Instruction Codes
● Theorganization of the computer is defined by its internal registers, the timing and
control structure, and the set of instructions that it uses.
● A computer instruction is a binary code that specifies a sequence of microoperations
for the computer.
● An instruction code is a group of bits that instruct the computer to perform a specific
operation.
● Instruction code is usually divided into two parts.
○ Operation part - Group of bits that define such operations as add, subtract,
multiply, shift, and complement.
○ Address part - Contains registers or memory words where the address of operand
is found or the result is to be stored.
● Each computer has its own instruction code format.
4.
Operation Code
● Theoperation code(op-code) of an instruction is a group of bits that define such
operations as add, subtract, multiply, shift, and complement.
● The number of bits required for the operation code of an instruction depends on the total
number of operations available in the computer.(n bits for 2n
operations)
● An operation code is sometimes called a macro-operation because it specifies a set of
micro-operations.
5.
Stored Program Organization
●Simplest way to organize computer is to have one processor register(Accumulator AC) and
an instruction code format with two parts.
○ First-Operation to be performed
○ Second – Address
● The memory address tells the control where to find an operand in memory.
● This operand is read from memory and used as the data to be operated on together with
the data stored in the processor register.
6.
● Instructions arestored in one section
of the memory and data in another.
● For a memory unit with 4096 words we
need 12 bits to specify an address
since 212
=4096.
● 4 bits are available for opcode to
specify one out of 16 possible
operations.
7.
Steps
● The controlreads a 16-bit instruction from the program portion of memory.
● It uses the 12-bit address part of the instruction to read a 16-bit operand from the data
portion of memory.
● It then executes the operation specified by the operation code.
● The operation is performed with the memory operand and the content of AC.
8.
Direct & IndirectAddressing Modes
● Following Addressing Modes are used for address portion of the instruction code.
○ Immediate- The address part specifies an operand.
Eg: ADD 5
○ Direct- The address part specifies the address of an operand.
○ Indirect- The address part specifies a pointer(another address) where the address of
the operand can be found.
● One bit of the instruction code(I) can be used to distinguish between a direct and an
indirect address.
9.
Effective Address
It isthe address of the operand in a
computation-type instruction or the target
address in a branch-type instruction.
The effective address in the instruction of Fig.(b)
is 457 and in the instruction of Fig(c) is 1350.
Computer Registers
Need ofRegisters?
● Computer instructions are normally stored in consecutive memory locations and are
executed sequentially one at a time.
● The control reads an instruction from a specific address in memory and executes it. It then
continues by reading the next instruction in sequence and executes it, and so on.
● This type of instruction sequencing needs a counter to calculate the address of the next
instruction after execution of the current instruction is completed.
● It is also necessary to provide a register in the control unit for storing the instruction code
after it is read from memory.
● The computer needs processor registers for manipulating data and a register for holding
a memory address.
12.
List of basicRegisters
Code Bits Name Purpose
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
13.
Common Bus System
Needof Common Bus System ?
● The basic computer has eight registers, a memory unit, and a control unit.
● Paths must be provided to transfer information from one register to another and between
memory and registers.
● The number of wires will be excessive if connections are made between the outputs of
each register and the inputs of the other registers.
● Hence more efficient scheme with a common bus is used.
14.
● The outputsof seven registers and
memory are connected to the
common bus.
● The specific output that is selected
for the bus lines at any given time
is determined from the binary
value of the selection variables S2,
S1, and S0.
● The number along each output
shows the decimal equivalent of
the required binary selection.
● For example, the number along the
output of DR is 3. The 16-bit
outputs of DR are placed on the
bus lines when S2S1S0 =011 since
this is the binary value of decimal 3.
15.
● The linesfrom the common bus are
connected to the inputs of each
register and the data inputs of the
memory.
● The particular register whose LD
(load) input is enabled receives the
data from the bus during the next
clock pulse transition.
● The memory receives the contents
of the bus when its write input is
activated.
● The memory places its 16-bit
output onto the bus when the read
input is activated and S2S1S0 =111.
16.
● INPR isconnected to provide
information to the bus but OUTR
can only receive information from
the bus.
● This is because INPR receives a
character from an input device
which is then transferred to AC.
● OUTR receives a character from AC
and delivers it to an output device.
● There is no transfer from OUTR to
any of the other registers.
17.
● The inputsof AC come from an
adder and logic circuit.
● This circuit has three sets of inputs.
● One set of 16-bit inputs come from
the outputs of AC. They are used to
implement register micro-
operations such as complement AC
and shift AC.
● Another set of 16-bit inputs come
from the data regisler DR. The
inputs from DR and AC are used for
arithmetic and logic
microoperations.
● A third set of 8-bit inputs come
from the input register INPR.
Instruction Format
● Thebasic computer has three instruction code formats each having 16 bits
○ Memory reference instructions
○ Register reference instructions
○ I/O instructions
● The opcode part of the instruction contains three bits and the meaning of the remaining
13 bits depends on the operation code encountered.
20.
Memory reference instructions
●Bits 0-11 for specifying address.
● Bits 12-14 for specifying address.
● 15th bit specifies addressing modes. (0 for direct and 1 for indirect)
● Opcode=000 through 110
● I=0 or 1
● Eg:
○ AND - 0xxx(direct) or 8xxx(indirect)
○ ADD - 1xxx or 9xxx
21.
Register reference instructions
●Recognized by the operation code 111 with a 0 in the 15th bit of the instruction.
● Specifies an operation on or a test of the AC register.
● An operand from memory is not needed.
● Therefore the 12 bits are used to specify the operation to be executed.
● Eg:
○ CLA - 7800 : Clear AC
○ CLE - 7400 : Clear E
22.
I/O instructions
● Theseinstructions are needed for transfering informations to and from AC register.
● Recognized by the opcode 111 and a 1 in the 15th bit.
● Bits 0-11 specify the type of I/O Operation performed.
● Eg:
○ INP - F800 : Input characters to AC
○ OUT - F400 : Output characters from AC
24.
Instruction Set Completeness
●The set of instructions are said to be complete if the computer includes a sufficient number
of instructions in each of the following categories
a. Arithmetic, logical, and shift instructions
b. Instructions for moving information to and from memory and processor registers
c. Program control instructions
d. Input and output instructions
Timing and Control
●The timing for all registers in the basic computer is controlled by a master clock generator.
● The clock pulses are applied to all flip-flops and registers.
● The clock pulses do not change the state of a register unless the register is enabled by a
control signal.
● Two major types of control organization:
○ hardwired control
○ microprogrammed control.
27.
Hardwired Control
● Thecontrol logic is implemented
with gates, flip-flops, decoders, and
other digital circuits.
● It can be optimized to produce a
fast mode of operation.
● Requires changes in the wiring
among the various components if
the design has to be modified or
changed.
Microprogrammed
Control
● The control information is stored in
a control memory.
● The control memory is
programmed to initiate the
required sequence of
microoperations.
● Required changes or modifications
can be done by updating the
microprogram in control memory.
28.
Hardwired control unit
●Consists of two decoders, a sequence
counter, and a number of control logic gates.
● An instruction read from memory is placed in
the instruction register(IR).
● The IR is divided into three parts:
○ I bit, opcode, and Address bits.
● Op-code in 12-14 bits are decoded with a 3x8
decoder.
● 8 outputs of the decoder are designated by
the symbols D0 through D7.
● Bit 15 is transferred to a flip-flop I.
● Bits 0-11 are applied to the control logic
gates.
29.
Hardwired control unit
●The 4-bit sequence counter can count in
binary from 0-15.
● The outputs of the counter are decoded into
16 timing signals T0-T15.
● The sequence counter SC can be incremented
or cleared synchronously.
● Mostly,SC is incremented to provide the
sequence of timing signals(T1,T2,...,T15)
● Once in a while, the counter is cleared to 0,
causing the next active timing signal to be T0.
● Eg: Suppose,at time T4, SC is cleared to 0 if
decoder output D3 is active. D3T4: SC ← 0.
30.
Hardwired control unit
●The SC responds to the positive transition of
the clock.
● Initially, the CLR input of SC is active.
● Hence it clears SC to 0, giving the timing
signal T0 out of the decoder.
● T0 is active during one clock cycle and will
trigger only those registers whose control
inputs are connected to timing signal T0.
● SC is incremented with every positive clock
transition, unless its CLR input is active.
● This produces the sequence of timing signals
T0, T1, T2, T3, T4 up to T15 and back to T0.
