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Some Microprocessor programs

2011-01-16T09:11:00.000+05:30

Exchange the contents of memory locations

Program 1:

LDA 2000H : Get the contents of memory location 2000H into accumulator

MOV B, A : Save the contents into B register

LDA 4000H : Get the contents of memory location 4000Hinto accumulator

STA 2000H : Store the contents of accumulator at address 2000H

MOV A, B : Get the saved contents back into A register

STA 4000H : Store the contents of accumulator at address 4000H

Program 2:

LXI H 2000H : Initialize HL register pair as a pointer to memory location 2000H.

LXI D 4000H : Initialize DE register pair as a pointer to memory location 4000H.

MOV B, M : Get the contents of memory location 2000H into B register.

LDAX D : Get the contents of memory location 4000H into A register.

MOV M, A : Store the contents of A register into memory location 2000H.

MOV A, B : Copy the contents of B register into accumulator.

STAX D : Store the contents of A register into memory location 4000H.

HLT : Terminate program execution.

Add contents of two memory locations

Sample problem:

(4000H) = 7FH

(400lH) = 89H

Result = 7FH + 89H = lO8H

(4002H) = 08H

(4003H) = 0lH

Source program:

LXI H, 4000H :HL Points 4000H

MOV A, M :Get first operand

INX H :HL Points 4001H

ADD M :Add second operand

INX H :HL Points 4002H

MOV M, A :Store the lower byte of result at 4002H

MVIA, 00 :Initialize higher byte result with 00H

ADC A :Add carry in the high byte result

INX H :HL Points 4003H

MOV M, A :Store the higher byte of result at 4003H

HLT :Terminate program execution

Pack the unpacked BCD numbers

Statement: Pack the two unpacked BCD numbers stored in memory locations 4200H and 4201H and store result in memory location 4300H. Assume the least significant digit is stored at 4200H.

Sample problem:

(4200H) = 04

(4201H) = 09

Result = (4300H) = 94

Source program:

LDA 4201H : Get the Most significant BCD digit
RLC
RLC
RLC
RLC : Adjust the position of the second digit (09 is changed to 90)
ANI FOH : Make least significant BCD digit zero
MOV C, A : store the partial result
LDA 4200H : Get the lower BCD digit
ADD C : Add lower BCD digit
STA 4300H : Store the result
HLT : Terminate program execution

Finding Two's complement of a number

Sample problem:

(4200H) = 55H

Result = (4300H) = AAH + 1 = ABH

Source program:

LDA 4200H : Get the number

CMA : Complement the number

ADI, 01 H : Add one in the number

STA 4300H : Store the result

HLT : Terminate program execution

Finding one's complement of a number

Sample problem:

(4400H) = 55H

Result = (4300B) = AAB

Source program:

LDA 4400B : Get the number

CMA : Complement number

STA 4300H : Store the result

HLT : Terminate program execution

Unpack a BCD number

Sample problem:

(4200H) = 58

Result = (4300H) = 08 and

(4301H) = 05

Source program:

LDA 4200H : Get the packed BCD number

ANI FOH : Mask lower nibble

RRC

RRC : Adjust higher BCD digit as a lower digit

STA 4301H : Store the partial result

LDA 4200H : .Get the original BCD number

ANI OFH : Mask higher nibble

STA 4201H : Store the result

HLT : Terminate program execution

Find the negative numbers in a block of data.

Sample problem 1:

(2200H) = 04H

(2201H) = 56H

(2202H) = A9H

(2203H) = 73H

(2204H) = 82H

Result = 02 since 2202H and 2204H contain numbers with a MSB of 1.

Source program :

LDA 2200H
MOV C, A : Initialize count
MVI B, 00 : Negative number = 0
LXI H, 2201H : Initialize pointer
BACK: MOV A, M : Get the number
ANI 80H : Check for MSB
JZ SKIP : If MSB = 1
INR B : Increment negative number count
SKIP: INX H : Increment pointer
DCR C : Decrement count
JNZ BACK : If count 0 repeat
MOV A, B
STA 2300H : Store the result
HLT : Terminate program execution

Separate even numbers from given numbers

Source program :

LXI H, 2200H : Initialize memory pointer l
LXI D, 2300H : Initialize memory pointer2
MVI C, 32H : Initialize counter
BACK:MOV A, M : Get the number
ANI 0lH : Check for even number
JNZ SKIP : If ODD, don't store
MOV A, M : Get the number
STAX D : Store the number in result list
INX D : Increment pointer 2
SKIP: INX H : Increment pointer l
DCR C : Decrement counter
JNZ BACK : If not zero, repeat
HLT : Stop

Evaluate the expression C=A2+B2

MVI B, Data#1 ; Data #1 is stored in register B

MOV C, B ; Copy of Data #1 is made in register C

MVI D, Data#2 ; Data #2 is stored in register D

MOV E,D ; Copy of Data #2 is made in register E

XRA A ; Accumulator content is cleared

Again: ADD B

DCR C : A2 is calculated by repeated Addition

JNZ Again

MOV H,A ; Calculated A2 value is stored in register H

XRA A ; Accumulator content is cleared

Loop: ADD D

DCR E :B2 is calculated by repeated Addition

JNZ Loop

ADD H ; A2+ B2 is determined, by adding result in A and register content H

STA 4200H ; Result is stored in memory location 4200H

2-Digit BCD to binary conversion

Sample problem 1:

(2200H) = 67H

(2300H) = 6 x OAH + 7 = 3CH + 7 = 43H

Source program :

LDA 2200H : Get the BCD number
MOV B, A : Save it
ANI OFH : Mask most significant four bits
MOV C, A : Save unpacked BCDI in C register
MOV A, B : Get BCD again
ANI FOH : Mask least significant four bits
RRC : Convert most significant four bits into unpacked BCD2
RRC
RRC
RRC
MOV B, A : Save unpacked BCD2 in B register
XRA A : Clear accumulator (sum = 0)
MVI D, 0AH : Set D as a multiplier of 10
Sum: ADD D : Add 10 until (B) = 0
DCR B : Decrement BCD2 by one
JNZ SUM : Is multiplication complete? i if not, go back and add again
ADD C : Add BCD1
STA 2300H : Store the result
HLT : Terminate program execution

2-Digit BCD to binary conversion

Statement: Convert a 2-digit BCD number stored at memory address 2200H into its binary equivalent number and store the result in a memory location 2300H.

Sample problem 1:

(2200H) = 67H

(2300H) = 6 x OAH + 7 = 3CH + 7 = 43H

Source program :

LDA 2200H : Get the BCD number
MOV B, A : Save it
ANI OFH : Mask most significant four bits
MOV C, A : Save unpacked BCDI in C register
MOV A, B : Get BCD again
ANI FOH : Mask least significant four bits
RRC : Convert most significant four bits into unpacked BCD2
RRC
RRC
RRC
MOV B, A : Save unpacked BCD2 in B register
XRA A : Clear accumulator (sum = 0)
MVI D, 0AH : Set D as a multiplier of 10
Sum: ADD D : Add 10 until (B) = 0
DCR B : Decrement BCD2 by one
JNZ SUM : Is multiplication complete? i if not, go back and add again
ADD C : Add BCD1
STA 2300H : Store the result
HLT : Terminate program execution

Generate and display BCD down counter

Source program 1:

LXI SP, 27FFH : Initialize stack pointer
MVI C, 99H : Initialize counter = 99
BACK:CALL Display : Call display subroutine
CALL Delay : Call delay subroutine
ADI 99H : See Addition below
DAA : Adjust for decimal
CPI 99H : Compare with last count
JNZ BACK :If no, repeat
HLT

Source Program 2:

LXI SP, 27FFH : Initialize stack pointer
MVI C, 99H : Initialize counter = 99
BACK: CALL Display : Call display subroutine
CALL Delay : Call delay subroutine
MOV A, C : Get count
ANI 0FH : Check for lower nibble
JNZ SKIP : If it is not 0FH go to skip
MOV A,C : Else get the count
SBI ,06 : Subtract 06
MOV C,A : Store the count
DCR C : Decrement count
CPI 99H : Check it for last count
JNZ BACK : If not, repeat
HLT : Stop

Generate and display BCD up counter with frequency 1Hz

Statement: Write a program for displaying BCD up counter. Counter should count numbers from 00 to 99H and it should increment after every 1 sec. Assume operating frequency of 8085 equal to 3MHz. Display routine is available

Source program:

LXI SP, 27FFH : Initialize stack pointer
MVI C, OOH : Initialize counter
BACK: CALL Display : Call display subroutine
CALL Delay : Call delay subroutine
MOV A, C
ADI A, 0 1 : Increment counter
DAA : Adjust it for decimal
MOV C,A : Store count
CPI ,00 : Check count is > 99
JNZ BACK : If not, repeat
HLT : Stop

Delay Routine:

Delay:MVI B, Multiplier-count : Initialize multiplier count
BACK 1 :LXI D, Initialize Count
BACK:DCX D : Decrement count
MOV A, E
ORA D : Logically OR D and E
JNZ BACK : If result is not a, repeat
DCR B : Decrement multiplier count
JNZ BACK 1 : If not zero, repeat
RET : Return to main program.

Generate and display binary up counter

Source program:

LXI SP, 27FFH : Initialize stack pointer
MVI C, OOH : Initialize counter
BACK: CALL Display : Call display subroutine
CALL Delay : Call delay subroutine
INR C : Increment counter
MOV A, C
CPI OOH : Check counter is > FFH
JNZ BACK : If not, repeat
HLT : Stop

Delay Routine:

Delay: LXI B, count : Initialize count
BACK: DCX D : Decrement count
MOV A, E
ORA D : Logically OR D and E
JNZ BACK : If result is not 0 repeat
RET : Return to main program

Find the factorial of a number

Statement: Program to calculate the factorial of a number between 0 to 8

Source program :

LXI SP, 27FFH ; Initialize stack pointer
LDA 2200H ; Get the number
CPI 02H ; Check if number is greater than 1
JC LAST
MVI D, 00H ; Load number as a result
MOV E, A
DCR A
MOV C,A ; Load counter one less than number
CALL FACTO ; Call subroutine FACTO
XCHG ; Get the result in HL
SHLD 2201H ; Store result in the memory
JMP END
LAST: LXI H, 000lH ; Store result = 01
END: SHLD 2201H
HLT

Subroutine Program:

FACTO:LXI H, 0000H
MOV B, C ; Load counter
BACK: DAD D
DCR B
JNZ BACK ; Multiply by successive addition
XCHG ; Store result in DE
DCR C ; Decrement counter
CNZ FACTO ; Call subroutine FACTO
RET ; Return to main program

Microprocessor Viva Questions and Answers

2011-01-15T16:13:00.000+05:30

What is a Microprocessor?
Microprocessor is a program controlled device, which fetches the instructions from memory, decodes and executes the instructions. Most Microprocessors are single chip devices.

What are the flags in 8086?
In 8086 Carry flag, Parity flag, Auxiliary carry flag, Zero flag, Overflow flag, Trace flag, Interrupt flag, Direction flag, and Sign flag.

In 8085 which is called as High order / Low order Register?
Flag is called as Low order register & Accumulator is called as High order Register.

What is Tri-state logic?
Three Logic Levels are used and they are High, Low, High impedance state. The high and low are normal logic levels & high impedance state is electrical open circuit conditions. Tri-state logic has a third line called enable line.

What happens when HLT instruction is executed in processor?
The Micro Processor enters into Halt-State and the buses are tri-stated.

Which Stack is used in 8085?
LIFO (Last In First Out) stack is used in 8085.In this type of Stack the last stored information can be retrieved first

What is Program counter?
Program counter holds the address of either the first byte of the next instruction to be fetched for execution or the address of the next byte of a multi byte instruction, which has not been completely fetched. In both the cases it gets incremented automatically one by one as the instruction bytes get fetched. Also Program register keeps the address of the next instruction.

What are the various registers in 8085?
Accumulator register, Temporary register, Instruction register, Stack Pointer, Program Counter are the various registers in 8085

What is 1st / 2nd / 3rd / 4th generation processor?
The processor made of PMOS / NMOS / HMOS / HCMOS technology is called 1st / 2nd / 3rd / 4th generation processor, and it is made up of 4 / 8 / 16 / 32 bits.

How many bit combinations are there in a byte?
Byte contains 8 combinations of bits.

Types of buses?
There are three types of buses.
Address bus: This is used to carry the Address to the memory to fetch either Instruction or Data.
Data bus: This is used to carry the Data from the memory.
Control bus: This is used to carry the Control signals like RD/WR, Select etc.

What is the Maximum clock frequency in 8086?
5 MHz is the Maximum clock frequency in 8086.

What is meant by Maskable interrupts?
An interrupt that can be turned off by the programmer is known as Maskable interrupt.

What is Non-Maskable interrupts?
An interrupt which can be never be turned off (ie. disabled) is known as Non-Maskable interrupt

What are the different functional units in 8086?
Bus Interface Unit and Execution unit, are the two different functional units in 8086.

What are the various segment registers in 8086?
Code, Data, Stack, Extra Segment registers in 8086.

Which Stack is used in 8086?
FIFO (First In First Out) stack is used in 8086.In this type of Stack the first stored information is retrieved first.

What is SIM and RIM instructions?
SIM is Set Interrupt Mask. Used to mask the hardware interrupts.
RIM is Read Interrupt Mask. Used to check whether the interrupt is Masked or not.

Give example for Non-Maskable interrupts?
Trap is known as Non-Maskable interrupts, which is used in emergency condition.

Give examples for Micro controller?
Z80, Intel MSC51 &96, Motorola are the best examples of Micro controller.

What is clock frequency for 8085?
3 MHz is the maximum clock frequency for 8085.

Give an example of one address microprocessor?
8085 is a one address microprocessor.

Give examples for 8 / 16 / 32 bit Microprocessor?
8-bit Processor - 8085 / Z80 / 6800; 16-bit Processor - 8086 / 68000 / Z8000; 32-bit Processor - 80386 / 80486

What is meant by a bus?
A bus is a group of conducting lines that carries data, address, & control signals.

What are the various registers in 8085?
Accumulator register, Temporary register, Instruction register, Stack Pointer, Program Counter are the various registers in 8085

Name 5 different addressing modes?
Immediate, Direct, Register, Register indirect, Implied addressing modes

In what way interrupts are classified in 8085?
In 8085 the interrupts are classified as Hardware and Software interrupts.

What is the difference between primary & secondary storage device?
In primary storage device the storage capacity is limited. It has a volatile memory. In secondary storage device the storage capacity is larger. It is a nonvolatile memory. Primary devices are: RAM / ROM. Secondary devices are: Floppy disc / Hard disk.

In 8085 name the 16 bit registers?
Stack pointer and Program counter all have 16 bits.

What is Stack Pointer?
Stack pointer is a special purpose 16-bit register in the Microprocessor, which holds the address of the top of the stack..

How many interrupts are there in 8085?
There are 12 interrupts in 8085.

Which interrupt has the highest priority?
TRAP has the highest priority

Can an RC circuit be used as clock source for 8085?
Yes, it can be used, if an accurate clock frequency is not required. Also, the component cost is low compared to LC or Crystal.