8085 microproceesor ppt

The 8085 Microprocessor
Architecture

The typical processor system consists of:
 CPU (central processing unit)
 ALU (arithmetic-logic unit)
 Control Logic
 Registers, etc…
 Memory
 Input / Output interfaces
Processor System Architecture
Interconnections between these units:
 Address Bus
 Data Bus
 Control Bus

The internal architecture of the 8085 CPU is capable of performing the following
operations:
 Store 8-bit data (Registers, Accumulator)
 Perform arithmetic and logic operations (ALU)
 Test for conditions (IF / THEN)
 Sequence the execution of instructions
 Store temporary data in RAM during execution
The 8085: CPU Internal Structure

Registers
 Six general purpose 8-bit registers: B, C, D, E, H, L
 They can also be combined as register pairs to
perform 16-bit operations: BC, DE, HL
 Registers are programmable (data load, move, etc.)
Accumulator
 Single 8-bit register that is part of the ALU !
 Used for arithmetic / logic operations – the result is always stored in the
accumulator.

• The Program Counter (PC)
– This is a register that is used to control the sequencing of the execution of
instructions.
– This register always holds the address of the next instruction.
– Since it holds an address, it must be 16 bits wide.
• The Stack pointer
– The stack pointer is also a 16-bit register that is used to point into memory.
– The memory this register points to is a special area called the stack.
– The stack is an area of memory used to hold data that will be retreived soon.
– The stack is usually accessed in a Last In First Out (LIFO) fashion.

The 8085 and Its Busses
• The 8085 is an 8-bit general purpose microprocessor that can
address 64K Byte of memory.
• It has 40 pins and uses +5V for power. It can run at a maximum
frequency of 3 MHz.
– The pins on the chip can be grouped into 6 groups:
• Address Bus.
• Data Bus.
• Control and Status Signals.
• Power supply and frequency.
• Externally Initiated Signals.
• Serial I/O ports.

The 8085 Bus Structure
The 8-bit 8085 CPU (or MPU – Micro Processing Unit) communicates with the other units
using a 16-bit address bus, an 8-bit data bus and a control bus.

The Address and Data Busses
• The 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.

The Control and Status Signals
• There are 4 main control and status signals. These are:
• ALE: Address Latch Enable. This signal is a pulse that
become 1 when the AD0 – AD7 lines have an address on
them. It becomes 0 after that. This signal can be used to
enable a latch to save the address bits from the AD lines.
• RD: Read. Active low.
• WR: Write. Active low.
• IO/M: This signal specifies whether the operation is a
memory operation (IO/M=0) or an I/O operation
(IO/M=1).
• S1 and S0 : Status signals to specify the kind of
operation being performed .Usually un-used in small
systems.

Frequency Control Signals
• There are 3 important pins in the frequency control group.
– X0 and X1 are the inputs from the crystal or clock generating
circuit.
• The frequency is internally divided by 2.
– So, to run the microprocessor at 3 MHz, a clock running
at 6 MHz should be connected to the X0 and X1 pins.
– CLK (OUT): An output clock pin to drive the clock of the rest of the
system.
• We will discuss the rest of the control signals as we get to them.

Microprocessor Communication and Bus Timing
• To understand how the microprocessor operates and uses these
different signals, we should study the process of communication
between the microprocessor and memory during a memory read
or write operation.
• Lets look at timing and the data flow of an instruction fetch
operation. (Example 3.1)

Steps For Fetching an
Instruction
• Lets assume that we are trying to fetch the instruction at memory
location 2005. That means that the program counter is now set to that
value.
– The following is the sequence of operations:
• The program counter places the address value on the address
bus and the controller issues a RD signal.
• The memory’s address decoder gets the value and determines
which memory location is being accessed.
• The value in the memory location is placed on the data bus.
• The value on the data bus is read into the instruction decoder
inside the microprocessor.
• After decoding the instruction, the control unit issues the
proper control signals to perform the operation.

Timing Signals For Fetching an Instruction
• Now, lets look at the exact timing of this sequence of events as that is
extremely important. (figure 3.3)
– At T1 , the high order 8 address bits (20H) are placed on the address
lines A8 – A15 and the low order bits are placed on AD7–AD0. The
ALE signal goes high to indicate that AD0 – AD8 are carrying an
address. At exactly the same time, the IO/M signal goes low to indicate
a memory operation.
– At the beginning of the T2 cycle, the low order 8 address bits are
removed from AD7– AD0 and the controller sends the Read (RD)
signal to the memory. The signal remains low (active) for two clock
periods to allow for slow devices. During T2 , memory places the data
from the memory location on the lines AD7– AD0 .
– During T3 the RD signal is Disabled (goes high). This turns off the
output Tri-state buffers in the memory. That makes the AD7– AD0
lines go to high impedence mode.

Demultiplexing AD7-AD0
– From the above description, it becomes obvious that the AD7– AD0
lines are serving a dual purpose and that they need to be
demultiplexed to get all the information.
– The high order bits of the address remain on the bus for three clock
periods. However, the low order bits remain for only one clock
period and they would be lost if they are not saved externally. Also,
notice that the low order bits of the address disappear when they
are needed most.
– To make sure we have the entire address for the full three clock
cycles, we will use an external latch to save the value of AD7– AD0
when it is carrying the address bits. We use the ALE signal to
enable this latch.

Demultiplexing AD7-AD0
– Given that ALE operates as a pulse during T1,
we will be able to latch the address. Then
when ALE goes low, the address is saved and
the AD7– AD0 lines can be used for their
purpose as the bi-directional data lines.
A15-A8
LatchAD7-AD0
D7- D0
A7- A0
8085
ALE

Cycles and States
• From the above discussion, we can define terms that will
become handy later on:
– T- State: One subdivision of an operation. A T-state lasts for
one clock period.
• An instruction’s execution length is usually measured in
a number of T-states. (clock cycles).
– Machine Cycle: The time required to complete one operation
of accessing memory, I/O, or acknowledging an external
request.
• This cycle may consist of 3 to 6 T-states.
– Instruction Cycle: The time required to complete the
execution of an instruction.
• In the 8085, an instruction cycle may consist of 1 to 6
machine cycles.

Generating Control Signals
• The 8085 generates a single RD signal. However, the signal
needs to be used with both memory and I/O. So, it must be
combined with the IO/M signal to generate different control
signals for the memory and I/O.
– Keeping in mind the operation of the IO/M signal we can
use the following circuitry to generate the right set of
signals:

A closer look at the 8085
Architecture
• Previously we discussed the 8085 from a
programmer’s perspective.
• Now, lets look at some of its features
with more detail.

The ALU
• In addition to the arithmetic & logic circuits, the ALU
includes the accumulator, which is part of every
arithmetic & logic operation.
• Also, the ALU includes a temporary register used for
holding data temporarily during the execution of the
operation. This temporary register is not accessible by
the programmer.

The Flags register
– There is also the flags register whose bits are affected by the arithmetic &
logic operations.
• S-sign flag
– The sign flag is set if bit D7 of the accumulator is set after an
arithmetic or logic operation.
• Z-zero flag
– Set if the result of the ALU operation is 0. Otherwise is reset. This
flag is affected by operations on the accumulator as well as other
registers. (DCR B).
• AC-Auxiliary Carry
– This flag is set when a carry is generated from bit D3 and passed
to D4 . This flag is used only internally for BCD operations.
(Section 10.5 describes BCD addition including the DAA
instruction).
• P-Parity flag
– After an ALU operation if the result has an even # of 1’s the p-
flag is set. Otherwise it is cleared. So, the flag can be used to
indicate even parity.
• CY-carry flag
– Discussed earlier

More on the 8085 machine
cycles
• The 8085 executes several types of instructions with each
requiring a different number of operations of different types.
However, the operations can be grouped into a small set.
• The three main types are:
• Memory Read and Write.
• I/O Read and Write.
• Request Acknowledge.
• These can be further divided into various operations (machine
cycles).

Opcode Fetch Machine Cycle
• The first step of executing any instruction is the Opcode fetch cycle.
– In this cycle, the microprocessor brings in the instruction’s Opcode
from memory.
• To differentiate this machine cycle from the very similar
“memory read” cycle, the control & status signals are set as
follows:
– IO/M=0, s0 and s1 are both 1.
– This machine cycle has four T-states.
• The 8085 uses the first 3 T-states to fetch the opcode.
• T4 is used to decode and execute it.
– It is also possible for an instruction to have 6 T-states in an opcode
fetch machine cycle.

Memory Read Machine Cycle
• The memory read machine cycle is
exactly the same as the opcode fetch
except:
– It only has 3 T-states
– The s0 signal is set to 0 instead.

The Memory Read Machine
Cycle
– To understand the memory read machine cycle, let’s study the
execution of the following instruction:
• MVI A, 32
– In memory, this instruction looks like:
• The first byte 3EH represents the opcode for loading a byte into
the accumulator (MVI A), the second byte is the data to be
loaded.
– The 8085 needs to read these two bytes from memory before it can
execute the instruction. Therefore, it will need at least two machine
cycles.
– The first machine cycle is the opcode fetch discussed earlier.
– The second machine cycle is the Memory Read Cycle.
– Figure 3.10 page 83.
2000H
2001H
3E
32

Machine Cycles vs. Number of bytes in the instruction
• Machine cycles and instruction length, do not have a direct relationship.
– To illustrate lets look at the machine cycles needed to execute the following
instruction.
• STA 2065H
• This is a 3-byte instruction requiring 4 machine cycles and 13 T-states.
• The machine code will be stored
in memory as shown to the right
• This instruction requires the following 4 machine cycles:
– Opcode fetch to fetch the opcode (32H) from location 2010H,
decode it and determine that 2 more bytes are needed (4 T-states).
– Memory read to read the low order byte of the address (65H) (3 T-
states).
– Memory read to read the high order byte of the address (20H) (3 T-
states).
– A memory write to write the contents of the accumulator into the
memory location.
2010H
2011H
2012H
32H
65H
20H

The Memory Write Operation
• In a memory write operation:
– The 8085 places the address (2065H) on the
address bus
– Identifies the operation as a memory write
(IO/M=0, s1=0, s0=1).
– Places the contents of the accumulator on the data
bus and asserts the signal WR.
– During the last T-state, the contents of the data
bus are saved into the memory location.

Memory interfacing
• There needs to be a lot of interaction between the
microprocessor and the memory for the exchange of
information during program execution.
– Memory has its requirements on control signals
and their timing.
– The microprocessor has its requirements as well.
• The interfacing operation is simply the matching of
these requirements.

Memory structure & its requirements
• The process of interfacing the above two chips is the
same.
– However, the ROM does not have a WR signal.
Address
Lines
Date
Lines
CS
RDOutput Buffer
ROM
Address
Lines
Data Lines
CS
RDOutput Buffer
RAM
WRInput Buffer
Data Lines

Interfacing Memory
– Accessing memory can be summarized into the following three
steps:
– Select the chip.
– Identify the memory register.
– Enable the appropriate buffer.
– Translating this to microprocessor domain:
– The microprocessor places a 16-bit address on the address
bus.
– Part of the address bus will select the chip and the other
part will go through the address decoder to select the
register.
– The signals IO/M and RD combined indicate that a
memory read operation is in progress. The MEMR signal
can be used to enable the RD line on the memory chip.

Address decoding
• The result of address decoding is the identification of
a register for a given address.
– A large part of the address bus is usually
connected directly to the address inputs of the
memory chip.
– This portion is decoded internally within the chip.
– What concerns us is the other part that must be
decoded externally to select the chip.
– This can be done either using logic gates or a
decoder.

The Overall Picture
• Putting all of the concepts together, we
get:
A15-A8
LatchAD7-AD0
D7- D0
A7- A0
8085
ALE
IO/MRDWR
1K Byte
Memory
Chip
WRRD
CS
A9- A0
A15- A10
Chip Selection
Circuit

8085 Instruction Set
 Data transfer operations
• Between registers
• Between memory location and a register
• Direct write to a register / memory
• Between I/O device and accumulator
 Arithmetic operations (ADD, SUB, INR, DCR)
 Logic operations
 Branching operations (JMP, CALL, RET)

Simple Data Transfer Operations
Examples:
 MOV B,A 47 From ACC to REG
 MOV C,D 4A Between two REGs
 MVI D,47 16 Direct-write into REGD
47

Simple Data Transfer Operations
Example:
 OUT 05 D3
05
Contents of ACC sent to output port number 05.

Simple Memory Access Operations

Branching Operations
Note: This is an unconditional jump operation.
It will always force the program counter to a fixed
memory address continuous loop !

Branching Operations
Conditional jump operations are very useful for
decision making during the execution of the program.

Direct Memory Access Operations

 Use a register PAIR as an address pointer !
 We can define memory access operations using the memory location (16 bit address)
stored in a register pair: BC, DE or HL.
 First, we have be able to load the register pairs.
LXI B, (16-bit address)
LXI D, (16-bit address)
LXI H, (16-bit address)
 We can also increment / decrement register pairs.
Indirect Memory Access Operations

In many real-time operations, the microprocessor should be able to receive an
external asynchronous signal (interrupt) while it is running a routine.
When the interrupt signal arrives:
 The processor will break its routine
 Go to a different routine (service routine)
 Complete the service routine
 Go back to the “regular” routine
Interrupts in 8085

In order to execute an interrupt routine, the processor:
 Should be able to accept interrupts (interrupt enable)
 Save the last content of the program counter (PC)
 Know where to go in program memory to execute
the service routine
 Tell the outside world that it is executing an interrupt
 Go back to the saved PC location when finished.
Interrupts in 8085

There are four other interrupt inputs in 8085 that
transfer the operation immediately to a specific address:
 TRAP: go to 0024
 RST 7.5: go to 003C
 RST 6.5 0034
 RST 5.5 002C
 RST 7.5, RST 6.5 and RST 5.5 are maskable interrupts, they are
acknowledged only if they are not masked !
Vectored Interrupts

8085 microproceesor ppt

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8085 microproceesor ppt