UNIT-IV

8255 PPI Various Modes Of Operation Interfacing To 8086. Interfacing Keyboard, Displays Stepper MotorD/A And A/D Converter Interfacing.

8255Programmable Peripheral Interface(PPI)

For most of the applications 8086 needs parallel input ,output data to/from the peripherals.

8255 is a device used to interface different real time peripherals to the microprocessors which can be initialized for different modes operations like simple I/O, Strobed I/O and handshaking I/O.

It has three 8-bit parallel ports :port A, port B, port C

Fig 4.2 8255 pin diagram

8255 has three operation modes: mode 0, mode 1, and mode 2

Fig 4.3 Control Word Register Bit Format of 8255 for I/O mode and BSR mode

Programming 8255 Mode 0:

— Ports A, B, and C can be individually programmed as input or output ports— Port C is divided into two 4-bit ports which are independent from each other

Mode 1:— Ports A and B are programmed as input or output ports— Port C is used for handshaking

PA[7:0]STBAIBFAINTRAPC3

PC5PC4

PB[7:0]STBBIBFBINTRBPC0

PC1PC2

PC6, 7

8255

PA[7:0]OBFAACKAINTRAPC3

PC6PC7

PB[7:0]OBFBACKBINTRBPC0

PC1PC2

PC4, 5

8255

Fig 4.4 Input and Output control signals of 8255 in mode1

Fig 4.5 Input and Output control signals of 8255 in mode1

Programming 8255 Mode 2:

— Port A is programmed to be bi-directional— Port C is for handshaking— Port B can be either input or output in mode 0 or mode 1

PA[7:0]

OBFAACKA

INTRA

PC4PC6PC7

STBA

IBFA

PC0

PC3PC58255

PC0PC0

PB[7:0]

In Out In OutIn Out

Mode 0

STBB OBFB IBFB ACKB

INTRB INTRB

Mode 1

Fig 4.6 Input and Output control signals of 8255 in mode2

Fig 4.7 Input and Output control signals of 8255 in mode2

Example: Mode 1 Input

BIT5 EQU 20HPORTC EQU 22HPORTA EQU 20H

READ PROC NEARRead:

IN AL, PORTC ; read portc TEST AL, BIT5 ;test IBF JZ Read ;if IBF=0 IN AL, PORTA ;Read Data

READ ENDP

keyboard

PA0

PA7

STBPC4 DAV

8255

Fig 4.8 Interfacing of 8255 to Keyboard for input in mode1

Example: Mode 1 output

Printer

PB0

PB7

ACKPC2 ACK

8255

PC4 DS

Fig 4.9 Interfacing of 8255 to Printer for output in

mode1

BIT1 EQU 2PORTC EQU 62HPORTB EQU 61HCMD EQU 63HPRINT PROC

NEAR; check printer ready?IN AL, PORTC ;get OBFTEST AL, BIT1 ;test OBFJZ PRINT ;if OBF=0 buffer

is full

;send character to printerMOV AL, AH ;get dataOUT PORTB, AL ;print

data; send data strobe to

printerMOV AL, 8 ;clear DSOUT CMD, ALMOV AL, 9 ;clear DSOUT CMD, AL;rising the data at the

positive edge of DSRET

PRINT ENDP

Example: Mode 1 output

Data bus

8086

D[7:0]

A0A1

RDWR

RESET

CS

Control port

PA[7:0]

PB[7:0]

PC[7:0]

A7A6A5A4A3A2

IO/M

A1 A0 Port

0 00 11 01 1

PAPBPCControl

Fig 4.1 Interfacing of 8255 PPI to 8086

Keyboard example

Fig 4.10 Interfacing of 8255 to 4x4 matrix Keyboard in mode1

Keyboard example

Fig 4.11 Flow chart of a keyboard scanning procedure

Bouncing Problem

Fig 4.12 Key bouncing problem

Fig 4.13 Key bouncing problem

Software Solution

Fig 4.14 Software Key debouncing solution

The time taken by the ADC from the active edge of SOC pulse till the active edge of EOC signal is called as the conversion delay of the ADC.

General algorithm for ADC interfacing contains the following steps

1. Ensure the stability of analog input, applied to the ADC

2. Issue start of conversion SOC pulse to ADC3. Read end of conversion EOC signal to mark

the end of conversion process4. Read digital data output of the ADC as

equivalent digital output.

ADC Interfacing

ADC 0808/0809

The analog to digital converter chips 0808 and 0809 are 8-bit CMOS, successive approximation converters. It is fastest technique.

The conversion delay is 100 µs at a clock frequency of 640 kHz, which is quite low as compared to other converters.

Block Diagram of ADC 0808/0809This converter internally has a 3:8 analog

multiplexer, so that at a time 8 different analog inputs can be connected to the chips.

Out of these 8 inputs only one can be selected for conversion by using 3 address lines A,B,C.

The CPU may drive these lines using output port lines in case of multichannel applications.

Fig 4.15 Pin diagram of ADC 0808/0809

•These are unipolar Analog to Digital (A to D) converters, they are able to convert only positive analog input voltages to their digital equivalents. This chips do not contain any internal sample & hold circuit.

Table 4.1 analog input selection

Interfacing between ADC to Microprocessor

Problem:-

Interface ADC 0808 with 8086 using 8255 ports. Use Port A of 8255 for transferring digital data output of ADC to the CPU & Port C for control signals. Assume that an analog input is present at I/P2 of the ADC and a clock input of suitable frequency is available for ADC. Draw the schematic & timing diagram of different signals of ADC0808.

Solution:-

•The analog input I/P2 is used & therefore address pins A,B,C should be 0,1,0 respectively to select I/P2.•The OE (Out put latch Enable) & ALE pins are already kept at +5v to select the ADC and enable the outputs. •Port C upper acts as the input port to receive the EOC signal while Port C lower acts as the output port to send SOC to ADC.•Port A acts as a 8-bit input data port to receive the digital data output from the ADC.

8255 Control Word:D7 D6 D5 D4 D3 D2 D1 D01 0 0 1 1 0 0 0 = 98HProgram:

MOV AL,98H ; Initialize 8255, send AL to control word (CWR) OUT CWR, ALMOV AL, 02H ;Select I/P2 as analog I/POUT Port B, AL ;Port B as outputMOV AL, 00H ; Give start of conversion pulse to the ADCOUT Port C, ALMOV AL, 01HOUT Port C, ALMOV AL, 00HOUT Port C, AL

WAIT: IN AL, Port C ; check for EOC by reading Port C upper & rotating

RCL ; through carry.JNC WAITIN AL, Port A ; if EOC, read digital equivalent in ACHLT ; stop.

Fig 4.16 Interfacing of ADC 0808 to 8086 through 8255

Interfacing D/A Converters

The Digital to Analog Converters (DAC) convert binary numbers into their analog equivalent voltages.

The DAC find applications in areas likeDigitally controlled gainsMotor speed controlsProgrammable gain amplifiers etc.AD 7523 8-Bit Multiplying DAC:--• Intersil’s AD 7523 is a 16 pin DIP, multiplying

digital to analog converter, containing R-2R ladder (R=10K) for digital to analog conversion.

Power supply +5v to +15v Vref -> -10v to +10v The maximum analog output voltage will be

+10v A Zener is connected between OUT1 & OUT2

to save the DAC from negative transients.An operational amplifier is used as a current – to –

voltage converter at the output of AD 7523.An external feedback resister acts to control the

gain.

Interfacing of AD 7523 with 8086Problem:--Interface DAC AD7523 with the 8086 running at 8MHz & write ALP to generate a saw tooth waveform of period 1ms with Vmax 5v.Solution:--Code segmentAssume cs:codeStart:MOV AL, 80H

OUT CWR, ALAGAIN: MOV AL, 00HBACK: OUT Port A, AL

INC ALCMP AL, 0F2HJB BACKJMP AGAIN

Code endsEnd Start

Fig 4.17 Pin diagram of AD7523

Fig 4.18 Interfacing of AD7523 to 8086 through 8255

Display interfaceInterface an 8255 with 8086 at 80h as an I/O address of port-

A. interface five 7 segment displays with the 8255. write a sequence of instructions to display 1,2,3,4 and 5 over the five displays continuously as per their positions starting with 1 at the least significant position. CWR address is 86h.

Number to be displayed

PA7dp

PA6

a

PA5

b

PA4

c

PA3

d

PA2

e

PA1

f

PA0

g

Code

1 1 1 0 0 1 1 1 1 CF

2 1 0 0 1 0 0 1 0 92

3 1 0 0 0 0 1 1 0 86

4 1 1 0 0 1 1 0 0 CC

5 1 0 1 0 0 1 0 0 A4

All these codes are stored in a look up table starting at 2000:0001.

Fig 4.19 Interfacing multiplexed 7-segment display to 8086 through 8255

ALP for display interface

again: mov cl,05h ;count for displaysmov bx,2000h ;initialize the data segment for mov ds,bx ; look-up tablemov ch,01h ;1st no. to be displayedmov al,80hout 86h,al ;load control word in the CWR

mov dl,01h ;enable code for least significant 7-seg display

nxtdgt: mov bx,0000h ;set pointer to look-up tablemov al,ch ;store number to be displayxlat ;find code from table

out 80h,almov al,dlout 82h,al ;enable the displayrol dl ;go for next digit displayinc chdec cl ;decrement counterjnz nxtdgt ;go for next digit displayjmp again

Stepper Motor InterfacingA stepper motor is a device used to obtain an accurate

position control of rotating shafts.It employs rotation of its shaft in terms of steps, rather

than continuous rotation as in case of AC or DC motors.In dot-matrix printer one small stepper motor which is

used to advance the paper to the next line position & another small stepper motor which is used to move the print head to the next character position.

In floppy disk stepper motor is used to position the read/write head over the desired track.

To rotate the shaft of the stepper motor, a sequence of pulses is needed to be applied to the windings of the stepper motor, in a proper sequence.

The no. of pulses required for one complete rotation of the shaft of the stepper motor are equal to its number of internal teeth on its rotor.

The stator teeth the rotor teeth lock with each other to fix a position of the shaft .

With a pulse applied to the winding input, the rotor rotates by one teeth position or an angle x. The angle x may be calculated as:

x = 3600 / no. of rotor teethThe stepper motors have been designed to work

with digital circuits. Binary level pulses of 0-5v are required at its winding inputs to obtain the rotation of shafts.

The sequence of pulses can be decided, depending upon the required motion of the shaft.

The count for rotating the shaft of the stepper motor through a specified angle may be calculated from the no. of rotor teeth

C = no. of rotor teeth / 3600 * θ0

Motion Step A B C D

Clockwise 1 1 0 0 0

2 0 1 0 0

3 0 0 1 0

4 0 0 0 1

5 1 0 0 0

Anticlockwise 1 1 0 0 0

2 0 0 0 1

3 0 0 1 0

4 0 1 0 0

5 1 0 0 0

Table 4.2 Excitation sequence for clockwise and anticlockwise rotation of a stepper motor

Problem

Design a stepper motor controller and write an ALP to rotate shaft of a 4-phase stepper motor:

i. In clockwise 5 rotationsii.In anticlockwise 5 rotations.The 8255 port A address is 0740h. The stepper

motor has 200 rotor teeth.The port A bit PA0 drives winding Wa, PA1 drives

winding Wb and so on.The stepper motor has an internal delay of 10msec.

Assume that the routine for this delay is already available.

Solution:

ALP:Assume cs:CodeCode segmentStart: MOV AL, 80H

OUT CWR, ALMOV AL, 88H; Bit

pattern 10001000MOV CX, 1000

Again1: OUT Port A, ALCALL DELAYROL AL, 01DEC CXJNZ Again1MOV AL, 88HMOV CX, 1000

Again2: OUT Port A, ALCALL DELAYROR AL, 01DEC CXJNZ Again2MOV AH, 4CHINT 21H

Code endsEnd start