Introduction to 8085_by_adi_ppt

Introduction
to
8085 Microprocessor
By
Prof. K. Adisesha (Ph. D)

Prof. K. Adisesha 2
8085 INTRODUCTION
Learning Objective:
Basic Concept and Ideas about Microprocessor.
Architecture of 8085
Addressing Modes and Instruction set of 8085
Interrupts of 8085
Peripheral Interfacings

Prof. K. Adisesha 3
Basic of Microprocessors
Differences between:
Microcomputer –a computer with a microprocessor as its
CPU. Includes memory, I/O etc.
Microprocessor –silicon chip which includes ALU, register
circuits & control circuits.
Microcontroller –silicon chip which includes microprocessor,
memory & I/O in a single package.

Prof. K. Adisesha 4
Definition of the Microprocessor:
 The microprocessor is a programmable device that takes in
numbers, performs on them arithmetic or logical operations
according to the program stored in memory and then produces
other numbers as a result.
 Programmable device: The microprocessor can perform different sets of operations
on the data it receives depending on the sequence of instructions supplied in the
given program. By changing the program, the microprocessor manipulates the data
in different ways.
 Instructions: Each microprocessor is designed to execute a specific group of
operations. This group of operations is called an instruction set. This instruction set
defines what the microprocessor can and cannot do.

Prof. K. Adisesha 5
Inside The Microprocessor:
 Internally, the microprocessor is made up of 3 main units.
 The Arithmetic/Logic Unit (ALU)
 The Control Unit.
 An array of registers for holding data while it is being manipulated.

Prof. K. Adisesha 6
Memory:
 Memory stores information such as instructions and data in binary
format (0 and 1). It provides this information to the microprocessor
whenever it is needed.
 Memory Map and Addresses.
 The registers inside the microprocessor
 Read Only Memory (ROM)
 Random Access Memory (RAM)

Prof. K. Adisesha 7
To execute a program:
 The user enters its instructions in binary format into the memory. The
microprocessor then reads these instructions and whatever data is
needed from memory, executes the instructions and places the results
either in memory or produces it on an output device.
 To execute a program, the microprocessor “reads” each instruction from
memory, “interprets” it, then “executes” it.
 To use the right names for the 3 cycles:
 Instruction Fetch Cycle
 Instruction Decode Cycles
 Instruction executes Cycle
 This sequence is continued until all instructions are performed.

Prof. K. Adisesha 8
Machine Language:
 The number of bits that form the “word” of a microprocessor is fixed for
that particular processor.
 These bits define a maximum number of combinations.
 For example an 4-bit microprocessor can have at most 24=16 different combinations.
 The 8085 (from Intel) is an 8-bit microprocessor.
 The 8085 uses a total of 246 bit patterns to form its instruction set.
 These 246 patterns represent only 74 instructions.
 The reason for the difference is that some (actually most) instructions have multiple
different formats.
 Because it is very difficult to enter the bit patterns correctly, they are
usually entered in hexadecimal instead of binary

Prof. K. Adisesha 9
Assembly Language:
 Entering the instructions using hexadecimal is quite easier than entering
the binary combinations. but it still is difficult to understand what a
program written in hexadecimal does.
 So, each company defines a symbolic code for the instructions.
 These codes are called “mnemonics”.
 The mnemonic for each instruction is usually a group of letters that suggest the
operation performed..
 Example:
 00111100 translates to 3C in hexadecimal (OPCODE)
 Its mnemonic is: “INR A”.
 INR stands for “increment register” and A is short for accumulator.

Prof. K. Adisesha 10
:

:

8085 INTRODUCTION
8085 MP Architecture:
 8-bit general purpose μp
 Capable of addressing 64 k of memory
 Has 40 pins
 Requires +5 v power supply
 Can operate with 3 MHz clock
 8085 upward compatible

8085 INTRODUCTION
The features of INTEL 8085 are :
It is an 8 bit processor.
It is a single chip N-MOS device with 40 pins.
It has multiplexed address and data bus.(AD0-AD7).
It works on 5 Volt dc power supply.
The maximum clock frequency is 3 MHz while minimum
frequency is 500kHz.
It provides 74 instructions with 5 different addressing modes.

The features of INTEL 8085 are :
It provides 16 address lines so it can access 2^16 =64K bytes of
memory.
It generates 8 bit I/O address so it can access 2^8=256 input ports.
It provides 5 hardware interrupts: TRAP, RST 5.5, RST 6.5, RST
7.5,INTR.
It provides Acc ,one flag register ,6 general purpose registers and
two special purpose registers(SP,PC).
It provides serial lines SID ,SOD. So serial peripherals can be
interfaced with 8085 directly.
8085 INTRODUCTION

PIN DIAGRAM -8085

 READY:This an output signal used to check the status of output
device.If it is low, µP will WAIT until it is high.
 TRAP:It is an Edge triggered highest priority , non mask able
interrupt. After TRAP, restart occurs and execution starts from
address 0024H.
 RST5.5,6.5,7.5:These are maskable interrupts and have low
priority than TRAP.
 INTR¯&INTA:INTR is a interrupt request signal after which µP
generates INTA or interrupt acknowledge signal.
 IO/M¯:This is output pin or signal used to indicate whether 8085
is working in I/O mode(IO/M¯=1) or Memory mode(IO/M¯=0 ).
PIN DIAGRAM -8085

 HOLD&HLDA:HOLD is an input signal .When µP receives HOLD
signal it completes current machine cycle and stops executing next
instruction. In response to HOLD µP generates HLDA that is HOLD
Acknowledge signal.
 RESET IN¯:This is input signal. When RESET IN¯ is low µp restarts
and starts executing from location 0000H.
 SID: Serial input data is input pin used to accept serial 1 bit data .
 X1X2 :These are clock input signals and are connected to external
LC,or RC circuit. These are divide by two so if 6 MHz is connected
to X1X2, the operating frequency becomes 3 MHz.
 VCC&VSS: Power supply VCC=+ -5Volt& VSS=-GND reference.
PIN DIAGRAM -8085

ARCHITECTURE-8085

Accumulator: It is 8 bit general purpose register.
◦ It is connected to ALU.
◦ So most of the operations are done in Acc.
Temporary register: It is not available for user
◦ All the arithmetic and logical operations are done in the temporary
register but user can’t access it.
Flag: It is a group of 5 flip flops used to know status of various
operations done.
◦ The Flag Register along with Accumulator is called PSW or
Program Status Word.
Arithmetic and Logical group

Flag Register is given by:
 S: Sign flag is set when result of an operation is negative.
 Z: Zero flag is set when result of an operation is 0.
 Ac: Auxiliary carry flag is set when there is a carry out of lower
nibble or lower four bits of the operation.
 CY: Carry flag is set when there is carry generated by an
operation.
 P: Parity flag is set when result contains even number of 1’s.
Rest are don’t care flip flops.
S Z X AC X P X CY
Arithmetic and Logical group

Register Group

 Temporary registers (W,Z):These are not available for user.
These are loaded only when there is an operation being
performed.
 General purpose:There are six general purpose registers in
8085 namely B, C, D, E, H, L. These are used for various
data manipulations.
 Special purpose :There are two special purpose registers in
8085:
 SP :Stack Pointer.
 PC: Program Counter.
Register Group

Stack Pointer: This is a temporary storage memory 16 bit register.
Since there are only 6 general purpose registers, there is a need to
reuse them .
 Whenever stack is to be used previous values are PUSHED on
stack and then after the program is over these values are POPED
back.
Program Counter: It is 16 bit register used to point the location from
which the next instruction is to be fetched.
 When a single byte instruction is executed PC is automatically
incremented by 1.
 Upon reset PC contents are set to 0000H and next instruction is
fetched onwards.
Register Group

INSTRUCTION REGISTER,DECODER & CONTROL
 Instruction register:When an instruction is fetched , it is
executed in instruction register.This register takes the Opcode
value only.
 Instruction decoder: It decodes the instruction from instruction
register and then to control block.
 Timing and control:This is the control section of µP.It accepts
clock input .
DECODER & CONTROL

INTERRUPT CONTROL
 It accepts different interrupts like TRAP INT5.5, 6.5, 7.5 and
INTR.
SERIAL IO CONTROL GROUP
 It is used to accept the serial 1 bit data by using SID and SOD
signals and it can be performed by using SIM & RIM
instructions.
DECODER & CONTROL

Classification of Instruction Set of 8085
The Instruction Set of 8085 can be categorized into five different
groups based on the nature of function of the instructions–
Data transfer operations
Arithmetic operations.
Logical operations
Branch operations and
Stack, Input / Output and Machine control operations
INSTRUCTIONS SET OF 8085

DATA TRANSFER GROUP
• Using mnemonics without any alteration in the content,
data can be transferred in four different cases –
From one register to another register
From the memory to the register
Move Immediate data to Register
From the register to the memory
INSTRUCTIONS SET OF 8085

From one register to another register
MOV Rd, Rs.(Move data from Rs to Rd).
Example:
MOV C,B. Move the content of register B to C.
Initially After execution
B=10H. B=10H.
C=20H. C=10H.
• Flags Affected :No flags affected.
• Addressing mode: Register.
DATA TRANSFER GROUP

From the memory to the register
MOV Rd, M (Move data from Memory to Rd).
Example:
MOV C,M. Move the content of Memory i.e. “H or L” to C.
Suppose the Data at memory pointed By HL pair at C200H is 10H.
H=C2,L=00,C=30H H=C2,L=00,C=10H.
Flags Affected :No flags affected.
Addressing mode: Indirect.
DATA TRANSFER GROUP

Move Immediate data to Register
MVI R, Data.(Move Immediate data to Register).
Example:
MVI B, 30H. (Move the data 30 H to Register B)
B=40H B=30H
◦ Flags Affected :No flags affected.
◦ Addressing mode: Immediate.
DATA TRANSFER GROUP

LXI Rp,16 bit .(Load 16 bit data to Register pair
Immediate).
Example:
LXI SP, C200H. (Load Stack pointer with C200H).
SP=C800H SP=C200H.
Flags Affected :No flags affected.
Addressing mode: Immediate.
DATA TRANSFER GROUP

From the register to the memory
STA address.(Store Acc data to address).
Example:
STA C200H. (Move the data from Acc to C200H).
Suppose in Acc the data is 10H.
A=10H, C200=20H C200=10H , A=10H
◦ Addressing mode: Direct.
DATA TRANSFER GROUP

LHLD address.(Load HL pair with data from address).
Example:
LHLD C200H. (Move the data from C200 to HL pair).
Suppose at C200 the data is 20H,30H .
H=10H,L=20H H=20H,L=30H.
C2=20H,00=30H C2=20H,00=30H
DATA TRANSFER GROUP

XCHG (Exchange the data from HL pair to DE pair)
Example : XCHG
H=20H,L=30H, H=40H,L=70H.
D=40H,E=70H. D=20H,E=30H.
◦ Addressing mode: Register.
DATA TRANSFER GROUP

IN 8 bit address (Move the data from address to Acc)
Example: IN 80H
Move the data from 80H port address to Accumulator.
Suppose data at 80H is 39H.
A=20H. A=39H
DATA TRANSFER GROUP

OUT 8 bit address (Move the data from Acc to address)
Example: OUT 80H
Move the data from Acc to port address 80H.
Suppose data at Acc is 39H.
A=39H, 80=10H. A=39H, 80=39H.
DATA TRANSFER GROUP

 Example:
 Write a program to exchange contents of memory location
D000H to D001H
LDA D000H Load Acc with data from D000
MOV B,A Move the data to B
LDA D0001H Load Acc with data from D001
STA 2000H Store Acc data at D000
MOV A,B Move B’s data to A
STA 2001H Store data from D000 to D0001
RST1 Stop.
DATA TRANSFER GROUP

Addressing modes in 8085 can be classified into 5 groups −
Immediate addressing mode
Register addressing mode
Direct addressing mode
Indirect addressing mode
Implied addressing mode
ADDRESSING MODES OF 8085

Immediate addressing:
Immediate data is transferred to address or register.
Example:
MVI A,20H. Transfer immediate data 20H to accumulator.
Number of bytes:
Either 2 or 3 bytes long.
1st byte is opcode.
2nd byte 8 bit data .
3rd byte higher byte data of 16 bytes.

Register addressing:
Data is transferred from one register to other.
Example:
MOV A, C :Transfer data from C register to accumulator.
Number of bytes:
Only 1 byte long.
One byte is opcode.

Direct addressing:
 Data is transferred from direct address to other register or
vice-versa.
Example:
LDA C200H .Transfer contents from C200H to Acc.
Number of bytes:
These are 3 bytes long.
1st byte is opcode.
2nd byte lower address.
3rd byte higher address.

Indirect addressing:
Data is transferred from address pointed by the data in a
register to other register or vice-versa.
Example:
MOV A, M: Move contents from address pointed by M to Acc.
Number of bytes:
These are 3 bytes long.
1st byte is opcode.
2nd byte lower address.
3rd byte higher address.

Implied addressing:
 These doesn’t require any operand. The data is specified in
Opcode itself.
Example: RAL: Rotate left with carry.
No. of Bytes:
These are single byte instruction or Opcode only.

ADD R (ADD register content with Acc and result in A ).
Example:
ADD C. (ADD the content of C with A).
Suppose the Data at C register is 10H.
C= 10H ,A=10H A=20H,C=10H.
◦ Flags Affected : All flags are modified.
◦ Addressing mode: Register
ARITHMETIC GROUP

ADD M(ADD H or L Reg content with Acc and result in A ).
Example:
ADD M. (ADD the content of HL with A).
Suppose the Data at memory pointed by HL register 1020H is 10H.
H= 10H ,L=20H . H=10H,L=20H.
A=20H,C=10H. A=30H.
o Flags Affected :All flags are modified.
o Addressing mode: Register Indirect.
ARITHMETIC GROUP

ADI Data (ADD immediate data with Acc and result in A ).
Example:
ADI 30H. (ADD 30H with A).
A=20H, A=50H.
◦ Flags Affected :All flags are modified.
ARITHMETIC GROUP

ADC R (ADD register content with Acc and carry and result in A ).
Example:
ADC C. (ADD the content of C with A with carry).
Suppose the Data at C register is 10H and carry is 01H.
. C= 10H ,A=10H A=21H,C=10H.
ARITHMETIC GROUP

Example: Write a program to perform 16 bit addition of 1234H &
4321H. Store answer at H & L registers.
MVI B,21H B=21H
MVI A,34H A=34H
MVI C,43H C=43H
MVI D,12H D=12H
ADD B A=34+21H
MOV L,A L=55H
MOV A,C A=43H
ADC D A=43+12H
MOV H,A H=55H
RST1 STOP.
ARITHMETIC GROUP

SUB R (Subtract register content from Acc and result in A ).
Example:
SUB B. (Subtract the content of B from A ).
Suppose the Data at B register is 10H .
B= 10H ,A=20H A=10H,B=10H.
ARITHMETIC GROUP

SBB R (Subtract register content from Acc with borrow and result in A ).
Example:
SBB B. (Subtract the content of B from A with borrow).
Suppose the Data at B register is 10H and borrow is 01H .
B= 0FH ,A=20H A=10H,B=0FH.
ARITHMETIC GROUP

SUI Data(Subtract immediate data from Acc and result in A ).
Example:
SUI 30H. (Subtract 30H from A).
A=80H, A=50H.
◦ Flags Affected : All flags are modified.
◦ Addressing mode: Immediate
ARITHMETIC GROUP

Example: Subtract data of C800 H from C200H.Store the result
at 2C00.
LDA C800H
MOV B,A
LDA C200H
SUB B
STA 2C00H
RST1
ARITHMETIC GROUP

DAD Rp (Add specified register pair with HL pair)
Example: DAD D.(Add the content of E with L and that of D with
H register and result in HL pair)
 Suppose the content of HL pair is H=20H ,L=40H and DE pair
is D=30H, E=10H.
H=20H ,L=40H H=50H ,L=50H
D=30H, E=10H D=30H, E=10H
◦ Flags Affected :Only carry flag is modified.
ARITHMETIC GROUP

DAA (Decimal adjust accumulator)
Example:
MVI A,12H
ADI 39H
DAA .
This instruction is used to store result in BCD form.If lower
nibble is greater than 9 ,6 is added while if upper nibble is
greater than 9,6 is added to it to get BCD result.
12+39=4B 12+39=51 in BCD form.
• Flags Affected : All flags are modified.
• Addressing mode: Register
ARITHMETIC GROUP

INR R (Increment register content by 1 ).
Example:
INR C. (Increment the content of C by 1).
C= 10H C=11H.
 Flags Affected :All flags are modified except carry flag.
 Addressing mode: Register.
ARITHMETIC GROUP

DCR R (Decrement register content by 1 ).
Example:
DCR C. (Decrement the content of C by 1).
C= 10H C=0FH.
 Flags Affected :All flags are modified except carry flag.
ARITHMETIC GROUP

INX Rp (Increment register pair content by 1 ).
Example:
INX SP (Increment the content of Stack pointer pair by 1).
INX B. (Increment the content of BC pair by 1).
Suppose the Data at BC register is 1010H and SP is C200H
BC= 1010H BC=1011H.
SP=C200H SP=C201H.
 Flags Affected :No flags are modified.
ARITHMETIC GROUP

ANA R (Logically AND register content with Acc and result in A ).
Example:
ANA C (AND the content of C with A).
C= 10H ,A=10H A=10H,C=10H.
 Flags Affected :S,Z,P are modified Cy=reset, AC=set.
LOGICAL GROUP

ANI Data (Logically AND immediate data with Acc and result in
A )
Example:
ANI 10H (AND 10H with A).
A=10H A=10H
◦ Flags Affected :S,Z,P are modified Cy=reset, AC=set.
LOGICAL GROUP

ORA R (Logically OR register content with Acc and result in A5 )
Example:
ORA C (OR the content of C with A).
C= 17H ,A=10H A=17H,C=17H.
◦ Flags Affected :S,Z,P are modified Cy=reset, AC=reset.
LOGICAL GROUP

ORI Data (Logically OR immediate data with Acc and result in A ).
Example:
ORI 10H (OR 10H with A).
A=30H A=30H
◦ Flags Affected :S,Z,P are modified Cy=reset,AC=set.
LOGICAL GROUP

XRA R (Logically XOR register content with Acc and result in A ).
Example:
XRA C (XOR the content of C with A).
C= 17H ,A=10H A=07H,C=17H.
◦ Flags Affected :S,Z,P are modified Cy=reset, AC=reset.
LOGICAL GROUP

CMP R (Compare register content with Acc and result in A ).
Example:
CMP C (Compare the content of C with A).
C= 10H ,A=17H A=17H,C=17H.
◦ Flags Affected :S=0,Z=0,P=0, Cy=reset, AC=reset.
LOGICAL GROUP

CPI Data (Compare immediate data with Acc ).
Example:
CPI 10H (Compare the content of C with A).
A=17H A=17H.
Flags Affected :S=0,Z=0,P=0, Cy=reset,AC=reset.
Addressing mode:Immediate.
LOGICAL GROUP

RLC (Rotate accumulator left ).
Example:
MOV A,03H.
RLC (Rotate accumulator left).
A=03H A=06H.
◦ Flags Affected :Only carry flag is affected.
◦ Addressing mode: Implied.
LOGICAL GROUP

RAL (Rotate accumulator left with carry ).
Example:
MOV A,03H.
RAL (Rotate accumulator left with carry).
A=03H , carry =01H A=07H.
Flags Affected :Only carry flag is affected.
Addressing mode: Implied.
LOGICAL GROUP

RRC (Rotate accumulator right ).
Example:
MOV A,03H.
RRC (Rotate accumulator right).
A=03H , A=81H.
◦ Flags Affected :Only carry flag is affected.
◦ Addressing mode: Implied.
LOGICAL GROUP

Write a program to reset last 4 bits of the number 32H
Store result at C200H.
MVI A, 32H A=32H
ANI F0H 00110010 AND 1111000
=00110000=30H
STA C200H. C200=30H
RST1 Stop
LOGICAL GROUP

JMP address(Unconditional jump to address)
Example:
JMP C200H.
 After this instruction the Program Counter is loaded with this
location and starts executing and the contents of PC are
loaded on Stack.
 Flags Affected :No Flags are affected.
 Addressing mode: Immediate.
BRANCH GROUP

CALL address(Unconditional CALL from address)
Example:
CALL C200H.
 After this instruction the Program Counter is loaded
with this location and starts executing and the contents
of PC are loaded on Stack.
◦ Flags Affected :No Flags are affected.
◦ Addressing mode: Immediate
BRANCH GROUP

Conditional Jump Instructions.
 JC (Jump if Carry flag is set)
 JNC (Jump if Carry flag is reset)
 JZ (Jump if zero flag set)
 JNZ (Jump if zero flag is reset)
 JPE (Jump if parity flag is set)
 JPO (Jump if parity odd or P flag is reset )
 JP (Jump if sign flag reset )
 JM (Jump if sign flag is set or minus)
BRANCH GROUP

Conditional Call Instructions.
 CC (Call if Carry flag is set)
 CNC (Call if Carry flag is reset)
 CZ (Call if zero flag set)
 CNZ (Call if zero flag is reset)
 CPE (Call if parity flag is set)
 CPO (Call if parity odd or P flag is reset )
 CP (Call if sign flag reset )
 CM (Call if sign flag is set or minus)
BRANCH GROUP

RET (Return from subroutine)
Example:
MOV A,C
RET
 After this instruction the Program Counter POPS PUSHED
contents from stack and starts executing from that address .
Flags Affected :No Flags are affected.
Addressing mode:Register indirect .
BRANCH GROUP

RST (Restart instruction)
Example:
MOV A,C
RST 1.
 After this instruction the Program Counter goes to address
0008H and starts executing from that address .
◦ Flags Affected :No Flags are affected.
◦ Addressing mode: Register indirect.
BRANCH GROUP

The addresses of the respective RST commands are:
Instruction Address
RST 0 0000H
RST 1 0008H
RST 2 0010H
RST 3 0018H
RST 4 0020H
RST 5 0028H
RST 6 0030H
RST 7 0038H
BRANCH GROUP

The stack is an area of memory identified by the programmer for
temporary storage of information.
 The stack is a LIFO structure.
– Last In First Out.
 The stack normally grows backwards into memory.
 In the 8085, the stack is defined by setting the
SP (Stack Pointer) register.
• LXI SP, FFFFH
STACK AND MACHINE CONTROL

Saving Information on the Stack.
 Information is saved on the stack by PUSHing it on.
 It is retrieved from the stack by POPing it off.
 The 8085 provides two instructions:
◦ PUSH
◦ POP
 for storing information on the stack and retrieving it back.
 Both PUSH and POP work with register pairs only.

PUSH Rp. (Push register pair contents on stack).
Example: LXI SP FFFFH.
PUSH H. (Move the content of HL pair on Stack).
 Suppose at HL pair the data is H= 20H,L= 30H & SP is
initialized at FFFFH
H=20H,L=30H H=20H,L=30H.
SP=FFFF H FFFD=30H,FFFE=20H

POP Rp. (Pop register pair contents from stack).
Example: POP D(POP the content of DE pair from Stack).
 Suppose at DE pair the data is H= 20H,L= 30H SP was
initialized at FFFFH
D=20H,E=30H D=10H,E=80H.
FFFD=80H,FFFE=10H
◦ Addressing mode: Register indirect

XTHL (Exchange HL register pair contents with top of stack).
Example: XTHL(Exchange top with HL pair).
 Suppose at HL pair the data is H= 20H,L= 30H & SP =FFFFH
& at locations FFFF=10H and at FFFE= 80H.
H=20H,L=30H H=10H,L=80H.
SP=FFFF =10H,FFFE=80H FFFD=20H,FFFE=30H

 Write a program to transfer a block of data from C550H to C55FH.
Store the data from C570H to C57FH .
LXI H ,C550H
LXI B ,C570H
MVI D,0FH
UP MOV A,M
STAX B
INX H
INX B
DCR D
JNZ UP
RST1
PROGRAM

 Find out errors in the following :
 MVI B,D =Immediate addressing doesn’t have register as operand
.Therefore, MVI B,80H.
 INX L=Increment operator always acts on the higher memory address
in register pair .Thus ,INX H.
 JP 80H = Conditional jump instructions doesn’t have any immediate
operand .Thus, JP UP.
If Flag contents are AB H, what is flag status
If flag contains AB H then it’s values from D7 to D0 are
10101011.
By comparing it with flag register we get S=1,Z=0,AC=0,
P=0,Cy=1.
PROGRAM

Write a program to add 10 data bytes. Data is stored from locations C200.
Store result at C300H.
LXI H,C200 H
MVI C, 0A H
UP MVI A,00 H
MOV B,M
ADD B
INX H
DCR C
JNZ UP
STA C300H
RST1.
PROGRAM

 The µP operates with reference to clock signal.The rise and
fall of the pulse of the clock gives one clock cycle.
 Each clock cycle is called a T state and a collection of
several T states gives a machine cycle.
 Important machine cycles are :
1. Op-code fetch.
2. Memory read.
3. Memory write.
4. I/Op-read.
5. I/O write.
TIMING AND STATE DIAGRAM

Op-code Fetch: It basically requires 4 T states from T1-T4
 The ALE pin goes high at first T state always.
 AD0-AD7 are used to fetch OP-code and store the lower byte of
Program Counter.
 A8-A15 store the higher byte of the Program Counter while IO/M¯
will be low since it is memory related operation.
 RD¯ will only be low at the Op-code fetching time.
 WR¯ will be at HIGH level since no write operation is done.
 S0=1,S1=1 for Op-code fetch cycle.

Op-code fetch cycle :

Memory Read Cycle: It basically requires 3T states from T1-T3
 AD0-AD7 are used to fetch data from memory and store the
lower byte of address.
 A8-A15 store the higher byte of the address while IO/M¯
will be low since it is memory related operation.
 RD¯ will only be low at the data fetching time.
 WR¯ will be at HIGH level since no write operation is done.
 S0=0,S1=1 for Memory read cycle.

Memory write Cycle: It basically requires 3T states from T1-T3 .
 AD0-AD7 are used to fetch data from CPU and store the lower
byte of address.
 A8-A15 store the higher byte of the address while IO/M¯ will
be low since it is memory related operation.
 RD¯ will be HIGH since no read operation is done.
 WR¯ will be at LOW level only when data fetching is done.
 S0=1,S1=0 for Memory write cycle.

Calculation of Delay using 8 bit counter:
 Consider following example:
MVI C, count(8 bit) H 7 T states
UP DCR C 4 T states
JNZ UP 10/7 T
RET 10T
 Here loop UP is executed (N-1) times.
 Thus delay is
Td=M+[(count)x N) -3.
 Where M= no.of T states outside loop.
N=no.of T states inside loop.
SUBROUTINE

 Here value of M= 17, N= 14.
 The maximum delay will occur if count is 255 or FF H.
 Thus Td max =17+[255x14]-3= 3584 T states.
 For 0.5 µsec delay for a T state, we get
 Td max=0.5 µsec x 3584= 1792 µsec or 1.792 m sec.
SUBROUTINE

 Generally µP 8085 can address 64 kB of memory .
 Generally EPROMS are used as program memory and RAM
as data memory.
 We can interface Multiple RAMs and EPROMS to single µP .
 Memory interfacing includes 3 steps :
1.Select the chip.
2.Identify register.
3.Enable appropriate buffer.
8085 Memory Interfacing

 Example: Interface 2Kbytes of Memory to 8085 with starting
address 8000H.
Initially we realize that 2K memory requires 11 address lines
(2^11=2048). So we use A0-A10 .
 Write down A15 –A0
A15
14 13 12 11 10 9 8 7 6 5 4 3 2 1 0
1
1
0
0
0
0
0
0
0
0
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
ADD
8000H
87FFH

 Address lines A0-A10 are used to interface memory while
A11,A12,A13,A14,A15 are given to 3:8 Decoder to provide an
output signal used to select the memory chip CS¯or Chip select
input.
 MEMR¯ and MEMW¯are given to RD¯and WR¯pins of Memory
chip.
 Data lines D0-D7 are given to D0-D7 pins of the memory chip.
 In this way memory interfacing can be achieved.

 The diagram of 2k interfacing is shown below:
A15-A8
Latch
AD7-AD0
D7- D0
A7- A0
8085
ALE
IO/M
RD
WR
2K Byte
Memory
Chip
WR
RD
CS
A10- A0
A15- A11
3:8DECODER

 In this example we saw that some address lines are used for
interfacing while others are for decoding.
 It is called absolute decoding.
 We sometimes don’t requires that many address lines. So
we ignore them. But this may lead to shadowing or multiple
address.
 This type of decoding is called linear decoding or partial
decoding.
 In partial decoding wastage of address takes place but it
requires less hardware and cost is also less as compared
with absolute one. Prof. K. Adisesha 97

PA0-PA7 I/O Port A Pins
PB0-PB7 I/O Port B Pins
PC0-PC7 I/O Port C Pins
D0-D7 I/O Data Pins
RESET I Reset pin
RD¯ I Read input
WR ¯ I Write input
A0-A1 I Address pins
CS ¯ I Chip select
Vcc , Gnd I +5volt supply
8255 PIN DIAGRAM

8255 BLOCK DIAGRAM

 Data Bus Buffer: It is an 8 bit data buffer used to interface
8255 with 8085. It is connected to D0-D7 bits of 8255.
 Read/write control logic:It consists of inputs
RD¯,WR¯,A0,A1,CS¯ .
 RD¯,WR¯ are used for reading and writing on to 8255 and
are connected to MEMR¯,MEMW¯ of 8085 respectively.
 A0,A1 are Port select signals used to select the particular
port .
 CS ¯ is used to select the 8255 device .
 It is controlled by the output of the 3:8 decoder used to
decode the address lines of 8085.
8255 BLOCK DIAGRAM

A1 A0 Selected port
0 0 Port A
0 1 Port B
1 0 Port C
1 1 Control Register
A0,A1 decide the port to be used in 8255.
8255 BLOCK DIAGRAM

 Group A and Group B Control:
 Group A control consists of Port A and Port C upper.
 Group B control consists of Port A and Port C lower.
 Each group is controlled through software.
 They receive commands from the RD¯, WR¯ pins to allow
access to bit pattern of 8085.
 The bit pattern consists of :
1. Information about which group is operated.
2. Information about mode of Operation.
8255 BLOCK DIAGRAM

 PORT A, B: These are bi-directional 8 bit ports each and are
used to interface 8255 with CPU or peripherals.
 Port A is controlled by Group A while Port B is controlled by
Group B Control.
 PORT C: This is a bi-directional 8 bit port controlled partially
by Group A control and partially by Group B control .
 It is divided into two parts Port C upper and Port C lower each
of a nibble.
 It is used mainly for control signals and interfacing with
peripherals.
8255 BLOCK DIAGRAM

 Mode 0 : Simple I/O
• Any of A, B, CL and CH can be programmed as input or output
 Mode 1: I/O with Handshake
• A and B can be used for I/O
• C provides the handshake signals
 Mode 2: Bi-directional with handshake
• A is bi-directional with C providing handshake signals
• B is simple I/O (mode-0) or handshake I/O (mode-1)
 BSR (Bit Set Reset) Mode
• Only C is available for bit mode access.
• Allows single bit manipulation for control applications
8255 MODES

 Here 8255 is interfaced in Memory Mapped I/O mode.
Initially we write down the addresses and then interface it .
A15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 Port
1 0 0 0 0 X X X X X X X X X 0 0 A
1 0 0 0 0 X X X X X X X X X 0 1 B
1 0 0 0 0 X X X X X X X X X 1 0 C
1 0 0 0 0 X X X X X X X X X 1 1 CW
INTERFACING 8085 & 8255

 Thus we get addresses ,considering don’t cares to be zero as
Port A =8000H
Port B =8001H
Port C =8002H
CWR =8003H
 We give A11,A12,A13 pins to A,B,C inputs of Decoder to
enable 8255 or Chip Select.
 A15 is logic 1 so it is given to active HIGH G1 pin& A14 ,IO/M
¯ are given to active low G2B ¯,G2A ¯ pins.
 Output from Latch is given as A0,A1 pins to 8255 while D0-D7
are given as data inputs.

8255
8085 3:8 decoder
74373
(AD0-AD7)
D7-D0
A0-A7
/CS
A0
A1
O0
O1
O7
A13
A12
A11
ALE
RD ¯
WR ¯
RD¯
WR¯
G2A G2B G1
A15
A14
IO/M
A
B
C
PA
PB
PC

Example: Take data from 8255 port B. Add FF H .Output result
to port A.
MVI A,82H Initialize 8255.
OUT 83H
LDA 81H Take data from port B
ADI FFH Add FF H to data
OUT 80H. OUT Result to port A.
RST1. STOP.

INTERFACING STEPPER MOTOR
with 8255

Serial Communications systems are of three types:
Simplex: This is a one way communication.
 Only one party can speak.
 The other party only hears to the first one but cant
communicate.
System A System B
unidirectional
Transmitter Receiver
SERIAL COMMUNICATION

System A System B
OR
Transmi
tter/Rec
eiver
Receiver
/Transm
itter
Half Duplex: It is a two way communication between two
ports provided that only party can communicate at a time.
• When one party stops transmitting the other starts transmitting.
• The first party now acts as a receiver.

OR/AND.
Full Duplex: It is a two way communication between two ports
and both parties can communicate at same time.
 Thus here efficient communication can be established.
Transmitter/
Receiver
Receiver/
Transmitter.

Asynchronous Synchronous
1. It transfers one character at a
time.
1. It transfers group of
characters at a time.
2. Used for transfer data rates
<20KBPS
2. Used for transfer data rates
>20KBPS
3. Start and stop bit for each
character which forms a frame.
3. No start and stop bit for
each character.
4. Two Clocks are used for Tx
and Rx
4. Single clock is used for both
Tx and Rx.
TRANSMISSION FORMATS

 Interrupt is a process where an external device can get the
attention of the microprocessor.
The process starts from the I/O device
The process is asynchronous.
 Classification of Interrupts
Interrupts can be classified into two types:
 Maskable Interrupts (Can be delayed or Rejected)
 Non-Maskable Interrupts (Can not be delayed or Rejected)
INTERRUPTS IN 8085

Interrupts can also be classified into:
◦ Vectored (the address of the service routine is hard-wired)
◦ Non-vectored (the address of the service routine needs to be
supplied externally by the device)
 An interrupt is considered to be an emergency signal that may
be serviced.
◦ The Microprocessor may respond to it as soon as possible.
INTERRUPTS IN 8085

The 8085 has 5 interrupt inputs.
 The INTR input.
◦ The INTR input is the only non-vectored interrupt.
◦ INTR is mask-able using the EI/DI instruction pair.
RST 5.5, RST 6.5, RST 7.5 are all automatically vectored.
◦ RST 5.5, RST 6.5, and RST 7.5 are all mask-able.
TRAP is the only non-mask-able interrupt in the 8085
◦ TRAP is also automatically vectored.
INTERRUPTS IN 8085

 Non vectored interrupts:
 The 8085 recognizes 8 RESTART instructions: RST0 - RST7
. Each of these would send the execution to a predetermined
hard-wired memory location:
Restart
Instruction
Equivalent to
RST0 CALL 0000H
RST1 CALL 0008H
RST2 CALL 0010H
RST3 CALL 0018H
RST4 CALL 0020H
RST5 CALL 0028H
RST6 CALL 0030H
RST7 CALL 0038H
INTERRUPTS IN 8085

Interrupt name Mask-able Vectored
TRAP No Yes
RST 7.5 Yes Yes
RST 6.5 Yes Yes
RST 5.5 Yes Yes
INTR YES NO
INTERRUPT PRIORITY

119
SOD
SDE
XXX
R7.5
MSE
M7.5
M6.5
M5.5
0
1
2
3
4
5
6
7
RST5.5 Mask
RST6.5 Mask
RST7.5 Mask
}0 - Available
1 - Masked
Mask Set Enable
0 - Ignore bits 0-2
1 - Set the masks according
to bits 0-2
Force RST7.5 Flip Flop to reset
Not Used
Enable Serial Data
0 - Ignore bit 7
1 - Send bit 7 to SOD pin
Serial Out Data
•SIM Instruction helps activate a particular interrupt.
•It can also mask a maskable interrupt.
SIM INSTRUCTION

 Example: Set the interrupt masks so that RST5.5 is
enabled, RST6.5 is masked, and RST7.5 is enabled.
 First, determine the contents of the accumulator.
- Enable 5.5 bit 0 = 0
- Disable 6.5 bit 1 = 1
- Enable 7.5 bit 2 = 0
- Allow setting the masks bit 3 = 1
- Don’t reset the flip flop bit 4 = 0
- Bit 5 is not used bit 5 = 0
- Don’t use serial data bit 6 = 0
- Serial data is ignored bit 7 = 0
SDO
SDE
XXX
R7.5
MSE
M7.5
M6.5
M5.5
0 1 0
0
0
0
0 1
EI ; Enable interrupts including INTR
MVI A, 0A ; Prepare the mask to enable RST 7.5, and 5.5, disable 6.5
SIM ; Apply the settings RST masks
SIM INSTRUCTION

Serial Data In
RST5.5 Interrupt Pending
0 - Available
1 - Masked
Interrupt Enable
Value of the Interrupt Enable
Flip Flop
SDI
P7.5
P6.5
P5.5
IE
M7.5
M6.5
M5.5
0
1
2
3
4
5
6
7
RST5.5 Mask
RST6.5 Mask
RST7.5 Mask
}
•Since the 8085 has five interrupt lines, interrupts may occur during an
ISR and remain pending.
•Using the RIM instruction, it is possible to can read the status of the
interrupt lines and find if there are any pending interrupts.
RIM INSTRUCTION

8253 BLOCK DIAGRAM

 Three independent 16 bit counters.
 24 pin Dual in line Package.
 Counting facility in Both BCD and Binary modes.
 Dc to 2 MHz operating Frequency.
 Can be used as a clock generator.
8253 Features

SC1 SC0 RL1 RL0 M2 M1 M0 BCD
D0
D7
0 0 Counter0
0 1 Counter1
1 0 Counter2
1 1 ILLEGAL
SC1 SC0 Select counter
0 0 Counter latching
0 1 Read/load LSB
1 0 Read/load MSB
1 1 R/L MSB 1st
then LSB.
RL1 RL0 Read/Load
CONTROL WORD

0 0 0 Mode 0
0 0 1 Mode 1
X 1 0 Mode 2
X 1 1 Mode 3
1 0 0 Mode 4
1 0 1 Mode 5
M2 M1 M0
BCD =0 Binary counter
BCD =1 BCD counter
CONTROL WORD

 Example: Use 8253 as a square wave generator with 1ms
period if the input frequency is 1MHz.
 We use counter 0 as a square wave generator and address of
counter 0 =10H and control register =13H.
 I/P frequency is 1MHz.So time is 1µsec.
 Count value = Required period /Input period = 1ms/1 µsec
 =1000(Decimal).
 Thus we use 8253 as a decimal counter.
8253 SQUARE WAVE

 Program:
MVI A,37H Initialize counter 0 mode 3
OUT 13H 16 bit count BCD
MVI A,00H Load LSB count to counter 0
OUT 10H
MVI A,10H Load MSB count to counter 0
OUT 10H.
 Thus, the output will be a square wave.
8253 SQUARE WAVE

 It is a 4 Channel DMA containing 4 individual I/P ,O/P
Channels.
CH0,CH1,CH2,CH3
 It is compatible with Intel processors.
 The maximum frequency is 3 MHz.
 It executes 3 cycles:
1.DMA read
2.DMA write.
3.DMA verify.
 The external device can terminate DMA Operation
8257 DMA

 Rotating priority mode: Each channel has equal priority.
Priority is shifted from one channel to other.
 Fixed priority mode: Each channel has a fixed priority and if
higher priority channels are busy then smaller priority will get
to serve.
 Extended write mode: This mode is used to interface slower
devices to the system.
 TC stop mode: If this bit is set the channel whose terminal
count is reached is disabled.
 Auto reload mode: If this bit is set data is transferred by
channel 2 only. All other channels are not used.
OPERRATING MODES OF 8257

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