Lecture 2

Chapter 4
Register Transfer and Microoperations
• 4-1 Register Transfer Language
• 4-2 Register Transfer
• 4-3 Bus and Memory Transfer
• 4-4 Arithmetic Microoperations
• 4-5 Logic Microoperations
• 4-6 Shift Microoperations
• 4-7 Arithmetic Logic Shift Units

EE-222 Computer Architecture
Muhammad Bilal Saif 33

CONNECTING REGISTERS
• In a digital system with many registers, it is impractical to
have dedicated connections between all registers.
• To completely connect n registers we require n(n-1) lines
– So this is not a realistic approach to use separate wiring for each register
in a large digital system

• Instead, take a different approach
• Have one centralized set of circuits for data transfer – the
bus
• Have control circuits to select which register is the source,
and which is the destination


BUS AND BUS TRANSFER
Bus is a path (of a group of wires) over which information is
transferred, from any of several sources to any of several destinations.
…
…

Dec
…
…

R1 R2 …… Rn

Bus

From a register to bus: BUS  R


Bus Implementation
• Bus can be implemented by
1. Mux
2. Tristate buffer


Bus implementation by using Mux
Register A Register B Register C Register D
0 1 2 3 0 1 2 3 0 1 2 3 0 1 2 3

B C D0 B 1 C1 D1 B2 C2 D2 B3 C3 D3
0 0

0 0 0 0
4 x1 4 x1 4 x1 4 x1
MUX MUX MUX MUX

0x
select
1
0y

Register A
4-line bus C 3rd bit of all the
0th
bit of all the 1st bit of all the Registers Registers
Registers
Muhammad Bilal Saif 2nd bit of all the Registers 37

General rules
• A bus system will multiplex k registers of n bits
each to produce an n-line common bus.
• The number of multiplexers needed to
construct the bus is equal to n (# of bits on
each register).
• The size of each multiplexer must be k x 1
since in multiplexes k data lines.


General rule (Simplified)
• # of Mux required is equal to # of bits in each
register.
• Size of Mux is dependent on total number of
registers


BUS TRANSFER IN RTL
• Depending on whether the bus is to be
mentioned explicitly or not, register transfer can
be indicated as either
R2 R1 Sout
n

Register Bus
or Clk
Bank
BUS R1, R2  BUS
Sin
n

• In the former case the bus is implicit, but in the
latter, it is explicitly indicated


Bus implementation by using Three state buffer

Three-State Bus Buffers

0
1 0
1
z Output Y=A if C=1
Normal input A High-impedance if C=0

Control input C
0
1

In high impedance the gate acts like an open circuit.


Three state Bus Buffer
• Tri-state gates can be used to implement the
functioning of any other conventional gate
such as AND, OR, NAND, NOR etc.

Assignment # 02:
a) Draw the circuit, which performs the functionality of OR gate, by using only
tri-state buffers and resistors.
b) Repeat the above for NOR gate

Three states gates in Verilog


Syntax of three state gates


Bus implementation by using Three state buffer

Bus line with three-state buffers
Bus line for bit 0
A0
B0
C0 Bus line for bit1
A1
D0
B1
S0 0
Select 1 C1
S1 2
Enable 3 D1

Bus line for bit2
Bus line for bit3 A2
A3
B2
B3
C2
C3
D2
D3


General Rule
• In case of three state buffers; a common bus
for k registers of n bits each, can be
constructed by using n circuits with k buffer in
each as shown in the previous slide.
• Only one decoder is required.


Surprise Quiz - 01
• Draw the diagram of a bus system by using tri-
state buffers, having
– 4 registers of 2 bits each.


Simple RAM design
Control Signals
Read/Write Control
8

8 bit Register
Address Bus 2
8 bit Register
Dec
8 bit Register

8 bit Register

Data Bus
In terms of RAM, each internal
register is called a “word”

MEMORY (RAM)
• Assume the RAM contains 2k words and each word
has n bits. It needs the following
– n data input lines
– n data output lines
data input lines
– k address lines
n
– A Read control line
– A Write control line address lines
k
RAM
Read
unit
Write

n
data output lines


RAM Calculation
In computer
210 = 1024 ≈ 1K
220 = 1048576 ≈ 1M
230 = 1073741824 ≈ 1G


MEMORY TRANSFER
• Collectively, the memory is viewed at the register level as
a device, “M”.
• Since it contains multiple locations, we must specify
which address in memory we will be using
• Memory is usually accessed in computer systems by
putting the desired address in a special register, the
Memory Address Register (MAR, or AR)
• When memory is accessed, the contents of the MAR get
sent to the memory unit’s address lines M
Memory Read
MAR / AR
unit Write
Further material regarding memory can be found at
article 2-7, p59 of text book Data out Data in

MEMORY READ
• To read a value from a location in memory and load it into a register,
the register transfer language notation looks like this:

R1  M[MAR]

• This causes the following actions to occur
– The contents of the MAR get sent to the memory address lines
– A Read (= 1) gets sent to the memory unit
– The contents of the specified address are put on the memory’s output data lines
– These get sent over the bus to be loaded into register R1


MEMORY WRITE

• To write a value from a register to a location in memory looks like
this in register transfer language:

M[MAR]  R1

• This causes the following to occur
– The contents of the MAR get sent to the memory address lines
– The values in register R1 get sent over the bus to the data input lines of the
memory
– A Write (= 1) gets sent to the memory unit
– The values get loaded into the specified address in the memory


SUMMARY OF R. TRANSFER MICROOPERATIONS

A B Transfer content of reg. B into reg. A
AR DR(AD) Transfer content of AD portion of reg. DR into reg. AR
A  constant Transfer a binary constant into reg. A
BUS  R1, Transfer content of R1 into bus and, at the same time,
R2 BUS transfer content of bus into R2
AR Address register
DR Data register
M[AR] Memory word specified by reg. AR
M Equivalent to M[AR]
DR  M Memory read operation: transfers content of
memory word specified by AR into DR
M  DR Memory write operation: transfers content of
DR into memory word specified by AR


Chapter 4
Register Transfer and Microoperations
• 4-1 Register Transfer Language
• 4-2 Register Transfer
• 4-3 Bus and Memory Transfer
• 4-4 Arithmetic Microoperations
• 4-5 Logic Microoperations
• 4-6 Shift Microoperations
• 4-7 Arithmetic Logic Shift Units


MICROOPERATIONS
• Computer system microoperations are of four
types:
– Register transfer microoperations
– Arithmetic microoperations We have covered this topic;
this microoperation does
– Logic microoperations not alter the register
contents while transfer
– Shift microoperations

These operations change
the register contents while
transfer

ARITHMETIC MICROOPERATIONS
• The basic arithmetic microoperations are
– Addition
– Subtraction
– Increment Can be implemented by
– Decrement 1. Binary up/down
Subtraction is achieved
by counter i.e.
addition
• The additional arithmetic microoperations are
– Add with carry 2. Adding/subtracting 1
R3R1 + R2’ + 1
– Subtract with borrow
– Transfer/Load
– etc. …

Summary of Typical Arithmetic Micro-Operations
R3  R1 + R2 Contents of R1 plus R2 transferred to R3
R3  R1 - R2 Contents of R1 minus R2 transferred to R3
R2  R2’ Complement the contents of R2
R2  R2’+ 1 2's complement the contents of R2 (negate)
R3  R1 + R2’+ 1 subtraction
R1  R1 + 1 Increment
R1  R1 - 1 Decrement

BINARY ADDER
For add microoperation we require
•Registers that hold the data
•Digital components that perform the addition (Full adder)

B3 A3 B2 A2 B1 A1 B0 A0

Binary Adder
FA C3 FA C2 FA C1 FA C0

C4 S3 S2 S1 S0

For n bit binary adder, we require n full adders.


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