it reaches its limit of FFFFH.
 When it rolls over from FFFF to 0000H, it sets high a flag bit called TF (timer flag).
 This is one of the most commonly used modes.

TIMERS / COUNTERS
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TIMERS / COUNTERS
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TIMER 0 MODE 2 (8-BIT AUTO-RELOAD TIMER)
 When a timer is in mode 2, THx holds the "reload value" and TLx operates as a timer.
 Thus, TLx starts counting up. When TLx reaches 255 and is subsequently incremented,
instead of resetting to 0 (as in the case of modes 0 and 1), it will be reset to the value
stored in THx.
 The auto-reload mode is very commonly used for establishing a baud rate.
 For example, suppose it is necessary to constantly count up 55 pulses generated by the
clock.
 In order to register each 55th pulse, the best solution is to write the number 200 to the
TH0 register and configure the timer to operate in mode 2.

TIMERS / COUNTERS
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TIMER 0 MODE 2 (8-BIT AUTO-RELOAD TIMER)

TIMERS / COUNTERS
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TIMER 0 MODE 3 (SPLIT TIMER)
 This mode is provided for applications requiring an additional 8-bit timer or counter.
 Mode 3 configures timer 0 so that registers TL0 and TH0 operate as separate 8-bit timers.
 In other words, the 16-bit timer 0 consisting of two registers TH0 and TL0 is split into two
independent 8-bit timers.
 The TL0 timer turns into timer 0, while the TH0 timer turns into timer 1.
 In addition, all the control bits of 16-bit Timer 1 (consisting of the TH1 and TL1 register),
now control the 8-bit Timer 1.
 While Timer 0 is in split mode, you may not start or stop the real timer 1 since the bits that
do that are now linked to TH0.

TIMERS / COUNTERS
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TIMER 0 MODE 3 (SPLIT TIMER)

TIMERS / COUNTERS
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STEPS TO PROGRAM 0 MODE 1 (16-BIT TIMER)
1. Load the TMOD value register indicating which timer (timer 0 or timer 1) and timer mode
(0 or 1) is selected. MOV TMOD,#01H.
2. Load registers TL and TH with initial count value. MOV TH0,#FFH; MOV TL0,#FCH.
3. Start the timer using SETB TRx. SETB TR0.
4. Keep monitoring the timer flag (TF). AGAIN: JNB TF0, AGAIN.
5. Stop the timer using CLR TRx. CLR TR0.
6. Clear the TF flag for the next round. CLR TF0.
7. Go back to Step 2 to load TH and TL again. SJMP STEP2.

TIMERS / COUNTERS
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LARGEST POSSIBLE DELAY USING TIMERS
Calculate TL and TH to get the largest time delay possible. Find the delay in ms. In your
calculation, exclude the overhead due to the instructions in the loop
Solution:
• To get the largest delay we make TL and TH both 0. This will count up from 0000 to
FFFFH and then roll over to zero
• Making TH and TL both zero means that the timer will count from 0000 to FFFF, and
then roll over to raise the TF flag.
• As a result, it goes through a total Of 65536 states. Therefore, we have delay
=(65536 - 0) ×1.085 us = 71.1065ms.

TIMERS / COUNTERS
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EXAMPLE-1
Write an 8051 assembly language program to blink an LED connected in P2.7 with a time
delay of 5ms by using timers. Assume that XTAL = 11.0592 MHz.
• Since XTAL = 11.0592 MHz, the counter counts up every 1.085 us.
• We must identify how many times 1.085 us intervals should be repeated to achieve
5 ms delay.
• To get that, perform 5 ms / 1.085 us = 4608 clocks.
• To Achieve that we need to load into TL and TH the value 65536 – 4608 = 60928.
• Convert 60928 into HEX i.e EE00. Therefore, we have TH = EE and TL = 00.

TIMERS / COUNTERS
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EXAMPLE-1
CLR P2.7 ;Clear P2.7
MOV TMOD,#01 ;Timer 0, 16-
bitmode
HERE: MOV TL0,#0 ;TL0=0, the low byte
MOV TH0,#0EEH ;TH0=EE, the
high byte
CPL P2.7 ;SET high P2.7
SETB TR0 ;Start timer 0
AGAIN: JNB TF0,AGAIN ;Monitor timer flag 0
CLR TR0 ;Stop the timer 0
CLR TF0 ;Clear timer 0 flag
SJMP HERE
Exercise: Assuming XTAL = 22MHz write a program to generate a square pulse of 200ms
period with 75% duty cycle on pin P2.4. Use timer 1 mode.

TIMERS / COUNTERS
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EXAMPLE-2
Generate a square wave with TON of 3ms and TOFF of 10ms on all pins of port 0. Assume
an XTAL of 22MHz.
For 22MHz , one timer cycle time is
• 22MHz / 12 = 1.83MHz
• T = 1/1.83M = 0.546 us
For OFF time calculation:
• 10ms/0.546 us = 18315 cycles
• 65536-18315 =47221 = B875H
For ON time calculation:
• 3ms/0.546us = 5494 cycles
• 65536 – 5494 = 60042=EA8AH

TIMERS / COUNTERS
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EXAMPLE-2
MOV TMOD, #01H
BACK: MOV TL0, #75H
MOV TH0, #0B8H
MOV P0, #00H
ACALL DELAY
MOV TL0, #8AH
MOV TH0, #0EAH
MOV P0, #0FFH
ACALL DELAY
SJMP BACK
ORG 300H
DELAY: SETB TR0
AGAIN: JNB TF0, AGAIN
CLR TR0
CLR TF0
RET
Exercise: Generate a square wave with TON of 5ms and TOFF of 2ms on all pins of port 0.
Assume an XTAL of 11.0592 MHz.

TIMERS / COUNTERS
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EXAMPLE-3
Assuming XTAL = 22MHz write a program to generate a square pulse of frequency 100kHz
on port pin P1.2 using timer-1 mode 2.
For 22MHz , one timer cycle time is
• 22MHz / 12 = 1.83MHz
• T = 1/1.83M = 0.546 μs
For 100kHz square wave,
• T=1/f = 0.01ms = 10μS
• TON = TOFF = 10uS/2 = 5μS
• 5 μs/ 0.546 μs = 9 cycles
• 256 – 9 = 247 = F7H
MOV TMOD, #20H
MOV TH1, #0F7H
SETB TR1
BACK: JNB TF1, BACK
CPL P1.2
CLR TF1
SJMP BACK
Exercise: Assume that XTAL = 11.0592 MHz, write a 8051 program to generate a square wave
of 2 kHz frequency on pin P1.7

TIMERS / COUNTERS
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COUNTERS
 Timers can also be used as counters counting events happening outside the 8051
 In counter, it is a pulse outside of the 8051 that increments the TH, TL registers
 TMOD and TH, TL registers are the same as for the timer discussed previously
 Programming the timer in the last section also applies to programming it as a counter
except the source of the frequency
 The C/T bit in the TMOD registers decides the source of the clock for the timer
 When C/T = 1, the timer is used as a counter and gets its pulses from outside the 8051
 The counter counts up as pulses are fed from pins 14 and 15, these pins are called T0
(timer 0 input) and T1 (timer 1 input).

TIMERS / COUNTERS
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EXAMPLE-4
Assuming that clock pulses are fed into pin T1(P3.5), write a 8051 assembly program for counter 1 in
mode 2 to count the pulses and display the state of the TL1 count on P1, which connects to 8 LEDs.
ORG 0000H
SETB P3.5 ; make T1 input
MOV P1,#00H ; Make Port 1 as output port
MOV TMOD, #60H ; counter 1, mode 2;C/T=1 external pulses
MOV TH1, #0 ; clear TH1
AGAIN: SETB TR1 ; start the counter
BACK: MOV A, TL1 ; get copy of TL
MOV P1, A ; display it on port 2
JNB TF1, BACK ; keep doing, if TF=0
CLR TR1 ; Stop the counter 1
CLR TF1 ; make TF=0
SJMP AGAIN ; keep doing it

TIMERS / COUNTERS
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EXAMPLE-5
Write an 8051 assembly language program to implement a counter for counting pulses of an input
signals. Assume the crystal frequency as 22 MHz. Configure TIMER 1 to generate a clock pulse for every one
seconds at P3.5 and TIMER 0 as a counter which receives input pulses at P3.4 from P3.5. Display final count
values in port P1 (TL0) & P2 (TH0).
ORG 000H
REPEAT: MOV TMOD, #15H
SETB P3.4
CLR P3.5
MOV TL0, #00
MOV TH0, #00
SETB TR0
MOV R0,#28
AGAIN: MOV TL1,#00
MOV TH1, #00
SETB TR1
………
………..
BACK: JNB TF1, BACK
CLR TF1
CLR TR1
DJNZ R0, AGAIN
MOV A, TL0
MOV P1,A
MOV A, TH0
MOV P2,A
SJMP REPEAT
END

SERIAL COMMUNICATION
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 In data transmission, serial communication is the process of sending data one bit at a time,
sequentially, over a communication channel or computer bus.
 To reduce the number of pins in a package, many ICs use a serial bus to transfer data
when speed is not important.
 Some examples of such low-cost serial buses include UART, USART, SPI, I²C etc.,
 Serial data communication uses two methods
 Synchronous method transfers a block of data at a time
 Asynchronous method transfers a single byte at a time
 There are special IC chips made by many manufacturers for serial communications
 UART (universal asynchronous Receiver-transmitter)
 USART (universal synchronous-asynchronous Receiver-transmitter)

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Wiring Connection between two UART
Serial communication message format

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 In UART, the start bit is always one bit (HIGH), the stop bit can be one or two bits (LOW).
 The rate of data transfer in serial data communication is stated in bps(bits per second).
 Another widely used terminology for bps is baud rate and is defined as the number of
signal changes per second. The baud rate and bps are the same, used inter changeably.
 8051 has two pins (11 and 10) that are used specifically for transferring (Tx) and
receiving(Rx) data serially as a part of the port 3 group (P3.1 and P3.0).
 To allow data transfer between two UART devices, we must make sure that the baud rate
of both devices matches.
 Different baud rate levels are, 150, 300, 600, 1200, 2400, 4800, 9600, 19200 etc.,

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 Dividing 1/12 of the crystal frequency by 32 is the default frequency (28800 Hz) for
timers to generate the baud rate.
BAUD RATE GENERATION IN 8051
 With XTAL=11.0592 MHz, find the TH1 value needed to set different baud rate.
 28800/3 = 9600 Where -3=FD (hex) is loaded into TH1
 28800/6 = 4800 Where -6=FA (hex) is loaded into TH1
 28800/12 = 2400 Where -12=F4 (hex) is loaded into TH1
 28800/24 = 1200 Where -24=E8 (hex) is loaded into TH1
XTAL
Oscillator
Divide by
12
Divide by
32 by UART
11.0592 MHz
MC Freq,
921.6 kHZ 28800 Hz
To timer 1 to set baud rate

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 Register required to work with 8051 serial communications are: SBUF and SCON
 SBUF is an 8-bit register used to hold a data during transmit and receive operation
MOV SBUF, #‘D’ ; load SBUF=44H, ASCII for ‘D’
MOV SBUF, A ; copy Accumulator into SBUF
MOV A, SBUF ; copy SBUF into Accumulator
 SCON is an 8-bit the special function register (bit-addressable).
 This register contain not only the mode selection bits but also the 9th data bit for transmit
and receive (TB8 and RB8) and the serial port interrupt bits (TI and RI).
SM0 SM1 SM2 REN TB8 RB8 TI RI

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BIT NAME DESCRIPTION
7 SM0 Serial port mode bit 0
6 SM1 Serial port mode bit 1
5 SM2 Multi processor communication enable bit
4 REN Receiver Enable. This bit is set in order to receive characters.
3 TB8 Transmit bit 8. The 9th
bit to transmit in mode 2 and 3.
2 RB8 Recieve bit 8. The 9th
bit to receive in mode 2 and 3.
1 TI Transmit Interrupt Flag. Set when a byte has been completely transmitted.
0 RI Receive Interrupt Flag. Set when a byte has been completely Received.
SM0 SM1 MODE DESCRIPTION BAUD RATE
0 0 0
Shift register: Serial data are transmitted and received through the RXD pin,
while the TXD pin output clocks
Fosc./12
0 1 1 8-Bit UART: 8-bit DATA, 1-bit for START, 1-bit for STOP Variable
1 0 2 9-Bit UART: 9-bit DATA, 1-bit for START, 1-bit for STOP Fosc./64 or Fosc./32
1 1 3 9-Bit UART: 9-bit DATA, 1-bit for START, 1-bit for STOP Variable

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STEP TO PROGRAM 8051 TO TRANSFER CHARACTER BYTES SERIALLY
1. Load TMOD with 20H, indicating the use of timer 1 in mode 2 (8-bit auto-reload) to set baud rate
2. The TH1 is loaded with one of the values to set baud rate for serial data transfer
3. Load SCON with 50H, indicating mode 1, where an 8-bit data is framed with start and stop bits
4. TR1 is set to 1 to start timer 1
5. The character byte to be transferred serially is written into SBUF register
6. The TI flag bit is monitored using JNB TI, xx to check the character is transferred completely or not
7. TI is cleared by CLR TI instruction
8. To transfer the next byte, go to step 5

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EXAMPLE-1
Write a program for the 8051 to transfer letter “A” serially at 4800 baud, continuously.
ORG 0000H
MOV TMOD,#20H ;timer1,mode 2(auto reload)
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 transfer
HERE: JNB TI,HERE ;wait for the last bit
CLR TI ;clear TI for next char
SJMP AGAIN ;keep sending A

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EXAMPLE-2
Write a program for the 8051 to transfer “YES” serially at 9600 baud, 8-bit data, 1 stop bit, do this
continuously MOV TMOD,#20H ;timer1,mode 2(auto reload)
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
;serial data transfer subroutine
TRANS: MOV SBUF,A ;load SBUF
HERE: JNB TI,HERE ;wait for the last bit
CLR TI ;get ready for next byte
RET

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EXAMPLE-3
Assume a switch is connected to pin P1.7. Write an 8051 program to monitor its status and send two messages to
serial port continuously as follows. Assume XTAL = 11.0592 MHz, 9600 baud, 8-bit data, and 1 stop bit.
* SW=0 send “NO”
* SW=1 send “YES” ORG 0000H ;starting position
MAIN: MOV TMOD, #20H ;timer1,mode 2(auto reload)
MOV TH1,#-3 ; 9600 baud rate
MOV SCON, #50H ;8-bit, 1 stop, REN enabled
SETB TR1 ; start timer
SETB P1.7 ; make SW an input
S1: JB P1.7, NEXT ; check SW status
MOV DPTR, #MESS1 ; if SW=0 display "NO“
FN: CLR A
MOVC A, @A+DPTR ; read the value; check for end of line
JZ S1 ; check for end of line
ACALL SENDCOM ; send value to serial port

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INC DPTR ; move to next value
SJMP FN ; repeat
NEXT: MOV DPTR, #MESS2 ; if SW=1 display "YES“
LN: CLR A
MOVC A, @A+DPTR ; read the value; check for end of line ;
JZ S1 ; check for end of line
ACALL SENDCOM ; send value to serial port
INC DPTR ; move to next value
SJMP LN ; repeat
SENDCOM: MOV SBUF, A ; place the value in SBUF
HERE: JNB TI, HERE ; wait until transmit complete
CLR TI ; clear TI
RET ; return
MESS1: DB ‘NO’,0
MESS2: DB ‘YES’, 0
END

51
STEP TO PROGRAM 8051 TO RECEIVE CHARACTER BYTES SERIALLY
1. Load TMOD with 20H, indicating the use of timer 1 in mode 2 (8-bit auto-reload) to set baud rate
2. The TH1 is loaded with one of the values to set baud rate for serial data transfer
3. Load SCON with 50H, indicating mode 1, where an 8-bit data is framed with start and stop bits
4. TR1 is set to 1 to start timer 1
5. The RI flag bit is monitored using JNB RI, xx to check the character is received completely or not
6. If RI is raised, SBUF register has a byte and transfer it to accumulator
7. RI is cleared by CLR RI instruction
8. To transfer the next byte, go to step 5

52
EXAMPLE-4
Write a program for 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.
ORG 0000H
MOV P1,#00H
MOV TMOD,#20H ;timer1,mode 2(auto reload)
HERE: JNB RI,HERE ;wait for char to come in
MOV A,SBUF ;saving incoming byte in A
MOV P1,A ;send to port 1
CLR RI ;get ready to receive next byte
SJMP HERE ;keep getting data

53
EXAMPLE-5
Assume that the 8051 serial port is connected to the COM port of 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 PC the message “We Are Ready”,
(b) receive any data send by PC and put it on LEDs connected to P1, and
(c) get data on switches connected to P2 and send it to PC serially.
The program should perform part (a) once, but parts (b) and (c) continuously, use 4800 baud rate.
ORG 0000H
MOV P1,#00H
MOV P2,#0FFH ;make P2 an input port
MOV TMOD,#20H ;timer 1, mode 2
MOV TH1,#0FAH ;4800 baud rate

54
MOVC DPTR,#MYDATA ;load pointer for message
H_1: CLR A
MOV A,@A+DPTR ;get the character
JZ B_1 ;if last character get out
ACALL SEND ;otherwise call transfer
INC DPTR ;next one
SJMP H_1 ;stay in loop
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

55
;----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 ;return to caller
;----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 ;return to caller
;-----The message---------------
MYDATA: DB “We Are Ready”,0
END
Exercise: Write a program to send the message “The Earth is but One Country” to serial port. Assume a SW is connected to
pin P1.2. Monitor its status and set the baud rate as follows:
• SW = 0, 4800 baud rate
• SW = 1, 9600 baud rate
Assume XTAL = 11.0592 MHz, 8-bit data, and 1 stop bit

56
PCON Register: Power control register

INTERRUPTS
57
57

INTERRUPTS
58
 A single microcontroller can serve several devices by two ways: (i) Interrupt (ii). Polling
 Polling can monitor the status of several devices and serve each of them as certain
conditions are met
 The polling method is not efficient, since it wastes much of the microcontroller’s time by
polling devices that do not need service, ex. JNB TF, target
 Interrupts: Whenever any device needs its service, the device notifies the microcontroller
by sending it an interrupt signal
 Upon receiving an interrupt signal, the microcontroller interrupts whatever it is doing and
serves the device
 The program which is associated with the interrupt is called the interrupt service routine
(ISR) or interrupt handler
 The advantage of interrupts is that the microcontroller can serve many devices (not all at
the same time).

INTERRUPTS
59
 Upon activation of an interrupt, the microcontroller goes through the following steps:
1. It finishes the instruction it is executing and saves the address of the next instruction (PC) on the
stack
2. It also saves the current status of all the interrupts internally (i. e: not on the stack)
3. It jumps to a fixed memory location called interrupt vector table, that holds the address of the ISR
4. The microcontroller gets the address of the ISR from the interrupt vector table and jumps to it
5. It starts to execute the interrupt service subroutine until it reaches the last instruction of the
subroutine which is RETI (return from interrupt)
6. Upon executing the RETI instruction, the microcontroller returns to the place where it was interrupted
7. First, it gets the program counter (PC) address from the stack by popping the top two bytes of the
stack into the PC
8. Then it starts to execute from that address

INTERRUPTS
60

INTERRUPTS
61
EA -- ET2 ES ET1 EX1 ET0 EX0
INTERRUPT ENABLE (IE) REGISTER
BIT NAME DESCRIPTION
7 EA Enable All must be set to 1 in order activate each interrupt given in the register
6 -- Reserved for future use
5 ET2 Enable/Disable Timer 2 overflow interrupt (for 8952)
4 ES Enable/Disable Serial port interrupt
3 ET1 Enable/Disable Timer 1 overflow interrupt
2 EX1 Enable/Disable eXternal interrupt 1
1 ET0 Enable/Disable Timer 1 overflow interrupt
0 EX0 Enable/Disable eXternal interrupt 0

INTERRUPTS
62
 The timer flag (TF) is raised when the timer rolls over
 In polling TF, we have to wait until the TF is raised
 The problem with this method is that the microcontroller is tied down while waiting for
TF to be raised, it can’t do anything else
 Using interrupts solves this problem and, avoids tying down the controller
 If the timer interrupt in the IE register

INTERRUPTS
63
EXAMPLE-1
Write a program using interrupts that continuously get 8-bit data from P0 and sends it to P1 while
simultaneously creating a square wave of 200 s period on pin P2.1. Use timer 0 to create the square
μ
wave. Assume that XTAL = 11.0592 MHz.
Delay calculation: Use timer 0 in mode 2 (auto reload). TH0 = 100/1.085 us = 92 =A4H
;--upon wake-up go to main, avoid using memory allocated to Interrupt Vector Table
ORG 0000H
LJMP MAIN ;by-pass interrupt vector table
;--ISR for timer 0 to generate square wave
ORG 000BH ;Timer 0 interrupt vector table
CPL P2.1 ;toggle P2.1 pin
RETI ;return from ISR

INTERRUPTS
64
;--The main program for initialization
ORG 0030H ;after vector table space
MAIN: MOV P0,#0FFH ;make P0 an input port
MOV P1,#00H ;make P0 an output port
CLR P2.1 ;make P2.1 as output pin
MOV TMOD,#02H ;Timer 0, mode 2
MOV TH0,#0A4H ;TH0=A4H for -92
MOV IE,#82H ;IE=10000010 (bin) enable Timer 0
SETB TR0 ;Start Timer 0
BACK: MOV A,P0 ;get data from P0
MOV P1,A ;issue it to P1
SJMP BACK ;keep doing it loop unless interrupted by TF0
END

INTERRUPTS
65
 The 8051 has two external hardware interrupts, Pin 12 (P3.2) and pin 13 (P3.3) of the
8051, designated as INT0 and INT1
 The interrupt vector table locations 0003H and 0013H are set aside for INT0 and INT1
 There are two activation levels for the external hardware interrupts
1. In Level trigged Mode, INT0 and INT1 pins are normally high, if a low-level signal is
applied to them, it triggers the interrupt
2. Edge trigged Mode:
• To make INT0 and INT1 edge-triggered (falling edge) interrupts, we must program
the bits of the TCON register bit TCON.2 (IT1) and TCON.0 (IT0).
• The falling edge of pins INT0 and INT1 are latched by the 8051 and are held by
the TCON.1 and TCON.3 bits of TCON register.

INTERRUPTS
66
TCON REGISTER
TF1 TR1 TF0 TR0 IE1 IT1 IE0 IT0
 TFx: Timerx Overflow Flag
occurred (i.e. Timerx goes to its
max and roll over back to zero).
not occurred.
 TRx: Timerx Run Control Bit
 TRx =1 means Timerx start.
 TRx =0 means Timerx stop.
 IEx: External Interruptx Edge Flag
interruptx occurred.
interruptx Processed.
 ITx: External Interruptx Trigger Type Select Bit
 ITx=1 means Interrupt occurs on falling
edge at INTx pin.
 ITx=0 means Interrupt occur on a low
level at the INTx pin.

INTERRUPTS
67
EXAMPLE-2
Assume that the INT1 pin is connected to a switch that is normally high. Whenever it goes low, it should
turn on an LED. The LED is connected to P1.3 and is normally off. As long as the switch is pressed low,
the LED should stay on. Simultaneously perform a toggle operation in P1.5 with the delay of 500ms.
ORG 0000H
LJMP MAIN
//ISR for INT1
ORG 0013H
SETB P1.3
RETI

INTERRUPTS
68
ORG 30H
MAIN: SETB P3.3
CLR P1.3
CLR P1.5
MOV IE,#10000100B
HERE: CLR P1.3
SETB P1.5
ACALL DELAY
CLR P1.5
ACALL DELAY
SJMP HERE
//Delay of 500ms
DELAY: MOV R2,#04H ;LOAD R2 WITH 07 HEX
HERE3: MOV R1,#0FFH ;LOAD R1 WITH 0FF HEX
HERE2: MOV R0,#0FFH ;LOAD R2 WITH 0FF HEX
HERE1: DJNZ R0,HERE1 ;DECREMENT R0
DJNZ R1,HERE2 ;DECREMENT R1
DJNZ R2,HERE3 ;DECREMENT R2
RET ;RETURN
END

INTERRUPTS
69
EXAMPLE-3
Assume that pin 3.3 (INT1) is connected to a pulse generator, write a program in which the falling
edge of the pulse will send a high to P1.3, which is connected to an LED (or buzzer). In other words, the
LED is turned on and off at the same rate as the pulses are applied to the INT1 pin.
ORG 0000H
LJMP MAIN
;--ISR for hardware interrupt INT1 to turn on LED
ORG 0013H ;INT1 ISR
SETB P1.3 ;turn on LED
;------MAIN program for initialization
ORG 30H
MAIN: SETB P3.3 ; make P3.3 as input
CLR P1.3 ; make p1.3 as output
SETB TCON.2 ;make INT1 edge-triggered.
MOV IE,#10000100B ;enable External INT 1
HERE: CLR P1.3 ;turn off the buzzer
SJMP HERE ;stay here until get interrupted
END
Exercise: Write a program in which the 8051 reads data from P1 and writes it to P2 continuously while giving a copy
of it to the serial COM port to be transferred serially. Assume that XTAL=11.0592. Set the baud rate at 9600.

INTERRUPTS
70
EXAMPLE-4
Write a program using interrupts to do the following:
(a) Receive data serially and sent it to P0,
(b) Have P1 port read and transmitted serially, and a copy given to P2,
(c) Make timer 0 generate a square wave of 5kHz frequency on P0.1.
Assume that XTAL-11,0592. Set the baud rate at 4800.
ORG 0
LJMP MAIN
ORG 000BH ;ISR for timer 0
CPL P0.1 ;toggle P0.1
ORG 23H
LJMP SERIAL ;jump to serial interrupt ISR

INTERRUPTS
71
ORG 30H
MAIN: MOV P0,#00H ;make P0 an output port
MOV P1,#0FFH ;make P1 an input port
MOV P2,#00H ;make P2 an output port
MOV TMOD,#22H ;timer 1,mode 2(auto reload)
MOV TH1,#0F6H ;4800 baud rate
MOV TH0,#-92 ;for 5kHZ wave
MOV IE,10010010B ;enable serial int.
BACK: MOV A,P1 ;read data from port 1
MOV SBUF,A ;give a copy to SBUF
MOV P2,A ;send it to P2
SJMP BACK ;stay in loop indefinitely

INTERRUPTS
72
ORG 100H
SERIAL: JB TI,TRANS ;jump if TI is high
MOV A,SBUF ;otherwise due to receive
MOV P0,A ;send serial data to P0
CLR RI ;clear RI since CPU doesn’t
TRANS: CLR TI ;clear TI since CPU doesn’t
END
Exercise: Write a 8051 assembly program using interrupts to do the following:
(a) Generate a 10 KHz frequency on P2.1
(b) Use timer 1 as an event counter to count and display it on P0. The pulse is connected to EX1.
Assume that XTAL = 11.0592 MHz. Set the baud rate at 9600.

INTERRUPTS
73
IP (Interrupt Priority) Register
Priority levels of the interrupts can be changed by changing the
corresponding bit in the Interrupt Priority (IP) register.

Microprocessor and microcontroller MODULE-4.pptx

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