Abc2

Microprocessor & MicrocontrollerMicroprocessor & Microcontroller -- II
T.ET.E SemSem V (Rev)V (Rev)
Prof. Nitin AhireProf. Nitin Ahire
XIE, MahimXIE, Mahim

1-Aug-13 Prof.Nitin Ahire 2
Overview of MicroprocessorOverview of Microprocessor
MICROPROCESSOR
( C P U )
MEMORY
INPUT OUTPUT (I/O)
DEVICE

Functional block DiagramFunctional block Diagram
• INPUT OUTPUT (I/O) DEVICE
I/P :Key board, scanner, card reader etc
O/P : Display, printer LED etc
• MEMORY
RAM, ROM
• MICROPROCESSOR
Central Processor Unit ( CPU ) include
ALU, Timing & control unit for synchronizations

Number SystemNumber System
• Decimal number system (DNS)(10)
0,1,2 ……,9,10
• Binary number system(2)
0,1,10,11,100
• Hexadecimal number system (16)
0,1,2,…..,9,A,B,C,D,E,F,10,11
• Advantages of Hex No over BCD No system
(1111 1111)2 (FF)16 (255)10

Review for Logic DevicesReview for Logic Devices
• Tri State Devices :
3 States are logic 1, logic 0 & high
impedances state ( Z )
Enable Enable
Active high Active Low

6
TriTri--State BuffersState Buffers
• An important circuit element that is used
extensively in memory.
• This buffer is a logic circuit that has three
states:
– Logic 0, logic1, and high impedance.
– When this circuit is in high impedance mode it
looks as if it is disconnected from the output
completely.
The Output is Low The Output is High High Impedance

7
The TriThe Tri--State BufferState Buffer
• This circuit has two inputs and one output.
– The first input behaves like the normal
input for the circuit.
– The second input is an “enable”.
• If it is set high, the output follows the proper
circuit behaviour.
• If it is set low, the output looks like a wire
connected to nothing.
Input Output
Enable
Input Output
Enable
OR

Review for Logic DevicesReview for Logic Devices
• Buffer e.g. 74LS244(unidirectionl) & 74LS245(Bidirection)
• Buffer is a logic CKT that amplifies the current or
power
• It has one I/P line and one O/P line
• The logic level of O/P is the same as that of the
I/P
• Basically used as to increase the driving capacity
of logic CKT
simple buffer Active low buffer

Introduction to 8085Introduction to 8085
• CPU built into a single semiconductor
chip is called as microprocessor
• The microprocessor work as a brain
of a computer
• It consist of ALU, registers and
control unit
• The microprocessor are usually
characterized by speed, word length
(bit), architecture, instruction set Etc

8085 Features8085 Features
• 8085 is a 8-bit processor
• Frequency of operation
a) 8085 --- 3Mhz
b) 8085-2 --- 5Mhz
c) 8085-1 --- 6Mhz
• 8085 has 16 bit address bus to access
memory
• 8 bit address bus to access I/O
location

8085 Features8085 Features
• It required only single +5V power supply
• 8085 has following registers
a) 8 bit accumulator
b) six 8- bit general purpose registers
c) 8-bit flag register
d) 16 –bit PC and SP
• It has 5 hardware and 8 software interrupt
• 8085 required 6 Mhz crystal
• It can transmit and receive serial data

8085
PIN DIG
1 X1 40 VCC
2 X2 39 HOLD
3 RESET OUT 38 HLDA
4 SOD 37 CLOCK (OUT)
5 SID 36 RESET IN
6 TRAP 35 READY
7 RST 7.5 34IO/M
8 RST 6.5 33 S1
9 RST 5.5 32 RD
10 INTR 31 WR
11 INTA 30 ALE
12 AD0 29 S0
13 AD1 28 A15
14 AD2 27 A14
15 AD3 26 A13
16 AD4 25 A12
17 AD5 24 A11
18 AD6 23 A10
19 AD7 22 A9
20 VSS 21 A8
8085
(3 MHz )
X1 Crystal 6 MHz
X2
PIN DIG
8085

Serial I/OSerial I/O
portsports
8085
Functional
Pin Diagram
A8-A15
AD0- AD7
RST 6.5RST 6.5
RST 5.5RST 5.5
INTRINTR
RESET INRESET IN
READYREADY
INTAINTA
HOLDHOLD
HLDAHLDA
SODSOD
TRAPTRAP
RST 7.5RST 7.5
ALEALE
X1X1
S0S0
S1S1
IO/MIO/M
RESET OUTRESET OUT
CLK OUTCLK OUT
WRWR
RDRD
ExternallyExternally
InitiatedInitiated
SignalSignal
ExternalExternal
AcknowledgeAcknowledge
SignalSignal
H.O.A.BH.O.A.B
MultiplexedMultiplexed
A/D BusA/D Bus
Control &Control &
StatusStatus
SignalSignal
SIDSID
X2X2 vccvcc CLK CKT &CLK CKT &
P.S.P.S.

Interrupt control Serial I/O Control
W 8 Z 8
D E
CB
H L
SP 16
PC 16
Internal latch
F/F 5
ALU
8
Temp. RegAccumulator 8 I.R. 8
Inst.
Decoder
&
M/C
Encoder
Timing and control unit Add. Buffer
A/D. Buffer
8 bit Internal BUS
AD0-AD7
A15-A8
SID SOD
INTR
INTA RST 7.5 to 5.5 TRAP
X1
X2
READY
WR RD ALE S0 S1 IO/M HLDA HOLD
CLK OUT RESET IN RESET OUT
P.S
+5V
GND
DECODRE
MUX

Interrupt control Serial I/O Control
W 8 Z 8
D E
CB
H L
SP 16
PC 16
Internal latch
F/F 5
ALU
8
Temp. Reg
8
Accumulator 8 I.R. 8
Inst.
Decoder
&
M/C
Encoder
Timing and control unit Add. Buffer
A/D. Buffer
8 bit Internal BUS
AD0-AD7
A15-A8
SID SOD
INTR
INTA RST 7.5 to 5.5 TRAP
X1
X2
READY
WR RD ALE S0 S1 IO/M HLDA HOLD
CLK OUT RESET IN RESET OUT
P.S
+5V
GND
DECODRE
MUX

RegistersRegisters
• The register contains a set of
binary storage cells/Flip Flop
• 6 general purpose 8 bit Reg.
B,C,D,E,H&L (or can be used as
pair of 16 bit reg. like BC,DE,HL)
• W & Z (Temp reg.)
• 16 bit Reg are PC And SP
• 8 bit flag register
B C
D E
H L
SP
PC
W Z
A F

InterruptsInterrupts
• Hardware interrupt
Trap (Non Mask able) (vectored)
RST 7.5(Mask able) (vectored)
RST 6.5 (Mask able) (vectored)
RST 5.5(Mask able) (vectored)
INTR (Mask able) (Non vectored)
• Software interrupt RST 0 to RST 7
All are vectored interrupt

InterruptsInterrupts
• 8085 has 5 hardware interrupts
8 software interrupts
• All software interrupt are vectored
• Out of 5 hardware interrupt 4 are
vector and 1 is non vector
also 4 are maskable and one is non
mask able

Flags Register ( 8 bit )Flags Register ( 8 bit )
D7 D6 D5 D4 D3 D2 D1 D0
• S –sign flag (for signed number)
if D7=1 the number in accumulator
will be –ve number
D7=0 the number in accumulator
will be +ve number
• Z – zero flag if D6=1The zero flag is set if the result
in accumulator is zero
S Z -- AC - P -- C

Flags Register ( 8 bit )Flags Register ( 8 bit )
D7 D6 D5 D4 D3 D2 D1 D0
AC – Auxiliary carry in the arithmetic operation, when the carry
is generated digit D3 and passed on digit D4 the AC flag is set
P – parity flag after an arithmetic and logical operation, if the
result has even number of ones the flag is set if it has odd
numbers of ones, the flag is reset
CY – Carry flag if an arithmetic operation results in carry, the
carry flag is set otherwise it is reset. The carry flag also serves as
a barrow flag for subtraction
S Z AC P C

Subtraction process in 8085Subtraction process in 8085
• 1 : find 1’s complement of the subtrahend
• 2 : find 2’s complement of the subtrahend
• 3 : Adds 2’s complement of the
subtrahend to the minuend
• 4 : complements the CY flag.
These steps are invisible to the user, only
the result is available to the user.
For unsigned number if CY is reset the result
is positive and if CY is set the result is
negative(2’complement)

Sign flag (used only for sign No.)Sign flag (used only for sign No.)
• Sign flag: This flag is used with signed
numbers in the arithmetic operation.
With sign number, bit D7 is reserved
for indicating the sign and the
remaining 7 bit are used to represent
the magnitude of a number
• Sign flag is irrelevant for unsigned
number

DD –– F/F (Latch)F/F (Latch)
D F/F
Latch
clk
I/P
O/P
clk
I/P
Q
Q

De multiplexing Of AD0De multiplexing Of AD0--AD7AD7
8085
Latch
AD0-AD7
D0-D7
ALE
A0-A7
A8-A15
IO/M

De multiplexing (AD0De multiplexing (AD0--AD7)AD7)

De multiplexing Of AD0De multiplexing Of AD0--AD7AD7
8085
LatchAD0-AD7
D0-D7
ALE
A0-A7
IO/M

Differentiate between IO/MDifferentiate between IO/M
8085
LatchAD0-AD7 A0-A7
D0-D7
IO/M Memory
IO device
ALE
A8-A15
A0-A7
1

1-Aug-13 Prof.Nitin Ahire 281-Aug-13 Prof.Nitin Ahire 28
Differentiate between IO/MDifferentiate between IO/M
8085
LatchAD0-AD7 A0-A7
D0-D7
IO/M Memory
IO device
ALE
A8-A15
A0-A7
0

Control & Status SignalsControl & Status Signals
S0S1

Generation of control SignalsGeneration of control Signals

Instruction, Data formatInstruction, Data format
and storageand storage
• Part of instruction each instruction has
two parts
1 opcode: one is the task to be perform
(operational code)
2 operand: data to be operated on
(data)
The data can be specified in the various
form it may in the memory or I/O or in
the instruction it self.

OpcodeOpcode
• Opcode : operational code
• Operand : Data
• Mnemonics : Instructions
Memory Locations Opcode Mnemonics Operand
2000 3E MVI A, 20
2001 20
2002 06 MVI B, 12
2003 12
2004 4F MOV C, A

ALU
(A)
INST.
DECODER
CONTROL
LOGIC
B C
D E
H L
SP
PC (2000)
2000
2001
2002
2003
2004
3E
MERD
MEMORY LOCATION
ADD BUS
Internal Data BUS
DECODER
DATA BUS
3E
20
06
12
4F

ALU
(A)
INST.
DECODER
CONTROL
LOGIC
B C
D E
H L
SP
PC (2001)
2000
2001
2002
2003
2004
20
MERD
MEMORY LOCATION
ADD BUS
Internal Data BUS
DECODER
DATA BUS
3E
20
06
12
4F
3E

ALU
(A)
INST.
DECODER
CONTROL
LOGIC
B C
D E
H L
SP
PC (2002)
2000
2001
2002
2003
2004
06
MERD
MEMORY LOCATION
ADD BUS
Internal Data BUS
DECODER
DATA BUS
3E
20
06
12
4F
20

ALU
(A)
INST.
DECODER
CONTROL
LOGIC
B C
D E
H L
SP
PC (2003)
2000
2001
2002
2003
2004
12
MERD
MEMORY LOCATION
ADD BUS
Internal Data BUS
DECODER
DATA BUS
3E
20
06
12
4F
20

ALU
(A)
INST.
DECODER
CONTROL
LOGIC
B C
D E
H L
SP
PC (2004)
2000
2001
2002
2003
2004
4F
MERD
MEMORY LOCATION
ADD BUS
Internal Data BUS
DECODER
DATA BUS
3E
20
06
12
4F
20

Instruction classificationInstruction classification
• “Instruction” is a command to the
microprocessor to perform a given task
on specified data”.
• The instruction can be classified into
following fundamental categories
1 Data transfer
2 Arithmetic & Logical operation
3 Branching operation
4 Machine control operation

• 1 Data transfer (copy)
basically used to copies data from source
to destination without modifying the
content of the source like,
Opcode operand
MOV rd, rs
MVI r, 8-bit
IN 8 bit port add.
OUT 8 bit port add.

• 1 Data transfer (copy)
• LXI Rp, 16-bit add.
• MOV R,M
• MOV M,R
• LDA 16-bit add.
• STA 16-bit add.
• LDAX R*
• STAX R*
• LHLD 16-bit add (1st memory location copy to L & 2nd memory
location to H)
• SHLD 16-bit add
*R – Register pair

LHLD 4000H & SHLD 4000HLHLD 4000H & SHLD 4000H
H L memory
» 4000
» 4001
03
70
0370

• Arithmetic operation
These instruction perform arithmetic
operation such as addition subtraction,
increment, decrement.
• ADD R
• ADI data
• ADC R
• ADC M
• ACI data
• DAD Rp

• SUB R
• SUB M
• SBB R
• SBB M
• SUI Data
• SBI Data
• DAA

• INR R
• DCR R
• INR M
• DCR M
• INX Rp
• DCX Rp

• Logical instruction.
These instruction perform various logical
operation with the content of the
accumulator
e.g. 1) AND,OR,EX-OR(ANA R,ANI Data,
XRA R)
2) Rotate (RAL,RAR,RLC,RRC)
3) Compare (CMP B,CPI Data)
4) Complement (CMC, CMA,STC)

RALRAL

More aboutMore about CMP RCMP R
instructioninstruction
• CMP R
• This instruction compare the contents of
accumulator with the contents of register
specified
• The operation of comparing is performed by
subtracting (Acc. – Reg.)
• The contents of Acc or Reg. are not altered
• The result of comparison is indicated by flags
When A>R ; CY=0 & Z=0
When A=R ; CY=0 & Z=1
When A<R ; CY=1 & Z=01-Aug-13 Prof.Nitin Ahire 47

• Branching operation
These group of instruction alter the
sequence of program execution
either conditionally or unconditionally
e.g. JUMP (conditionally or unconditionally)
CALL & RET (conditionally or unconditionally)

• Machine control instruction
These instruction control the machine
function such as Halt (HLT),
interrupt (RST 1) or do noting (NOP)

More aboutMore about DAADAA
instructioninstruction
• Decimal Adjust Accumulator
• If lower nibble of A > 9 or AC =1
then, lower nibble of A = lower
nibble of A + 06H.
• If Higher nibble of A > 9 or CY =1
then, Higher nibble of A = Higher
nibble of A + 06H.

• E.g. 1) 55 h + 06 h = 5B h
55 + 06 = 61 D

ProgrammingProgramming
• Write a generalized program for
addition of two 8 -bit numbers
and get the result in BCD
• 1st No. is located at memory location
C200h
• 2ndNo. is located at memory location
C201h
• Save the result at next location C202

Addition of two 8 bitAddition of two 8 bit numbernumber
located at C200h & C201hlocated at C200h & C201h
LXI H, C200H; load HL pair by C200h
MOV A,M; Move 1st No. in the Reg. A
INX H ; increment the HL pair by 1
ADD M; Add A+(M)=A
DAA
INX H; increment the HL pair by 1
MOV M,A; move the result in M
HLT; Stop

• Write a generalized program for addition
of two 8 -bit numbers located at (with carry)
• 1st No. is located at memory location C200h
• 2ndNo. is located at memory location C201h
• Save the result at next memory location C202h
• And save the status of carry at next memory
location C203h

Addition of two 8Addition of two 8--bit numbersbit numbers
• MVI C,00H ; use reg. C to know the status of CY flag
• LXI H,C200h ; set memory pointer at C200h
• MOV A,M ; transfer the 1st No. from memory to Acc.
• INX H; go to the next memory location point to 2nd No.
• ADD M; Add A+M =A
• JNC Down (Put 16-bit memory address of target instruction)
• INR C ; If there is carry increment the reg. C by 1
• Down: STA C202H ; load the Acc. Data at C202h
• MOV A,C ; load Carry to Acc
• STA C203H ; load Acc. Data (carry) at C203h
• RST 1/ HLT ; Stop

• Write a generalized program for addition
of two 16-bit numbers located at (with carry)
• 1st No. is located at memory location C200h & C201h
• 2ndNo. is located at memory location C202h & C203h
• Save the result at next memory location C204h & C205h
• And save the status of carry at next memory location
C206h

Addition of two 16 bit numberAddition of two 16 bit number
MOV C,00h ; Reg. C for to save the status of carry flag
LHLD C200H ; Here C200h-L=34, C201h H=12
XCHG ; exchange HL with DE
LHLD C202H ; Here L= 21, H=43
DAD D ; DE+HL = HL
JNC down
INR C
Down: SHLD C204H ; store the result
MOV A,C
STA C206H ; store the carry
HLT ; stop

Find the largest number from the series of 10Find the largest number from the series of 10
numbersnumbers
• The 10 numbers are saved from memory
location 8500h to 8509h
• Save the largest number at memory location
8550h
8500 8501 8502 8503 8504 8505 8506 8507 8508 8509
99 66 AA DF DD 97 F3 44 FE 67
8550
XX

Find the largest number from the series of 10 numbersFind the largest number from the series of 10 numbers
• MVI C,O9H ; counter
• LXI H,8500H ; memory pointer
• MOV A,M; move 1ST No. to Acc.
• UP:INX H ; Increment HL pair point to 2nd No.
• CMP M; Compare 1st No. with 2nd No.
• JNC DOWN; Jump on no carry ( A > M)
• MOV A,M ; If carry change the No. ( A < M)
• DOWN:DCR C ; Decrement the counter by 1
• JNZ UP; Check for zero flag ( counter = zero)
• STA 8550H ; Save the result ( Acc to 8850h)
• HLT; Stop

Memory
location
Opcode,
Operand
Mnemonics Remark
8000 0E ,09 MVI C,09H COUNTER
8002 21,00,85 LXI H,8500H MEMORY POINTER
8005 7E MOV A,M 1ST No. to Acc.
8006 23 UP:INX H Increment HL pair
8007 BE CMP M Compare
8008 D2,0C,80 JNC DOWN Jump on no carry
800B 7E MOV A,M If carry change the No.
800C 0D DOWN:DCR C Decrement the counter
800D C2,06,80 JNZ UP Check for zero flag
8010 32,50,85 STA 8550H Save the result
8013 76 HLT Stop

8550
FE
• After the execution of program
status of memory location

Block TransferBlock Transfer
• Write a program to transfer a block
of data from 8500H to 8509H. Store
the data from 8570H to 8579H .
8500 8501 8502 8503 8504 8505 8506 8507 8508 8509
99 66 AA DF DD 97 F3 44 FE 67
8570 8571 8572 8573 8574 8575 8576 8577 8578 8579
xx xx xx xx xx xx xx xx xx xx

ProgramProgram
LXI H ,8500H ; source locationLXI H ,8500H ; source location
LXI BLXI B ,8570H ; Destination location,8570H ; Destination location
MVIMVI D,0AH ; CounterD,0AH ; Counter
UPUP MOVMOV A,M ; 1A,M ; 1stst No. transfer in Acc.No. transfer in Acc.
STAX B;STAX B; Save Acc. Content at memory location pointed by BCSave Acc. Content at memory location pointed by BC RegReg
INXINX HH
INX BINX B
DCR DDCR D
JNZJNZ UPUP
HLTHLT

• After the execution of program
status of memory location
8500 8501 8502 8503 8504 8505 8506 8507 8508 8509
99 66 AA DF DD 97 F3 44 FE 67
8570 8571 8572 8573 8574 8575 8576 8577 8578 8579
99 66 AA DF DD 97 F3 44 FE 67

Draw a minimum mode system of 8085Draw a minimum mode system of 8085
• The minimum number of components
required to make a system using
8085 is called minimum system or
minimum mode system.

8
0
8
5
74
LS
373
74
LS
138
Memory I/O
MERD
MEWR
IORD
IOWR
A0-A15
D0-D7A0-A7
ALE
AD0-AD7
A8-A15
RD
IO/M
WR
TRAP
RST 7.5
RST 6.5
RST 5.5
INTR
INTA
READY
SOD
SID
RESET OUT
RESET IN
VCC
Minimum mode systemMinimum mode system(8085(8085))
X1
X2

• The 8085 instruction set is classified
into the following 3 group according
to word size or byte size
1) 1- byte instruction
2) 2- byte instruction
3) 3 –byte instruction

11-- byte instructionbyte instruction
• A 1- byte instruction includes the
opcode and the operand in the same
byte
e.g. Opcode operand hex code
1 MOV C, A (4F) (opcode)
2 ADD B (80) (Data)
each instruction required 1 memory
location ( 8-bit)

• In the 2- byte instruction the first byte
specifies the operation code and the
second byte specifies the operand
MVI A, 12H 3E (opcode)
12 (Data)
These instruction required 2 memory
location

• In 3-byte instruction the first byte specifies the
opcode and the following 2 bytes specify the 16-
bit address
LDA 2050 3A (Opcode)
50 (Data)
20 (Data)
Note the second byte is the lower address and the
third byte is the high order address
These instruction required 3 memory location

AddressingAddressing modemode
• The different methods (mode) to select
the operands or address are called
addressing mode
• For 8085 they are
1 Immediate addressing
2 Register addressing
3 Direct addressing
4 Indirect addressing
5 Implied addressing

Addressing modeAddressing mode
1 Immediate addressing
• In the immediate addressing mode the
data is specified in the instruction it self.
• The immediate addressing mode
instruction are either 2- byte or 3- byte
long.
• The instruction contain the letter “I”
indicate the immediate addressing mode.
e.g. 1 MVI A,12h
2 LXI H,2000h

2 Register addressing mode
• In register addressing mode the source
and destination operands are general
purpose registers
• The register addressing mode
instructions are generally of 1 –byte
e.g. 1 MOV A,B
2 ADD B
3 PCHL
4 XRA A

3 Direct addressing
• In the direct addressing mode the
16 bit address of the data or operand
is directly specified in the instruction
• These instruction are 3 –byte
instruction.
e.g. 1 LDA 2000h
2 STA 2060h

4 Indirect addressing
• In the Indirect addressing mode the
instruction reference the memory
through the register pair i.e. the
memory address where the data is
located is specified by the register
pair
e.g.1 MOV A,M
2 LDAX B

5 Implied addressing
• The Implied mode of addressing does not
required any operand
• The data is specified within the opcode it-
self
• Generally these instructions are 1-byte
instruction
• The data is supposed to be present in the
Accumulator
e.g. 1 RAL
2 CMC

Timing diagramTiming diagram
• For better understanding of each
instruction it is very essential to
understand the Timing diagram of
each instruction.
• The graphical representation of each
instruction with respective to time i.e.
CLOCK is called “Timing Diagram”

• Instruction cycle (IC) : The essential step
required by CPU to fetch and execute an
instruction is called IC
IC=FC+EC
• Machine cycle (MC) : Time required by
microprocessor to complete the operation
of accessing memory or I/O device is
called MC.
• T –state :Each clock cycle is called T-state

Timing diagram
• The P operates with reference to clock signal.
• 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/O-read.
5. I/O write.

• Timing diagram for the 1- byte
instruction
• single MC = opcode fetch
Memory
location /(PC)
Opcode Instruction
2005 4F MOV C,A

Control & Status SignalsControl & Status Signals
S0S1

• Step 1 (State T1) In the state, 8085 sends the status
signals, IO/M=0, S1=1 and S0=1
• The 8085 send a 16 bit address on A8-A15 and AD0-
AD7
• The high order bytes of PC (20) is placed on the A8-A15
lines, and it remain there upto T3 state.
• The low order bytes of PC (05) placed on the AD0-
AD7,lines which remain there only for T1
• During this state, ALE gives a positive pulse signal is
used to latch the add A0-A7.
• No control signal is generated in state. ( RD & WR )

• Step2 (T2): The content of PC lower byte will
disappear on AD0-AD7 lines, so the same line can be
used as data line . The contents of A0-A7 are still
available for memory from external Latch.
• The control signal RD is made low by the processor
which enables the read ckt of addressed memory
device.
• Then the memory device send the content on the data
bus D0-D7 (4F)
• In addition to these the processor increments PC
content by one

• Step3 (T3): during this cycle the data from
memory (opcode) is transfer in the instruction
Reg. and RD control signal made HIGH

Step 4 (T4): the microprocessor perform only
internal operation.
The opcode decoded by the CPU and 8085 decide
1) Whether it should enter T5 and T6 states or not
2) How my bytes of instruction it is ?
If instruction doesn’t required T5 &T6 states, it
go to the next MC

• Step 5 (T5 &T6): T5 and T6 states, states are
required to complete decoding and some
operations inside the 8085 it depend on the
type of instruction

• Following instruction required T5 & T6 states for the opcode
fetch MC
1 CALL
2 CALL conditional
3 DCX Rp
4 INX Rp
5 PCHL
6 SPHL
7 PUSH Rp
8 RET conditional
All other instruction except the above instruction required opcode
fetch of T1 to T4 states only.

• Machine cycle (type)
Few more MC
1 opcode fetch
2 operand fetch
3 Memory read
4 Memory write
5 I/O read
6 I/O write
7 Interrupt Ack M-cycle
8 Ideal M-cycle

• Timing diagram for the 2- byte
instruction
• 2 MC = 1 opcode fetch & 2 Data fetch
Memory
location /(PC)
Opcode Instruction
2000 06 MVI B , 43 h
2001 43

Stack control and branching groupStack control and branching group
• Stack is the reserved area of the
memory in RAM where temporary
information may be stored
• Stack pointer (SP): an 16-bit SP is
used to hold the address of the most
recent stack entry. It work on the
principle of LIFO or FILO.

STACK MEMORYSTACK MEMORY
STACK
Memory
(Reserve area)
Total Memory
Size
Reserve
area

Stack Related InstructionsStack Related Instructions
• LXI SP,16-bit address
• LXI SP, 2000 h
• LXI SP,FFFF h
• PUSH Rp(PUSH B, PUSH H, PUSH D,PUSH PSW)
• POP RP (POP B, POP D,POP H,POP PSW)
• SPHL ( HL SP)
• XTHL ( HL SP)
• PCHL ( HL PC)

• PUSH B
Let BC=3010, B=30h, C=10h
suppose SP initialized at FFFF h
after execution of instruction PUSH B
SP=SP-1=FFFF-1=FFFE
B [FFFE] =30h
again SP=SP-1=FFFE-1=FFFD
C [FFFD]=10h
SP=[FFFD] New location of SP

PUSH H
Contents on the stack & in the register
after the PUSH instruction
4242 F2F2
F
C
E
L
A
B
D
H
SP
F2
42
X
2097
2098
2099
2097
8085 Register
Memory
Initially SP at 2900 & H=42 , L=F2
1 2

• POP B
initially B=20, C=40h
SP at=[FFFD]=10h
at=[FFFE]=30h
After execution of POP B
SP=[FFFD]=10h [C]
SP=SP+1=[FFFE]=30h [B]
Again SP=SP+1=[FFFF]
Now B=30h, C=10h and SP=[FFFF]

POP HPOP H
Initially SP at 2097 After POP H ,
H= 42
L= F2
Contents on the stack and in the registers
after the POP instruction
42 F2
2099
F2 2097
42 2098
X 2099
F
C
E
L
A
B
D
H
SP
8085 Register
12

SubroutinesSubroutines
• Whenever we need to use a group of instruction
several times throughout a program there is a way we
can avoid having to write the group of instructions each
time we want to use them.
• One way is to write the group of instruction
separately, Called Subroutines
• Whenever we want to execute that group of
instruction we can call that Subroutine.
• The return address has to be stored back on the stack
memory

SubroutinesSubroutinese.g.
6FFD 31 LXI SP, FFFF h
FF
FF
7000 CD CALL C200h
7001 00
7002 C2
7003 Next instruction
When this instruction is executed PC contents 7003h
(next instruction) will stored on to the stack and
microprocessor will load PC with C200h and start
executing instruction from C200h

• If SP= FFFF h
• CALL C200. PC stack (memory)
(SP-1)=Pc H FFFF
(SP-2)=Pc L FFFE
SP=SP-2 FFED
PC=new C200
70
03
0370

• Conditional Call instructions
When condition is true, then CALL at NEW address else execute
the next instruction of the program
1) CZ Add
2) CNZ Add
3) CP Add
4) CM Add
5) CPO Add
6) CPE Add
7) CC Add
8) CNC Add If condition false 9T True16T

• RET unconditionally
1 SP (PC L) PC Stack
2 SP+1 (PC H)
3 SP+2=SP
Initially SP at FFFD
After execution of RET SP=FFFF
70
03
0370
FFFD
FFFE
FFFF

• RET conditionally
When condition is true, then RET at the main
program
1) RZ
2) RNZ
3) RP
4) RM
5) RPO
6) RPE
7) RC
8) RNC

Nested SubroutinesNested Subroutines
• Whenever one subroutine calls another
subroutine in order to complete a specific
task, the operation is called as nesting.
• The First subroutine may call the second
subroutine and in turn the second
subroutine may called first or third
subroutine such routines called NESTED
subroutines

Nested subroutinesNested subroutines
. sub 1 sub 2
. . .
call sub 1 call sub 2 .
. . .
. . .
. RET RET

• There are two kinds of subroutines
1) Re-entrant subroutines.
2) Recursive subroutines.
Nested SubroutinesNested Subroutines

Re entrantRe entrant SubroutinesSubroutines
1)Re-entrant Subroutine
It may happened a Subroutine ‘1’ is called
from main program and ‘2’ Subroutine is
called from Subroutine ‘1’ and Subroutine
‘2’ may called Subroutine ‘1’ then the
program re entrant in Subroutine ‘1’ this is
called re-entrant Subroutine

Re entrantRe entrant subroutinesubroutine
• Main Program
• Next
• line
SUB 1 SUB 2
CALL
CALL
RET
RET

RecursiveRecursive subroutinesubroutine
• A procedure which called it self is
called a recursive subroutine
• The recursive subroutine loop takes
long time to execute
• In this type of subroutine we
normally define N ( recursive depth)
it is decrement by one after each
subroutine is call until N=0

RecursiveRecursive subroutinesubroutine
• Main Program N=3 N=2 N=1
• Next
• line
SUB 1 SUB 2
CALL
CALL
RET
RET
RET
SUB 3
CALL

SoftwareSoftware DelayDelay
• Delay : operating after an some
time interval.
• Microprocessor take fixed amount of
time to execute each instruction
• Microprocessor driven by fixed
frequency (crystal)
• So using instruction we can generate
a Delay.

Software DelaySoftware Delay
• E.g. delay using NOP instruction
NOP ( 1-byte instruction- 4T state)
assume crystal freq= 4Mhz..
CLK freq = 2Mhz
(T=0.5 microsecond)
Delay using NOP = 4 X 0.5 microsecond
= 2.0 microsecond

Software DelaySoftware Delay
• If we want the more delay than
4T,then we go on increasing NOP after
NOP.
• Impractical (size of program increase)

Delay using 8Delay using 8 ––bit counterbit counter
Delay
Initializes 8-bit counter
Decrement counter
RET
yes
No

•
MVI C, Count 7T
up: DCR C 4T
JNZ :up 10T/7T
RET 10T
The loop is executed ( count-1 ) times

• Formula for delay value
Td=[ M + {(count) X N} -3 ] T
M=No. of T state out side the loop
N=No. of T state inside the loop
M=10+7=17 T; N=10+4=14T

JNZ instruction required 10 or 7 T state
based on the condition ( Z=0 or 1)
(when condition is satisfied it take
10T state and if not satisfied it take 7T
state)
so 3 T must be subtracted from total
value

• Td max= [17+[ {255} X 14 ] -3] T
= 3584 T
= 3584 X 0.5 microsecond
= 1792 microsecond
(FF)=(255)

Delay using 16 bit counterDelay using 16 bit counter
• LXI B, (count)H 10T
up: DCX B 6T
MOV A,C 4T
ORA B 4T
JNZ :up 10/7T
RET 10T
DCX not affect the zero flag.

Td=[ M + {(count) X N} -3 ] T
M=No. of T state out side the loop
N=No. of T state inside the loop
M=20 T; N=24T

• Td max= [20+[ {65535}X 24 ] -3] T
= 1572857 T
=1572857 X 0.5 microsecond
= 78642 microsecond
(FFFF)=(65535) largest count.

Memory and I/O interfacingMemory and I/O interfacing
• Types of memory
1 ROM (EPROM)
2 RAM

Memory structure & it’sMemory structure & it’s
requirement'srequirement's
• The read write
memories consist
of an array of
registers, where in
each register has a
unique address
• M=No. of register
• N=No. of bits
MXN
I/Pdecoder
I/P Buffer
O/P Buffer
A1
A0
AM
Data i/Ps
Data o/p s
WR
RD
CS

• If the memory having 13 address
line and 8 data lines, then it can
access 213 address lines = 8192
and N= 8 bit or 1-byte
• No of address lines of the ‘up’ is to
be used to find how much memory
array can be access by that
processor.

No. of address lineNo. of address line
required to accessed the memoryrequired to accessed the memory
• No of lines size of memory
1 21= 2
2 22= 4
3 23= 8
4 24 =16
10 210= 1K= 1024
11 211 =2K= 2048
16 216 =64K= 65536

EPROM IC available in theEPROM IC available in the
marketmarket
IC NO size
2716 2k X 8
2732 4k X 8
2764 8k X 8
27128 16k X 8
27256 32K x 8
27512 64k X 8

RAM IC available in theRAM IC available in the
marketmarket
IC NO size
6116 2k X 8
6264 8k X 8
62512 64k X 8
2114 1k X 4

Comparison between fullComparison between full
and partial decodingand partial decoding
• full decoding
1) Also referred to be as
absolute decoding
2) All address lines are
consider
3) More hardware
required
• partial decoding
1) Also referred to be
as liner decoding
2) Few address lines are
ignored
3) Decoder hardware is
simple

8085 Memory Interfacing

Memory and I/O interfacingMemory and I/O interfacing
• Types of memory
1 ROM (EPROM)
2 RAM

• The read write
memories consist
of an array of
registers, where in
each register has a
unique address
• M=No. of register
• N=No. of bits
MXN
I/Pdecoder
I/P Buffer
O/P Buffer
A1
A0
AM
Data i/Ps
Data o/p s
WR
RD
CS

• If the memory having 13 address
line and 8 data lines, then it can
access 213 address lines = 8192
and N= 8 bit or 1-byte
• No of address lines of the ‘up’ is to
be used to find how much memory
array can be access by that
processor.

No. of address lineNo. of address line
required to accessed the memoryrequired to accessed the memory
• No of lines size of memory
1 21= 2
2 22= 4
3 23= 8
4 24 =16
10 210= 1K= 1024
11 211 =2K= 2048
16 216 =64K= 65536

EPROM IC available in theEPROM IC available in the
marketmarket
IC NO size
2716 2k X 8
2732 4k X 8
2764 8k X 8
27128 16k X 8
27256 32K x 8
27512 64k X 8

RAM IC available in theRAM IC available in the
marketmarket
IC NO size
6116 2k X 8
6264 8k X 8
62512 64k X 8
2114 1k X 4

Comparison between fullComparison between full
and partial decodingand partial decoding
• full decoding
1) Also referred to be as
absolute decoding
2) All address lines are
consider
3) More hardware
required
• partial decoding
1) Also referred to be
as liner decoding
2) Few address lines are
ignored
3) Decoder hardware is
simple

8085 Memory Interfacing8085 Memory Interfacing
• 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.
4. linkmemory_interfacing.ppt

• 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/MRDWR
2K Byte
Memory
Chip
WRRD
CS
A10- A0
A15- A11
3:8DECODER

Microprocessor & MicrocontrollerMicroprocessor & Microcontroller
-- II
T.E Sem V (Rev)T.E Sem V (Rev)

Connection of I/O devices.Connection of I/O devices.
• Polling method
• Interrupt method

Interrupt system of 8085Interrupt system of 8085
• Definition: “It is a mechanism by
which an I/O device ( Hardware
interrupt) or an instruction (software
interrupt) can suspend the normal
execution of the processor and get it
self serviced.”

Types of interruptTypes of interrupt
• 1) Hardware interrupt
• 2) Software interrupt

Hardware interruptHardware interrupt
• Interrupt : “It is an external
asynchronous input that inform the
‘up’ to complete the instruction that
it is currently executing and fetch a
new routine in order to offer a
service to that I/O devices. Once the
I/O device is serviced, the ‘up’ will
continue with execution of its normal
program.”

Hardware interruptHardware interrupt
• 8085 has ‘5’ hardware interrupt
1)Trap
2)RST 7.5
3)RST 6.5
4)RST 5.5
5)INTR

Types of Hardware interruptTypes of Hardware interrupt
• NMI( non maskable)
1) It can’t be masked or
made pending
2) Highest priority
3) This interrupt disable
all maskable interrupts
4) Used for emergency
purpose like power
failure, smoke detector
e.g. TRAP
• Maskable
1) It can be masked or
made pending
2) Lower priority
3) These interrupt dose
not disable non
maskable interrupt
4) Used to interface
peripherals.
e.g. RST 7.5

Hardware InterruptHardware Interrupt
Priority interrupt ISR address trigger
1 TRAP 0024h edge +level
2 RST 7.5 003Ch edge
3 RST 6.5 0034h level
4 RST 5.5 002Ch level
5 INTR No specific level
location

Software interruptSoftware interrupt
• 8085 has ‘8’ software interrupt
1)RST0
2)RST1
3)RST2
4)RST3
5)RST4
6)RST5
7)RST6
8)RST7

• These instruction ( RST0-RST7) allow
the ‘up’ to transfer the program
control from main program to the
subroutine program (i.e. ISR)
ISR: interrupt service routing

Interrupt Restart locations
RST 0 0 X 8 = 0000h
RST 1 1 X 8 = 0008h
RST 2 2 x 8 = 0010h
RST 3 3 X 8 = 0018h
RST 4 4 X 8 = 0020h
RST 5 5 X 8 = 0028h
RST 6 6 X 8 = 0030h
RST 7 7 X 8 = 0038h

Software interrupt / hardwareSoftware interrupt / hardware
interruptinterrupt
• Software interrupt
1)It is as synchronous event
2)This interrupt is requested
by executing instruction
3)PC is incremented
4)The priority is highest
5)It can’t be ignored
6)It is not used to interface
the peripheral
Used in debugging
• Hardware interrupt
1)It is an asynchronous
event
2)This interrupt is
requested by external
device
3)PC is not incremented
4)The priority is lower
than softer interrupt
5)Can be masked
6)It is used to interface
peripheral devices

Interrupt relatedInterrupt related
instructionsinstructions
1) EI : it is used to enable the all
maskable interrupt. It required 1-
byte, one MC (4T). It does not
affect on TRAP
2) DI : it is used to disable all
maskable interrupt. 1-byte (4T). It
does not affect on TRAP

SIM : (set interrupt mask)
1-byte (4T) state.
• Used to enable or disable RST 7.5, RST
6.5, RST 5.5 interrupts.
• It does not affect on TRAP & INTR .
• It is used in serial data transmission
• It also transfer serial data bit ‘D7’of ‘A’
to the SOD pin
• Hence the CWR format must be load in
the ‘A’ before execution of SIM
instruction.

SIM (bit pattern)SIM (bit pattern)
• SOD pin
• D7= SOD
• D6= serial data enable 1=enable, 0=disable
• D5= Don’t care
• D4= Reset R7.5 F/F, 1=Reset 0=no effect
• D3=MSE Mask set enable 1=D2,D1,D0 bit are effective
0=D2,D1,D0 bit are ignored
• D2= M’7.5 Mask RST 7.5 1= Mask or disable R7.5
0= Enable RST 7.5
• D1=M’6.5 Mask RST 6.5 1= Mask or disable R6.5
0= Enable RST 6.5
• D0=M’5.5 Mask RST 5.5 1= Mask or disable R5.5
0= Enable RST 5.5
SOD SDE M’ 6.5M’ 7.5MSER 7.5X M’ 5.5

RIM : ( read interrupt mask)
1-byte (4T) state.
• It gives the status of the pending maskable
interrupt (RST 7.5 – RST 5.5)
• It does not affect on TRAP & INTR
• It can also transfer the contents of the serial
input data on the SID pin into the
accumulator (‘D7’ bit.)
• Hence after execution of this instruction
serial data get load in to the accumulator

RIM (bit pattern)RIM (bit pattern)
• SID pin
• D7= SID
• D6=
• D5= if 1 respective interrupt is pending
• D4= 0 respective interrupt is not pending
• D3=IE interrupt enable
• D2=
• D1= if 1 respective interrupt is Masked
• D0= 0 respective interrupt is unmasked
SID I 7.5 M 6.5M 7.5IEI 5.5I 6.5 M 5.5

8155 (PPI)8155 (PPI)

Features of 8155 (PPI)Features of 8155 (PPI)
• It is a multifunction device designed to use in
minimum mode system
• It contain RAM, I/O ports and Timer
• Features
1) 2k static RAM cell organized as 256 bytes
2) 2 programmable 8 bit I/O ports (A,B)
3) 1 programmable 6 bit I/O port (c)
4) 1 programmable 14 bit binary down
counter/timer
5) An internal address latch to de multiplex AD0-
AD7 using ALE
6) Internal selection logic for memory and I/O.
using command register

81558155
AD0-AD7
IO/M
CE
ALE
RD
WR
VCC GND
RESET TIMER IN TIMER OUT
PA0-PA7
PB0-PB7
PC0-PC5
A
B
C
256x8
Static RAM
Timer
CE – 8155
CE - 8156

Pin outPin out
• ADo-AD7 : address and data lines
internally de multiplex by using
internal latch and ALE signal address
lines are used to access the memory
or I/O port depending on the status
of IO/M^ pin i/p
• D0-D7 lines act as data bus

Pin outPin out
• ALE : used to de multiplex the AD0-
AD7
• IO/M^ : used to differentiate
between IO or memory
• CE^ : used to select the 8155
• RD^ : used for to read the data from
memory or I/O
• WR^: used for to write the data from
memory or I/O

Pin outPin out
• Reset : used to reset the 8155 IC
• PA0-PA7,PB0-PB7 : Port A and Port B I/P 8 bit
pins they can be programmed either i/p or o/p
port using command register
• PC0-PC5 : these are 6 bit I/O pins they can be
used as simple Input output port or control port
when PA and/or PB are used in handshake mode
• Timer in: this is an i/p to the timer
• Timer out :This is an o/p pin depending on the
mode of the timer o/p can be either a square
wave or pulse.
• VCC, GND : +5 v resp. to GND

I/O port or timer selectionI/O port or timer selection
IO/M^ A2 A1 A0
1 0 0 0 CWR
1 0 0 1 PORT A
1 0 1 0 PORT B
1 0 1 1 PORT C
1 1 0 0 Timer LSB
1 1 0 1 Timer MSB
0 Memory option

Control Word of 8155Control Word of 8155
• D0=Port A 0=Input
• D1=Port B 1=Output
• D2 &D3 used with port C
• D4 (IEA=interrupt Enable Port A) 1=Enable
• D5 (IEB=interrupt Enable Port B) 0=Disable
• D6&D7 used in timer mode
D7 D6 D1D2D3D4D5 D0

D7 D6 timer commands
0 0 NOP
0 1 Stop counting if timer
is running
1 0 Stop after TC (stop after
at the count)
1 1 Start timer if is not
running

Timer modeTimer mode
• MSB
• LSB
• Mode 0 In this mode the timer o/p remains high for half the
count and goes low for the remaining
count, thus providing the single square wave. The
pulse width is determined by count and clk freq
• Mode 1 In this mode the initial count is automatically reloaded
at the end of each count. Provide the continuous square
wave.
• Mode 2 In this mode single clock pulse is provided at the end
of count
• Mode 3 This is similar to mode 2 except the initial count is
reloaded to provided a continuous wave form
M2 M1 D9D10D11D12D13 D8
D7 D6 D1D2D3D4D5 D0

• For port C
D3 D2
0 0 = ALT 1(port C as Input mode)
1 1 = ALT 2(Port C as Output Mode)
0 1 = ALT 3 used in handshake mode
1 0 = ALT 4 along with Port A &B

Example 1Example 1
• Design a square wave generator with
a pulse width of 100 us by using the
8155 timer if clock freq is 3MHz.
Sol : T=1/F=1/3MHz=330ns
Timer count=pulse period/CLK period
= 200us/330ns=606
count = 025Eh

• Assuming the port addresses
CWR=20h
Timer LSB=24h
Timer MSB=25h
Count =025Eh
Therefore 5Eh must be load in the LSB timer
Select mode 1 for square wave.
Therefore 42h (01000010)must be load in the
MSB timer

• Control word
To start the timer D7 and D6 bit must
be 1
set the bit of CWR and send to
address 20h
therefore C0h (11000000) must be
load in CWR register.

Program for square waveProgram for square wave
• MVI A,5Eh
OUT 24H
MVI A,42H
OUT 25H
MVI A,C0H
OUT 20H
HLT

Feature of 8255Feature of 8255
• It is programmable parallel I/O device
• It has 3,8 bit I/O Ports: Port A, Port B,
Port C, which are arranged in two
groups of 12 pins.
• TTL compatible
• Direct bit set/reset capability is
available for Port C

8255 PIN DIAGRAM8255 PIN DIAGRAM
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 BLOCK DIAGRAM8255 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.

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.

Group A and Group B Control:
Group A control consists of Port A and Port C
upper.
Group B control consists of Port B 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.

• 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 Operating Mode8255 Operating Mode
• 8255 provides one control word register
• It is selected when
A0=1,A1=1,CS^=0,WR^=0
• The read operation is not allowed for CWR
• There are two CWR formats (mode)
1)BSR mode 2)I/O mode
• The two basic modes are selected by D7
bit of control register
• when D7=1 it is a I/O mode & D7=0 it is
BSR mode

8255 MODES8255 MODES
• BSR (Bit Set Reset) Mode
• Only C is available for bit mode access.
• Allows single bit manipulation for control
applications
• 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)

• Bit D7 must be zero for BSR mode
• The BSR mode is a port C bit set/reset mode.
• The indivisible bit of port C can be set or reset
by writing a control word in CWR.
• Port C bit set/reset ; if D0=0 reset
D0=1 set
• The port pin of port C is selected using bit
D3,D2,D1
• The BSR mode affect only one bit of port C at a
time
BSR modeBSR mode
0 X bbbxx S/R
D7 D6 D1D2D3D4D5 D0

Port C bit selectionPort C bit selection
D3/b D2/b D1/b Bit
0 0 0 Bit 0
0 0 1 Bit 1
0 1 0 Bit 2
0 1 1 Bit 3
1 0 0 Bit 4
1 0 1 Bit 5
1 1 0 Bit 6
1 1 1 Bit 7

ExampleExample
• Write a set of instruction to perform the
following
1)Set bit 4 of port C
2)Reset bit 4 of port C( Assume Port C Address
=12 h)
Sol: 1) MVI A, 09h
OUT 12h
2) MVI A,08h
OUT 12h

I/O mode (CWR)I/O mode (CWR)
• When the bit D7=1 then I/O mode is selected
• The bit D6 and D5 used for mode selection of Group A
• If the D4 =1port A act as I/P port
D4 = 0 port A act as O/P Port
• If the D3 = 1 Port C Upper act as I/P Port
D3 = 0 port C Upper act as O/P Port
• The bit D2 used for mode selection of Group B
• If the D1 = 1 port B act as I/P Port
• D1 = 0 port B act as O/P Port
• If the D0 = 1 Port C Lower act as I/P Port
D0 = 0 Port C Lower act as O/P Port
I/O,BSRMode A PBMode BPCUPAMode A PCL
D7 D6 D1D2D3D4D5 D0

ExampleExample
• Write a program to initialize 8255 in
the configuration given below:
1 Port A : Simple I/P
2 Port B : Simple O/P
3 Port CL: O/P
4 Port CU: I/P
Assume CWR address is 83h

solutionssolutions
• MVI A,98h
OUT 83h
I/O,BSRMode A PBMode BPCUPAMode A PCL
1 0 00110 0

INTERFACING 8085 & 8255INTERFACING 8085 & 8255
• Here 8255 is interfaced in I/O 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

• Thus we get addresses ,considering don’t cares to
be zero as
Port A =80H
Port B =81H
Port C =82H
CWR =83H
• Then, 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.

82558085 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.
SolutionSolution

INTERFACING STEPPERINTERFACING STEPPER
MOTOR with 8255MOTOR with 8255

Stepper motorStepper motor
• Hardware : A stepper motor is a
digital motor. It can be driven by
digital signals motor shown in the
figure ( ckt ) has two phases, with
center tap winding.
• The center taps of these winding are
connected to the +5Volt supply.
• Due to this, motor can be excited by
grounding 4 terminals of the 2
winding.

Locking system for stepperLocking system for stepper
motormotor
• In the stepper motor it is not
desirable to excite both the ends of
the same winding simultaneously.
• This cancel the flux and motor
winding may damage.
• To avoid this digital clocking system
must be design.

Stepper Motor ProgrammingStepper Motor Programming
• MVI A, 80H (CWR)
• OUT 43H (CWR ADDRESS)
• UP:MVI A,88H
• OUT 40H (PORT A ADDRESS)
• CALL 6666H (CALL Delay)
• MVI A,11H
• OUT 40H
• CALL 6666H
• MVI A,22H
• OUT 40H
• CALL 6666H
• MVI A,44H
• OUT 40H
• JMP UP1-Aug-13 Prof.Nitin Ahire 198

Data bit pass to the port AData bit pass to the port A
PA0 PA1 PA2 PA3
1 0 1 0 = 0A
1 0 0 1 = 09
0 1 0 1 = 05
0 1 1 0 = 06

Data bitsData bits
• 5000H 0AH
• 5001H 09H
• 5002H 05H
• 5003H 06H

Initialized Port A as O/P PortInitialized Port A as O/P Port
• Program for stepper motor
LXI SP,FFFFH
MVI A,80H ; to make port A as o/p port
OUT CWR
BACK: LXI H,5000H ;HL act as memory pointer
MVI C,04H ;counter for the steps
UP: MOV A,M ;data bits transfer on the lower nibble of
port A
OUT PORT A
CALL DELAY ; delay for the steps
INX H ; increment the HL pair for the next
data bits
DCR C ; decrement the counter
JNZ UP ;check zero flag
JMP BACK ; jump back for continuous loop ( motor
rotation)

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