ECE 2211 Microprocessor and Interfacing Chapter 8 The 8088/8086 Microprocessors and their memory and I/O interfaces Br. Athaur Rahman Bin Najeeb Room 2.105.

ECE 2211 Microprocessor and Interfacing

Chapter 8

The 8088/8086 Microprocessors and their memory and I/O interfaces

Br. Athaur Rahman Bin NajeebRoom 2.105

Email: [email protected]: http://eng.iiu.edu.my/~athaur

Consultation : Tuesday 10.00 am ( appointment)

Comparision 8088 - 8086

Difference : Min and Max mode

8088 Minimum-Mode Signals

8086 Minimum-Mode Signals

ModeSelect MN/MX’

Interruptinterface

8086 MPU

Power supply

Vcc GND

INTR

_____INTA

_____TEST

NMI

RESET

HOLD

HLDA

Address / data bus

AD0-AD15,A16/S3-A19/S6

ALE____BHE/S7

M/IO’

DT/R’___RD___WR____DEN

READY

CLK

DMAinterface

Memory/IOcontrols

BLOCK DIAGRAM OF THE 8288

SYSTEM CLOCK

Clock (CLK) : input signal which synchronize the internal and external operations of the microprocessor.

• The clock source is generated by 8284 ( clock generator and Driver IC )

• CLK ( 8) of 8284 is connected to pin 19 8088/8086

• 8284 also supplies it with 2 of it's control lines – RESET and READY. The RESET signal does resets the 8088. This line can also be used by other peripherals on the computer so that they reset when the 8088 resets.

•READY used to slow down the 8088 ; Dfrom IO circuit thru RD1 and RD2

• A crystal oscillator is connected between X1 and X2 which provides a FUNDAMENTAL CRYSTAL FREQUENCY. ( FCF)

• 33% duty cycle the FCF is divided by 3 internally by 8244 to provide the necessary CLK

• output pin pclk provide 50% of duty cycle to drive periperal devices

CLOCK GENERATOR IC

http://en.wikipedia.org/wiki/Crystal_oscillator

What is the frequency

Complete connection from 8244 - 8088

8.7 BUS CYCLE AND TIME STATES

• A bus cycle defines the basic operation that a microprocessor performs to communicate with external devices.

•Examples of bus cycle are memory read, memory write, input/output read and input/output write.

•A bus cycle corresponds to a sequence of events that starts with an address being output on the system bus followed by a read or write data transfer.

•During these operations, a series of control signal are also produced by the MPU to control the direction and timing of the bus.

•Each bus cycle consists of at least four clock periods, T1, T2, T3 and T4.

•These clock period are also called T-state.

• These 4 clock states gives a bus cycle duration of ( 125 ns * 4 ) = 500 ns in a 8-Mhz 8088

• Idle State: no bus activity ; one clock period

• Wait state : controlled by READY signal ; inserted between T3 and T4 when READY = 0 . Bus cycle will complete when READY = 1

BUS CYCLE AND TIME STATES

• T1

– Address placed on bus

– ALE active

• T2

– Change direction of Data bus for READ instructions

• T3-4

– Data transfer occurs

Timer States

Bus Cycle and Time StatesT1 - start of bus cycle. Actions include setting control signals to give the

required values for ALE, DTR, IO/M putting a valid address onto the address bus.

T2 - the RD or WR control signals are issued, DEN is asserted and in the case of a write, data is put onto the data bus. The DEN turns on the data bus buffers to connect the CPU to the external data bus. The READY input to the CPU is sampled at the end of T2 and if READY is low, a wait state TW (one or more) is inserted before T3 begins.

T3 - this clock period is provided to allow memory to access the data. Ifthe bus cycle is a read cycle, the data bus is sampled at the end of T3.

T4 - all bus signals are deactivated in preparation for the nextclock cycle. The 8088 also finishes sampling the data (in a readcycle) in this period. For the write cycle, the trailing edge of theWR signal transfers data to the memory or I/O, which activatesand write when WR returns to logic 1 level.

System Timing Diagrams

T-State:— One clock period is referred to as a T-State

T-State— An operation takes an integer number of T-States

CPU Bus Cycle:

— A bus cycle consists of 4 or more T-States

T1 T2 T3 T4

Wait and Idle States

• Idle State

– No bus activity required

– Each is 1 clock period long

– Occurs when instruction queue is full or the MPU does

not need to read/write to memory

• Wait State

– Triggered by events external to MPU

– Buffer full will trigger a wait state

– Triggered by READY pin

– Inserted between T3 and T4

What is the duration of the bus cycle in the 8088 based microcomputer if the clock is 8MHz and two wait states are inserted

The duration of the bus cycle is in an 8MHz system is given in general by

Tcyc = 500 ns + N x 125ns

Tcyc = 500 ns +2 x 125ns

= 750 ns

What is the duration of the bus cycle in the 8086 based microcomputer if the clock is 5MHz

a)No wait state ?

b)with three wait states are inserted.

8.8 HARDWARE ORGANIZATION OF THE

MEMORY ADDRESS SPACE

Hardware Organization of the memory Address Space

8088 8086

Low / EvenBankHigh / Odd

Bank

Bank – select signals

8088 byte transfer 8088 word transfer

Byte / Word Transfer 8088

QuestionA memory cycle for an 8088 running at 5Mhz has no wait / idle state. What is the duration

for

A) to write a byte into memory

B) to write a word into memory

8086 memory access

8086 byte access on even address ( low)

Address busA19 – A1

D15 – D8

D7 – D0

____BHE ( HIGH )

Y Y + 1X + 1 X

A0 (LOW)

A0 is set to 0 to enable low bank

BHE is set to logic 1 to disable high bank

Data is transferred via D0 ( LSB ) – D7 ( MSB )

8086 byte access on odd address ( high bank)

Address busA19 – A1

D15 – D8 D7 – D0

____BHE ( LOW )

Y Y + 1

X + 1 X

A0 (HIGH)

A0 is set to 1 to disable low bank

BHE is set to logic 0 to enable high bank

Data is transferred via D8 ( LSB ) – D15 ( MSB )

8086 word access on even address ( lowbank) - aligned

Address busA19 – A1

D15 – D8 D7 – D0

____BHE ( LOW )

Y Y + 1

X + 1 X

Both A0 and BHE is enabled ; data transferred from both banks at same time

Data is transferred via D0 ( LSB ) – D15 ( MSB )

Aligned and occurs in 1 bus cycle

8086 word access on odd address ( high bank)

A word starting at ODD address : Misaligned

The LSB is located at lower address in High Bank,( Example : 00003(h) and 00004(h)

Requires 2 bus cycles, where X+1 address in high bank is accessed during the first Bus cycle ( A0=1, BHE=0) and data is transferred using D8 to D15

In second bus cycle ( A0 = 0, BHE = 1 , data transferred via D0 – D7 )

The next two slaids demonstrated this activity

Address busA19 – A1

D15 – D8 D7 – D0

____BHE ( LOW )

X + 2X + 3

X + 1 X

A0 (HIGH)

ODD ADDRESS WORD TRANSFER BY THE 8086

First bus cycle

Address busA19 – A1

D15 – D8 D7 – D0

____BHE ( HIGH )

X + 2X + 3

X + 1 X

A0 (LOW)

ODD ADDRESS WORD TRANSFER BY THE 8086

Second bus cycle