Unit. 3 coa.ppt

1
Basic Computer Organization & Design
Computer Organization Computer Architectures Lab
BASIC COMPUTER ORGANIZATION AND DESIGN
• Instruction Codes
• Computer Registers
• Computer Instructions
• Timing and Control
• Instruction Cycle
• Memory Reference Instructions
• Input-Output and Interrupt
• Complete Computer Description
• Design of Basic Computer
• Design of Accumulator Logic

2
INTRODUCTION
• Every different processor has its own design
(different registers, buses, micro-operations, machine instructions, etc)
• Modern processor is a very complex device
• It contains
– Many registers
– Multiple arithmetic units, for both integer and floating point calculations
– The ability to pipeline several consecutive instructions to speed execution
– Etc.
• However, to understand how processors work, use a simplified
processor model
• This is similar to what real processors were like ~25 years ago

3
THE BASIC COMPUTER
• The Basic Computer has two components, a processor and
memory
• The memory has 4096 words in it
– 4096 = 212, so it takes 12 bits to select a word in memory
• Each word is 16 bits long
CPU RAM
0
4095
0
15

4
INSTRUCTIONS
Instruction codes
• Program
– A sequence of (machine) instructions
• (Machine) Instruction
– A group of bits that tell the computer to perform a specific operation
(a sequence of micro-operation)
• The instructions of a program, along with any needed data
are stored in memory
• The CPU reads the next instruction from memory
• It is placed in an Instruction Register (IR)
• Control circuitry in control unit then translates the
instruction into the sequence of microoperations
necessary to implement it

5
INSTRUCTION FORMAT
Instruction codes
• A computer instruction is often divided into two parts
– An opcode (Operation Code) that specifies the operation for that
instruction
– An address that specifies the registers and/or locations in memory to
use for that operation
• In the Basic Computer, since the memory contains 4096 (=
212) words, we needs 12 bit to specify which memory
address this instruction will use
• In the Basic Computer, bit 15 of the instruction specifies
the addressing mode (0: direct addressing, 1: indirect
addressing)
• Since the memory words, and hence the instructions, are
16 bits long, that leaves 3 bits for the instruction’s opcode
Opcode Address
Instruction Format
15 14 12 0
I
11
Addressing
mode

6
ADDRESSING MODES
Instruction codes
• The address field of an instruction can represent either
– Direct address: the address in memory of the data to use (the address of the
operand), or
– Indirect address: the address in memory of the address in memory of the data
to use
• Effective Address (EA)
– The address, that can be directly used without modification to access an
operand for a computation-type instruction, or as the target address for a
branch-type instruction
0 ADD 457
22
Operand
457
1 ADD 300
35
1350
300
Operand
1350
+
AC
+
AC
Direct addressing Indirect addressing

7
PROCESSOR REGISTERS
Instruction codes
• A processor has many registers to hold instructions,
addresses, data, etc
• The processor has a register, the Program Counter (PC) that
holds the memory address of the next instruction
– Since the memory in the Basic Computer only has 4096 locations, the PC
only needs 12 bits
• In a direct or indirect addressing, the processor needs to keep
track of what locations in memory it is addressing: The
Address Register (AR) is used for this
– The AR is a 12 bit register in the Basic Computer
• When an operand is found, using either direct or indirect
addressing, it is placed in the Data Register (DR). The
processor then uses this value as data for its operation
• The Basic Computer has a single general purpose register –
the Accumulator (AC)

8
PROCESSOR REGISTERS
Instruction codes
• The significance of a general purpose register is that it can be
used for loading operands and storing results
– e.g. load AC with the contents of a specific memory location; store the
contents of AC into a specified memory location
• Often a processor will need a scratch register to store
intermediate results or other temporary data; in the Basic
Computer this is the Temporary Register (TR)
• The Basic Computer uses a very simple model of input/output
(I/O) operations
– Input devices are considered to send 8 bits of character data to the processor
– The processor can send 8 bits of character data to output devices
• The Input Register (INPR) holds an 8 bit character gotten from an
input device
• The Output Register (OUTR) holds an 8 bit character to be send
to an output device

9
BASIC COMPUTER REGISTERS
List of Registers
DR 16 Data Register Holds memory operand
AR 12 Address Register Holds address for memory
AC 16 Accumulator Processor register
IR 16 Instruction Register Holds instruction code
PC 12 Program Counter Holds address of instruction
TR 16 Temporary Register Holds temporary data
INPR 8 Input Register Holds input character
OUTR 8 Output Register Holds output character
Registers
Registers in the Basic Computer
11 0
PC
15 0
IR
15 0
TR
7 0
OUTR
15 0
DR
15 0
AC
11 0
AR
INPR
0 7
Memory
4096 x 16
CPU

10
COMMON BUS SYSTEM
Registers
• The registers in the Basic Computer are connected using a
bus
• This gives a savings in circuitry over complete
connections between registers

11
COMMON BUS SYSTEM
Registers
S2
S1
S0
Bus
Memory unit
4096 x 16
LD INR CLR
Address
Read
Write
AR
LD INR CLR
PC
LD INR CLR
DR
LD INR CLR
AC
ALU
E
INPR
IR
LD
LD INR CLR
TR
OUTR
LD
Clock
16-bit common bus
7
1
2
3
4
5
6

12
COMMON BUS SYSTEM
Registers
AR
PC
DR
L I C
L I C
L I C
AC
L I C
ALU
E
IR
L
TR
L I C
OUTR L
INPR
Memory
4096 x 16
Address
Read
Write
16-bit Common Bus
7 1 2 3 4 5 6
S0 S1 S2

13
COMMON BUS SYSTEM
Registers
• Three control lines, S2, S1, and S0 control which register the
bus selects as its input
• Either one of the registers will have its load signal
activated, or the memory will have its read signal activated
– Will determine where the data from the bus gets loaded
• The 12-bit registers, AR and PC, have 0’s loaded onto the
bus in the high order 4 bit positions
• When the 8-bit register OUTR is loaded from the bus, the
data comes from the low order 8 bits on the bus
0 0 0 x
0 0 1 AR
0 1 0 PC
0 1 1 DR
1 0 0 AC
1 0 1 IR
1 1 0 TR
1 1 1 Memory
S2 S1 S0 Register

14
BASIC COMPUTER INSTRUCTIONS
Instructions
• Basic Computer Instruction Format
15 14 12 11 0
I Opcode Address
Memory-Reference Instructions (OP-code = 000 ~ 110)
Register-Reference Instructions (OP-code = 111, I = 0)
Input-Output Instructions (OP-code =111, I = 1)
15 12 11 0
Register operation
0 1 1 1
15 12 11 0
I/O operation
1 1 1 1

15
BASIC COMPUTER INSTRUCTIONS
Hex Code
Symbol I = 0 I = 1 Description
AND 0xxx 8xxx AND memory word to AC
ADD 1xxx 9xxx Add memory word to AC
LDA 2xxx Axxx Load AC from memory
STA 3xxx Bxxx Store content of AC into memory
BUN 4xxx Cxxx Branch unconditionally
BSA 5xxx Dxxx Branch and save return address
ISZ 6xxx Exxx Increment and skip if zero
CLA 7800 Clear AC
CLE 7400 Clear E
CMA 7200 Complement AC
CME 7100 Complement E
CIR 7080 Circulate right AC and E
CIL 7040 Circulate left AC and E
INC 7020 Increment AC
SPA 7010 Skip next instr. if AC is positive
SNA 7008 Skip next instr. if AC is negative
SZA 7004 Skip next instr. if AC is zero
SZE 7002 Skip next instr. if E is zero
HLT 7001 Halt computer
INP F800 Input character to AC
OUT F400 Output character from AC
SKI F200 Skip on input flag
SKO F100 Skip on output flag
ION F080 Interrupt on
IOF F040 Interrupt off
Instructions

16
INSTRUCTION SET COMPLETENESS
• Instruction Types
Set of instructions using which user can construct machine
language programs to evaluate any computable function.
Functional Instructions
- Arithmetic, logic, and shift instructions
- ADD, CMA, INC, CIR, CIL, AND, CLA (other than ADD/AND?)
Transfer Instructions
- Data transfers between the main memory
and the processor registers
- LDA, STA
Control Instructions
- Program sequencing and control
- BUN, BSA, ISZ
Input/Output Instructions
- Input and output
- INP, OUT
Instructions

17
CONTROL UNIT
Instruction codes
• Control unit (CU) of a processor translates from machine
instructions to the control signals (for the microoperations)
that implement them
• Control units are implemented in one of two ways
• Hardwired Control
– CU is made up of sequential and combinational circuits to generate the
control signals
• Microprogrammed Control
– A control memory on the processor contains microprograms that
activate the necessary control signals
• We will consider a hardwired implementation of the control
unit for the Basic Computer

18
TIMING AND CONTROL
Control unit of Basic Computer
Timing and control
Instruction register (IR)
15 14 13 12 11 - 0
3 x 8
decoder
7 6 5 4 3 2 1 0
I
D0
15 14 . . . . 2 1 0
4 x 16
decoder
4-bit
sequence
counter
(SC)
Increment (INR)
Clear (CLR)
Clock
Other inputs
Control
signals
D
T
T
7
15
0
Combinational
Control
logic

19
TIMING SIGNALS
Clock
T0 T1 T2 T3 T4 T0
T0
T1
T2
T3
T4
D3
CLR
SC
- Generated by 4-bit sequence counter and 416 decoder
- The SC can be incremented or cleared.
- Example: T0, T1, T2, T3, T4, T0, T1, . . .
Assume: At time T4, SC is cleared to 0 if decoder output D3 is active.
D3T4: SC  0
Timing and control

20
INSTRUCTION CYCLE
• In Basic Computer, a machine instruction is executed in the
following cycle:
1. Fetch an instruction from memory
2. Decode the instruction and calculate effective address (EA)
3. Read the EA from memory if the instruction has an indirect address
(Fetch operand)
4. Execute the instruction
• After an instruction is executed, the cycle starts again at
step 1, for the next instruction
• Note: Every different processor has its own (different)
instruction cycle

21
FETCH and DECODE
• Fetch and Decode T0: AR PC (S0S1S2=010, T0=1)
T1: IR  M [AR], PC  PC + 1 (S0S1S2=111, T1=1)
T2: D0, . . . , D7  Decode IR(12-14), AR  IR(0-11), I  IR(15)
S2
S1
S0
Bus
7
Memory
unit
Address
Read
AR
LD
PC
INR
IR
LD Clock
1
2
5
Common bus
T1
T0
Instruction Cycle

22
DETERMINE THE TYPE OF INSTRUCTION
= 0 (direct)
D'7IT3: AR M[AR]
D'7I'T3: Nothing
D7I'T3: Execute a register-reference instr.
D7IT3: Execute an input-output instr.
Instrction Cycle
Start
SC  0
AR  PC
T0
IR  M[AR], PC  PC + 1
T1
AR  IR(0-11), I  IR(15)
Decode Opcode in IR(12-14),
T2
D7
= 0 (Memory-reference) =>opcode ≠ 111
(Register or I/O) = 1
I
I
Execute
register-reference
instruction
SC  0
Execute
input-output
instruction
SC  0
M[AR]

AR Nothing
= 0 (register)
(I/O) = 1 (indirect) = 1
T3 T3 T3 T3
Execute
memory-reference
instruction
SC  0
T4

23
REGISTER REFERENCE INSTRUCTIONS
r = D7 IT3 => Register Reference Instruction
Bi = IR(i) , i=0,1,2,...,11
- D7 = 1, I = 0
- Register Ref. Instr. is specified in b0 ~ b11 of IR
- Execution starts with timing signal T3
Instruction Cycle
Register Reference Instructions are identified when
r: SC  0
CLA rB11: AC  0
CLE rB10: E  0
CMA rB9: AC  AC’
CME rB8: E  E’
CIR rB7: AC  shr AC, AC(15)  E, E  AC(0)
CIL rB6: AC  shl AC, AC(0)  E, E  AC(15)
INC rB5: AC  AC + 1
SPA rB4: if (AC(15) = 0) then (PC  PC+1)
SNA rB3: if (AC(15) = 1) then (PC  PC+1)
SZA rB2: if (AC = 0) then (PC  PC+1)
SZE rB1: if (E = 0) then (PC  PC+1)
HLT rB0: S  0 (S is a start-stop flip-flop)

24
MEMORY REFERENCE INSTRUCTIONS
AND to AC
D0T4: DR  M[AR] Read operand
D0T5: AC  AC  DR, SC  0 AND with AC
ADD to AC
D1T4: DR  M[AR] Read operand
D1T5: AC  AC + DR, E  Cout, SC  0 Add to AC and store carry in E
- The effective address of the instruction is in AR and was placed there during
timing signal T2 when I = 0, or during timing signal T3 when I = 1
- Memory cycle is assumed to be short enough to complete in a CPU cycle
- The execution of MR instruction starts with T4
MR Instructions
Symbol
Operation
Decoder
Symbolic Description
AND D0 AC  AC  M[AR]
ADD D1 AC  AC + M[AR], E  Cout
LDA D2 AC  M[AR]
STA D3 M[AR]  AC
BUN D4 PC  AR
BSA D5 M[AR]  PC, PC  AR + 1
ISZ D6 M[AR]  M[AR] + 1, if M[AR] + 1 = 0 then PC  PC+1

25
Memory, PC after execution
21
0 BSA 135
Next instruction
Subroutine
20
PC = 21
AR = 135
136
1 BUN 135
Memory, PC, AR at time T4
0 BSA 135
Next instruction
Subroutine
20
21
135
PC = 136
1 BUN 135
Memory Memory
LDA: Load to AC
D2T4: DR  M[AR]
D2T5: AC  DR, SC  0
STA: Store AC
D3T4: M[AR]  AC, SC  0
BUN: Branch Unconditionally
D4T4: PC  AR, SC  0
BSA: Branch and Save Return Address

26
MR Instructions
BSA:
D5T4: M[AR]  PC, AR  AR + 1
D5T5: PC  AR, SC  0
ISZ: Increment and Skip-if-Zero
D6T4: DR  M[AR]
D6T5: DR  DR + 1
D6T4: M[AR]  DR, if (DR = 0) then (PC  PC + 1), SC  0

27
FLOWCHART FOR MEMORY REFERENCE INSTRUCTIONS
MR Instructions
Memory-reference instruction
DR  M[AR] DR  M[AR] DR  M[AR] M[AR]  AC
SC  0
AND ADD LDA STA
AC  AC DR
SC  0
AC  AC + DR
E  Cout
SC  0
AC  DR
SC  0
D T
0 4 D T
1 4 D T
2 4 D T
3 4
D T
0 5 D T
1 5 D T
2 5
PC  AR
SC  0
M[AR]  PC
AR  AR + 1
DR  M[AR]
BUN BSA ISZ
D T
4 4 D T
5 4 D T
6 4
DR  DR + 1
D T
5 5 D T
6 5
PC  AR
SC  0
M[AR]  DR
If (DR = 0)
then (PC  PC + 1)
SC  0
D T
6 6


28
INPUT-OUTPUT AND INTERRUPT
• Input-Output Configuration
INPR Input register - 8 bits
OUTR Output register - 8 bits
FGI Input flag - 1 bit
FGO Output flag - 1 bit
IEN Interrupt enable - 1 bit
- The terminal sends and receives serial information
- The serial info. from the keyboard is shifted into INPR
- The serial info. for the printer is stored in the OUTR
- INPR and OUTR communicate with the terminal
serially and with the AC in parallel.
- The flags are needed to synchronize the timing
difference between I/O device and the computer
A Terminal with a keyboard and a Printer
I/O and Interrupt
Input-output
terminal
Serial
communication
interface
Computer
registers and
flip-flops
Printer
Keyboard
Receiver
interface
Transmitter
interface
FGO
OUTR
AC
INPR FGI
Serial Communications Path
Parallel Communications Path

29
PROGRAM CONTROLLED DATA TRANSFER
/* Input */ /* Initially FGI = 0 */
loop: If FGI = 1 goto loop
INPR  new data, FGI  1
loop: If FGO = 1 goto loop
consume OUTR, FGO  1
-- CPU -- -- I/O Device --
loop: If FGI = 0 goto loop
AC  INPR, FGI  0
/* Output */ /* Initially FGO = 1 */
loop: If FGO = 0 goto loop
OUTR  AC, FGO  0
I/O and Interrupt
Start Input
FGI=0
AC  INPR
More
Character
END
Start Output
FGO  1
FGO=1
More
Character
END
consume OUTR
yes
no
yes
no
FGI=0 FGO=1
yes
yes
no
no

30
INPUT-OUTPUT INSTRUCTIONS
D7IT3 = p
IR(i) = Bi, i = 6, …, 11
p: SC  0 Clear SC
INP pB11: AC(0-7)  INPR, FGI  0 Input char. to AC
OUT pB10: OUTR  AC(0-7), FGO  0 Output char. from AC
SKI pB9: if(FGI = 1) then (PC  PC + 1) Skip on input flag
SKO pB8: if(FGO = 1) then (PC  PC + 1) Skip on output flag
ION pB7: IEN  1 Interrupt enable on
IOF pB6: IEN  0 Interrupt enable off
CPU Side

31
PROGRAM-CONTROLLED INPUT/OUTPUT
• Program-controlled I/O
- Continuous CPU involvement
I/O takes valuable CPU time
- CPU slowed down to I/O speed
- Simple
- Least hardware
I/O and Interrupt
Input
LOOP SKI DEV
BUN LOOP
INP DEV
Output
LDA DATA
LOOP SKO DEV
BUN LOOP
OUT DEV

32
INTERRUPT INITIATED INPUT/OUTPUT
- Open communication only when some data has to be passed --> interrupt.
- The I/O interface, instead of the CPU, monitors the I/O device.
- When the interface founds that the I/O device is ready for data transfer,
it generates an interrupt request to the CPU
- Upon detecting an interrupt, the CPU stops momentarily the task
it is doing, branches to the service routine to process the data
transfer, and then returns to the task it was performing.
* IEN (Interrupt-enable flip-flop)
- can be set and cleared by instructions
- when cleared, the computer cannot be interrupted

33
FLOWCHART FOR INTERRUPT CYCLE
R = Interrupt f/f
- The interrupt cycle is a HW implementation of a branch
and save return address operation.
- At the beginning of the next instruction cycle, the
instruction that is read from memory is in address 1.
- At memory address 1, the programmer must store a branch instruction
that sends the control to an interrupt service routine
- The instruction that returns the control to the original
program is "indirect BUN 0"
I/O and Interrupt
Store return address
R
=1
=0
in location 0
M[0]  PC
Branch to location 1
PC  1
IEN  0
R  0
Interrupt cycle
Instruction cycle
Fetch and decode
instructions
IEN
FGI
FGO
Execute
instructions
R  1
=1
=1
=1
=0
=0
=0

34
REGISTER TRANSFER OPERATIONS IN INTERRUPT CYCLE
Register Transfer Statements for Interrupt Cycle
- R F/F  1 if IEN (FGI + FGO)T0T1T2
 T0T1T2 (IEN)(FGI + FGO): R  1
- The fetch and decode phases of the instruction cycle
must be modified Replace T0, T1, T2 with R'T0, R'T1, R'T2
- The interrupt cycle :
RT0: AR  0, TR  PC
RT1: M[AR]  TR, PC  0
RT2: PC  PC + 1, IEN  0, R  0, SC  0
After interrupt cycle
0 BUN 1120
0
1
PC = 256
255
1 BUN 0
Before interrupt
Main
Program
1120
I/O
Program
0 BUN 1120
0
PC = 1
256
255
1 BUN 0
Memory
Main
Program
1120
I/O
Program
256
I/O and Interrupt

35
FURTHER QUESTIONS ON INTERRUPT
How can the CPU recognize the device
requesting an interrupt ?
Since different devices are likely to require
different interrupt service routines, how can
the CPU obtain the starting address of the
appropriate routine in each case ?
Should any device be allowed to interrupt the
CPU while another interrupt is being serviced ?
How can the situation be handled when two or
more interrupt requests occur simultaneously ?
I/O and Interrupt

36
COMPLETE COMPUTER DESCRIPTION
Flowchart of Operations
Description
=1 (I/O) =0 (Register) =1(Indir) =0(Dir)
start
SC  0
R
AR  PC
R’T0
IR  M[AR], PC  PC + 1
R’T1
AR  IR(0~11), I  IR(15)
D0...D7  Decode IR(12 ~ 14)
R’T2
AR  0, TR  PC
RT0
M[AR]  TR, PC  0
RT1
PC  PC + 1, IEN  0
R  0, SC  0
RT2
D7
I I
Execute
I/O
Instruction
Execute
RR
Instruction
AR <- M[AR] Idle
D7IT3 D7I’T3 D7’IT3 D7’I’T3
Execute MR
Instruction
=0(Instruction =1 (interrupt
Cycle) Cycle)
=1(Register or I/O) =0(Memory Ref)
D7’T4
IEN
FGI
FGO
=1
=1
=1
=0
=0
=0
R  1

37
Microoperations
Description
Fetch
Decode
Indirect
Interrupt
Memory-Reference
AND
ADD
LDA
STA
BUN
BSA
ISZ
RT0:
RT1:
RT2:
D7IT3:
RT0:
RT1:
RT2:
D0T4:
D0T5:
D1T4:
D1T5:
D2T4:
D2T5:
D3T4:
D4T4:
D5T4:
D5T5:
D6T4:
D6T5:
D6T6:
AR  PC
IR  M[AR], PC  PC + 1
D0, ..., D7  Decode IR(12 ~ 14),
AR  IR(0 ~ 11), I  IR(15)
AR  M[AR]
R  1
AR  0, TR  PC
M[AR]  TR, PC  0
PC  PC + 1, IEN  0, R  0, SC  0
DR  M[AR]
AC  AC  DR, SC  0
DR  M[AR]
AC  AC + DR, E  Cout, SC  0
DR  M[AR]
AC  DR, SC  0
M[AR]  AC, SC  0
PC  AR, SC  0
M[AR]  PC, AR  AR + 1
PC  AR, SC  0
DR  M[AR]
DR  DR + 1
M[AR]  DR, if(DR=0) then (PC  PC + 1),
SC  0
T0T1T2(IEN)(FGI + FGO):

38
Register-Reference
CLA
CLE
CMA
CME
CIR
CIL
INC
SPA
SNA
SZA
SZE
HLT
Input-Output
INP
OUT
SKI
SKO
ION
IOF
D7IT3 = r
IR(i) = Bi
r:
rB11:
rB10:
rB9:
rB8:
rB7:
rB6:
rB5:
rB4:
rB3:
rB2:
rB1:
rB0:
D7IT3 = p
IR(i) = Bi
p:
pB11:
pB10:
pB9:
pB8:
pB7:
pB6:
(Common to all register-reference instr)
(i = 0,1,2, ..., 11)
SC  0
AC  0
E  0
AC  AC
E  E
AC  shr AC, AC(15)  E, E  AC(0)
AC  shl AC, AC(0)  E, E  AC(15)
AC  AC + 1
If(AC(15) =0) then (PC  PC + 1)
If(AC(15) =1) then (PC  PC + 1)
If(AC = 0) then (PC  PC + 1)
If(E=0) then (PC  PC + 1)
S  0
(Common to all input-output instructions)
(i = 6,7,8,9,10,11)
SC  0
AC(0-7)  INPR, FGI  0
OUTR  AC(0-7), FGO  0
If(FGI=1) then (PC  PC + 1)
If(FGO=1) then (PC  PC + 1)
IEN  1
IEN  0
Description
Microoperations

39
DESIGN OF BASIC COMPUTER(BC)
Hardware Components of BC
A memory unit: 4096 x 16.
Registers:
AR, PC, DR, AC, IR, TR, OUTR, INPR, and SC
Flip-Flops(Status):
I, S, E, R, IEN, FGI, and FGO
Decoders: a 3x8 Opcode decoder
a 4x16 timing decoder
Common bus: 16 bits
Control logic gates:
Adder and Logic circuit: Connected to AC
Control Logic Gates
- Input Controls of the nine registers
- Read and Write Controls of memory
- Set, Clear, or Complement Controls of the flip-flops
- S2, S1, S0 Controls to select a register for the bus
- AC, and Adder and Logic circuit
Design of Basic Computer

40
CONTROL OF REGISTERS AND MEMORY
Scan all of the register transfer statements that change the content of AR:
LD(AR) = R'T0 + R'T2 + D'7IT3
CLR(AR) = RT0
INR(AR) = D5T4
Address Register; AR
R’T0: AR  PC LD(AR)
R’T2: AR  IR(0-11) LD(AR)
D’7IT3: AR  M[AR] LD(AR)
RT0: AR  0 CLR(AR)
D5T4: AR  AR + 1 INR(AR)
AR
LD
INR
CLR
Clock
To bus
12
From bus
12
D'
I
T
T
R
T
D
T
7
3
2
0
4

41
CONTROL OF FLAGS
pB7: IEN  1 (I/O Instruction)
pB6: IEN  0 (I/O Instruction)
RT2: IEN  0 (Interrupt)
p = D7IT3 (Input/Output Instruction)
IEN: Interrupt Enable Flag
D
I
T3
7
J
K
Q IEN
p
B7
B6
T2
R

42
CONTROL OF COMMON BUS
x1 for placing AR onto bus
D4T4: PC  AR
D5T5: PC  AR
x1 = D4T4 + D5T5
x1
x2
x3
x4
x5
x6
x7
Encoder
S 2
S 1
S 0
Multiplexer
bus select
inputs
x1 x2 x3 x4 x5 x6 x7 S2 S1 S0
selected
register
0 0 0 0 0 0 0 0 0 0 none
1 0 0 0 0 0 0 0 0 1 AR
0 1 0 0 0 0 0 0 1 0 PC
0 0 1 0 0 0 0 0 1 1 DR
0 0 0 1 0 0 0 1 0 0 AC
0 0 0 0 1 0 0 1 0 1 IR
0 0 0 0 0 1 0 1 1 0 TR
0 0 0 0 0 0 1 1 1 1 Memory

43
DESIGN OF ACCUMULATOR LOGIC
Circuits associated with AC
All the statements that change the content of AC
Design of AC Logic
16
16
8
Adder and
logic
circuit
16
AC
From DR
From INPR
Control
gates
LD INR CLR
16
To bus
Clock
D0T5: AC  AC  DR AND with DR
D1T5: AC  AC + DR Add with DR
D2T5: AC  DR Transfer from DR
pB11: AC(0-7)  INPR Transfer from INPR
rB9: AC  AC Complement
rB7 : AC  shr AC, AC(15)  E Shift right
rB6 : AC  shl AC, AC(0)  E Shift left
rB11 : AC  0 Clear
rB5 : AC  AC + 1 Increment

44
CONTROL OF AC REGISTER
Gate structures for controlling
the LD, INR, and CLR of AC
AC
LD
INR
CLR
Clock
To bus
16
From Adder
and Logic
16
AND
ADD
DR
INPR
COM
SHR
SHL
INC
CLR
D0
D1
D2
B11
B9
B7
B6
B5
B11
r
p
T5
T5
Design of AC Logic

45
ALU (Adder & Logic Circuits)
Design of AC Logic

46

Unit. 3 coa.ppt

More Related Content

What's hot (20)

Similar to Unit. 3 coa.ppt (20)

Recently uploaded (20)

Unit. 3 coa.ppt