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MICROPROCESSORS-8086 PowerPoint PPT Presentation


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MICROPROCESSORS-8086. Architecture Programming Interfacing. FOR CONVINIRNT STUDY OF MICROPROCESSORS TWO TYPES OF MODELS ARE USED :. PROGRAMMER’S MODEL :- THIS MODEL SHOWS FEATURES , SUCH AS INTERNAL REGISTERS, ADDRESS ,DATA & CONTROL BUSES ; THAT WE NEED TO PROGRAM THE DEVICE.

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Microprocessors 8086 l.jpg

MICROPROCESSORS-8086

  • Architecture

  • Programming

  • Interfacing


Slide2 l.jpg

FOR CONVINIRNT STUDY OF MICROPROCESSORS TWO TYPES OF MODELS ARE USED :

  • PROGRAMMER’S MODEL :- THIS MODEL SHOWS FEATURES , SUCH AS INTERNAL REGISTERS, ADDRESS ,DATA & CONTROL BUSES ; THAT WE NEED TO PROGRAM THE DEVICE.

  • THE HARDWARE MODEL:- THIS MODEL SHOWS THE PIN DIAGRAM AND THE SIGNALS TO/FROM THIS PINS TO UNDERSTAND HOW A MOCROCOMPUTER SYSTEM IS BUILT AROUND.


Microcomputer l.jpg

MICROCOMPUTER

  • A MICROCOMPUTER SYSTEM IS ONE WHICH USES A MICROPROCESSOR AS ITS CPU

  • IN ADDITION THE MICROCOMPUTER ALSO HAS A MEMORY UNIT,INPUT/OUTPUT DEVICES AND SYSTEM BUSES.

  • THE SYSTEM BUSES ARE OF THREE TYPES: 1.ADDRESS BUS 2.DATA BUS 3.CONTROL BUS

  • PHYSICALLY BUSES ARE GROUP OF WIRES


8086 buses l.jpg

8086 BUSES

  • THE 8086 HAS

  • 20 ADDRESS LINES

  • 16 DATA LINES

  • 4-10 CONTROL LINES.

  • WITH THIS THE 8086 IS ABLE

  • TO ADDRESS 1,048,,576 (220 ) MEMORY LOCATIONS/PORTS.

  • TO MANIPULATE AND/OR OPERATE ON 16-BITS(2-BYTES) OF DATA AT A TIME.

  • TO GENERATE NECESSARY CONTROL SIGNALS.


Architecture l.jpg

  • THE INTERNAL ARCHITECTURE OF 8086 CAN BE MAINLY DIVIDED INTO TWO UNITS:

  • BUS INTERFACE UNIT (BIU)

  • EXECUTION UNIT (EU)

ARCHITECTURE

  • THE BIU CONTAINS :

  • CODE SEGMENT REGISTER (CS)

  • DATA SEGMENT REGISTER (DS)

  • EXTRA SEGMENT REGISTER (ES)

  • STACK SEGMENT REGISTEER (SS) AND

  • INSTRUCTION POINTER (IP)


Slide6 l.jpg

  • THE EU CONTAINS THE FOLLOWING 8-BIT REGISTERS:

  • AH & AL (AX-16 BIT)

  • BH & BL (BX-16 BIT)

  • CH & CL (CX-16 BIT)

  • DH 7 DL (DX-16 BIT)

  • IT ALSO INCLUDES THE FOLLOWING 16-BIT REGISTERS:

  • STACK POINTER (SP)

  • BASE POINTER (BP)

  • SOURCE INDEX (SI)

  • DESTINATION INDEX (DI)


8086 system connections timing troubleshooting l.jpg

8086 SYSTEM CONNECTIONS,TIMING & TROUBLESHOOTING


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40

VCC

AD15

GND

1

A16-A19 & S3-S6

2

35

34

BHE/S7

33

MN/MX

32

RD

AD14 TO AD0

31

RQ/GTO(HOLD)

30

RQ/GT1(HLDA)

8086 CPU

29

LOCK (WR)

28

S2 (M/IO)

27

S1 (DT/R)

16

26

NMI

S0 (DEN)

17

INTR

18

25

QS0 (ALE)

CLK

19

24

QS1(INTA)

GND

20

TEST

23

RESET

21

22

READY


Slide9 l.jpg

BASIC 8086 MINIMUM MODE SYSTEM

MN/MXM/IO

INTA

RD

WR

DT/R

DEN

ALE

AD0-AD15

A16-A19

CLK READYRESET

8284A CLOCKGENE-RATOR

8282

LATCH

ADDR

WAIT STATE GENERATOR

8286

TRAN- CEIVER

ADDR/DATA

DATA

RAM 2142

PERI-

PHERAL

2716

PROM


Timing sequence l.jpg

TIMING SEQUENCE

  • AN EXTERNAL CLOCK GENERATOR DEVICE IS CONNECTED TO 8086 TO PROVIDE CLOCK SIGNALS THROUGHOUT THE SYSTEM.

  • ONE CYCLE OF CLOCK IS CALLED A STATE OR T-STATE.

  • EACH BASIC OPERATION SUCH AS READING A MEMORY LOCATION OR WRITING TO A PORT REQUIRES SEVERAL STATES.THIS GROUP OF STATES IS CALLED A MACHINE CYCLE.

  • THE TOTAL TIME REQUIRED TO FETCH AND EXECUTE AN INSTRUCTION IS CALLED AN INSTRUCTION CYCLE. AN INSTRUCTION CYCLE CONSISTS OF ONE OR MORE MACHINE CYCLE.


Basic signal flow on 8086 buses l.jpg

BASIC SIGNAL FLOW ON 8086 BUSES

BASICALLY THERE ARE TWO OPERATIONS TO SEE:

1.READ OPERATION

2. WRITE OPERATION

WILL SEE WHAT IS GOING ON DURING THIS TWO CYCLES OF OPERATION.


Read cycle l.jpg

READ CYCLE

  • HERE WE WILL SEE THE ACTIVITIES CARRIED OUT ON 8086 BUSES AT VARIOUS TIME INSTANTS WHEN IT READS FROM A MEMORY LOCATION OR FROM A PORT.

  • HERE WE WILL ASSUME THAT THE 8086 IS OPERATED IN IS MINIMUM MODE.


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T1

T2

T3

TW

T4

CLK

M/IO

ALE

MEMORY ACCESS TIME

ADDR/ DATA

RESERVED FOR DATA

VALID D15-D0

A15-A0

ADDR/ STATUS

A19-A16

RD/INTA

READY

DT/R

DEN


Write cycle l.jpg

WRITE CYCLE

  • HERE WE WILL SEE THE ACTIVITIES CARRIED OUT ON 8086 BUS AT VARIOUS TIME INSTANTS WHEN IT WRITES TO A PORT OR A MEMORY LOCATION.

  • HERE WE WILL ASSUME THAT THE 8086 IS OPERATED IN IS MINIMUM MODE.


Slide15 l.jpg

T1

T2

T3

TW

T4

CLK

M/IO

ALE

ADDR/ DATA

A15-A0

DATA OUT (D15-D0)

ADDR/ STATUS

A19-A16

WR

READY

DT/R

DEN


Addressing l.jpg

ADDRESSING

1. ADDRESSING MEMORY

2. ADDRESSING PORTS

  • DECODER IS THE CIRCUITRY USED FOR ADDRESING. IT SERVES TWO PURPOSES:

  • TO ENABLE RAM,ROM OR PORT.

  • TO MAKE SURE THAT ONLY ONE DEVICE IS ENABLED AT A TIME.


A system rom decoder l.jpg

A SYSTEM ROM DECODER

  • TO UNDERSTAND THE CONCEPT A GENERAL DIGITAL SYSTEM WITH 8 DATA LINES AND 16 ADDRESS LINES IS CONSIDERED.

  • THE HARDWARE USED CONSIST

  • 2732 ROM-8(EACH-4 KB)

  • 74LS138 DECODER/DEMULTIPLEXER

  • 16 ADDRESS LINES

  • 8 DATA LINES


Slide18 l.jpg

A0

A1

A2

A11

ROM 1

2732

ROM 7

2732

ROM 0

2732

CS

CS

CS

D0

D5

D6

D7

Y0 Y1 Y7

A12

74LS138

A13

G2A

G2B

G1

A14

A15

RD

+5V


Address decoder worksheet l.jpg

ADDRESS DECODER WORKSHEET


A system ram decoder l.jpg

A SYSTEM RAM DECODER

TO THE SAME SYSTEM WE WANT TO ADD 16KB RAM.SO ADDITIONAL HARDWARE WE REQUIRE:

  • 2142 RAM-8(EACH 2KB)

  • ONE MORE 74LS138 DECODER/DEMULTIPLEXER

  • SINCE WE HAVE EACH RAM OF 2KB(2048 BYTES) ,WE NEED 11 ADDRESS LINES TO ADDRESS EACH MEMORY LOCATION WITHIN A RAM.AS WE HAVE OCCUPIED ADDRESSES 0000 THROUGH 7FFF FOR ROM.WE MUST START RAM ADDRESSES AFTER 7FFF.

  • SO WE HAVE 5 MORE ADDRESS LINES.WHICH WE WILL USE TO PROVIDE ADDRESS DECODING.


Slide21 l.jpg

A0

A1

A2

A10

RAM 1

2142

RAM 7

2142

RAM 0

2142

CS

CS

CS

D0

D5

D6

D7

Y0 Y1 Y7

A11

74LS138

A12

G2A

G2B

G1

A13

A14

GND

A15


A system port decoder l.jpg

A SYSTEM PORT DECODER

A SYSTEM PORT CAN BE ADDRESSED IN TWO WAYS :

  • USING MEMORY MAPPED I/O- HERE A PORT IS TREATED AS IF IT IS A MEMORY LOCATION.

  • IT USES MEMORY RELATED INSTRUCTIONS.

  • THE OPERATION IS FASTER.

  • A MAXIMUM OF 1MB INPUT AND 1MB(220) OUTPUT DEVICES CAN BE ADDRESSED.

  • IT USES MEM.READ AND MEM.WRITE CONTROL SIGNALS.

  • IT CONSUMES THE ADDRESS RANGE USED BY PROGRAM MEMORY.

  • DECODING IS COMPLEX.


Slide24 l.jpg

  • USING DIREC I/O-HERE AN INPUT OR AN OUTPUT DEVICE IS TREATED AS A DISTINCT I/O DEVICE.

  • IT DOES NOT CONSUME PROGRAM MEMORY.

  • THE DECODING IS SIMPLE.

  • IT USES IN AND OUT INSTRUCTIONS.

  • A MAXIMUM OF 64K BYTE TYPE INPUT AND OUTPUT DEVICES CAN BE ADDRESSED OR 32K WORD TYPE INPUT AND OUTPUT DEVICES CAN BE ADDRESSED.

  • IT USES IORD AND IOWRT CONTROL SIGNALS.

  • OPERATION IS SLOWER.


Slide25 l.jpg

A15

A14

G1

Y0

Y1

Y2

Y3

Y4

Y5

Y6

Y7

CS (CHIP SELECT) SIGNAL FOR PORT DEVICES

A13

G2B

74LS138

A12

G2A

A5

A4

A3


8086 physical memory l.jpg

8086 PHYSICAL MEMORY

THE TOTAL MEMORY (1MB) OF 8086 IS ARRANGED IN TWO BANKS. AN ODD BANK AND AN EVEN BANK. BOTH THE BANKS HAVE EQUAL NO. OF LOCATIONS.

THE ODD BANK CONTAINS ODD NUMBERED MEM. LOCATIONS.IT IS KNOWN AS UPPER BANK.

THE EVEN BANK CONTAINS ONLY EVEN NUMBERED MEM. LOCATIONS.IT IS KNOWN AS LOWER BANK.

THIS ARRANGE MENT IS DONE IN ORDER TO SPEED UP THE OPERATION.

THE ARRANGEMENT AND THE SIGNAL FOLLOWED, EXPLAINS THE SAME.


The 8086 memory bank l.jpg

THE 8086 MEMORY BANK

UPPER BANK

LOWER BANK

ODD

EVEN

CS

CS

A1---A19

A0

BHE

D15-D8

D7-D0


Addressing with 8086 l.jpg

ADDRESSING WITH 8086

  • PROBLEM: TWO 16K ROM AND TWO 32K RAM ARE REQUIRED TO BE INTERFACED WITH 8086 CPU.THE RAM ADDRESS MUST START AT 00000H.THE ROM ADDRESS RANGE MUST INCLUDE FFFF0H IN ITS RANGE.


Address map l.jpg

ADDRESS MAP

THE RAM ADDRESS STARTS AT 00000H.

TOTAL RAM IS 2*32K.

SO RAM ADDRESS RANGE IS FROM 00000H TO 0FFFFH.(FFFF-0000)H=216 =64K.

SINCE THE ROM ADDRESS MUST INCLUDE FFFF0H. WE TAKE LAST ADDRESS OF ROM AS FFFFFH.

AS TOTAL SPACE FOR ROM IS 2*16K,THE FIRST ADDRESS FOR ROM IS F8000H. (FFFFF-F8000)H=215=32K.

ADDRESS LINES A1-A14 ARE CONNECTED TO ROM.ADDRESS LINES A1-A15 ARE CONNECTED TO RAM.AND REMAINING LINES ARE USED FOR CHIP SELECTION.(NOTE: A0 IS RESERVED FOR BANKS.)


Slide31 l.jpg

TO GENERATE THE CHIP SELECT SIGNAL THE FOLLOWING LOGIC IS USED:

THE CHIPSELECT SIGNAL IS ACTIVE LOW.

SO A PARTICULAR CHIP CAN BE SELECTED ONLY WHEN THIS SIGNAL IS LOW.

SECONDLY AT A TIME ONLY ONE CHIP SHOULD BE SELECTED.

FURTHER ,THE ODD BANK WILL BE ENABLED ONLY IF BHE SIGNAL IS ACTIVATED.AND THE EVEN BANK WILL BE ENABLED ONLY IF AO SIGNAL IS LOW.


Slide34 l.jpg

  • HERE WE CONSIDER AS AN EXAMPLE A SYSTEM WITH FEW PORTS.

  • AN 8251(UNIVERSAL SYNCHRONOUS ASYNCHRONOUS RECEIVER TRANSMITTER)

  • TWO 8255A(PROGRAMMABLE PARALLEL PORTS.)

  • AN 8279 (DISPLAY AND KEYBOARD INTERFACING.)

  • AN 8275 (CRT CONTROLLER)

  • AN 8272A(FLOPPY DISK CONTRLLER)


Slide35 l.jpg

BHE OFFBOARD

P

O

R

T

S

E

L

E

C

T

A15

G2A G2B

G

A5

Y0

Y1

Y2

Y3

Y4

Y5

Y6

Y7

8255-1

8255-2

A4

8251

74LS138

8279

8275

8272

A3

RESERVED FOR FUTURE USE.

A2


Map for ports l.jpg

MAP FOR PORTS


Slide37 l.jpg

  • LINES A1 AND A0 ARE USED FOR SETTING INTERNAL REGISTERS OF THE ADDRESSED PORT.

  • NOTE THAT THIS IS AN ABSOLUTE ADDRESSING.i.e. A PORT CAN BE SELECTED FOR ONLY ONE ADDRESS.

  • NON-ABSOLUTE ADDRESING ALSO EXISTS.THERE A PORT CAN BE SELECTED FOR MORE THAN ONE ADDRESSES.


The off board decoder l.jpg

THE OFF-BOARD DECODER

  • THE SDK-86 USES AN OFF-BOARD CIRCUITRY .

  • THE PURPOSE OF THIS CIRCUITRY IS TO PRODUCE AN ACTIVE LOW OFF BOARD SIGNAL WHENEVER THE 8086 ADDRESSES A PORT OR MEMORY WHICH IS NOT DECODED ON THE SDK-86 BOARD.

  • USUALLY WHEN WE CONNECT ADDITIONAL PORT USING DIRECT I/O, THIS DEVICE IS CONSIDERED AS OFF-BOARD(OUT OF THE DECODING MAP OF THE SYSTEM).

  • SO WHENEVER THIS PORT IS CALLED UPON,THE 8086 ASSERTS THE OFF BOARD SINAL.

  • TO COMMUNICATE WITH OFF BOARD DEVICES,EXTRA HARDWARE SUCH AS 74LS138 OR AN EPROM DECODER IS REQUIRED FOR ADDRESSING THIS DEVICE.


8088 memory and port addressing l.jpg

8088 MEMORY AND PORT ADDRESSING

  • IN 8088 THERE ARE ONLY 8 DATA LINES AND 20 ADD.LINES.

  • THE 8088 MEMORY IS NOT DEVIDED INTO ODD AND EVEN BANKS.BUT THERE IS ONLY A SINGLE BANK.

  • THERE ARE 1MB LOCATIONS EACH 1-BYTE LONG.SO TO READ OR WRITE A WORD 8088 REQUIRES TWO MACHINE CYCLES ALWAYS.


The 8086 timing parameters l.jpg

THE 8086 TIMING PARAMETERS

  • THE 8086 TIMING PARAMETERS ARE REQUIRED TO FIND WHETHER A PARTICULAR DEVICE IS FAST ENOUGH TO WORK IN OUR SYSTEM. IF NOT WE CAN USE WAIT STATES TO ALLOW THE SLOW DEVICE TO RESPOND.

  • THE 8086 RECEIVES TWO DIFFERENT CLOCK SIGNALS FRON 8284 CLOCK GENERATOR: 4.9MHz AND 2.45MHz USING A DIVIDE BY 2 NETWORK.


Slide41 l.jpg

  • NOW SUPPOSE THAT 8086 USES A FREQ. OF 4.9 MHz AND WE WANT TO DETERMINE WHETHER 2716 EPROMs ARE ABLE TO WORK CORRECTLY WITH OUR SYSTEM.

  • TO READ FROM EPROM, WE NEED TO PROVIDE ITS ADDRESS,WE ALSO NEED TO PROVIDE ITS CHIP SELECT SIGNAL AND IN ADDITION WE ALSO NEED TO ENABLE ITS OUTPUT.

  • FOR THIS ,THE 8086 PROVIDES ON AD0-AD15 THE ADDRESS OF THE PARTICULAR LOCATION,THE CE (CHIP ENABLE-ACTIVE LOW) SIGNAL AND OE (OUTPUT ENABLE-AVTIVE LOW SIGNAL).


Slide42 l.jpg

  • FROM A DATA BOOK THE TYPICAL ACCESS TIMES FOR EPROM 2716 ARE:

  • Tacc : 450 ns

  • THIS MEANS THAT IF 2716 HAS ITS CE AND OE SIGNALS ENABLED,THAN IT WILL TAKE AT THE MAX. 450 ns,AFTER THE ADDREES IS PUT ON AD0-AD15 LINES.

  • Tce: 450 ns

  • THIS MEANS THAT THE ADDRESS AND OE SIGNALS ARE ALREADY ENABLED THAN IT WILL TAKE AT THE MAX. ,450 ns AFTER CE SIGNAL IS ENABLED.

  • Toe: 120 ns

  • THIS THE MAX. TIME REQUIRED WHEN AD0-AD15 AND CE SIGNALS ARE ENABLED BUT OUTPUT BUFFERS OF 2716 ARE NOT ENABLED.


Slide43 l.jpg

NEXT, WE CHECK THAT EACH OF THESE TIMES ARE SHORT ENOUGH TO WORK WITH 8086 AT 4.9MHz.

  • LOOKING AT THE WAVE FORMS OF READ CYCLE THERE IS A TIME GAP BETWEEN THE STARTING OF STATE-T1 AND THE TIME AT WHICH VALID ADDRESS IS AVAILABLE.

  • THIS TIME INTERVAL IS SYMBOLISED AS TCLAV TIME FROM CLOCK LOW TO ADDRESS VALID. AS PER 8086 DATA SHEET THIS TIME IS 110 ns MAX.

  • AGAIN LOOKING AT READ CYCLE ,WE FIND THAT VALID DATA MUST BE AVAILABLE BEFORE THE END OF T3.THERE ALSO EXISTS A TIME GAP BETWEEN ,THE LINE IS AVAILABLE FOR PUTTING VALID DATA AND THE END OF STATE-T3.

  • THIS TIME INTERVAL IS SYMBOLISED AS TDVCLTIME DATA MUST BE VALID BEFORE CLOCK GOES LOW. AS PER 8086 DATA SHEET THIS TIME IS 30 ns.


Slide44 l.jpg

  • THE TIME BETWEEN THE END OF TCLAV AND THE START OF TDVCL IS THE TIME AVAILABLE FOR GETTING THE ADDRESS TO THE MEMORY AND TO Tacc THE MEMORY DEVICE.

  • THE TIME BETWEEN THE START OF TCLAV AND THE END OF TDVCL IS 3 T-STATES.

  • AT 4.9 MHz ,TIME FOR ONE T-STATE IS 2.04 ns. SO TOTAL TIME FOR 3 STATES IS 612 ns.

  • FROM THIS WE SUBTRACT TCLAV AND TDVCL.

  • =612-110-30=472 ns.


Slide45 l.jpg

  • SO, 472 ns AVAILABLE FOR CALLING 2716 AND ACCESSING ITS DATA.

  • NOW,LOOKING AT THE MINIMUM MODE SYSTEM,WE SEE THAT THE ADDRESS INFORMATION FOR 2716 GOES THROUGH LATCHES 74S373. SO, THE PROPOGATION DELAY FOR THIS MUST BE SUBTRACTED FROM 472 ns.THE PROPAGATION DELAY (CALLING TIME OR GETTING ADDRESS TO 2716 TIME) IS 12 ns. THIS LEAVES 472-12=460 ns AS Tacc.

  • SO FROM THIS WE SEE THAT THE Tacc NEEDED FOR 2716 IS 450 ns max. WHILE THE 8086 PROVIDES 460 ns, AT LEAST.

  • SO, WE CONCLUDE THAT WE DO NOT NEED ANY WAIT STATE TO BE INSERTED. AS FAR AS Tacc. IS CONCERN.


Slide46 l.jpg

  • ANOTHER PARAMETER TOBE CHECKED IS Tce.

  • THE CE SIGNAL REQUIRES EITHER A0 OR BHE AS ITS ENABLE SIGNAL. SO,IT REQUIRES THE SAME CALCULATION AS THAT OF Tacc.

  • SO,Tce ALSO DOESN’T NEED WAIT STATE.

  • THE LAST TO BE SEEN IS Toe.

  • THE OE SIGNAL WILL BE GENERARED WHEN RD SIGNAL IS APPLIED.

  • NOW, LOOKING AT THE READ CYCLE RD SIGNAL WILL NOT BE ASSERTED TILL THE STATE T2.THE RD SIGNAL WILL BE ENABLED WITHIN A TIME TCLRL AFTER T2 STARTS.THE MAX. VALUE FOR THIS SIGNAL IS 165 ns. AGAIN,Toe MUST BE ACTIVE BEFORE VALID DATA IS PLACED ON AD0-AD15. i.e. BEFORE STARTING OF TDVCL.


Slide47 l.jpg

  • SO, SUM OF THESE TIMINGS MUST BE SUBTRACTED FROM THE TOTAL TIME AVAILABLE.

  • THE TOTAL TIME AVAILABLE IS STATE-T2 & T3.=408 ns.

  • 408-165-30=183 ns.

  • SO, 8086 PROVIDES 183 ns FOR ENABLING OUTPUT BUFERS.

  • WHILE THE TIME Toe REQUIRED BY 2716 IS 120 ns.

  • SO, FOR Toe ALSO WE NEED NO WAIT STATE.


Troubleshooting an 8086 based microcomputer l.jpg

TROUBLESHOOTING AN 8086 BASED MICROCOMPUTER

  • TROUBLESHOOTING IS TO FIND FAULTS IN A SYSTEM THAT WAS WORKING WELL, ONCE.

  • THE FOLLOWING STEPS ARE RECOMMENDED FOR A SYSTEMATIC AND TIME-EFFECTIVE TROUBLESHOOTING.


Slide49 l.jpg

  • IDENTIFYING THE SYMPTOMS.

  • CAREFULLY OBSERVE THE WORKING OF BAD SYSTEM AND TRY TO IDENTIFY THE SYMPTOMS.

  • 2. MAKING A CAREFUL VISUAL AND TACTILE INSPECTION.

  • CHECK FOR COMPONENTS THAT ARE EXCESSIVELY HOT.

  • DO NOT TOUCH FIRMLY,BECAUSE IT MAY RESULT IN SKIN BURN.

  • MAKE IT SURE THAT ALL ICs ARE FIRMLY SEATED IN THEIR SOCKETS AND NO PIN IS BENT.

  • CHECK FOR BROCKEN WIRES AND LOOSE CONNECTORS.


Slide50 l.jpg

  • 3. CHECKING POWER SUPPLY.

  • FROM THE MANUAL DETERMINE THE REQUIRED OPERATING VOLTAGE AND COMPARE IT WITH THE VOLTAGES AVAILABLE. ALSO CHECK THAT THE SAME VOLTAGE AVAILABLE AT PIN Vcc.

  • 4. SIGNAL ROLL CALL.

  • MAKE A CHECK AT SOME KEY SIGNALS e.g. CLOCK SIGNAL ON A CRO.OTHER SIGNAL SUCH AS ALE , RD AND WR SHOULD ALSO BE CHECKED.


Slide51 l.jpg

  • 5. SYSTEMATICALLY SUBSTITUTING ICs.

  • THIS STEP CAN BE USED ONLY IF THERE ARE TWO IDENTICAL SYSTEMS. ONE IS WORKING AND THE OTHER IS NOT WORKING.AND ALSO ALL ICs ARE IN SOCKETS.

  • FIRST TURN THE POWER OFF. MAKE MARKS ON THE ICs OF THE BAD SYSTEM.NOW EXCHANGE THE CPUs OF THE TWO SYSTEM AND SEE IF IT WORKS. IF SO, IT SURE THAT THE CPU IS BAD. IF NOT ,RESTORE THE ORIGINAL CPUs.IN THIS WAY GO ON TESTING ICs .

  • IF ROM IS ACCESIBLE AND RAM IS NOT,THEN TRY TO CHECK FOR RAM DECODER IC.

  • THIS WAY YOU CAN LOCATE BAD IC/ICs .


Slide52 l.jpg

  • 6.TROUBLE SHOOTING A SYSTEM WITH SOLDERED-IN ICs.

  • MOST OF THE TIME ALL ICs EXCEPT CPU AND ROM ICs ARE SOLDERED.

  • TRY TO RUN THE BASIC MONITOR PROGRAM WHICH IS IN ROM,AND COMPARE THE SIGNALS ON THE TWO.THIS WILL PROBABLY LOCATE THE FAULTY IC.

  • TRY TO EXCHANGE ROM OF THE TWO SYTEMS. AND/OR EXCHANGE CPU.

  • THEN RUN A RAM TEST ROUTINE .THE ROUTINE WILL WRITE ALL 1’S TO SELECTED LOCATIONS.THEN, READ THIS DATA BACK IF THEY ARE NOT 1’S ,THEN THE PROBLEM MAY BE WITH SYTEM BUS,(AS IT CAN NOY WRITE TO DESIRED LOCATION ) OR WITH RAM OR WITH RAM DECODER.


Slide53 l.jpg

  • 7.USING A LOGIC ANALYZER.

  • A LOGIC ANALYZER SHOWS YOU SIGNALS ON 16 TO 64 LINES AT A TIME. IT GIVES YOU A SNAPSHOT OF LOGIC LEVELS WHICH ARE ON ITS INPUT LINES AT THE TIME OF TAKING SAMPLE OR “PHOTO”.

  • SO, WE CAN CONNECT ALL THE DATA,ADDRESS AND CONTROL LINES TO THE INPUTS OF A LOGIC ANALYZER AND WE CAN SEE THE ACTIVITIES ON THIS LINES. THIS WILL SURELY TELL IF THE FAULT IS THERE WITH YOUR 8086.

  • USING A LOGIC ANALYZER IS A VERY EFFECTIVE AND THE MOST SOPHISTICATED SYSTEM OF TROUBLE SHOOTING.BUT IT REQUIRES EXPERTISE IN USING THE ANALYZER.


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  • 8.OTHER EQUIPMENT.

  • A LOGIC ANALYZER IS A BIG INSTRUMENT AND CAN NOT BE CARRIED WITH.TO USE THE SAME YOU ALSO NEED TO KNOW THE PROGRAMMING AND THE HARDWARE OF THE SYSTEM IN DETAIL.

  • BUT SUPPOSE YOU ARE TO REPAIR SEVERAL DIFFERENT TYPES OF MICROPROCESSORS AND YOU DO NOT KNOW MUCH ABOUT THE PROGRAMMING OF THE SYSTEM WHICH YOU ARE TROUBLE SHOOTING ,THEN YOU NEED TO USE EQUIPMENT SUCH AS FLUKE 9010A.

  • SUCH EQUIPMENT KEEPS YOU FREE FROM THE STRESS OF KNOWING ALL THE SOFTWARE AND

  • HARDWARE DETAILS OF THE SYSTEM AT HAND.


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