1. EKT 221 : Chapter 4 Computer Design Basics

2. Chapter Overview Part 1 – Datapaths Part 2 – A Simple Computer
Introduction
Datapath Example
Arithmetic Logic Unit (ALU)
Shifter
Datapath Representation
Control Word
Part 2 – A Simple Computer
Instruction Set Architecture (ISA)
Single-Cycle Hardwired Control
Instruction Decoder
Sample Instructions
Single Cycle Computer Issues
Multiple Cycle Hardwired Control
Sequential Control Design
Digital 2 will focus on Part 1
Part 2 will be covered in
Computer Architecture & Microprocessor. however prior reading is encourage.

3. Part 1 : Datapath Computer Specification
Instruction Set Architecture (ISA)
The specification of a computer's appearance to a programmer at its lowest level.
It describe all the available instruction set in the computer, where it is kept (address) and how to use it (read).
Computer Architecture
A high-level description of the hardware implementing the computer derived from the ISA

4. Part 1 : Datapath
The architecture usually includes additional specifications such as speed, cost, and reliability.
Simple computer architecture comprise of:
Datapath for performing operations
Control unit for controlling datapath operations
A datapath is specified by:
A set of registers
The microoperations performed on the data stored in the registers
A control interface

5. Datapaths : Guiding principles for basic datapaths (Typical):
The set of registers
Collection of individual registers
A set of registers with common access resources called a register file
A combination (individual & set of reg.) of the above
Microoperation implementation
One or more shared resources for implementing microoperations
Buses - shared transfer paths
Arithmetic-Logic Unit (ALU) - shared resource for implementing arithmetic and logic microoperations
Shifter - shared resource for implementing shift microoperations

6. The combination of: Datapath
A set of registers with a shared ALU, and interconnecting paths.

7. Block Diagram of a Generic Datapath
Four parallel-load registers
• Two mux-based register selectors
• Register destination decoder
• Mux B for external constant input
• Buses A and B with external address and data outputs
• ALU and Shifter with Mux F for output select (Function Unit)
• Mux D for external data input
• Logic for generating status bits V, C, N, Z

8. Block Diagram of a Generic Datapath
Example:
R1  R2 + R3
A Select
Place contents of R2 into Bus A 10
B Select
Place contents of R3 into the input of MUX B 11
MB Select
Place the 0 input of MUX B into Bus B
G Select
Provide the arithmetic operation A + B
???? (4bits)
MF Select
Place the ALU o/p on MUX F o/p
MD Select
Place the MUX F o/p onto Bus D
Destination Select
To select R1 as the destination of the data on Bus D 01
Load enable
To enable a register
R1 = HIGH
Note : G Select must refer to Function Table of Arithmetic Circuit (refer next 3 slides)

9. The Arithmetic/ Logic Unit
(ALU)

10. Arithmetic Logic Unit (ALU)
ALU Comprise of:
An arithmetic circuit (add, subtract)
A logic circuit (bitwise operation)
A selector to pick between the two circuits 1 2 3

11. Arithmetic Logic Unit (ALU)
ALU Comprise of:
An arithmetic circuit
An n-bit parallel adder
A block of input logic with 2 selectors S1 and S0 1
* Mode Select (S2) distinguishes between arithmetic and logic operations which actually construct item 3
G Select
(4-bits)

12. ALU
Is a combinational circuit that performs a set of basic microoperations on:
Arithmetic, and
Logic
Has a number of selection lines used to determine the operations to be performed
e.g.
n selection lines can specify up to 2n types of operations.

13. An n-bit ALU
n data inputs of A are combined with n data inputs of B, to generate the result of an operation at the G outputs.
S2=0  Arithmetic operations (8). Which one – is specified by S1, So and Cin.
S2=1  Logic operations (4). Which one – specified by So and Cin.

14. Question What are the 8 arithmetic operations?
What are the 4 logic operations?

15. How to design the ALU? Design the arithmetic section
Design the logic section
Combine both sections

16. To be designed… One Stage of ALU S2 = 0 for Arithmetic
S2 = 1 for Logic

17. Arithmetic Circuit Design
Given …

18. Arithmetic Circuit Design1
Analyse the Circuit:
Use
Note : X = A
G = A + Y + Cin
For S1 and S0 = 00, then
G = A
G = A
Eg. We can verify for n = 4 bit:
A = 1010
B = 0101

19. Building the B input Logic
Input = S1, S0 and B
Output = Y
Obtain the K-Map
Get the Boolean Expression
Y = BS0 + BS1

20. Building the B input Logic
Y = BS0 + BS1 Y
Example of a 4-bit Arithmetic Circuit
Any other alternative?

21. Use Multiplexer B B 1

22. Building the Logic Circuit2
The Logic Circuit performs bitwise operation
Commonly : AND, OR, XOR and NOT
Note : if 4 bit is wanted, then we have to arrange it in array
One Stage of Logic Circuit

23. Building the Selector for choosing Arithmetic or Logic Unit3
S2 = 0 for Arithmetic
S2 = 1 for Logic
One Stage of ALU
Refer also Fig. 10.2

24. Exercise Draw the complete diagram of the ALU, which consists of:
The Arithmetic Circuit
The Logic Circuit
The Selector circuit
For:
A bit-slice (one stage) circuit
A 4-bits circuit

25. S2 = 0 for Arithmetic operation S1 = 0 S0 = 1 Cin = 0
Example:
R1  R2 + R3
G Select
Provide the arithmetic operation A + B
???? (4bits)
Therefore;
S2 = 0 for Arithmetic operation
S1 = 0
S0 = 1
Cin = 0
LSB
G Select
(4-bits)
0010
0010
Answer :
MSB

26. The Shifter
The Shifter Shifts the value on Bus B, placing the result on an input of MUX F
The Shifter can:
Shift Right
Shift Left
It is obvious that the shifter would be a bidirectional shift register with parallel load

27. 4 Bit Basic Shifter S Operation 00
Parallel Load B (B to be passed thru the shifter unchanged) 01
Shift Right 10
Shift Left

28. Barrel Shifter
Sometimes the data need to be shifted ore rotated more than one bit position or in a single clock cycle
A Barrel Shifter can perform this by using MUX
2n input requires 2n MUX

29. 4-Bit Barrel Shifter
A rotate is a shift in which the bits shifted out are inserted into the positions vacated
The circuit rotates its contents left from 0 to 3 positions depending on Selector S.
Note that a left rotation by three (3) position the same as a right rotation by one position in this 4 bit barrel shifter

30. Datapath Representation
Register File
Simplified
Block
Diagram
Function Unit

31. Datapath Representation
Notice that the G, H and MF Select are combined as FS
Boolean Equations:
MF = F3.F2
Gi = Fi
Hi = Fi

32. Control Word The datapath has many control inputs
The signals driving these inputs can be defined and organized into a control word
To execute a microinstruction, we apply control word values for a clock cycle. For most microoperations, the positive edge of the clock cycle is needed to perform the register load
The datapath control word format and the field definitions are shown on the next slide

33. The Control Word Fields
Total : 16-bit
(3-bit)
(3-bit)
(3-bit)
(4-bit)
(1-bit)
Fields :
DA – D Address
AA – A Address
BA – B Address
MB –Mux B
FS – Function Select
MD –Mux D
RW – Register Write
The connections to datapath are shown in the next slide

35. Control Word Encoding

36. Example 1
Given the 16-bit Control Word as
Field : DA AA BA MB FS MD RW
Binary :
Symbolic :
R1 R2 R3 Reg F=A+B+1 Function Write
Refer to Control Word Encoding Table
Answer :
R1  R2 + R3 + 1

37. Example 2 1000101100111001 R4  sl R6 Given the 16-bit Control Word as
Field : DA AA BA MB FS MD RW
Binary :
Symbolic :
R4 R2 R6 Reg F=sl B Function Write
Refer to Control Word Encoding Table
Answer :
R4  sl R6

38. Example 3
Given the 16-bit Control Word as
Field : DA AA BA MB FS MD RW
Binary :
Symbolic :
R2 R0 R3 Reg F = A-1 Data_In No_Write
Refer to Control Word Encoding Table
Answer :
Data Out  R3
Address B is selected because MB = 0 (Refer to block Diagram on Slide 7)

39. Example 4
Given the 16-bit Control Word as
Field : DA AA BA MB FS MD RW
Binary :
Symbolic :
R4 R0 R3 Reg F = A-1 Data_In Write
Refer to Control Word Encoding Table
Answer :
R4  Data In
Address D is selected because MD = 1 (Refer to block Diagram on Slide 7)

40. Example 5
Given the 16-bit Control Word as
Field : DA AA BA MB FS MD RW
Binary :
Symbolic :
R5 R0 R3 Reg F = A+B Function Write
Refer to Control Word Encoding Table
Answer :
R5  R0 + R3

41. Example 5 continue…… 1010000110001001 R5  R0 + R3
Given the 16-bit Control Word as
Answer :
R5  R0 + R3
Assume : Registers are 8-bit and the value given are in hex, Find:
The new value of register content if R0 = 5 and R3 = 3
The new value of register content if R0 = 7 and R3 = 1C
Answers in 8-bit binary. 1. 2.

42. Other Examples of Control Word

43. Assignment Folder Questions
10 – 3, 4, 5, 6, 7

44. THANK YOU