Microprocessor and Programming

Chapter No.1Basics of Microprocessor8085

1/15/16 BY UBAID SAUDAGAR 2

0000 1111 (0FH)

0010 0100 (24H)

1010 1110 (AEH)-

0001 1010 (1AH)

0101 0111 (57H)D ATABUS

Data input lines used for

writing operation

Address lines

Control inputs

Data output lines used for

reading operation

Every location consist of N bit data (in

general 8 bit data)

No. of locations =M

To address any location or to

select any location we

require address lines.No of address

lines =

To select the chip, whether read or write operation

Eg: 3 bit address lines are therefore = 8

locations and address varies as:

000001010

-111

and at every location 8 bit data present

DATA BUS

Memory basics

What is a Microprocessor (µp) ?As we know that to perform logical operations like AND,OR,NOT etc. we have digital IC’s like IC 7400,7408 etc.

Simple addition operation we require half adders and full adders, for subtraction we require half and full subtractor, which can be constructed using digital IC’s.

In general for performing any arithmetic and logical operation we require we require n number of IC’s and bulky circuitry.

Now integrating all this circuitry onto a single chip is called a microprocessor.

Integrating all this circuitry on a single chip is done with the help of a technology called as VLSI technology i.e. Very large scale integration.

The added advantage is that we can control all this circuitry with the help of commands.

Eg. For adding two 8 bit numbers we require is one half adder and seven full adders and we get the o/p by switching on and off the switches but now we will give the instruction to the microprocessor as ADD A,B which will add content of A and B which are both 8 bits.


What is a Microprocessor (µp) ? Cont.…

Therefore,

A microprocessor is an integrated circuit on a tiny silicon chip that contains thousands or millions of tiny on/off switches, known as transistors.

The transistors are laid out such that they store or manipulate data. These circuits manipulate data in certain patterns, patterns that can be programmed by software to make machines do many useful tasks.

One of the biggest tasks microprocessors perform is acting as the brains inside a personal computer. The number of transistors on a chip has grown from 2,300 on the 4004 in 1971 to 5.5 million on today's Pentium Pro processor. These advances have made today's microprocessors 1,000 faster than their predecessors.


Microprocessor Based System (Microcomputer)

A microprocessor can itself do nothing on its own, it requires some supporting devices do complete a particular task called as peripherals.

Eg. Our brain can thinks of picking up a book but it is not the job of the brain to pick up a book, it has a supporting device i.e. its hand which when gets an impulse from the brain picks up the book.

Similarly µp has peripherals attached to it like memory, i/p devices, o/p devices, io ports etc.


Microprocessor

i/p o/p devices

Ports

Memory

8085 Microprocessor


Features of 8085 microprocessorIt is a 8 bit microprocessor.

It is manufactured with N-MOS technology.

It has 16-bit address bus and hence can address up to 2^16 = 65536 bytes (64KB) memory locations through A0 -A15

The first 8 lines of address bus and 8 lines of data bus are multiplexed AD0 – AD7

Data bus is a group of 8 lines D0 – D7

It supports external interrupt request.

A 16 bit program counter (PC)

A 16 bit stack pointer (SP)

Six 8-bit general purpose register arranged in pairs: BC, DE, HL.

It requires a signal +5V power supply and operates at 3.2 MHZ single phase clock.

It is enclosed with 40 pins DIP (Dual in line package).1/15/16 BY UBAID SAUDAGAR 7

ARCHITECTURE OF 8085

The internal hardware design of the µp is called the Architecture of that µp.


Register OrganizationIn the architecture above we see many registers. The basic use of the register is temporary storage of data.

But there are also some registers which are used for special purpose as well which will be defined later.

There are some registers which are accessible to the user and some are those which are not accessible to the user.

Since the microprocessor is an 8 bit microprocessor hence all the general purpose registers are 8 bit in size. It means that it can store 8 bit data.

These are six general purpose registers namely B,C,D,E,H,L all 8 bit in size.

They are used in general to store 8 bit data temporarily. Sometimes during calculations to

store the intermediate value, sometimes to store the result of any calculations etc.

These six registers can also be used to store 16 bit data by using them in pairs as BC, DE, HL.

When used in pairs they are also used to store addresses of memory locations which are 16 bits.


Register organization cont.. Accumulator is also an 8 bit register which is used to store temporary data. But it has also other tasks

It is the most important register of a microprocessor and every microprocessor consists of an accumulator.

It is used by the ALU (arithmetic logic unit). During addition or subtraction operation the result is stored in the accumulator. After which we store it anywhere we want.

Eg. ADD B instruction will add the content of register B (where temporary data is stored) and accumulator and store the result in the accumulator itself.

Also when the microprocessor is sending some data to the io devices the data must be stored first in the accumulator and then transferred.

This is an 8 bit temporary register which is not used by the user but only used by the microprocessor itself to store temporary data.1/15/16 BY UBAID SAUDAGAR 11

Register organization cont.. This an 8 bit flag register. Flag register is normally used to reflect the result of some operations or to

control some operations. Out of the 8 bits only 5 bits are used. Each bit is called as a flag which reflects the result of some operation or controls some operations.

These are two registers which actually act as pointers .

Program counter points to the next memory address while current instruction is in execution stage.

Eg. 2000H: MOV A,B

2001H: MVI B,32H

When the instruction at the address 2000H is being executed the program counter points to the next memory address i.e. 2001H.

Stack pointer is also a 16 bit pointer which points to the stack memory. What is stack memory we will see later


Basic concepts about Opcode and Operand

Note: An instruction is actually a command which we give to the microprocessor which it follows and performs the tasks as instructed.

Opcode indicates what operation is to be performed.

Operand indicates on which the operation is to be performed.

How the operation is to be performed we come to know by addressing mode.

Let’s understand by an example:

MVI A,32H is an instruction which the microprocessor does not understand directly. Hence what we do is we write its hex codes.

The above is a two byte instruction and hex codes are written as

2000 : 3EH

2001 : 32H

Where 2000 and 2001 are the addresses where the hex codes are stored.


Basic concepts about Opcode and Operand

As we can see above there are 2 hex codes namely 3EH and 32H

Now this 3EH is actually 1 byte i.e. 0011 1101 B

Now this byte is the first byte which will enter the microprocessor

Now this byte when decoded will give two information's viz. what operation is to be performed and the size of the instruction.

Hence we call this first byte as opcode or operational code which tells the microprocessor what operation is to be performed.

The second thing which the microprocessor understands is that it is a 2 byte instruction hence when the next byte i.e. 32H i.e. 0011 0010 B goes into the microprocessor, instead of decoding it will treat is as data and move into the accumulator as directed by the opcode.


Working overview of the microprocessor

Let's understand the working by considering a memory attached to the microprocessor.

Consider a read operation takes place. Now a control signal goes low controlled by the timing and control unit i.e. RD

Note: The timing and control unit is used to generate timing signals for proper working of the microprocessor and also it generates control signals to control the different operations and peripherals of the microprocessor.

It is not that the operations are working randomly in any manner, they are working in a sequence. Eg. Lectures and practical's are going in a sequence, if there is no time table the department would not run properly. Similarly the timing and control unit gives a proper sequence to the processor and controls different operations with the help of control signals.


A bus

AD bus

Control bus

Microprocessor

2000: 32H2001: 3EH2002: 47H

Memory

Working overview of the microprocessor cont..

RD is connected to the memory chip at OE pin which is also active low. The memory is now ready for read operation. Now this RD signal from the timing and control unit goes low indicating read operation. IO/M also goes low indicating memory operation.

Now we require the address of the memory location from where we need to access data which is available in the program counter.

Note: The program counter is a 16 bit register which holds the next address of the memory location to be accessed or in other words it indirectly points to the next memory address as shown below.



Now this 16 bit address available in the PC is given to the address buffers. Both these address buffers are 8 bit in size, therefore the address is divided into two parts i.e. upper address and lower address.

For Eg: consider address 2001H, now this address is divided as 20H

and 01H. Now this 20H is called as upper address or higher order

address and 01H is called as lower order address.

The higher order address from the address buffer (1) is transferred to the higher order address bus and the lower order address from the address buffer (2) is transferred to the multiplexed lower order address/data bus.

The ALE signal from the timing and control unit goes high making address available on the multiplexed AD bus.

Both the addresses gets transferred on the address bus and the particular memory location is selected.

After that the data on that particular location is transferred on the multiplexed AD bus which is 8 bit wide. The ALE signal now goes low i.e. ALE = 0 and the data from the memory now gets transferred from the memory to the microprocessor.


1 2


Since the size of the AD bus is 8 bits hence the microprocessor can process only 8 bit data at a time.

This 8 bit data which enters the microprocessor gets loaded to the instruction register and then goes to the instruction decoder where it is decoded.

Once it is decoded the microprocessor comes to know two things viz. what task is to be performed and the size of the instruction.

Eg. MVI A,32H is an instruction which is stored in the memory as 3EH, 32H

3EH = 0011 1101 = 1 byte

32H = 0011 0010 = 1 byte

Always the first byte is opcode and then is the operand (data).

Opcode: Operational code is a binary code which has a specific meaning understood by the microprocessor. Opcode gives two information viz. what task is to performed and what is the size of the instruction.

Eg. Consider two instructions MOV A,B and MVI A,32H



The first instruction is one byte instruction for which the opcode is 78H i.e. 0111 0100 when this opcode is decoded by the instruction decoder, the µp comes to know that it is a 1 byte instruction, and it has to move the content present in B register to the A register.

But when the second instruction is executed, the first byte of that instruction i.e. 3EH = 0011 1101 when decoded by the µp, it comes to know that it is a 2 byte instruction and it has to move the immediate value 32H in the accumulator.

After which it goes to the next memory address where the second byte is present i.e. 32H and moves this byte into the µp. Now this byte is not treated as opcode but treated as operand (data) which will we added to the contents of the accumulator and the result will be stored in the accumulator itself.


About Address And Data BusThe address bus has 8 signal lines A8 – A15 which are unidirectional.

The other 8 address bits are multiplexed (time shared) with the 8 data bits.So, the bits AD0 – AD7 are bi-directional and serve as A0 – A7 and D0 – D7 at the same time.During the execution of the instruction, these lines carry the address bits during the early part, then during the late parts of the execution, they carry the 8 data bits.In order to separate the address from the data, we can use a latch to save the value before the function of the bits changes.

There is a control signal called as ALE which when goes high, the address is available on the multiplexed AD bus and it gets latched otherwise this address gets lost

When ALE goes low data is available on the AD bus and this data either goes to the io device or to the memory as per the address


How address and data are separated ?


Interrupt control Interrupts are those which break the normal sequence of operation of the microprocessor.

There are 6 hardware interrupts out of which 5 are input and one output interrupt pin.

There are maskable and nonmaskable interrupts.

Maskable interrupts are those which can be avoided.

Eg. Mobile phone, when the phone bell rings, if I am busy I can avoid the phone call by not picking it up.

Similarly when an interrupt which is maskable arrives at the microprocessor, if the microprocessor is busy it can avoid this interrupt and respond to it at any other time.

Non maskable interrupts are those which cannot be avoided.

Eg. Smoke detector mounted on the ceiling of a classroom. It has to detect the smoke anyhow and cannot avoid it.

Similarly when a non maskable interrupt arrives at the microprocessor It cannot be avoided, the microprocessor stops its current task and responds to the interrupt.


Serial IO controlTo transfer the data and receive data serially to and from the microprocessor, serial io control is used with two pins namely SID and SOD.

These pins are used when microprocessor needs to communicate with any device serially or in other words during serial communication where data is transferred is transferred to and from the microprocessor in serial form (Bit by bit).

Serial i/p data and Serial o/p data for receiving and transferring data serially.


Binary number formatsNibble : 4 bits form a nibble

Eg : 1000

Byte : 8 bits form a byte

Eg : 1000 1011

Word : 16 bits form a word

Eg : 1001 1000 1011 1010

Quad word : 64 bits form a quad word

Eg : 1000 10001 1010 1011 1100 1101 0001 0010 1000 10001 1010 1011 1100 1101 0001 0010


Lower nibble

Higher nibble

Lower byte

Higher byte

Pin diagram of 8085


Limitation Of 8085 Microprocessor i. The 8085 is an 8-bit microprocessor hence it can perform operation only on two 8-bit numbers.

ii. The 8085 has 16 bit address lines and 8-bit data lines.

iii. The 8085 can address maximum memory of 64 Kb (2^16).

iv. The 8085 cannot perform any operation on ASCII number.

v. The 8085 cannot perform multiplication and division of numbers.

vi. The 8085 cannot perform subtraction, multiplication and division of BCD number.

vii. The 8085 cannot transfer the data directly from one memory to another memory.

viii. The 8085 cannot perform the pipelining of instruction.


Chapter No.216 bit Microprocessor8086

What do you mean by 16 bit processor?

A processor is said to be 16 bit processor, it means that:

1. The size of the data bus is 16 bits

2. Since the size of the data bus is 16 bit, therefore 16 bit data can be accessed and processed at a time (i.e. in parallel i.e. simultaneously).

3. Size of the ALU (arithmetic logic unit) is 16 bits.

4. Size of the registers is 16 bits.

Note: This is because when 16 bit data enters the µp it needs to be stored temporarily and processed simultaneously. For Eg. If the size of the ALU is 8 bit and 16 bit enters the µp, then if the µp was taking 1 clock cycle to process the data it will take 2 clock cycles to process the data and will work as an 8 bit processor.

5. Size of the microprocessor never depends on the address bus.

Note: Whenever we open the My computer in windows and right clicking on the C drive, then properties, there we see 32 bit operating system, it means that the operating system is working on a 32 bit µp.


Salient features of 8086 µpIt requires +5V power supply

It is a 40 pin IC

Available in three clock rates: 5,8 and 10 MHz

It is a 16 bit microprocessor

Size of data bus is 16 bits

Size of address bus is 20 bits

Direct memory access upto 1Mb =

Note: Size of the memory always depends on the number of address lines.

Two separate units: Bus Interface Unit and Execution Unit

External Co processor (8087) supported

6 byte instruction queue present to help speed up instruction execution

Has multiplexed address and data bus

It consist two memory banks viz. odd memory bank and even memory bank


Salient features of 8086 µp


Memory Size of the data bus is 16 bits

This is indicating bus (no. of lines together)

Two address lines

No. of combinations possibleon two address lines are:00011011Therefore we can access 4 memory locations =

Concept of memory addressing

Architecture of 8086


The internal hardware design of the µp is called the Architecture of that µp.As shown in the block diagram 8086 is divided into two functional units : BUS INTERFACE UNIT (BIU) and EXECUTION UNIT. Dividing the work between these two units speeds up processing.

The BIU :- Sends out addresses - Fetches instructions and data from memory- Reads the portsIn other words all operations related to outside world (outside the µp) are handled by the BIU i.e. all data transfers and address transfers are handled by this unit for BIU.The word Bus interface unit itself indicates that this unit is used to interface the buses with the outside world.

The EU - Tells BIU where to fetch instructions and data from - Decodes the instruction - Executes the instruction

ARCHITECTURE


AH AL

BH BL

CH CL

DH DL

SP

BP

SI

DI

ARITHMETIC LOGIC UNIT (ALU)

TEMPORARY REGISTER

FLAG REGISTER

INSTRUCTION DECODER AND CONTROL UNIT

6

5

4

3

2

1

DS

ES

CS

SS

IP

Σ20

INTERNAL SYSTEM BUSES- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

- - - - - - - -

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Memory Interface

6 byte instruction

queue registerSegment Registers

Instruction pointer

General purpose Registers

Pointers and index Registers

BIU

EU

BIU (Bus Interface Unit)The BIU consist of the following modules:

1. Physical address calculation unit (Generates 20 bit physical address)

2. Segment registers (Used to store the base address to access a segment)

3. Instruction pointer (Used to store the offset address, used along with the code segment)

4. Queue register, 6 bytes long (Used to store 6 prefetch instruction bytes from the memory)

With the help of these modules it performs the following action:

-.It sends out addresses to access memory and i/o ports

-.It fetches instructions and data from the memory and also performs write operation with the help of system bus

-.In other words it interfaces with the external world

-.Instruction queuing in the queue register


EU (Execution Unit)The execution unit consist of the following modules:

1. General purpose registers (They are used for multipurpose)

2. Pointers and index registers (As offset registers to select a particular location within a segment)

3. Arithmetic logic unit (ALU)

4. Operand register or temporary register ()

5. Flag register

6. Instruction decoder

7. Timing and control unit

With the help of these modules the following actions are performed:

-.The instruction bytes coming from the queue are decoded by the instruction decoder (here we come to know two things : (1). How many bytes instruction (2). What operation is to be performed (by decoding the opcode))

-.After decoding execution takes place and data is placed in proper registers or send to the BIU to save in the memory


Instruction Decoder and Control UnitAs we have seen that a decoder is a combinational circuit which has binary inputs and it converts these binary inputs to corresponding code depending upon the circuit which it is going to drive.

Eg : A seven segment decoder driver i.e. IC 7447 had 4 inputs and 7 outputs.

These 4 were inputs in binary form and o/p combination of 7 o/p’s was depending upon the number to be displayed.

Similarly in the execution unit the decoder which is present, what it does, whenever any byte arrives from the BU into the decoder, it decodes in such a way that it comes to know two information viz. what operation is to be performed and the size of instruction in bytes.

Note: Always the first byte which arrives at the decoder is taken as opcode from which we come to know what operation is to be performed and there after it comes to know that the instruction is of how many bytes.

After decoding it drives the control unit which generates proper control signals to execute those bytes which have been decoded.1/15/16 BY UBAID SAUDAGAR 35

General purpose registersThere are in all 4 general purpose registers viz. AX, BX, CX, DX each of 16 bits i.e. 2 bytes i.e. 1 word.

These 4 registers are actually are pairs of 8 registers viz. AH, AL, BH, BL, CH, CL, DH, DL each of 8 bits i.e. 1 byte.

The letter X specify complete 16 bit register. The letters L and H specify lower and higher bytes of a register.

To store every bit 1 f/f is used hence an 8 bit register is used to store 8 bits and to store 16 bits, 16 bit register is used.

Importance of each register is given on next slide:1/15/16 BY UBAID SAUDAGAR 36

AH AL

BH BL

CH CL

DH DL

15 8

7 0

General purpose registers1. AL/AX (Accumulator): It is intermediate register between the ALU and the memory. Mostly, but not always the results of arithmetic and logical operations are stored in the accumulator and then from there it is transferred to wherever required.

It is the most important general purpose register of the µp.

Eg: When the instruction DIV BL is executed what the µp does is that it divide content of AX register with the content of BL register by default and after dividing the quotient is by default stored in the AL register and remainder is stored in the AH register.

2. BX (Base Register): To select any memory location particularly in the data segment we require two addresses : base address (stored in the segment register) and offset address which can be stored in the base register in certain addressing modes.

3. CL/CX (Counter Register): It is used as a default counter register during loop, shift, rotate, repeat instructions. It can be used as 8 bit counter as well as 16 bit counter.

4. DX (Data register): During multiplication and division operations when the result goes beyond 16 bits i.e. 32bits, then the upper 16 bits are stored in the DX register.


Segment registersThere are 4 segment registers in the Bus interface unit, size of each segment register is 16 bits.

Code Segment Register (CS register)

Data Segment Register (DS register)

Extra Segment Register (ES register)

Stack Segment Register (SS register)

The entire memory can be divided into four segments

- Code segment (Used to store instruction codes of the program)

- Data segment (Used to store data bytes/words)

- Extra segment (It is an additional segment to store data)

- Stack segment (Its used to store stack of data and addresses of main program during subroutine call)

Segment registers give the base address (starting address) to select any memory segment.


Code segment register (CS)

Data segment register (DS)

Extra segment register (ES)

Stack segment register (SS)

15 0

Segment registers--

Code segment 0000H

FFFFH

Data segment 0000H

FFFFH

Extra segment 0000H

FFFFH

Stack segment 0000H

FFFFH

--

--


1 MB memory

2500H

4000H

5000H

7654H

Segment registersCS

DS

ES

SS

Base addrese

s

Operand register or Temporary register

It is a 16 bit register which can store 16 bit data.

This register cannot be used by the programmer/user.

It is used by the µp to store intermediate data or results therefore it is called as temporary register.


Pointers and Index RegistersThe 16 bit register from where the µp will take 16 bit effective address is called memory pointer.

There are three pointer registers as follows:

- Instruction pointer (IP)

- Stack Pointer (SP)

- Base Pointer (BP)

Also there are 2 index registers which have multiple purposes such as:

- Can be used for pointer addressing of data

- Used as source in some string processing instructions

- Used to store Offset address

Note: The index registers are also a type of memory pointers but having multiple purpose

There are two index registers as follows:

- Source Index register (SI) - Destination Index Register (DI)1/15/16 BY UBAID SAUDAGAR 41

Instruction pointer (IP) registerThe different instruction codes are stored in the memory in the code segment. Now to execute these instructions the µp needs to access these codes. Hence it generates 20 bit physical address on the address bus.

Now to form this 20 bit physical address we require base address and offset address.

µp will always take the base address from CS register and offset address from IP register.


8086

-

-

Code segment

-

-

-Microprocessor

Memory

1000H

2FACH

CS

IP

PA = BA + EA = 20 bits = 12FACH

00000H

FFFFFH

Stack Pointer (SP) registerThe stack memory segment is used to save useful data in the form of stack of data and addresses when the µp transfers the control from the main program to the subroutine.

To access the stack memory segment again we require physical address of 20 bits. Now we require base address and offset address. µp will always obtain the base address from SS register and offset address from SP register.

The stack segment register and the stack pointer register together address the stack top.


SS

SP

Microprocessor

BA

OA

Stack segment

-

-

25

00

98

34

PA = BA + EA = 10000H + FFFCH = 1FFFCH

0000H

FFFDH

FFFCH

FFFEH

FFFFH

Top of stack

1000H

--

Initial stack top

Stack memory

The stack in the 8086/8088 microprocessor, like that in many microprocessors, is a region of memory that can store information for later retrieval. It is called a stack, because you "stack" things on it. The philosophy is that you retrieve (pop) things in the opposite order of storing (push) them.

While executing main program you require some instructions which need to be executed again and again within the same program, so instead of writing the same set of instructions again and again we write a sub program or subroutine and then we call that subroutine from the main program.

So when we call the subroutine the address on the address bus changes as the control now gets transferred to a different program. Therefore to avoid loss of address of main program we store this address in a memory called as stack memory.

Now when the control gets transferred from main program to sub program the address on the address bus changes. After executing the subroutine the address of the main program is required which is stored in the stack memory.

This address is then again popped from the stack memory, and control gets transferred to main program.


Stack concept

3AH

-

5DH

6FH

8BH

-

3EH

2CH


SS = BA = 1000H

0000H

1000H

FFFFH

0FFFH

0FFEH

-

-

Offset address varies from 0000H

to FFFFH

STACK SEGMENT64 Kb

STACK SEGMENT64 Kb

Initially SP = 0000HWhen PUSH inst. Is executed storage of data starts from 1 less i.e. FFFFH and then points to the

data last entered i.e. FFFEH

FFFEH

But if SP is loaded initially = 1000H and

then PUSH inst. Is executed then data storage starts from

0FFFH and then points to data last entered i.e. 0FFEH

PUSH AX when SP = 0000H

AX = 2C3EH

PUSH AX when SP = 1000H

AX = 6F5DH

data segment

a dw 2500h

b dw 3498h

data ends

code segment

start:

mov ax, data

mov ds, ax

mov ax,0000h

mov ax, a

mov bx, b

push ax

push bx

mov ax,0000

mov bx,0000

pop cx

pop dx

int 3

code ends

end start


Let’s understand Stack with help of a program

Base pointer (BP) registerThe stack segment is basically used to store stack of data and addresses during subroutine calls or Call instruction. The size of the stack segment is 64kb but the entire 64kb is not used as stack data starts from the last addresses i.e. FFFFH and then moves up. The upper region i.e. from 0000H is mostly empty and user can store its normal data over there.

We know that SP register is used along with SS register to form the physical address and the SP register points to the top of stack i.e. last byte entered into the stack memory as stack.

Now if the user wants to store normal data into the stack i.e. from the address 0000H, the µp by default uses the BP register along with the SS register to form the physical address.

Note: Therefore the stack segment (SS) register can be used along with BP and SP


Base pointer (BP) register

0 0 1 0 0 1 0 1

- - - - - - - -

0 0 1 1 0 1 1 1

- - - - - - - -

- - - - - - - -

0 1 0 1 0 1 1 0

- - - - - - - -

1 0 0 1 1 0 0 0


SS

BP

SP

Microprocessor

EA

OA

BA

PA = SS + BP = 00000H + 0010H = 00010H

0010H

0000H

FFFFH

FFFCH

0000H

Stack top pointed by

SP

Source index (SI) and Destination index (DI) registers

We know that data segment is used to store data. To access data segment we require 20 bit physical address.

Now this 20 bit physical address is formed using base address an offset address.

µp will take the Base address from the DS register and the offset address from either SI register or DI register or even BX register which is called as base register.


DS

SI

DI

BX

µprocessor

BA

OA

EA

EA

FFFFH

0000H

PA = DS + (SI or DI or BX)= 10000H + 2000H = 12000H

or = 10000H + 3000H = 13000H

or= 10000H + 4000H = 14000H

1000H

2000H

3000H

4000H

Flag register of 8086

X X X X OF DF IF TF SF ZF X AC X PF X CF


X – Indicates Intel reserved (for future use)

CF – Carry FlagPF – Parity FlagAC – Auxiliary Carry FlagZF – Zero FlagSF – Sign FlagTF – Trap FlagIF – Interrupt FlagDF – Direction FlagOF – Overflow Flag

8 LSB’s8 MSB’s

Total 9 flags

A 16 bit register used out of which 9 are used by the µp and the rest are reserved by Intel for future use.Since it is a register it can either be 0 or 1 Therefore a flag may be reset or set The flags may be status flags or control flagsStatus flags are those which give the current status of the µpControl flags are those which control the operations of the µp

Bit 16

Bit 15

Bit 14

Bit 13

Bit 12

Bit 11

Bit 10

Bit 9 Bit 8 Bit 7 Bit 6 Bit 5 Bit 4 Bit 3 Bit 2 Bit 1

Flag Register of 8086 cont..Flags are basically used to reflect the status of the operations of the microprocessor or to control the operations of the microprocessor.

Flags are a modified kind of register that record the condition of a microprocessor's calculation.

Flag Register determines the current state of the processor.

Let’s understand by an example:

Let AL = 95H = 1001 0101

BL = 6FH = 0110 1111

We ADD the contents of AL and BL registers with the help of the instruction ADD AL , BL.

When we add the two contents: 1001 0101

+ 0110 1111

0001 0000 0100 1/15/16 BY UBAID SAUDAGAR 51

Carry generated which is reflected by

the carry flag

This 8 bit value is stored in the AL register

Flag Register of 8086 cont..This instruction will add the contents of AL and BL and the result i.e. 104H has to be stored in the register AL.

But what we notice is that the size of AL register is 8 bits only and it cannot accumulate 104H as it is 12 bits.

So what the microprocessor will do is that, it will store 04H in AL register and set the carry flag to indicate that a carry is generated due to addition.


Carry flag (Bit 1)Type: Status flag

Operation: During an addition operation if a carry is generated from the MSB or during a subtraction operation a borrow is generated carry flag sets indicating a carry or borrow respectively.

Eg: 8 bit addition

We use the instruction ADD AL,BL (the contents of AL and BL are added and result is by default stored in AL. Eg. 95H and 6FH ; the result is 104H). Now 104H is 12 bits and size of AL register is 8 bits. Hence it indicates that a carry is generated which needs to be reflected somewhere, so we are reflecting in the carry flag.

Therefore when CF = 1; carry is generated

CF = 0; carry not generated

Similarly for 16 bit addition we use the instruction ADD AX,BX (the contents of AX and BX will be added and result will be stored in AX by default). If the result goes beyond 16 bits it means that a carry is generated which will be reflected by the carry flag, it means that it will set.


Carry flag cont..The carry flag is also used to reflect the result of subtraction.

Eg: 8 bit subtraction

We use the instruction SUB AL,BL (the operation performed is AL – BL and result is stored in register AL)

Now if AL >= BL then there is no borrow Eg. 35H – 34H therefore CF = 0; reset

Now if AL < BL then a borrow is required Eg. 34H – 35H therefore CF = 1; set

Similarly for 16 bit subtraction we use the instruction SUB AX,BX (the operation performed is AX - BX and result is stored in AX).

Now if AX >= BX then there is no borrow Eg. 3500H – 3400H therefore CF = 0; reset

Now if AX < BX then a borrow is required Eg. 3400H – 3500H therefore CF = 1; set

Therefore the carry flag reflects the result on subtraction operation.1/15/16 BY UBAID SAUDAGAR 54

Auxiliary carry flag (Bit 5)Type: Status Flag

Operation: During an addition operation if a carry is generated from the lower nibble to the higher nibble, then AC flag sets. In other words that carry is stored in the Auxiliary carry flag bit i.e. Bit 5, or we can say that the carry which is generated from the lower nibble to the higher nibble is reflected in Bit No.5 of the flag register.

Eg: 8 bit addition

Instruction ADD AL,BL is executed

AL = 45H = 0100 0101

BL = 5BH = 0101 1011

ADD = 1010 0000


Lower nibbles

111 Carry

1 11

This carry which is generated from the lower nibble to higher is called as AC and it is stored in BIT no. 5 of flag register

Auxiliary carry flag cont..Eg: 8 bit subtraction

SUB AL,BL instruction is executed.

Let AL = 10H = 0001 0000

BL = 18H = 0001 1000

SUB = 1000

As we see above that a borrow is taken from the higher nibble. As soon as the borrow is taken from the higher nibble the AC flag sets i.e. AC = 1


1

Borrow taken from the

higher nibble

Parity flag (Bit 3)Type: Status flag

Operation: When an addition, subtraction, multiplication, division or any logical operation takes place then depending on the result the parity flag gets affected.

µp checks the 8 LSB’s of the result:

• If the number of one’s in the 8 LSB’s are 0/2/4/6/8 then the parity flag sets i.e. PF = 1 indicating even parity

• If the number of one’s in the 8 LSB’s are 1/3/5/7 then the parity flag resets i.e. PF = 0 indicating odd parity

• Eg: Let AX = 1002H

BX = 2008H

ADD AX,BX instruction is executed. The result obtained is 300AH (0011 0000 0000 1010) which is stored in AX register.

Now µp will check only the last 8 bits of the result

Here It came to know that No. of 1’s are 2 which is even hence PF = 1


µp checks these 8 bits

Zero flag (Bit 7)Type: Status Flag

Operation: Whenever any arithmetic or logical operation takes place and the result is 0 then zero flag sets to indicate that the result is 0 i.e. ZF = 1

Eg: Let AL = 80H

BL = 80H

SUB AL,BL instruction is executed. The result obtained is 00H in AL register. When µp checks this it has to reflect it somewhere that the result is zero, so it reflects in bit 7 which is reserved for zero flag by setting its value to 1 i.e. ZF = 1

Similarly for 16 bit operations, when the result is 0000H. ZF sets i.e. ZF = 1

Else it is always 0 i.e. ZF = 0


Sign flag (Bit 8)Type: Status flag

Operation: Whenever the result of any arithmetic operation is negative the sign flag sets i.e. SF = 1

Eg: 8 bit subtraction

SUB AL,BL instruction is executed

AL = 94H = 1 0 0 1 0 1 0 0

BL = 96H = 1 0 0 1 0 1 1 0

SUB = 1 1 1 1 1 1 1 0 = FEH


1

1

1

1

11

11

1

11

This borrow comes from

sign flag

Sign flag cont..As we see above that AL < BL answer is negative which is indicated by the sign flag. Whenever a borrow is

required by the MSB bit it indicates that answer is negative, hence sign flag sets and borrow to the MSB is given

from the sign flag bit i.e. bit 8 in the flag register.


Overflow flag (Bit 12)Type: Status flag

Operation: Whenever an arithmetic or logical operation takes place and the result which gets stored in any of the registers goes out of range then the Overflow Flag sets to 1 i.e. OF = 1

Eg: 8 bit addition

ADD AL,BL instruction is executed and the result will be by default stored in the AL register (size of AL = 8 bits)

AL = 89H = 1000 1001

BL = 89H = 1000 1001

ADD = 0001 0001 0010


The result is 12 bits and AL = 8 bits hence when a carry is generated form the MSB indicating that the result is long enough to fit in the AL

register hence OF sets

Interrupt flag (Bit 10)Type: Control flag

Operation: Interrupt flag is used to control the maskable interrupts of the µp 8086 which arrive at the INTR pin. Let’s see how it controls the interrupts.

Shown above is the internal circuitry to control the maskable interrupts. When Bit 10 of Flag register is 0 i.e.

IF = 0, input to the AND gate is 0 therefore o/p of AND gate goes low. Therefore any input to the AND gate won’t be recognized hence any interrupt occurring on the INTR pin won’t be acknowledged as it is not recognized.

When IF = 1 and interrupt on the INTR pin occurs it is recognized and given to the internal INTR circuitry through AND gate and then acknowledged.


IF

INTR pinTo INTR circuit

Trap Flag (Bit 9)Type: Control Flag

Operation: This flag is used to detect any error in the program (debugging) by executing the program in single stepping mode.

- If logic 0 is stored in TF, then the µp will execute all the instructions of a program in one operation (free run operation)

Eg: When we execute the program and press key F9, the entire program gets executed in one stroke. The imp. Point to understand is that when we press the key F9, value 0 gets placed bit 9 of the flag register therefore Trap flag resets i.e. TF = 0.

- If logic 1 is stored in TF, then the µp will execute one instruction of the program at a time, after executing each instruction µp will execute INT1 (software interrupt), so µp will branch from main program to subroutine.

The subroutine has a program which displays the result in different registers of µp on the screen, so after each instruction the programmer can verify the result.

Eg: We use F8 function key to execute the program in single stepping mode. When we press key F8 value 1 gets placed into bit 9 of flag register, hence trap flag sets i.e. TF = 11/15/16 BY UBAID SAUDAGAR 63

Direction Flag (Bit 11 )Type: Control Flag

Operation: To understand the operation let’s take an example:

In the above example we see two memory blocks. The offset address of first memory block is placed in the SI register and offset address of the second memory block is placed in the DI register.

Now we need to manually increment SI and DI after each byte transfer from data segment to extra segment.


40H

53H

87H

98H

DS : 0000

DS : 0001

DS : 0002

DS : 0003

ES : 0000

ES : 0001

ES : 0002

ES : 0003

Memory

block

Offset address placed in

SI register

Memory

block

Offset address

placed in DI register

Direction Flag cont..To avoid all this what we do is after moving one byte from Data segment to extra segment we clear the direction flag (CLD instruction is used to clear the direction flag) which puts SI and DI in auto increment mode.

In other words whenever DF = 0, SI and DI increment automatically.

If we are copying the data from the last location we set the direction flag (STD instruction is used to set the direction flag) which puts the SI and DI in auto decrement mode.

In other words whenever DF = 1, SI and DI decrement automatically.


Source memory block Destination memory block

OA

OA OA

OA

Arithmetic Logic Unit (ALU)In digital techniques we have seen that to add two single bit numbers we used half adder, then to add two single bit numbers along with carry we used full adder.

Similarly we designed half and full subtractor and so on. Hence for different arithmetic calculations different ckts have to be designed.

Similarly logic operations such as AND, OR, NOT etc. we required different IC’s like for AND gate we require IC 7408 and so on for other IC’s.

Therefore for performing arithmetic and logical operations different ckts were designed and used which is very much bulky.

Microprocessor has a unit called as ALU (arithmetic and logic unit) which does all the arithmetic and logical operations which we do with the help of so many ckts.

The different operations performed by the ALU are: Addition, Subtraction, Multiplication, Division etc. and ANDing, ORing etc.

The size of the ALU is 16 bits which means that the data coming from the memory to the µp which is in general 16 bits, that 16 bit data can be processed at a time i.e. in one clock cycle that data can be processed.

Hence the size of the ALU defines how many bit processor. Size of ALU = n ; Processor = n bits


6 byte instruction queue registerWhy is the 8086 queue only six bytes long?

As we know that the longest instruction in the instruction set of 8086 is six bytes long.

Hence with a six bytes long queue it is possible to prefetch even the longest instruction in the instruction set.


6 byte instruction queue registerIn general what happens in 8085 is that there is a microprocessor and memory.

Memory contains opcodes and operands.

With the help of Opcodes we come to know two things: how many byte instruction and what operation is to be performed.

Eg: MVI A, 32H

Now this instruction is a two byte instruction, first byte is opcodes and second is operand which is stored in the memory as follows:


3EH

32H

-

-

Memory

1000H1001H Random

address

6 byte instruction queue registerNow the µp fetches 3EH first from the location 1000H, and brings it into the µp where it is decoded.

When 3EH is decoded, what it comes to know that it is a two byte instruction and second that it needs to move immediate data 32H into the accumulator.

So after decoding it again goes back to memory and fetches 32H and this 32H after being decoded as data is moved to the Accumulator.

We notice that when it was being decoded at that the data bus, address bus was idle, it has no work, and further what we notice Is that, after decoding again the µp has to go to the memory and fetch data.

So this wastes a lot of time of the µp.

This is a simple example of a car manufacturer which defines a person to make a car. Now that person starts making the engine, then on its completion, he starts preparing the outer body and continues.

We need to understand is that the person is completing one task and then starting another, hence to manufacture one car will take a lot of time.


6 byte instruction queue registerThe manufacturer came up with an idea, that why not increase my staff. Instead of one staff I’ll make it 5.

First : Engine manufacture

Second : Body manufacture

Third : Seat manufacture

Fourth : Painting

Fifth : Tyre manufacture

Now what happens that the work got divided and each person is busy. After completion of first module it goes to second and so on till the fifth.

So every module is continuously busy, and the work got divided.


6 byte instruction queue register

0.2sec (T1)

0.2sec (T2)

0.2sec (T1)

0.2sec (T3)

0.2sec (T2)

0.2sec (T1)

0.2sec (T4)

0.2sec (T3)

0.2sec (T2)

0.2sec (T1)

0.2sec (T5)

0.2sec (T4)

0.2sec (T3)

0.2sec (T2)

0.2sec (T1)

0.2sec (T6)

0.2sec (T5)

0.2sec (T4)

0.2sec (T3)

0.2sec (T2)

0.2sec (T7)

0.2sec (T6)

0.2sec (T5)

0.2sec (T4)

0.2sec (T3)1/15/16 BY UBAID SAUDAGAR 71

TOTAL TIME = 1sec

TOTAL TIME = 0.2 sec

TOTAL TIME = 0.2 sec

M1 M2 M3 M4 M5

1 person can do the work in 1 sec, now the work got divided, but first time all the modules are in wait state and waiting for the previous one to complete i.e. M2 is waiting for M1 to complete. So time taken is same i.e. 1s. But after the entire cycle is complete once, now the time taken is only 0.2 sec.

This is called as parallel processing

6 byte instruction queue registerSimilarly in the microprocessor 8086 it has created a queue register of 6 bytes.

This queue is based on FIFO (first in first out) basis which means that the byte which comes first also goes out first from this register.


From memory

First in first out

To execution unit (For Decoding and Execution)

Byte 1Byte 2

Byte 3

Byte 4

Byte 5

Byte 6

6 byte instruction queue registerµp 8086 has created 2 logical units within its architecture one is the bus interface unit (BIU) and the other is the execution unit (EU).

While the EU is decoding an instruction or executing an instruction which does not require the use of buses, the BIU fetches upto six bytes from the memory.

The BIU stores these prefetch bytes in FIFO register called as queue register.

When the EU is ready for the next instruction it simply takes (fetches) the next byte from the instruction queue in the BIU.

This is much faster than sending out address to the system memory and waiting for memory to send back the next instruction bytes.

Drawback: The only drawback of the queue is when a branch instruction comes in the queue:

Let’s understand with an example:1/15/16 BY UBAID SAUDAGAR 73

6 byte instruction queue registerSuppose an instruction is at the address 3000H

3000H : JMP 8000 The BIU fetches inst. from the memory and keeps in the queue, it fetches 6 bytes and keeps in the queue.


Address at which it is going to jump

4CH

00H

80H

56H

67H

7AH

69H

Memory

3000H3001H

3002H

3004H

3005H

3003H

--

7AH

67H

56H

80H

00H

4CH

Address at which instruction lies

TO EXECUTION UNIT

Assuming that 4CH is opcode for JUMP instr.

6 byte instruction queue registerWhen 4CH goes to the execution unit, there it is decoded where the µp comes to know that it is a three byte instruction and second thing that it has to do jump operation.

Since when 4CH was decoded and µp came to know that it is a three byte instr. and jump instruction hence µp wants the address where to jump. So it fetches more two bytes i.e. 00H and 80H from the queue and decodes them as address.

During their decoding process queue is again full with 6 bytes.

Now the IP jumps to address 8000H to fetch instructions from that location.

The bytes which had been prefetched in the queue are now of no use and are thus flushed.

Therefore the same process starts again as explained before.


What is pipelining ?When one instruction is getting decoded and executed and simultaneously next instruction is being fetched from the memory is called as pipelining process.

In 8086 with the help of queue register 6 instruction bytes are prefetched in the queue register, whenever the EU requires bytes for decoding and execution purpose it just has to fetch from the queue register instead of going to the main memory and waist its time.

By the time it is decoding and executing the byte fetched from the queue, µp does not allow the buses to remain idle, the external bus goes to the memory and fetches the next bytes and places in the queue.

This is actually parallel processing and similar to a water pipe where the water continues to remain in motion. Similar way the instruction bytes are in continuous motion i.e. from the memory to the BIU then into the queue then to the execution unit for execution.

In other words the motion of bytes does not stop anywhere ideally unless and until a branch instruction arises in the queue.

Note: The µp does not perform the next fetch operation till at least two bytes of the instruction queue are emptied.


Pipelining process


Memory

BIUExecution unit

6

5

4

3

2

16 bytes queue

Here we can see that there Is a pipeline structure

It has a two stage pipeline1. Fetch stage (prefetches upto 6 bytes) 2. Decode and Execute stage (where)Instructions are decoded and executed

Signal description (Pin description)


Pin diagram of 8086 µp


8086µp

VccGNDGND

40120

CLK RESET READY

192122

NMI INTR

1718

TEST MN/MX

2333

BHE/ RD

3432

- 16 -2, 39

/ - / 38 - 35

3130292627282524

HOLD /HLDA /WR LOCKDENDT/RM/ IOALEINTA

Minimum mode pins

Maximum mode pins

Pin configuration8086 is 16 bit µp available in three clock rates 5, 8, and 10 MHz and is a 40 pin ic.

It works in two modes minimum mode and maximum mode.

The pins on the µp 8086 can be categorized into 3 types:

1. Pins common to minimum and maximum mode

2. Minimum mode pins

3. Maximum mode pins


From pin number 16 – 2 are used by the multiplexed address data bus.

Multiplexing address and data lines means that the same lines are used for address and data. Therefore the same lines are used to carry address information as well as data.

During T1 state of every machine cycle ALE signal goes high and enables the latches, during this T1 state address which is generated by the µp is available on this multiplexed bus.

After T1 state gets over ALE signal goes low which disables the latches and DEN signal also goes low which enables the Transrecievers. Now data is available on the multiplexed bus.

Refer concept of latch


(Pin 16 - 2) - Multiplexed Address Data Bus

- (Pin 38 – 35) These lines work as Address Bus (A16 – A19) during T1 state of every machine cycle i.e. when ALE signal goes high.

T2 onwards these work as status signals S3 to S6

S3 and S4 gives the status of the memory segment currently accessed.

S5 gives the status of Interrupt Flag.

S6 remains low in general.


Segment accessed

0 0 Extra segment

0 1 Stack segment

1 0 Code segment

1 1 Data segment

Segment accessed

0 0 Extra segment

0 1 Stack segment

1 0 Code segment

1 1 Data segment

Interrupt basicsWhat is an interrupt?

An interrupt is an external signal to the µp which breaks the normal sequence of the program running in the µp.

For Eg. A person is reading a book and while reading he hears the door bell so what he does is that he puts a bookmark in the book and goes and opens the door, receives the request from the person standing on the door and then comes back.

Now it depends who has arrived on the door, if a non important person is there then he can tell him to wait or ignore. But if an imp. person is there whom he cannot avoid, then he has to compulsory respond.

Similarly there are two types of interrupts one maskable (can be avoided) and the other non – maskable (cannot be avoided)

Another Eg. Is of gas detector (at any how it is going to detect the gas (non - maskable)), and a mobile phone (if a non important call is there user can avoid it and if an important call is there user will accept it)

µp 8086 consist of two interrupt pins i.e. NMI and INTR and one acknowledge pin INTA

Instead of being wired directly to the CPU, the interrupts from the external devices are channeled through Programmable interrupt controller (PIC 8259).


Interrupt pins (INTR, NMI, INTA)


INTR (Pin 18):

This is input pin to the microprocessor.

It is an interrupt pin and the input is level triggered, which means that when logic high arrives on this pin it gets activated.

An interrupting device such as keyboard, printer etc. is not directly connected to this pin but through a PIC (IC 8259).

Interrupts arriving on this pin can be masked with the help of IF flag.

Initially the interrupt flag (IF) is in the set state so that any interrupt which arrives to the microprocessor gets response i.e. the interrupt is unmasked (allowed).

When the instruction CLI is executed, IF flag gets cleared and interrupts arriving on this pin are masked (they don’t get any response).

When one interrupt is getting response from the processor (i.e. processor is executing ISR), CLI instruction is executed so that no other device may interrupt the processor


NMI (Pin 17)This is an input pin to the µp .

It is a positive edge triggered input.

It is a non maskable interrupt.

For any interrupt to be recognized the i/p at this pin should be high for at least 2 clock cycles

Whenever there is a power failure this pin comes into use.


INTA (Pin 24)This is an acknowledge signal going out from the µp to the PIC (IC 8259).

Whenever the µp is interrupted on the INTR pin with a high signal, if the µp accepts the interrupt which means that the IF is set, it sends an acknowledgement signal to the PIC from where the interrupt came, indicating that it has accepted the interrupt.


RD (Pin 32) ReadThis is an active low pin

It is an output pin. Whenever the µp performs memory read or i/o read operation, this pin goes low to indicate read operation. This pin is always used along with M/IO to indicate memory read or I/O read.


RD

Microprocessor

OE (o/p enable)

Memory

Data from the memory goes to the µp

READY (Pin 22)This is an input acknowledgement signal from the slow peripheral devices to the microprocessor to indicate that data transfer is complete.

Eg. Printer connected to microprocessor.

The main use of this pin is to insert wait states into the timing of the microprocessor.

If the READY pin is at logic 0 level, the µp enters wait states into its timing cycle and remains idle.

The need for this is because the speed of the µp is very fast, it operates in MHz and speed of peripheral device like printer is slow as compared to that of the µp. Hence µp sends data at a faster rate continuously to the peripheral, which may cause loss of data, hence the peripheral sends a low signal on the READY input to send the data at a slower rate by adding wait states to its timing cycle.

Once one slot of data is received by the peripheral then it sends an acknowledgement high signal on its input to indicate that data transfer is complete and send the next slot of data.


CLK (Pin 19) – Clock Clock pin provides the basic timing signal to the microprocessor.


CLK

Microprocessor

T f/f

Schmitt trigger converting sine wave into square wave

X1 X2

Crystal oscillator

IC 8284Clock generator

The crystal oscillator produces sinusoidal oscillations.We assume that the crystal frequency = 10 MHz Schmitt trigger converts this sine wave into square wave T f/f divides the crystal frequency by 2This frequency is applied to the µp which will be the Operating frequency of the µp Therefore OF of µp = 5 MHz

Vcc (Pin 40) ; GND (Pin 1 and 20)Vcc i.e. +5V is applied at pin 40 and is used as power supply

There are two layers of 8086, hence two separate GND pins for two layers of 8086 chip are used

This helps in improve noise rejection


RESET (Pin 21)It causes the processor to immediately terminate its present activity .

The 8284 clock generator provides this signal. This signal must be active high for at least 4 clock cycles

It clears all the flag register, the instruction queue, DS,SS, ES and IP register and sets the bits of the CS register

Hence whenever the processor boots the reset vector address of 8086 is FFFF0H (as CS = FFFFH and IP = 0000H)

Note: The reset vector is the default location a central processing unit will go to find the first instruction it will execute after a reset. That is to say, the reset vector is a pointer or address where the CPU should always begin as soon as it is able to execute instructions.

The system designer will store the monitor program starting from the address FFFF0H. The monitor program should be executed to initialize system peripherals whenever the system is switched ON. To enable automatic execution of monitor program, Whenever the system is switched ON, the EPROM should be mapped at the end of memory space in 8086-based system. Monitor program is a permanent program written by the system designer to take care of system initializations. The system initializations includes the following.

1. Programming 8279 for keyboard scanning and display refreshing.

2. Programming peripheral ICs 8259,8257,8255,8251,8254,etc.,.

3. Initializing stack.

4. Display a message on display (output) device.


Concept of latchAs we have learned the concept of f/f . What we have noticed that in some f/f’s that whatever we give them the input they store that input at their o/p.

A simple example is of a D f/f, where when we apply a logic 1 at its input that logic 1 appears at the o/p and doesn’t change unless and until next clock appears.


Flip Flop or Latch

D QLogic ‘0’ at i/p Logic ‘0’ at o/p

CLK

Concept of latchSo in general what we can say that it stores that 1 bit at its o/p or it holds this value at its o/p, therefore we can also term it as a D latch.

A latch is nothing but a f/f but basically used to hold values at its o/p for other devices.

Now an Octal (Eight) Latch is a device with eight D f/f’s.

It’s IC number is 8282 . It has an i/p pin STB which is active high and when logic 1 is applied to it activates the IC and i/p’s are applied to all the latches and correspondingly o/p’s appear at the o/p.

Now let’s understand why we require octal latch:

No. of address lines = 20

No. of data lines = 16

Now these 16 lines are multiplexed with 16 address lines and these are time multiplexed.


Concept of latchMultiplexing address and data lines means that the same lines are used for address and data.

Therefore the same lines are used to carry address information as well as data.

The remaining address lines are multiplexed with 4 status signals (S3 – S6)

Let’s understand this multiplexing with the help of a simple multiplexer diagram :


2 : 1 MUX

Address

Data

When ALE = 1, it selects address and makes it available at the o/pWhen ALE = 0, it selects data and makes it available at the o/p

Address or Data depending upon ALE

Concept of latch and machine cycleALE is an o/p signal from the µp which goes to the octal latch to make it active.

Since we know that the multiplexing is time multiplexing hence ALE becomes high for a short duration and then becomes low.

Now we need to understand when does it become high and low.

- Crystal frequency = 10 MHz (assumption)

- Operating frequency = 5 MHz

- Time period of each clock pulse = 1/(5 * ) = 0.2 µs = 1T state


- - - - - - -

- - - - - - -

1 T state= 0.2 sec

Concept of latch and machine cycleNow many T states together form 1 machine cycle and one instruction consist of many machine cycle.

For Eg. Consider the instruction MVI A, 32H

This instruction is of 2 bytes viz. first byte is the opcode and second is the operand.

For this instruction 2 machine cycles are required viz. opcode fetch machine cycle and memory read machine cycle.

To fetch the opcode from the memory it requires 4 T states (T1, T2, T3, T4) = 0.8 µs

To fetch the operand from the memory it requires 3 T states (T1, T2, T3) = 0.6 sec

Therefore total time to execute an instruction = 1.4 µs

The main aim to clear the concept of T state and machine cycle is that during T1 state of every machine cycle ALE goes high and that moment of time address is available on the multiplexed address and data bus.


Concept of latch and machine cycleNow this ALE is connected to STB pin of the octal latch, and the multiplexed 8 lines are connected to octal latch

Since its an octal latch hence only 8 lines can be connected to it, therefore we require 3 octal latches.

Three octal (8 bit) latches are required, the connection is as follows:


Microprocessor

8282

8282

8282

ALE .

.

STB

STB

STB -

Multiplexed Address and Data bus

-

-

-

20 bit address generated to access Memory or I/O device

/ - /

How address gets latched?In the above figure we see that there are three external octal latches.

The ALE signal is connected with the strobe input (STB i/p) of the latches which enables the latches and loads the address into the latches.

Since each latch is of 8 bits hence we have to use 3 latches to latch 20 bit address.


-

Microprocessor

3EH

32HOctal latche

s

/ - /

20 bit addressALE

Data bus

This is the opcode fetched

during first machine

cycleThis is the

data fetched during second

machine cycle

How address gets latched?Consider the instruction MVI A,32H is executed.

This instruction is a 2 byte instruction and has two machine cycles. First machine cycle has 4 T states and second machine cycle has 3 T states.

On T1 state of every machine cycle ALE goes high and enables the latches. The addresses is available on the multiplexed bus for this T1 state and then disappears.

Therefore in order to save our address so that it doesn't get lost we use latches which latch the address at its o/p (which means that it holds the address at its o/p).

This is because when T2 begins, ALE becomes low, latches gets disabled and data is available on the multiplexed bus.

Since latches are disabled they do not respond to data appearing on the bus, and address is available on their o/p with which the proper memory location is accessed.


ALE (Pin 25)ALE stands for Address Latch Enable.

As we studied in the previous slides about latch. Latch has an input pin STB which is connected to the o/p pin of the µp i.e. ALE

As the name suggests, it is used to enable the latch by generating a high signal during T1 state of every machine cycle.

As soon as ALE pin goes high, latches gets enabled, address which is available on the address bus goes to the input of the latches where it is latched (it holds the address at its o/p).

Now ALE goes low after T1 state gets over, hence disabling the latches (as STB also goes low).

Now data is available on the multiplexed A/D bus.

Since latches are disabled hence data which arrives at latches get no response.1/15/16 BY UBAID SAUDAGAR 101

Concept of Numeric Processor 8087A system having two microprocessor will require only lesser time to complete the task.

Using a general-purpose microprocessor such as the 8086 to perform mathematical functions such as log, sine, and others is very time consuming, not only for the CPU but also for programmers writing such programs.

In the absence of a math coprocessor, programmers must write subroutines using 8086 instructions for mathematical functions.

You may take a normal example of simple calculator and a scientific calculator.

In a simple calculator normal calculations takes place, but in scientific calculator apart from the normal calculations, log, sine, cos and many other calculations are present.

For performing those complicated calculations numeric processor is designed and Operation of 8087 does not require any software support from the system software or operating system.

They are Unable to fetch the code from the memory so they work under the control of main processor .

The 8087 instruction may lie interleaved in the 8086 program, but it is the task of 8086 to identify the 8087 instructions from the program, send it to 8087 for further execution & after the completion of execution cycle the result may be referred back to CPU.


TEST (Pin 23)It is an active low input line dedicated for 8087 math co – processor(numeric processor).

When this pin is high in the maximum mode, it indicates to the microprocessor that it is busy.

Whenever those instructions arrive in the program which are dedicated to the co – processor then they are sent to the co – processor for execution.

While the co – processor is busy executing those instructions it sends a high signal on the TEST pin of the µp indicating that it is busy executing those instructions, till then the µp remains in idle state.


MN/MX (Pin 33)This is an input pin to the µp which will indicate the processor will work in which mode.

If input to this pin is high i.e. if the pin is connected to Vcc (+5V) then µp will enter in minimum mode

If input to this pin is low i.e. if the pin is connected to GND (0V) then µp will enter in maximum mode


Concept of Memory BankingNEED :

We know that our microprocessor is 16 bit microprocessor, hence it has 16 data lines, it means that 16 bit data can be accessed and processed from the memory in 1 machine cycle.

But we have seen that a memory chip has n locations and each location has 8 bits, therefore only 8 bit data bus can be connected to the memory chip.

Since only 8 bit data bus can be connected hence only 8 bit data can be accessed at a time. Therefore in 1 machine cycle only 8 bit data can be accessed and processed.

Therefore it will behave as an 8 bit processor.

So to avoid this the entire 1 Mb memory is divided into two banks viz. odd bank and even bank.

Note: The 8086 microprocessor has 20 address lines and with the help of these 20 lines we can address 1 Mb of memory.

Each bank will have a capacity of 512 Kb (odd + even bank = 512 Kb + 512 Kb = 1 Mb)


Concept of Memory Banking1010 1011

1011 1100

0101 1010

0011 0101

0011 1100


1000 1001

0111 1000

0101 0110

0010 0011

0001 0010Even bank Odd bank

00000H

00002H----

FFFFAH

FFFFCH

FFFFEH

00001H

00003H----

FFFFBH

FFFFDH

FFFFFH

All even addressesIn even bank

All odd addresses In odd bank

8 bit data at every location

8 bit data at every location

OVERVIEW :

Concept of Memory Banking

Since the memory size reduces to 512 Kb each, hence address bus size also reduces as follows:

No. of address lines required to access 512 Kb = ; where x is the no. of address lines

We replace x by 19 therefore = 524288 locations and at each location 1 byte = 524288 bytes ==̃ 512 Kb


Even Memory Bank

Odd Memory Bank

Address Bus

Address Bus

Data Bus Data Bus

CONNECTION OF ADDRESS AND DATA BUS :


As we know that only 19 address lines are required to access memory banks, 1 address line is left.

With the help of the previous diagram we understood how address and data bus is connected to the memory banks. Now we need to understand how this memory banks are enabled and disabled.


A0BHE

Microprocessor 8086

Even Memory

Bank

CS

Odd Memory

Bank

CS

SELECTION OF MEMORY BANKS :

Concept of Memory BankingIn the above figure we see that the address line A0 is not used for addressing purpose but used for chip selection. The chip select on the memory is active low hence when A0 is low, lower memory bank or even bank gets selected.

To select the odd bank or higher memory bank an o/p signal BHE is used. When this pin goes low odd memory bank or higher memory bank gets selected.

We need to now understand how addressing takes place.

Refer next slide




A19

A18

A17

A16

A15

A14

A13

A12

A11

A10

A9 A8 A7 A6 A5 A4 A3 A2 A1 A0 BHE

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0

A0BHE

A1 – A19

8086

DO – D7

D8 – D15

34H

4FH

5CH

-

6BH

3EH

4AH

9CH

-

54H

00000H00002H00004H

FFFFEH

00005H

00003H

FFFFFH

00001H

ADDRESS BUS

ADDRESS BUSDATA BUS

DATA BUS


From the above diagram and table what we see is that, the address lines are connected to both the banks simultaneously.

The lower order data lines (D0 – D7) are connected to lower memory bank and higher order data lines (D8 – D15) are connected to higher memory bank.

Consider the 1st row in the table, where we see that A0 and BHE, both are low hence both the banks are enabled.

Now the address from A19 – A1 = 0000000000000000000, this address will select the 0th location of both the banks, since both the banks are enabled data will be loaded to both the memory banks in case of write operation or data will be loaded from both the memory banks to the data buses in case of read operation.

Consider the 2nd row in the table, where we see that A0 is 0 and BHE is 1, therefore lower bank is enabled and higher bank is disabled.


ADDRESSING THE MEMORY BANKS :

Concept of Memory BankingNow the address from A19 – A1 = 0000000000000000001, this address will select the 1st location of both the banks, but since lower bank is enabled and higher bank is disabled therefore data will be loaded only from the lower memory bank onto the lower order data bus or vice versa in case of read or write operation.

Consider the 3rd row in the table, where we see that A0 is 1 and BHE is 0, therefore lower bank is disabled and higher bank is enabled.

Now the address from A19 – A1 = 0000000000000000010, this address will select the 2nd location of both the banks, but since lower bank is disabled and higher bank is enabled therefore data will be loaded only from the higher memory bank onto the higher order data bus or vice versa in case of read or write operation.



As we discussed before that our main aim of memory banking is that the microprocessor should be able to access 16 bits i.e. 2 bytes i.e. 1 word at a time and process it in 1 machine cycle.

But depending upon instruction and address it accesses bytes or words.

Consider the table on next slide


ACCESSING WORDS AND BYTES FROM BANKS :



INSTRUCTION ADDRESS DATA TYPE BHE A0 MACHINE CYCLE

DATA LINES USED

MOV AL,[0000H]

0000H BYTE 1 0 ONE D0 – D7

MOV AX,[0000H]

0000H WORD 0 0 ONE D0 – D15

MOV AL,[0001H]

0001H BYTE 0 1 ONE D8 – D15

MOV AX,[0001H]

0001H WORD 01

10

FIRSTSECOND

D8 – D15D0 – D7 Byte is accessed in one

machine cycle as this instruction is accessing a byte from even memory

location. Hence even bank is enabled and odd

bank is disabled and data is available on the

lower order data bus D0 - D7

Word is accessed in one machine cycle as this

instruction is accessing a word from even memory location. Hence even and

odd bank both are enabled to and data is available on the lower

order data bus D0 - D7

Byte is accessed in one machine cycle as this

instruction is accessing a byte from odd memory

location. Hence odd bank is enabled and

even bank is disabled and data is available on the higher order data

bus D8 – D15

Word is accessed in two machine cycles. This inst.

Is accessing a word starting from the odd

memory location hence in the first machine cycle 1

byte is accessed from odd bank, in the 2nd machine cycle 2nd byte is accessed

from even bank

BHE/ (Pin 34) is used for future development.

BHE is active low o/p signal and stands for Bus High Enable.

This o/p pin is used to activate the odd memory bank, when the o/p of this pin is low it activates the odd memory bank and data is available on the higher order data bus (D8 – D15)

If this pin goes high then odd memory bank gets disabled and any type of data transfer from the higher order data bus is disabled.


WR – Write (Pin 29)This is an o/p pin is used in minimum mode, i.e. MN/MX = 1

It is connected to the memory or i/o device. When this pin goes low it indicates write operation.

It is used along with M/IO pin to indicate whether memory write or io write operation.


WR

8086 Microprocess

or

WR

Memory

Data from the microprocessor goes to memory

M/IO (Pin 28) This is an o/p pin used in minimum mode, to distinguish between a memory operation and io operation.

When this pin is low it indicates io operation and is used with RD and WR.

For Eg. M/IO is high, and RD is also low whereas WR is high, then memory read operation takes place.

For Eg. M/IO is high, and WR is also low whereas RD is high, then memory write operation takes place.

In similar manner IO read and IO write operation takes place, only M/IO is low.


WR RD

8086 Microprocessor

WROE

Memory

Data Bus

Concept of DMA (Direct Memory Access)

Consider that an io device and memory are connected to the µp.

Lets consider that a pen drive is connected which will be treated as an io device.

Now bulk data is to be transferred from the main memory to the pen drive.

The procedure will be that the pen drive will access the µp, which in turn will perform a read operation from the memory, then a write operation to the pen drive.


8086Microprocessor

Memory

I/O deviceSystem Bus

System Bus

Data transfer takes place betweenI/O device and memory

Reads data from the memory

Writes data to I/O device


If the data is small, then going this way is not a big issue . But if data is more, then going through this method is quite tedious and will take a lot of time.

Secondly µp’s are not designed just for data transfer from one device to another, going this way we are wasting the time of the µp.

Hence our requirement is that data should be transferred directly between the two devices without any intervention from the µp.

To achieve our requirement Intel designed a controller called as DMA controller which meets our requirement.

Now what is going to happen that the I/O device gets connected to the DMA controller which in turn gets connected to the µp.

Whenever there is a requirement of data transfer between I/O device and memory, I/O device sends a request signal to the DMA controller, DMA will send a request signal (HOLD) to the µp to release the system bus.



µp will send an acknowledge signal to the DMA that it has released the system bus it actually means that the system bus from the µp side is tri stated.

DMA also sends an acknowledge signal to the I/O device indicating that it has the control over the system bus. Now data transfer takes place directly between the I/O device and the memory under DMA control, hence by passing the µp and this data transfer takes place at a higher rate.

By the time data transfer takes place outside the µp, µp may perform some internal calculation part as external system bus is tristated.


HOLDHLDA

Microprocessor8086

REQDACK

DMA Controller

8257

REQ

I/O device

Memory

System bus tri stated from external devices

System bus tri stated from external devices

System Bus

System Bus

HOLD (Pin 31)It is input pin used in the minimum mode. The i/p comes from the DMA controller which is acting another bus master (as it also has data and address lines) requesting the µp to release the system bus.

In response 8086 completes the current machine cycle and releases the system bus.

In other words it tristates its external system bus.


HLDA (Pin 30)It is acknowledge signal generated by the µp to the DMA controller after releasing the system bus, indicating that it has released the system bus.

After which DMA operation takes place.


Concept of Octal Bus TransrecieversAs we have seen that the octal latch is used to latch the address obtained from the multiplexed A/D bus during the T1 state.

After T1 state ALE goes low and data is available on the multiplexed A/D bus. This data cannot be latched as the latches are disabled.

Therefore for driving this data coming from the data bus we require Octal Transrecievers (IC 8286)

Now these Transrecievers contain 8 bidirectional buffers. Let’s see what a buffer is.

Shown in the above diagram is the basic diagram of a buffer.


Input Output

Concept of Octal Bus TransrecieversIt is a single input, single output, non inverting device with gain =1.

As we have seen that, an inverter is a device which is same as a buffer but having a bubble at the end which inverts the input.

But in case of a buffer if there is 0 at the input there will be 0 at the o/p.

The main important feature of a buffer is that it is used to amplify the current or power.

This is because the o/p current of the microprocessor is very low say few mA like 5mA.

Now the device which it is driving gets activated or gets drived at 50mA. So we require a device which will amplify the current to drive that device.

Hence buffers are used.

Now there are 8 such buffers present in an octal bus Transreciever.


Concept of Octal Bus TransrecieversAlso the Buffers used are tristated buffers which means that it has three states viz. high, low and hi – z (hi - impedance).

High and low is understood but hi – z state i.e. high impedance state effectively removes the device's influence from the rest of the circuit.

In other words infinite resistance state.

In the above diagram we see an active high tri state buffer. When the enable i/p is high the buffer acts normally i.e. allows i/p to appear at the o/p with an increase in current as current is amplified.


Enable

Input Output

Concept of Octal Bus TransrecieversBut when the enable i/p is 0, the buffer enters in hi – z state i.e. it isolates the input from the o/p

Now, an octal Transreciever has bidirectional buffers as follows:

As we know that data bus is bidirectional, hence we see that buffers are also bidirectional.


Enable Enable

Input A

Input B

Output A

Output B

A

B

Concept of Octal Bus TransrecieversNow this octal bus Transreciever consist of such 8 bidirectional buffers.

When data gets transferred between the µp and the memory, ALE goes high during T1 state, address gets latched with the help of an 8282 octal latch.

After T1 state gets over, T2 starts, ALE goes low, latches gets disabled, data is now available on the multiplexed bus.

8086 has an o/p pin DEN which is connected to the OE pin of the Transreciever.

When DEN pin is high Transreciever is inactive, when it becomes low Transreciever gets activated.

During the T1 state, this pin is high as no data is available on the multiplexed A/D bus, therefore 8286 is inactive.

As soon as T1 state gets over, DEN pin goes low activating the 8286 Transreciever

As we know that Data bus is bidirectional hence there are two possibilities viz. data transfer from the µp to the memory, data transfer from memory to the µp.


Concept of Octal Bus TransrecieversNow with the help of DEN and OE of µp and 8286 Transreciever, the IC gets enabled but we need to decide the direction of data flow.

Hence there is a pin labelled T on the IC 8286 . When T = 1, flow is from µp to memory and when T = 0, flow is from memory to µp.

8086 µp has a pin labelled DT/R, when this pin goes high, it indicates data transmit, therefore data flows from µp to memory by making T = 1.

When this pin goes low, it indicates receive, thereby making T = 0, therefore data flows from memory to µp.

We now understand that why we require bi directional buffers.

When T = 1, buffer A gets activated and buffer B is in hi – z state.

When T = 0, buffer B gets activated and buffer A is in hi – z state.


DEN (Pin 26) – Data EnableThis pin is used to activate the octal bus Transreciever IC 8286 . It is connected to the OE pin of the Transreciever which consist of 8 bidirectional buffers. Actually when DEN goes low OE goes low and all the internal buffers get activated.


DEN

Microprocessor 8086

OE

8286 Transreciev

er

OE

8286 Transreciev

er

Active low o/p

Active low i/p

DT/R (Pin 27) – Data transmit/ ReceiveThis pin is used to decide the direction of data flow i.e. either data will flow from the µp to memory or io device or data will flow from the memory or io device to the µp.

When this pin goes high i.e. when DT/R = 1, which indicates data transmit from µp to memory or io device

In other words whenever write operation takes place, DT/R goes high, hence flow of data is from the µp.

When this pin goes low i.e. when DT/R = 0, which indicates receive i.e. µp receives data from memory or io device.

In other words whenever read operation takes place, DT/R goes low, hence flow of data is towards the µp.


DT/R

8086 Microprocess

or

T

8286 Transreciev

er

(Pin 28 – 26) – Status lines These are the active low status lines used in maximum mode which reflect the type of operation being carried out by the processor. It indicates which machine cycle is currently in operation.

Refer table below:


Indication

0 0 0 Interrupt acknowledge

0 0 1 Read I/O port

0 1 0 Write I/O port

0 1 1 Halt

1 0 0 Code Access

1 0 1 Read memory

1 1 0 Write memory

1 1 1 Passive

Indication

0 0 0 Interrupt acknowledge

0 0 1 Read I/O port

0 1 0 Write I/O port

0 1 1 Halt

1 0 0 Code Access

1 0 1 Read memory

1 1 0 Write memory

1 1 1 Passive

LOCK (Pin 29)This pin is used in the maximum mode.

When a programmer is executing some important instructions during which no interrupt can be tolerated (such as when we restart our computer and there are some updates which the windows is taking at that time if any other command is given to the PC it will not react to it) at that time what the programmer does it writes a LOCK prefix alongside the instruction, due to which this pin goes low indicating to the external bus masters such as DMA that it cannot take control over the system bus.

When the next instruction arrives without the LOCK prefix then this pin goes high and allows other bus masters to access gain over the system bus.


(Pin 24 – 25) QS stands for queue status. It means that outside the chip we come to know what is the status of the queue register inside the microprocessor.

Let’s understand from the table below:


Queue status

0 0 No operation (All the six registers of the queue are full)

0 1 It indicates that the opcode is being executed

1 0 The instruction queue is empty, all registers are cleared due to branching operation

1 1 It indicates that operands are being executed

Queue status

0 0 No operation (All the six registers of the queue are full)

0 1 It indicates that the opcode is being executed

1 0 The instruction queue is empty, all registers are cleared due to branching operation

1 1 It indicates that operands are being executed

// This signal works in the maximum mode when more than one processor is connected in the system.

RQ stands for bus request. GT stands for bus grant.

0 has the higher priority than 1. It means that when two processors simultaneously asks the 8086 to release the system bus, the processor which is connected at pin 31 will get bus grant first then the one which is connected at pin 30.

The external processor will send a low signal as a request to the µp to release the system bus, the µp will complete its current machine cycle and release the bus i.e. it will grant the bus to the bus master by sending a low signal.

After using the system bus again the external bus master will send a low signal thereby releasing back the system bus to the µp.


Minimum mode configuration


Minimum mode operationIn minimum mode configuration MN/MX = 1. It consists of 8284 clock generator and reset circuit.

Three octal latches (IC 8282) are used to demultiplex the address/data lines and 2 octal Transrecievers (IC 8286) are used as data amplifiers for the transmission of data to the memory or i/o from the microprocessor and vice versa.

In this mode all the control signals are given out by the microprocessor itself. Only a single processor is present on the system.

The latches gets activated when the ALE signal from the microprocessor goes high, which is connected to STB pin of the Octal latch. This allows address to be available on the output of latch after T1 state.

The Transrecievers have two pins OE and T. The Transrecievers gets activated when it receives a low signal on the OE pin from the microprocessor pin DEN.

To select the direction of data DT/R is connected to T pin of the Transrecievers. When DT/R = 1, direction of data is from the microprocessor and when DT/R = 0, direction of data is towards the microprocessor.


Minimum mode operationIC 8284 clock generator is used to provide clock signals (timing signals) to the processor so that all the operations of the processor are properly synchronized and work according to proper timing signals.

The wait state generator is used to add wait states when a slow peripheral is connected to the processor.


IC 8284 (CLOCK GENERATOR)


IC 8284 (CLOCK GENERATOR)The 8284 clock generator consist of a reset circuitry which is used to reset the microprocessor.

When a low signal is applied to the RES pin of the clock generator, it sends a high signal to the RESET pin of the processor thereby restarting the processor (reboot the processor).

Once booted a logic 1 appears on the RES pin as the capacitor gets charged to +5V.

PCLK i.e. peripheral clock is a timing signal to the peripherals to synchronize their operations with the processor.

The EFI i.e. external frequency input is used to determine the oscillator frequency.

RDY1 and RDY2 are input signals from the peripherals to the clock generator with RDY1 having higher priority.


74LS373 (Octal LATCH)


74LS373 (Octal LATCH)


An octal latch has 8 D-latches and 8 tri-state buffers.

The 74LS373 octal latch has two control inputs ‘Enable’ (G) and ‘Output Control’ (OC).

The ‘Enable’ control is active high and is connected to CLK inputs of all 8 latches. Logic I in CLK input will store the

logic levels present at D inputs in respective latches.

The ‘Output Control’ signal is active-low and is connected to control inputs of all 8 tri-state buffers.

Logic O will enable the buffers to output data from respective latches.

The latch Intel 8282 provides the same functionality of 74LS373

74LS245 (Bidirectional Buffer)


74LS245 (Bidirectional Buffer)We know that 74LS245 is a bi-directional 8-bit buffer.

It has DIR signal to control the direction of data flow.

When DIR signal is low, data flow is from B to A and when it is high data flow is from A to B.

It also has G signal which when activated enables the buffer.


8288 Bus controller


8288 Bus controllerThe input to the bus controller are the status signals from the microprocessor in maximum mode.

The output of the bus controller are the control signals which are required to drive the latches and Transrecievers.

Also control signals such as MEMR, MEMW, IORD, IOWR are also generated from the bus controller.

Which signals should be generated depends on the status of the status signals to the bus controller.

Refer S2,S1,S0 status signals section.


Maximum mode operation


Maximum mode operation As we can see in the block diagram, maximum mode is the mode in which more than one processor is present on the system.

Therefore the control signals required to drive the latches and Transrecievers are generated by the bus controller.

Also control signals required for memory selection and io selection are generated by the bus controller.

The bus controller gets status signals from the microprocessor which gives the status of the operation to be performed.

The remaining pins which were used in minimum mode but now their functionality changed in maximum mode are used for request and grant signals to other processors on the system.


Real mode memory addressing8086 operates in the real mode.

It is said to be real mode as it can access only 1 Mb of memory.

Assuming that we have 4 Gb address space, to address 4 Gb locations we require 32 addressing lines therefore = 4294967295 locations = 4 Gb locations

Therefore the address varies from:

00000000H to FFFFFFFFH i.e.

0000 0000 0000 0000 0000 0000 0000 0000 to

1111 1111 1111 1111 1111 1111 1111 1111

Now the first 1 Mb of memory is called as real memory.

8086 is therefore accessing the real memory and the way we will access the real memory is called as real mode addressing.


4 Gb address space

First 1 Mb memory called as real

memory

Memory segmentationThe µp 8086 has 20 address lines therefore the number of memory locations it can access is = 1048576 = 1 Mb.

Therefore consider the memory below:


0000 0001 (01H)

0010 0100 (24H)

-

-

-

-

1000 1111 (8FH)

1001 1010 (9AH)

Address varies from 0000 0000 0000 0000 0000 1111 1111 1111

1111 1111

We see that, at every location only 8 bit data is available and it is stored in hexadecimal form and has a 20 bit address.

This is the normal way of addressing a memory location

Memory segmentation cont.. Hence our main aim is to access a memory location and from there access data.

The main point to understand is that the size of the registers is 16 bit and address is of 20 bits.

Therefore 8086 uses the memory segmentation scheme to access 1 Mb of memory or 1M locations.

The memory is divided into segments or blocks. The length of the segment is 64 Kb. Two registers are used, one to access a segment and one to move within the segment (select any location within the segment).

Consider the memory given on next slide:

Two steps to access any location within a memory:

Select the segment with the help of base address or segment address (starting address (indicating start of any segment). It means that the segment register contains the base address with the help of which we are able to select any segment.

Now to move within the segment or in other words to select any location within the segment we use offset address or effective address which we get from any pointer or index register or any general purpose register.

Assume base address = 1000H and offset address = 0FFFH .


Memory segmentation cont..


0th segment

1st segment

2nd segment

--

8th segment

--

15th segment

Memory is divided into 16 segments and size of every segment is 64 Kb

= 65536 = 64 KbIt means that in every segment there are 65536 locations.Now we have such 16 segments and every segment has 65536 locations it means that the total size of the memory is = 65536 * 16 = 1048576 = 1 Mb

To move within the segment i.e. to select any

location within the segment first is to select the segment, second is to select any location within

the segment.

Base address = 1000H which selects the segment

64 Kb

Offset address = 0FFFHWhich selects the location 0FFFHAnd now from this location data will be accessed

Memory segmentationNow these segments are logical segments, it means that there is no physical division within the memory.

The segments can either be Code Segment, Data Segment, Extra Segment, Stack segment.

Code segment : when a segment is assigned as code segment, it is used to store the codes or instructions

Data Segment : when a segment is assigned as data segment, it is used to store data used in the programs

Extra Segment : it is also another segment used to store data

Stack Segment : it is used to store stack of data and address of main program during subroutine call


Physical address calculation


The No. of address lines in 8086 = 20Memory accessible = = 1Mb Scheme of accessing = Segmentation schemeMemory is divided into 64 Kb logical segments

Whenever a memory is accessed it is accessed using the physical addressIn segmentation scheme physical address is calculated using two addresses viz. base address or segment address and offset address or effective address.

Since address bus is of 20 bits = Physical address is of 20 bits

The calculation is as follows:Physical address (PA) = BA + EA

But addition is a little different, let’s understand with an example:

Physical address calculation cont..Let BA = 1000H = 0001 0000 0000 0000 = 16 bit

EA = 3467H = 0011 0100 0110 0111 = 16 bit

Now base address is always appended with 0H i.e. 0000B at the end internally by the µp.

Therefore BA now becomes = 0001 0000 0000 0000 0000 = 10000H

Now,

PA = 0001 0000 0000 0000 0000 = 10000H

+ 0011 0100 0110 0111 = 3467H

---------------------------------------------

0001 0011 0100 0110 0111 = 13467H

---------------------------------------------1/15/16 BY UBAID SAUDAGAR 154

20 bit physical address

Physical address calculation cont..Now this 20 bit physical address is used to access the memory as:


0000HFFFFH

-

-

0000HFFFFH

0000HFFFFH

0000HFFFFH

F000H

6000H

5000H

0000H

-

-

Physical address

varies from F0000H to FFFFFH

64 Kb

Physical address

varies from 00000H to 0FFFFH

Physical address varies from 00000H

To FFFFFH

Starting addresses

of segments

The base and the offset address can

also written as 4000 : 0023

Physical address calculation cont..In the above diagram we see that the base address are the starting address of any segment. Now that segment may be code segment, data segment, extra segment, stack segment.

Hence the base address may vary from 0000H to F000H.

The offset address is also called as displacement. This address is required to select any location within the segment which has been selected with the help of base address.

The offset address varies from 0000H to FFFFH which is equal to 64Kb.

0000 0000 0000 0000 = 0000H

0000 0000 0000 0001 = 0001H

-

1111 1111 1111 1111 = FFFFH

Imp: Since these are logical segments hence it is not necessary that base address should start at fixed locations like 1000H, 2000H…….F000H etc., it may start at any random location where memory is empty.


Between the first address and the last address there are

65536 locations = approx. 64K locations within a segment

Physical address calculation cont..Consider,

Base address : 1238H

Offset address : 01FFH

Therefore Physical address = 12380H

+ 01FFH

-------------- 1257FH ---------------One important point to notice is that the offset address varies as 0000H to FFFFH in every segment but the base address of every segment is different. Hence physical address varies due to base address.


Physical address calculation cont..Segment Register Starting address of segment Ending address of segment

2000H 20000H 2FFFFH

2001H 20010H 3000FH

2100H 21000H 30FFFH

AB00H AB000H BAFFFH

1234H 12340H 2233FH


Chapter No. 38086 Instruction Set


About assembly language programming

Assembly language is a low level language which uses mnemonics (opcodes) and operands (data) to write an instruction.

It does not consist of powerful statements like C, C++ which are said to be high level language.

To convert C codes to machine codes (also called as machine level language which only the machine understands) we require a compiler.

To convert an assembly language program to machine language which the machine i.e. µp understands an assembler is required.

An assembler is a software which converts assembly language program to machine language which the microprocessor understands.

Several assemblers are available for developing assembly language programs.

Turbo assembler and macro assembler are widely used tools.1/15/16 BY UBAID SAUDAGAR 160

Program development toolsEditor:

It is used to type the assembly language statements for the program and save in a file.

The file containing the text of assembly language statements is called as source file and has the extension .asm

Assembler:

Assembler is a software which reads the text from the source file, translates into respective binary codes and saves into two files, one with the extension .obj called as object file and one with the extension .lst called as list file.

The object file contains binary codes and addresses of the instructions.

The list file contains assembly language statements, binary codes, and offset address for each instruction.

Assembler indicates syntax errors if any in the source file.1/15/16 BY UBAID SAUDAGAR 161

Program development toolsLinker:

Large assembly language programs are usually divided into many small modules.

The code for each module are separately developed, tested and finally linked into a single large executable program.

The linker is used to join many object files into a single large object file.

Usually linker produces files with the extension .exe which can be directly loaded in the memory and executed.

Debugger:

It enables the program to be loaded into the system memory, execute it and then debug it. It also enables to insert breakpoints at any desired locations, change the content of register, memory location and rerun the program.


How does microprocessor differentiate between data and instruction and code?- As soon as µp is turned on it begins execution of code in the memory sequentially

- An instruction consists of two parts viz. opcode and operand

- Opcode tells the microprocessor what operation is to be performed.

- Operand is the one on which the operation is to performed.

- How the operation is to be performed we come to know from addressing mode.

- Both opcode and data (operand) present in the instruction are in binary numbers, so how to differentiate?

- The µp always interprets the first byte it fetches as opcode and second as operand

- Eg. We tell the processor that our program begins at 2000h. The first code it fetches is 3Eh, when it decodes it knows that it is a 2 byte instruction, hence the second code i.e. 32h is a data byte. If we enter 06h instead of 32h then it will load 06 in the accumulator instead of 32


Concept about instructionWhat is word length of a microprocessor?

The word length is defined as the no. of bits, the microprocessor recognizes and processes at a time.

As we know that our microprocessor is 16 bit microprocessor hence it recognizes 16 bits at a time.

What actually it means is that the number of binary combinations possible to pass on the data bus.

Since data bus is of 16 bits hence 16 binary bits can pass on it from :

0000 0000 0000 0000 Between these two limits the number of binary combinations are = 65536

1111 1111 1111 1111

Microprocessor understands only binary codes, now each 16 bit binary pattern is assigned a special task, or we can say that the design engineer assigns a special meaning to each 16 bit binary combination using logic ckts, this is called an instruction.

For Eg. 1111 1110 1100 1001 is an instruction which tells the microprocessor to decrement the content of CL register by one


Concept about instructionWhat is an instruction ?

An instruction is a binary pattern designed inside a microprocessor to perform a specific function.

The entire group of instructions that a microprocessor supports is called Instruction Set.

As explained above the different binary combinations are different instructions and a set of these binary combinations are called as the instruction set of that microprocessor.

Instructions tell us what operation is to be performed.

Now how to perform the operation, we come to know from addressing modes.


Addressing modesThe different addressing modes are as follows:

1. Implicit addressing mode

2. Register addressing mode

3. Immediate addressing mode

4. Direct addressing mode

5. Register Indirect addressing mode

6. Register relative addressing mode

7. Based indexed addressing mode

8. Relative based indexed addressing mode


Immediate addressing modeIn this addressing mode the data (operand) is present in the instruction itself.

Eg.

Mov bl,02h ;this instruction when executed will move immediate data 02h into the bl register

Assume bl = 9fh before execution

After execution bl = 02h

Mov ax,0000h ;this instruction when executed will move immediate data 0000h to the ax register

Assume ax = 121ah before execution

After execution ax = 0000h

Note:

Mov bl,1000h ;is not valid as 1000h is 16 bit and bl is a register which is 8 bits

Always in an instruction the source and the destination size should be same, if source is 8 bits in size, then destination should also be 8 bits.


Direct addressing modeIn this addressing mode, the data is present in the memory i.e. data segment, and the effective address (offset address) where the data is present is given in the instruction.

Eg.

Mov al, [1234h] ;this instruction moves the data present at the effective address 1234h

Assume data at the address ds : 1234 = 78h; al = 90h

After execution al = 78h

Note:

If we write : mov ax,1234h, this instruction will move 1234h into the ax register, hence to denote address in the instruction we write the address in square bracket [].


Register addressing modeIn this addressing mode the data is present in any of the general purpose registers.

The name of the register is mentioned in the instruction itself.

Eg:

Mov al,bl ;when this instruction is executed the contents of bl register is copied to al register

If bl = 55h; al = 00h

After instruction is executed

Bl = 55h; al = 55h

Note:

Mov ax,bl ;this instruction is not valid as the size in bits of source and destination is not same


Register indirect addressing modeIn this addressing mode the data is present in the memory location and the address of the memory location is not given directly in the instruction but instead is stored in some register pair or index register.

For this purpose three registers are used viz. BX, SI, and DI

Eg.

Mov al,[si] ;it moves the contents at the memory location whose offset address is given by si register

Assume al = 00h, si = 145Ah, now square brackets in the instruction indicates address

Let data at the location DS : 145AH = 34h

After execution al = 34h

Similarly we can write

Mov al,[bx] or mov al,[di] ;here bx and di hold offset address where data is present 1/15/16 BY UBAID SAUDAGAR 170

Register relative addressing modeIn this addressing mode, data is present in the memory and to access this data we require its address.

The address is not given directly in the instruction but instead is obtained by adding the contents of the base register and displacement or index register and displacement.

Eg.

Mov al,08[si] ; si = 1002h, displacement = 08, therefore effective address = 1002h + 08h = 100Ah

The above instruction will move the content at the memory location 100Ah which is formed by adding si and displacement in al register.

Assume data at location DS : 100Ah = 56h

Therefore al = 56h after execution

Note: The displacement may be 8 bit or 16 bit1/15/16 BY UBAID SAUDAGAR 171

Based indexed addressing modeIn this addressing mode the data is present in the memory whose address is not given directly in the instruction but instead is calculated by adding the contents in the base register and index register.

BX and BP are used as base registers and SI and DI are used as index registers.

Therefore effective address may be calculated using {BX or BP} + {SI or DI}

Eg.

Mov al,[bx][si] ;let bx = 1087h, si = 2004h, si+bx = 308Bh = effective address or offset address

This instruction will mov the data present at the memory location 308Bh into al register.

Let’s assume al = 09h before execution and address 308Bh contains 8Ch data.

After execution al = 8Ch


Relative based indexed addressing mode

In this addressing mode the data is present in the memory and to access this data, we require its address which is not present directly in the instruction.

The address is calculated by adding the contents in the base register and index register and displacement

Therefore {BX or BP} + {SI or DI} + {displacement}

Note: The displacement may be 8 bit or 16 bit

Eg.

Mov al,1234[si][bp] ;let si = 1000h, bp = 1200h, displacement = 1234h, EA = 1000h+1200h+1234h = 3434h

When the above instruction is executed the data at the location 3434h will be copied in the al register.

Assume al = 23h before execution, data at the location 3434h = 6Fh

After execution al = 6Fh1/15/16 BY UBAID SAUDAGAR 173

Implicit addressing modeIn the addressing mode the data is neither present in the instruction, nor any register, nor at any memory location.

There are some instructions which work on some registers directly such addressing scheme is called as implicit addressing mode.

Eg.

STI ;set interrupt flag

CLD ;clear direction flag

STD ;set direction flag

Note: In the above instructions there is no concern with data.


Assembler directivesAssembly language contains two types of statements

- Instructions

- Directives

Instructions are translated into machine code by the assembler which the processor understands by decoding them. In other words instructions, instruct the processor to do a particular task.

Directives are instructions to the assembler which help the assembler in the assembly process and these directives are not translated into machine code.

Or

Statements that direct the assembler to do some special task


DB, DW, DD, DQ, DT directives Cont..D B : define byte

The DB directive is used to reserve a byte or bytes of memory location in the available memory.

Eg: array DB 20h {assembler reserves 1 byte of memory for the variable named sum and initializes its value to 20}

array DB 0 {assembler reserves 1 byte of memory for the variable named sum and initializes it to 0}

D W : define word

The DW directive is used to reserve a word or words of memory location in the available memory.

Eg: array DW 1234h, 2345h, 3456h, 5678h {this makes the assembler reserve four words in memory(8 bytes), and initializes the words the specified values in the statements}


DB, DW, DD, DQ, DT directives cont..D Q : define Quad word

This directive is used to direct the assembler to reserve four words(8 bytes) of memory for the specified variable.

DD: define double word

This directive is used to direct the assembler to reserve two words(4 bytes) of memory for the specified variable.

D T : define ten bytes

This directive directs the assembler to define the specified variable requiring 10 bytes for its storage and initialize 10 bytes with the specified values.


DB, DW, DD, DQ, DT directives cont..Let's understand with the help of an example

Data segment

A db 20h,45,00101101B

B dw 2786h,34h,76,10011010B

C dd 12345678h,32h,6ABFh

D dq 12345678ABCD9876h

E dt 123456789ABCDE874590h

Data ends


Assume directive A S S U M E : assume logical segment name

The ASSUME directive is used to inform the assembler the names of the logical segments to be assumed for different segments used in the program.

Eg: the statement ASSUME CS : CODE directs the assembler that the machine code(executable program) are available in a segment named CODE and hence CS register is to be loaded with the segment address for the label CODE, while loading.

Eg: the statement ASSUME DS : DATA indicates to the assembler that the data items related to the program are available in the logical segment named DATA, and DS register is to be initialized by the segment address for data segment while loading


Segment, ends and end directiveS E G M E N T :

The segment directive is used to indicate the start of a logical segment.

Eg: CODE SEGMENT statement indicates to the assembler the start of a logical segment called CODE

Eg: DATA SEGMENT statement indicates to the assembler the start of a logical segment called DATA

E N D S : end of segment

This directive marks the end of a logical segment. The logical segments are assigned with the names using ASSUME directive. The names appear with the ENDS directive as prefix to mark the end of those particular segments.

Eg: DATA SEGMENT

.

.

DATA ENDS


Segment, ends and end directive cont.. E N D : end of program

The END directive marks the end of an assembly language program. When the assembler comes across the END directive, it ignores the source lines available later on.


Org (originate) directiveThis assembler directive is used to assign starting memory location of program.

If org is not given then program code will be stored from the effective address 0000h by default.

Eg: Org 1000h will start storing the program from the address 1000h.


EQU (equate) directiveThis directive tells the assembler to assign a value to the corresponding label.

This directive is used in general when we use some value ,multiple times in a program. So when we want to change the value. We need to go everywhere in the program where we entered the value.

Instead of that what we will do is that we will use equ directive and assign that value to a label through this directive as follows:

Count equ 100h

Now the above directive will assign value 100h to count label.

Now we use this count label many times in the main program. If we want to change its value, we will just change the value above once and not go to every instruction like others.


Proc and endp directiveThe subprogram called as subroutine is also called as procedure.

The proc and endp directive are used to indicate the start and of the subprogram or procedure.

Eg.

Add proc far/near

The above statement identifies the start of a procedure named add and tells the assembler that the procedure is far (different code segment)

Endp

The above directive indicates the end of procedure.


Even directive This assembler directive informs the assembler to increment the memory location counter to the next even address if it is not already at the even address.

As we know that mp 8086 is a 16 bit processor, therefore it can access a word in one machine cycle.

But the condition is that the word is located at an even address.

Eg:1005h is a word located at 2000h and 2001h

Which means that at:

2000h – 10h

2001h – 05h

Hence in such case we don’t require an even directive.

But if the word is located at 2001h and 2002h

Then it means that:

2001h – 10h

2002h – 05h

Now to access this word it will take 2 machine cycles, hence to reduce the access time we use even directive


Chapter No.3 Instruction set of 8086

Data transfer instructionsThis type of instructions are used to transfer 8/16 bit data from source to destination without any changes in the source operand and only changing the source operand.

Always in such cases size of source and destination should be same.


XSource

XDestinati

on

8/16 bits 8/16 bits

Arithmetic instructions Arithmetic instructions are used to perform different arithmetic operations in ALU.


XCHG D,S (Exchange the content of destination and source)

This instruction exchanges the content of source and destination. The size of the source and destination must be equal.

The possible combinations of source and destinations are as follows:

It is to be noted that immediate address is also not allowed

Eg: XCHG cl,[0004h]

Where 0004h is the offset address, but it is not allowed.

For examples refer class notes.


Destination Source

Register Register

Memory Register

DAA (decimal adjust AL after addition)This instruction is used to adjust the result and get the final result in packed bcd form.

Operation performed:

To add two 8 bit numbers and obtain the result in AL register.

Then execute DAA instruction

This instruction will check lower nibble of AL register (if > 9 or AC = 1, then it will add 06h to AL register)

Also it will check higher nibble of AL register (if > 9 or CF = 1, then it will add 60h to AL register)

If both the nibbles are greater than 9 then AL = AL + 66h

Eg: ADD AL, BL

DAA

Refer testing.asm


DAS (Decimal adjust AL after subtraction)

This instruction is used to adjust the result and get the final result in packed bcd form.

Operation performed:

To subtract two 8 bit numbers and obtain the result in AL register.

Then execute DAS instruction

This instruction will check lower nibble of AL register (if > 9 or AC = 1, then it will subtract 06h from AL register)

Also it will check higher nibble of AL register (if > 9 or CF = 1, then it will subtract 60h from AL register)

If both the nibbles are greater than 9 then AL = AL - 66h

Eg: SUB AL, BL

DAS

Refer testing.asm


AAA (ASCII adjust after addition) Need for ASCII adjust instructions: numerical data coming into a computer from a terminal is usually in ASCII code.

In this code numbers 0 to 9 are represented by the ASCII codes 30H to 39H.

The 8086 allows you to add two ASCII code for two decimal digits without masking off 3 in the upper nibble of each.

After the addition, the AAA instruction is used to make sure that the result is in correct unpacked BCD form.

Operation, assume AL = 0011 0101 = ASCII 5, BL = 0011 1001 = ASCII 9

Add al, bl = 35h + 39h = 6eh which is incorrect bcd form

AAA when executed – AL = AL AND 0fh = 6eh AND 0fh = 0eh

Since lower nibble of AL > 9, AL = AL + 06 = 14h

AH = AH + 1 = 00h + 01h = 01h (note that AH should be zero to get correct result)

AL = AL AND 0FH = 14h AND 0fh = 04h

Therefore AH = 01h AL = 04h i.e. correct unpacked bcd result


AAS (ASCII adjust after subtraction)Operation:

Sub al, bl ; let bl = 0011 1001 – 39h = ascii 9 ; let al = 0011 0101 – 35h = ascii 5

aas

If lower nibble of AL > 9 or AF = 1, then AL = AL – 06

AH = AH – 1

AF = CF =1

AL = AL AND 0F

This instruction corrects the result in AL register after subtracting two unpacked ASCII operands.

The result is unpacked bcd format.

Refer testint.asm


AAM (ASCII adjust for multiplication) This instruction can be used to convert the result of the multiplication of two valid unpacked BCD numbers

This instruction must be used after multiplication instruction. The multiplication of two unpacked BCD numbers is always less than 100, hence the result will be in AL and that number is finally then converted into unpacked BCD form.

Eg:

If AL = 06 (ASCII value is 30), BL=08 (ASCII value is 38)

MUL BL [ AX = 30H {48 decimal} ]

AAM [ AH = 04 AL = 08 ]


AAD (ASCII adjust before division)Syntax: AAD

This instruction is used before division instruction.

Eg:

If AX = 0205 and BL = 07

AAD ; after execution AX = 0019 (25 decimal)

DIV BL ; content of AX is divided by BL, after execution AL = 03 (quotient), AH = 04 (Remainder)


PUSH source, POP destinationPUSH:

This instruction pushes the content of the specified register or memory location on to the stack memory.

The PUSH instruction decrements the stack pointer by 2 and copies a word (16 bit only) into the stack segment which is pointed by the stack pointer.

Eg. PUSH BX

The above instruction decrements the SP by 2 and copies the contents of the BX register onto the stack memory

POP:

This instruction removes the contents from the stack memory i.e. from the stack segment and loads into the specified register or memory location

Eg. POP CX

The above instruction when executed, increments the stack pointer by 2 and removes a word from the stack segment and loads in the CX register

Refer chapter 2 for stack concept


PUSHF and POPFPUSHF:

The above instruction copies the flag register contents i.e. 16 bits into the stack segment pointed by the stack pointer.

So whenever the above instruction is executed the stack pointer decrements by 2 and the flag register contents are stored on to the stack.

POPF:

The above instruction copies the contents (16 bits) from the stack segment pointed buy the stack pointer into the flag register.

So whenever the above instruction is executed the stack pointer is incremented by 2 and contents on the stack memory are moved to the flag register.


Inc d (Increments the contents of destination)

This instruction increments the contents of destination by one and stores the result in the same destination.

The destination may be a register or any memory location but not any immediate data.

Eg:

Inc ch

Inc si ;where si consist of offset address

Refer testint.asm


Dec d (decrements the contents of the destination)

This instruction decrements the contents of the destination by one and stores the result in the destination again.

The destination may be a register or memory, but not an immediate data.

Eg.

Dec ch

Dec si

Refer testint.asm


XLATThe XLAT instruction is used to translate a byte from one code to another code

The byte to be translated is put in the AL register.

There is a lookup table where a number of bytes are present

Eg. Data segment

Table db 00h,01h,04h,09h,16h,25h

a db 04

Data ends

Code segment

Mov al, a

Lea bx, table

Xlat

Code ends


TEST D,S (AND destination and source data)

This instruction is used to AND the destination content and source content and the result is copied in the ALU, so the old value of destination is not lost.


S

D

ALU

Flag controlled instructionsSTC – Set carry flag

STD – Set direction flag

When direction flag sets, it sets the SI and DI flag in auto decrementing mode

STI – Set interrupt flag

When interrupt flag is set, the interrupts appearing on the INTR pin are enabled

CLC – Clear carry flag

CLD – Clear direction flag

When direction flag clears, it sets the SI and DI flag in auto incrementing mode

CLI – Clear interrupt flag

When interrupt flag is cleared, the interrupts appearing on the INTR pin are disabled

CMC – Complement the carry flag


MOV D,S (Move source data into destination)

In this instruction, content of the source is transferred or copied into the destination. The source value does not changes.

The possible combination of source and destination are:


Destination Source

Register Register

Memory Register

Register Memory

Register/memory Data

Register/memory Segment register

Segment register except CS

Register/memory

MOV D,S (Move source data into destination)

Register can be any 8/16 bit register.

But important is to notice that source and destination should be of same size.

Memory can be indicated by any method as discussed in the addressing modes.

For examples refer class notes


LAHF (Load AH register with 8 LSB’s of flag register)

This instruction copies the content of 8 LSB’s of flag register into AH register, so the old value of flag register is not lost but the old content of AH register is lost.

X may be 0 or 1

As reserved by Intel

Refer testint.asm


8 LSB’s of flag register

AH

SAHF (Store contents of AH register into 8 LSB’s of flag register)

This instruction copies the content of AH register into the 8 LSB’s of flag register, the content of AH register does not change but the content in the flag register changes.

X may be 0 or 1

As reserved by Intel

Refer testint.asm

Note: The value which is reserved by Intel cannot be changed or manipulated

Eg: if at bit no.6, Intel has reserved a 0, if we try to set that bit, it cannot be set.


AH 8 LSB’s of flag

register

LDS R,Memory (Load DS and given register {16 bit register} with memory contents )

This instruction copies the contents of four memory locations in the given register and DS register.

The register R should be of 16 bits. Refer testint.asm

Note: immediate addressing is also not allowed over here

Memory


R

DS

LES R,Memory (Load ES and given register {16 bit register} with memory contents )

This instruction copies the contents of four memory locations in the given register and ES register.

The register R should be of 16 bits. Refer testint.asm

Note: immediate addressing is also not allowed over here

Memory


R

ES

LEA R,EA(OA) (Load effective address of memory into given register R)

This instruction copies the effective address of memory into the given register R.

The register R should be of 16 bits, as EA or OA is always of 16 bits.

Refer testing.asm and class notes.


REA of

memory16 bit register

ADD D,S (Add destination and source data)

This instruction adds the source and destination and the result is stored in the destination itself.

The size of the source and destination is the same. The possible combinations of source and destination are given below.

Eg: add ax, bx

Add al, bl

Add dx, ds: [1234h]

Refer testing.asm


Destination Source

Register Register

Register Memory

Memory Register

Register/memory

Data

ADC D, S (Add destination and source data with carry)

This instruction adds the content of source and destination with carry and result is stored in the same destination.

The size of the source and destination should be same.

The possible combinations are same as add instruction.

Eg: adc ax, bx

adc al, bl


SUB D, S (subtract source from destination)

The subtract instruction subtracts the source from the destination and the result is stored in the destination.

The source operand and destination operand should be same.

The possibility of source and destination are:

Eg: sub ax, bx

sub al, bl

Refer testing.asm


Sbb d, s (subtract the content of source with borrow from the content of destination)

This instruction subtracts the contents of source with borrow from the content of destination and the result is stored in the destination itself.

The possible combinations are same as sub instruction

Eg: Sbb ax, bx

Sbb al, bl


Neg d (negate the contents of destination)

This instruction finds the 2’s complement of the contents of the destination and stores the result in the same destination.

The destination may be a register or memory but not an immediate data

Eg.

Neg ah

Neg cx

Neg si ; si register contains the memory address


Cbw (convert sign byte into sign word)This instruction converts the sign byte present in register AL into sign word and stores the result in register AX.

Eg.

Content of AL = 94h

So after cbw instruction is executed, the result in AX = FF94H

Note: FF is indicating the number is negative


Cwd (convert sign word into sign double word)

This instruction converts the sign word present in the ax register into sign double word and stores the result in register DX – AX

Eg.

Let AX = 1004H

Therefore after execution of cwd instruction

The result in the DX – AX register after execution = 0000H - 1004 H

Note that 0000h here indicates the number is positive


Mul s (8/16 bits) (unsigned multiplication of source data and accumulator)

This instruction is used multiply two 8/16 bit numbers and store the result in AX if result is 16 bit or DX-AX in case result is 32 bits

The source may be a memory or register, but not an immediate data.

Eg.

Mul bl

This instruction will multiply content in al (by default) and bl and store the result in ax register.

Mul bx

This instruction will multiply content in ax (by default) and bx and store the result in dx-ax register.

Refer testint.asm


IMUL S (8/16 bits) (integer multiplication of source data with accumulator)

This instruction is used for signed multiplication which multiplies source operand with the contents of accumulator and the result is stored in AX if result is 16 bits.

It is stored in DX-AX register if result is 32 bits.

Refer testint.asm


DIV S (Division)This instruction is used for division operation.

To understand this instruction let’s take some examples.

Eg1.

DIV BL

The above instruction will divide AX/BL by default (where AX is dividend and BL is divisor)

In other words, whenever the source operand is a byte, the dividend is a word i.e. AX register

After division we get a quotient and a remainder, quotient is stored in AL register and remainder is stored in AH register by default

Eg2.

DIV BX

The above instruction will divide DX-AX/BX by default (where DX-AX is dividend and BX is divisor)


Call operand (unconditional call)When the microprocessor executes the CALL instruction, then the microprocessor will branch from the main program to the subprogram (procedure (subroutine)).

If the main program and the sub program are present in the same code memory segment, then it is called as intra segment branching in which only the IP is changed i.e. the offset address changes but base address remains same.

But if the main program and the sub program are present in different code memory segments, then it is called as inter segment branching in which the contents of CS and IP are changed i.e. both base and offset address are changed.

There are two types of call i.e. Near CALL and Far CALL.

Near CALL is associated with intra segment branching and Far CALL is associated with Inter segment branching.

When the CALL instruction is executed, the control gets transferred to the sub program.


Call operand (unconditional call) cont..When the control gets transferred to the sub program the contents of the Instruction Pointer (IP) are changed only during Near CALL

When the control gets transferred to the sub program the contents of the Code Segment register (CS) and Instruction Pointer (IP) are changed only during Far CALL

When the control gets transferred to the sub program, the sub program gets executed and at the end of the program there exist a RET instruction which returns the control to the main program

Near CALL

Main program and Sub program = same code segment

When control transfers to sub program, next instruction’s address (offset present in IP) are saved on the stack and CS contents remain unchanged

The control gets transferred to subprogram by loading starting address of sub program in IP register

At the end of subprogram RET instruction gets executed due to which the address of the next instruction in the main program is loaded from the stack into the IP register and execution of main program continues.


Call operand (unconditional call) cont..Far CALL

Main program and Sub program = different code segment

When control transfers to sub program, next instruction’s address (offset present in IP and base address present in CS) are saved on the stack.

The control gets transferred to subprogram by loading base and offset of sub program in CS and IP register

At the end of subprogram RET instruction gets executed due to which the base and the offset address of the next instruction in the main program is loaded from the stack into the CS and IP register and execution of main program continues.


Call operand (unconditional call) cont..Main program Sub program (Procedure)

CALL

Y

HLT RET


Next instructio

n

String instructionsR E P

This is an instruction prefix, which can be used in string instructions.

It can be used with string instructions only.

It causes the instruction to be repeated CX no. of times.

After each execution the SI and DI registers are incremented or decremented by 1.

It is important that before we use the REP instruction prefix the following steps must be carried out:

1. CX must be initialized to count value.

2. If auto incrementing is required, DF must be cleared using CLD instruction else set using STD instruction.


String instructionsMOVSB/MOVSW{move string}

It is used to transfer a word/byte from data segment to extra segment

The offset of the source in the data segment is in SI

The offset of the destination in extra segment is in DI

Eg: MOVSB {ES:[DI] DS:[SI]…..byte transfer}

Eg: MOVSW {ES:[DI], ES:[DI+1] DS:[SI],DS:[SI+1]}


String instructions2 . L O D S B / L O D S W { L o a d s t r i n g }

It is used to load AL(or AX) register with a byte(or word) from data segment.

The offset of the source in data segment is in SI

Eg: LODSB { AL DS:[SI]….byte transfer}

Eg: LODSW {AL DS:[SI]; AH DS:[SI+1]}


String instructions3 . S T O S B / S T O S W { S t o r e s t r i n g }

It is used to store AL(or AX) in the extra segment.

The offset of the destination in extra segment is in DI

Eg: STOSB {ES:[DI] AL ….byte transfer}

Eg: STOSW {ES:[DI] AL; ES:[DI+1] AH}


String instructions4.CPMSB/CPMSW {Compare string}

It is used to compare a byte(or word) in the data segment with a byte or a word in the extra segment

The offset of the byte(or word) in extra segment is in DI, the offset of the byte(or word) in data segment is in SI

Comparison is done by subtracting the byte(or word) in extra segment from the byte(or word) in data segment

Flags are affected but result is not stored anywhere

Eg: CMPSB {compare DS[SI]with ES:[DI]….byte operation}

Eg: CMPSW {compare DS:[SI],DS:[SI+1] and ES:[DI],ES:[DI+1]}



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