VTU 4th Semester ECE dept Microcontroller lecture slides module 3

SHRI MADHWA VADIRAJA INSTITUTE OF
TECHNOLOGY AND MANAGEMENT
Microcontroller
18EC46
Venugopala Rao A S
Dept. of CSE
venugopalrao.cs@sode-edu.in

8051 Stack, I/O Port Interfacing and Programming:
• Module-3: 8051 Stack, I/O Port Interfacing and Programming:
• 8051 Stack [T1: 2.7],
• Stack and Subroutine instructions [T2: 8.3],
• Assembly language program examples on subroutine and involving loops.[T1:
Ex:3-11, 3-12],
• Assembly language program examples on subroutine and involving loops.[T1:
Ex:4-7],
• Interfacing simple switch and LED to I/O ports to switch on/off LED with respect
to switch status.
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• 8051 Stack
• The stack is a section of RAM used by the CPU to store information temporarily
• This information could be data or an address
• The register used to access the stack is called the SP (stack pointer) register
• The stack pointer in the 8051 is 8 bit wide, hence it can take value of 00 to FFH
• When the 8051 is powered up, the SP register contains value 07
• RAM location 08 is the first location begin used for the stack by the 8051
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• The storing of a CPU register in the stack is called ___
• PUSH
• SP is pointing to the __
• last used location of the stack
• As we push data onto the stack, the SP is ____
• incremented by one
• Loading the contents of the stack back into a CPU register is called a ___
• POP
• With every pop, the top byte of the stack is copied to the register specified by the
instruction and the stack pointer is decremented once
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• Show the stack and stack pointer from the following. Assume the default stack
area.
• MOV R6, #25H
• MOV R1, #12H
• MOV R4, #0F3H
• PUSH R6
• PUSH R1
• PUSH R4
• Solution:
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• Examining the stack, show the contents of the register and SP after execution of
the following instructions. All value are in hex.
• POP R3 ; POP stack into R3
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• The CPU also uses the stack to save the address of the instruction just below the
CALL instruction
• This is how the CPU knows where to resume when it returns from the called
subroutine.
• Why to increment SP after push?
• To make sure that the stack is growing from lower to upper addresses (to 7FFH)
• If the stack pointer were decremented after push ???
• We would be using RAM locations 7, 6, 5, etc. which belong to R7 to R0 of bank
0, the default register bank
• When 8051 is powered up, register bank 1 and the stack are using the same
memory space
• We can reallocate another section of RAM to the stack
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• Call and RET
• Already covered
• Interrupts and Returns
• An interrupt is a hardware-generated call.
• Similar to a call opcode which can automatically access a subroutine, certain pins
on the 8051 can cause a call when external electrical signals on them go to a low
state.
• Internal operations of the timers and the serial port can also cause an interrupt call
to take place.
• The subroutines called by an interrupt are located at fixed hardware addresses and
are called as Interrupt Service Routine (ISR).
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• When an interrupt call takes place, hardware interrupt disable flip-flops are set.
• This is to prevent another interrupt of the same priority level from taking place.
• Disabled Hardware interrupts can be enabled back by executing an interrupt
return instruction. (Generally written at the end of the interrupt subroutine.)
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INTERRUPT ADDRESS (HEX) CALLED
IE0 0003
TF0 0008
IEl 0013
TFl 0018
SERIAL 0023

• RETI
• Pops two bytes from the stack into the program counter(PC) and reset the interrupt
enable Flip-Flops
• The only difference between the RET and RETI instructions is the enabling of the
interrupt logic when RETI is used.
• RET is used at the ends of subroutines called by an opcode. RETI is used by
subroutines called by an interrupt.
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• TIME DELAY CALCULATION IN 8051:
• Machine Cycle in 8051 :
• The CPU takes a certain number of clock cycles to execute an instruction.
• These clock cycles are referred to as machine cycles.
• The length of the machine cycle depends on the frequency of the crystal
oscillator connected to the 8051 system.
• The crystal oscillator frequency can vary from 4MHz to 30MHz.
• To make the 8051 system compatible with the serial port of the personal computer
PC, 11.0592MHz crystal oscillators is used.
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• In the 8051, one machine cycle lasts 12 oscillator periods.
• So to calculate the machine cycle, we take 1/12 of the crystal frequency, then take
the inverse of it results in time period. i.e frequency = 1/time period.
• E.g.: Find the time period of the machine cycle in each case for the following
crystal frequency of different 8051 based systems: 11.0592 MHz, 16 MHz, 20
MHz
• Solution:
• Crystal Frequency: 11.0592 MHz:
• 11.0592/12 = 921.6 KHz
• Machine cycle = 1/921.6 KHz = 1.085us [us=microsecond]
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• Crystal Frequency: 16 MHz:
• 16MHz/12 = 1.333 MHz
• Machine cycle = 1/1.333 MHz = 0.75us [us=microsecond]
• Crystal Frequency: 20MHz:
• 20MHz/12 = 1.66 MHz
• Machine Cycle = 1/1.66 MHz = 0.60us [us=microsecond]
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• Now let us find how long it takes to execute each of the following instructions, for
a crystal frequency of 11.0592 MHz. The machine cycle of a system of 11.0592
MHz is 1.085 us.
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INSTRUCTION MACHINE CYCLE TIME TO EXECUTE
MOV R2,#55H 1 1x1.085 us = 1.085 us
DEC R2 1 1x1.085 us = 1.085 us
DJNZ R2,target 2 2x1.085 us = 2.17 us
LJMP 2 2x1.085 us = 2.17 us
SJMP 2 2x1.085 us = 2.17 us
NOP 1 1x1.085 us = 1.085 us
MUL AB 4 4x1.085 us = 4.34 us

• How to Calculate Exact Time Delay in 8051 microcontroller?
• The delay subroutine consists of 2 parts: Setting a counter and creating a loop.
• Let us assume that the crystal frequency of 11.0592 MHz is connected
• Consider the following code
• MOV A,#55H
• AGAIN: MOV P1,A
• ACALL DELAY
• CPL A
• SJMP AGAIN
• ;-----------------Time Delay
• DELAY: MOV R3,#225
• HERE: DJNZ R3,HERE
• RET
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• Machine cycles for each instruction of the DELAY subroutine is as shown below.
• Therefore, we have a time delay of [(255 x 2) + 1 + 1] x 1.085 us = 555.52 us
• Very often we used to calculate the time delay based on the instructions inside the
loop and ignore the clock cycles associated with the instructions outside the loop.
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DELAY: MOV R2,#255 Machine Cycle = 1
HERE: DJNZ R2,HERE Machine Cycle = 2
RET Machine Cycle = 1

• NOP Instruction:
• NOP instruction is used to increase the delay in the loop. NOP means "No
Operation" simply wastes time.
• Loop Inside a Loop Delay:
• This method is used to get a larger delay i.e. is used to loop inside a loop, which is
also called a nested loop.
• E.g.: Lets find the time delay for the following subroutine with 11.0592 MHz
crystal frequency is connected to the 8051 system.
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• The time delay inside the HERE loop [255(1+1+1+1+2)] x 1.085 us = 1660.05 us
• The time delay of the two instructions outside the loop is:
• 1660.05 us+[1+1]x1.085 us = 1662.67 us
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DELAY: MOV R2,#255 Machine Cycle = 1
HERE: NOP Machine Cycle = 1
NOP Machine Cycle = 1
DJNZ R2,HERE Machine Cycle = 2
RET 1

• E.g.: For a crystal frequency of 11.0592 MHz, find the time delay in the following
subroutine. The machine cycle is 1.085 us.
• 'HERE' Loop Calculations: 1+1+2, so [(1+1+2)x250] x 1.085 us = 1085 us.
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DELAY: MOV R2,#200
AGAIN: MOV R3,#250
HERE: NOP
NOP
DJNZ R3,HERE
DJNZ R2,AGAIN
RET
Machine Cycle = 1
Machine Cycle = 1
Machine Cycle = 1
Machine Cycle = 1
Machine Cycle = 2
Machine Cycle = 2
Machine Cycle = 1

• 'AGAIN' Loop Calculations:
• In this loop "MOV R3,#250" and "DJNZ R2,AGAIN" at the begining and end of
the AGAIN loop give delay of [(1+2)x200]x1.085 us = 651us
• The AGAIN loop repeats the HERE loop 200 times so 200x1085 us = 217000 us.
• The total time delay is 217000 us + 651 us = 217651 us or 217.651milliseconds.
• The time is approximate as we have ignored the first and the last instructions in
the subroutine i.e. DELAY: "MOV R2,#200" and "RET".
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• I/O Port Interfacing and Programming
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• The four 8-bit I/O ports P0, P1, P2 and P3 each uses 8 pins
• All the ports upon RESET are configured as input, ready to be used as input ports
• When the first 0 is written to a port, it becomes an output
• To reconfigure it as an input, a 1 must be sent to the port
• Port 0
• Port 0 can be used for input or output, 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
• Open drain is a term used for MOS chips in the same way
that open collector is used for TTL chips
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• The following code will continuously send out to port 0 the alternating value 55H
and AAH
• BACK: MOV A,#55H
• MOV P0,A
• ACALL DELAY
• MOV A,#0AAH
• MOV P0,A
• ACALL DELAY
• SJMP BACK
• In order to make port 0 an input, the port must be programmed by writing 1 to all
the bits
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ORG 300H ;put DELAY at address 300H
DELAY:MOV R5,#0FFH ; R5=255 counter
AGAIN:DJNZ R5,AGAIN ; stay here until R5 = 0
RET ;return to caller (when R5 =0)
END ;end of asm file

• E.g.: Program to configure Port 0 as an input port and then receive data from that
port and sent to P1
• MOV A,#0FFH ;A=FF hex
• MOV P0,A ;make P0 an i/p port by writing it all 1s
• BACK: MOV A,P0 ;get data from P0
• MOV P1,A ;send it to port 1
• SJMP BACK ;keep doing it
• Port 1
• Port 1 can be used as input or output
• This port does not need any pull-up resistors since it already has pull-up resistors
internally
• Upon reset, port 1 is configured as an input port
• To make port 1 an input port, it must be programmed by writing 1 to all its bits
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• Port 1 to Port 3 Structures and their operation:
• All the ports have the following components in them
• D latch
• Output driver
• Input buffer
• Port 1 Structure:
• In the figure, Internal FET pull-up is the internal load for P1 which is present in
P2 and P3 as well but not found in P0.
• 8051 ports have both latch and buffer.
• While reading the data, it can be either reading from the pin or from the latch
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• Reading the input pin:
• To make any port pin as input, we must write 1(Logic High) to that pin.
• Writing High (1) to the port bit means that the High is given to the Latch (D flip-
flop) input.
• This makes Q to go high and Q` to go low
• Since Q` is low and connected to gate of Transistor M1, M1 goes to OFF state.
• When M1 is OFF, it blocks any path to ground and hence input signal directly
goes to the tri-state buffer.
• When we read the input port through instructions such as MOV A, P1 we are
reading the data present at the input pin. This instruction activates the read pin of
the tri-state buffer and lets the data at the pin to go into CPU’s internal bus
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• Writing zero to port:
• To make any bits of any ports as output port we must write ZERO onto it.
• If we write a ZERO to any port bit, then Q=0 and Q`=1 which makes transistor to
go to ON state.
• This provides a path to ground for both load and input pin.
• So when we try to read from input pin it always gives a LOW irrespective of the
status of the input bit
•
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• Port 2 and Port 3
• Just like P1, port 2 also does not need any pull-up resistors
• Upon reset, port 2 is configured as an input port
• To make port 2 an input port, it must be programmed by writing 1 to all its bits.
• Port 3 does not need any pull-up resistors
• Port 3 is configured as an input port upon reset.
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• E.g.: Program to continuously send out to port 1 the alternating value 55H and
AAH
• MOV A,#55H
• BACK: MOV P1,A
• ACALL DELAY
• CPL A
• SJMP BACK
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• E.g.: Program to configure Port 1 first as an input port then receive data from that
port and save it in R7 and R5
• MOV A,#0FFH ;A=FF hex
• MOV P1,A ;make P1 an input port by writing it all 1s
• MOV A,P1 ;get data from P1
• MOV R7,A ;save it to in reg R7
• ACALL DELAY ;wait
• MOV A,P1 ;another data from P1
• MOV R5,A ;save it to in reg R5
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• I/O Programming- Different ways of Accessing Entire 8 Bits
• Program to access the entire 8 bits of Port 1
• BACK: MOV A,#55H
• MOV P1, A
• ACALL DELAY
• MOV A,#0AAH
• MOV P1,A
• ACALL DELAY
• SJMP BACK
• Can we rewrite the code in a more efficient manner ?
• Access the port directly without going through the accumulator
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BACK:MOV P1,#55H
ACALL DELAY
MOV P1,#0AAH
ACALL DELAY
SJMP BACK
MOV A,#55H
BACK:MOV P1,A
ACALL DELAY
CPL A
SJMP BACK

• Write a 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 P2.3
• SETB P1.2 ;make P1.2 an input
• MOV A,#45H ;A=45H
• AGAIN: JNB P1.2,AGAIN ;get out when P1.2=1
• MOV P0,A ;issue A to P0
• SETB P2.3 ;make P2.3 high
• CLR P2.3 ; make P2.3 low for H to L
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• Write a program to toggle the bits of port1 with delay which depends on the value
of a number in R0.
• ORG 0
• BACK:MOV A,#0H
• MOV P1,A ; Send 0h to port P1
• MOV R0, #30H ; count value for generating delay
ACALL DELAY ; Call the subroutine DELAY
• CPL A ; Complement A to toggle
• MOV P1,A ; Send AAh to port1
• MOV R0, #0FFH ; count value for generating delay
ACALL DELAY ; Call the subroutine DELAY
• SJMP BACK ; Keep doing this indefinitely
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• ------Delay Subroutine---------
• ORG 300H
• DELAY: NOP ;Do nothing
• AGAIN: DJNZ R0,AGAIN ;Stay here until R0 becomes 0
• RET ;Return to the Called program
• END
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• Assume that P2.3 is an input port pin and represents the condition of an oven.
Monitor the bit continuously. If it goes high, it means that the oven is hot.
Whenever it goes high, send a high-to-low pulse to port P1.5 to turn on a buzzer.
• Solution:
• ORG 00H
• 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
• END
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• A switch is connected to pin P1.7. Write a 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
• Solution:
• AGAIN: JB P1.7,SENDY ;jump if P1.7=1
• MOV P2,#’N’ ;SW=0, issue ‘N’ to P2
• SJMP AGAIN ;keep monitoring
• SENDY: MOV P2,#’Y’ ;SW=1, issue ‘Y’ to P2
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• A switch is connected to pin P1.0 and an LED to pin P2.7. Write a program to get
the status of switch and send it the LED.
• Solution:
• AGAIN: MOV C, P1.0 ;Read the switch status in to CF
• MOV P2.7,C ;Send the switch status on to LED
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• A switch is connected to pin P1.0 and an LED to pin P2.7. Write a program to get
the status of switch and send it the LED.
• AGAIN: MOV C,P1.0 ;Read the switch status in to CF
• MOV P2.7,C ;Send the switch status on to LED
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