Microprocessors 1-8086

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.

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
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
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
THE INTERNAL ARCHITECTURE OF 8086 CAN BE
MAINLY DIVIDED INTO TWO UNITS:
BUS INTERFACE UNIT (BIU)
EXECUTION UNIT (EU)

THE BIU CONTAINS :
CODE SEGMENT REGISTER (CS)

 DATA SEGMENT REGISTER (DS)
 EXTRA SEGMENT REGISTER (ES)
STACK SEGMENT REGISTEER (SS) AND
 INSTRUCTION POINTER (IP)

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

40 VCC
AD15
GND 1 A16-A19
2 & S3-S6
35
34 BHE/S7
33 MN/MX
32 RD
AD14 TO 31 RQ/GTO(HOLD)
AD0 30 RQ/GT1(HLDA)
8086 CPU 29 LOCK (WR)
28 S2 (M/IO)
27 S1 (DT/R)
16
NMI 26 S0 (DEN)
17
INTR 18 25 QS0 (ALE)
CLK 19 24 QS1(INTA)
GND 20
23 TEST
RESET 21 22 READY

BASIC 8086 MINIMUM MODE SYSTEM

MN/MX
CLK M/IO
8284A
READY
CLOCK INTA
RESET
GENE-
RD
RATOR
WR
8282
DT/R
LATCH
DEN
ALE ADDR
WAIT STATE
GENERATOR AD0-
AD15
A16-A19

8286
TRAN-
ADDR/DATA CEIVER
DATA

RAM 2142 PERI-
2716
PHERAL
PROM

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
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
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.

T1 T2 T3 TW T4

CLK

M/IO

ALE
MEMORY ACCESS TIME
ADDR/ RESERVED VALID
A15-A0
DATA FOR DATA D15-D0

ADDR/ A19-A16
STATUS
RD/INTA
READY
DT/R
DEN

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.

T1 T2 T3 TW T4

CLK

M/IO

ALE

ADDR/
A15-A0 DATA OUT (D15-D0)
DATA
ADDR/ A19-A16
STATUS
WR
READY
DT/R
DEN

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
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

A0
A1
A2

A11

ROM 0 ROM 1 ROM 7

CS 2732 CS 2732 CS 2732

D0

D5
D6
D7

Y0 Y1 Y7 A12
74LS138 A13
G2A G2B G1 A14

A15 RD +5V

ADDRESS DECODER WORKSHEET
A15 & A11-A8 A7-A4 A3-A0 HEX EQUI.
A14-A12 ADDRESS
ROM0 ST. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 =0 0 0 0
END 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 =0 F F F
ROM1 ST. 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 =1 0 0 0
END 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 =1 F F F
ROM2 ST. 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 =2 0 0 0
END 0 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 =2 F F F
ROM3 ST. 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 =3 0 0 0
END 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 =3 F F F
ROM4 ST. 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 =4 0 0 0
END 0 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 =4 F F F
ROM5 ST. 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 =5 0 0 0
END 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 =5 F F F
ROM6 ST. 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 =6 0 0 0
END 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 =6 F F F
ROM7 ST. 0 1 1 1 0 0 0 0 0 0 0 0 0 =7 0 0 0
END 0 1 1 1 1 1 1 1 1 1 1 1 =7 F F F

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.

A0
A1
A2

A10

RAM 0 RAM 1 RAM 7

CS 2142 CS 2142 CS 2142

D0

D5
D6
D7

Y0 Y1 Y7 A11
74LS138 A12
G2A G2B G1 A13

A14 GND A15

A SYSTEM PORT DECODER
A SYSTEM PORT CAN BE ADDRESSED IN TWO WAYS :
1. 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.

2. 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.

A15
A14 G1 Y0
Y1 CS (CHIP
SELECT)
Y2 SIGNAL
A13 G2B
Y3 FOR PORT
74LS138 DEVICES
Y4

A12 Y5
G2A
Y6
Y7

A5 A4 A3

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
UPPER BANK LOWER BANK

ODD EVEN

CS CS

BHE A1---A19 A0
D15-D8 D7-D0

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
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.)

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.

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)

BHE OFFBOARD

P
A15 G2A G2B O
A5 G R
T
Y0 8255-1
S
Y1 8255-2
E
A4 Y2 8251
L
Y3 8279
74LS138 E
Y4 8275
8272 C
Y5
A3 T
Y6
Y7 RESERVED
A2 FOR FUTURE
USE.

MAP FOR PORTS
A15-A5 A4 A3 A2 A1 A0 EQUI HEX
ADDRESS
8255-1 1 0 0 0 =F F E O
=F F E 3
8255-2 1 0 0 1 =F F E 4
=F F E 7
8251 1 0 1 0 =F F E 8
=F F E B
8279 1 0 1 1 =F F E C
=F F E F
8275 1 1 0 0 =F F F 0
=F F F 3
8272 1 1 0 1 =F F F 4
=F F F 7

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
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
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

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.

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).

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.

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 TDVCL TIME
DATA MUST BE VALID BEFORE CLOCK GOES LOW. AS PER 8086
DATA SHEET THIS TIME IS 30 ns.

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.

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.

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.

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

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.

1. 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.

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.

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 .

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.

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.

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.

Microprocessors 1-8086

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Microprocessors 1-8086