Instructor: Alexander Stoytchev

Instructor: Alexander Stoytchev
CprE 281: Digital Logic Instructor: Alexander Stoytchev

Simple Processor CprE 281: Digital Logic
Iowa State University, Ames, IA Copyright © Alexander Stoytchev

Administrative Stuff Final Project (7% of your grade)
Posted on the class web page (Labs section) This is due this week (during your lab)

Administrative Stuff The FINAL exam is scheduled for
Friday Dec 12:00 – 2:00 PM It will be in this room.

Final Exam Format The exam will cover: Chapter 1 to Chapter 6, and Sections Emphasis will be on Chapters 5, 6, and 7 The exam will be open book and open notes You can bring up to 5 pages of handwritten notes, plus your textbook.

Final Exam Format The exam will be out of 130 points
You need 95 points to get an A It will be great if you can score more than 100 points. but you can’t roll over your extra points 

Topics for the Final Exam
K-maps for 2, 3, and 4 variables Multiplexers (circuits and function) Synthesis of logic functions using multiplexers Shannon’s Expansion Theorem 1’s complement and 2’s complement representation Addition and subtraction of binary numbers Circuits for adding and subtracting Serial adder Latches (circuits, behavior, timing diagrams) Flip-Flops (circuits, behavior, timing diagrams) Counters Registers

Topics for the Final Exam
Synchronous Sequential Circuits FSMs Moore Machine Mealy Machines State diagrams, state tables, state-assigned tables State minimization Designing a counter Arbiter Circuits Reverse engineering a circuit ASM Charts Register Machines Bus structure and Simple Processors Something from Star Wars

How to Study for the Final Exam
Form a study group Go over the slides for this class Go over the problems at the end of Ch 5 & 6 Exercise Get some sleep

Digital System Datapath circuit Control circuit To store data
To manipulate data To transfer data from one part of the system to another Comprise building blocks such as registers, shift registers, counters, multipliers, decoders, encoders, adders, etc. Control circuit Controls the operation of the datapath circuit Designed as a FSM

A Simple Processor [ Figure 7.9 from the textbook ]

What are the components?
[ Figure 7.9 from the textbook ]

Register R0 [ Figure 7.9 from the textbook ]

All Registers [ Figure 7.9 from the textbook ]

2-Bit Register IN1 IN0 OUT1 OUT0 1

4-Bit parallel-access shift register
(this one is not used in this example) [ Figure 5.18 from the textbook ]

Tri-State Driver [ Figure 7.9 from the textbook ]

All Tri-State Drivers [ Figure 7.9 from the textbook ]

Tri-state driver (see Appendix B for more details)

Bus Structure A way to transfer data from any register (device) to any other register (device) A set of n wires to transfer an n-bit data What if two registers write to the bus at the same time? n Bus

Tri-State Driver (or Tri-State Buffer)
Z: High impedance state Figure 7.1. Tri-state driver. device3 device2 device1 device0 Then several devices can be connected to a single wire Note that at any time, at most one of e0, e1, e2, and e3 can be set to 1 e2 e3 e1 e0

An n-bit Tri-State Driver
An n-bit tri-state buffer can be constructed using n 1-bit tri-state buffers. Input e e e e e Output

How to connect registers to a bus (using tri-state drivers)
This shows only two 2-bit registers, but this design scales to more and larger registers. [ Figure 7.3 from the textbook ]

An alternative approach uses multiplexers to implement a bus

An alternative approach uses multiplexers to implement a bus
(eight 5-to-1 multiplexers) This requires one multiplexer per bit. Assuming 8-bit registers, we need eight 5-to-1 multiplexers. [ Figure 7.4 from the textbook ]

Adder/Subtractor [ Figure 7.9 from the textbook ]

Adder/Subtractor unit
y y y n – 1 1 Add Sub control x x x n – 1 1 c n -bit adder c n s s s n – 1 1 [ Figure 3.12 from the textbook ]

The first two stages of a carry-lookahead adder
x 1 y g p s c 2 [ Figure 3.15 from the textbook ]

A hierarchical carry-lookahead adder
Block x 15 8 – y 7 3 1 Second-level lookahead c s P G 31 24 16 32 [ Figure 3.17 from the textbook ]

Control Circuit [ Figure 7.9 from the textbook ]

Designing The Control Circuit

The function register and decoders
Clock X w En y 1 2 3 2-to-4 decoder Function Register Y I FR in f Rx Ry Function [ Figure 7.11 from the textbook ]

Clock X w En y 1 2 3 2-to-4 decoder Function Register Y I FR in f Rx Ry Function All three decoders are always enabled [ Figure 7.11 from the textbook ]

Operations performed by this processor
Function Performed Load Rx, Data Rx  Data Move Rx, Ry Rx  [Ry] Add Rx, Ry Rx  [Rx] + [Ry] Sub Rx, Ry Rx  [Rx] - [Ry] [ Table 7.1 from the textbook ]

Function Performed Load Rx, Data Rx  Data Move Rx, Ry Rx  [Ry] Add Rx, Ry Rx  [Rx] + [Ry] Sub Rx, Ry Rx  [Rx] - [Ry] Where Rx and Ry can be one of four possible options: R0, R1, R2, and R3 [ Table 7.1 from the textbook ]

Function Load 1 Move Add Sub Rx1 Rx0 Register R0 1 R1 R2 R3 Ry1 Ry0 Register R0 1 R1 R2 R3

 Move R3, R0 f1 f0 Function Load 1 Move Add Sub Rx1 Rx0 Register R0 1 R1 R2 R3 Ry1 Ry0 Register R0 1 R1 R2 R3

Clock X w En y 1 2 3 2-to-4 decoder Function Register Y I FR in f Rx Ry  Move R3, R0

Clock X w En y 1 2 3 2-to-4 decoder Function Register Y I FR in f Rx Ry 0 1 1 1 0 0  Move R3, R0

Clock X w En y 1 2 3 2-to-4 decoder Function Register Y I FR in f Rx Ry 0 1 1 1 0 0 0 1 1 1 0 0  Move R3, R0

 Add R1, R3 f1 f0 Function Load 1 Move Add Sub Rx1 Rx0 Register R0 1 R1 R2 R3 Ry1 Ry0 Register R0 1 R1 R2 R3

 Sub R0, R2 f1 f0 Function Load 1 Move Add Sub Rx1 Rx0 Register R0 1 R1 R2 R3 Ry1 Ry0 Register R0 1 R1 R2 R3

x x  Load R2, Data f1 f0 Function Load 1 Move Add Sub Rx1 Rx0 Register R0 1 R1 R2 R3 Ry1 Ry0 Register R0 1 R1 R2 R3

Similar Encoding is Used by Modern Chips
[

Sample Assembly Language Program For This Processor
Move R3, R0 Add R1, R3 Sub R0, R2 Load R2, Data

Machine Language vs Assembly Language
Meaning / Interpretation 011100 100111 110010 001000 Move R3, R0 Add R1, R3 Sub R0, R2 Load R2, Data R3  [R0] R1  [R1] + [R3] R0  [R0] – [R2] R2  Data

Meaning / Interpretation 011100 100111 110010 001000 Move R3, R0 Add R1, R3 Sub R0, R2 Load R2, Data R3  [R0] R1  [R1] + [R3] R0  [R0] – [R2] R2  Data For short, each line can be expressed as a hexadecimal number

Meaning / Interpretation 1C 27 32 08 Move R3, R0 Add R1, R3 Sub R0, R2 Load R2, Data R3  [R0] R1  [R1] + [R3] R0  [R0] – [R2] R2  Data

Intel 8086 [

Intel 8086 Memory Address [

Intel 8086 Machine Language [

Intel 8086 Assembly Language [

Intel 8086 Comments [

Another Part of The Control Circuit

A part of the control circuit for the processor
Reset Up-counter Clear w En y 1 2 3 T 2-to-4 decoder Q Clock [ Figure 7.10 from the textbook ]

What are the components?

2-Bit Up-Counter T y 2-to-4 decoder w En Q Clock Up-counter Clear

2-bit Synchronous Up-Counter
Q Clock 1 Clear_n

2-bit Synchronous Up-Counter with Enable
Q Clock Enable Clear_n

2-to-4 Decoder with Enable Input

2-to-4 Decoder with an Enable Input
[ Figure 4.13c from the textbook ]

2-to-4 Decoder with an Enable Input
1 (always enabled in this example) [ Figure 4.13c from the textbook ]

So How Does This Work? T y 2-to-4 decoder w En Q Clock Up-counter
Reset Up-counter Clear w En y 1 2 3 T 2-to-4 decoder Q Clock

So How Does This Work? 0 0 T y 2-to-4 decoder w En Q Clock Up-counter
Reset Up-counter Clear w En y 1 2 3 T 2-to-4 decoder Q Clock 0 0

So How Does This Work? 0 0 1 0 0 0 T y 2-to-4 decoder w En Q Clock

Meaning/Explanation This is like a FSM that cycles through its four states one after another. But it also can be reset to go to state 0 at any time. The implementation uses a counter followed by a decoder. The outputs of the decoder are one-hot-encoded. This is like choosing a state assignment for an FSM in which there is one Flip-Flop per state, i.e., one-hot encoding (see Section in the textbook)

Deriving the Control Signals

Control Signals [ Figure 7.9 from the textbook ]

Another Control Signal

Yet Another Control Signal
Clock X w En y 1 2 3 2-to-4 decoder Function Register Y I FR in f Rx Ry [ Figure 7.11 from the textbook ]

Control signals asserted in each time step
(Load): I0 Extern Rin = X Done (Move): I1 Rin = X Rout = Y (Add): I2 Rout = X Ain Gin AddSub = 0 Gout (Sub): I3 AddSub = 1 [ Table 7.2 from the textbook ]

(Load): I0 Extern Rin = X Done (Move): I1 Rin = X Rout = Y (Add): I2 Rout = X Ain Gin AddSub = 0 Gout (Sub): I3 AddSub = 1 These come from the outputs of the 2-to-4 decoder in Figure 7.10. They are also one-hot encoded. These are the outputs of the first 2-to-4 decoder that is connected to the two most significant bits of the function register. They are one-hot encoded so only one of them is active at any given time (see Fig 7.11). [ Table 7.2 from the textbook ]

Different Operations Take Different Amount of Time
(Load): I0 Extern Rin = X Done (Move): I1 Rin = X Rout = Y (Add): I2 Rout = X Ain Gin AddSub = 0 Gout (Sub): I3 AddSub = 1 1 clock cycle 3 clock cycles [ Table 7.2 from the textbook ]

Expressing the 'Extern' signal
(Load): I0 Extern Rin = X Done (Move): I1 Rin = X Rout = Y (Add): I2 Rout = X Ain Gin AddSub = 0 Gout (Sub): I3 AddSub = 1 Extern = I0 T1

Expressing the 'Done' signal
(Load): I0 Extern Rin = X Done (Move): I1 Rin = X Rout = Y (Add): I2 Rout = X Ain Gin AddSub = 0 Gout (Sub): I3 AddSub = 1 Done = (I0 + I1)T1 + (I2 + I3)T3

In Chapter 6 we looked at the following Register Swap Example

Register Swap Controller

Register Swap Controller
Design a Moore machine control circuit for swapping the contents of registers R1 and R2 by using R3 as a temporary. [ Figure 6.10 from the textbook ]

If an output is not given, then assume that it is 0.
State Diagram D: R 3 out 1 = in Done , w C: 2 B: A: No Transfer Reset If an output is not given, then assume that it is 0. [ Figure 6.11 from the textbook ]

Animated Register Swap

0xFF 0x42 0xC9 These are the original values of the 8-bit registers

1 0xFF 0x42 0xC9 For clarity, only inputs that are equal to 1 will be shown.

1 0xFF 1 0x42 1 0xC9

1 0xFF 1 0x42 0x42 1 0xC9

1 0xFF 1 0x42 1 0x42

1 1 0xFF 1 0x42 0x42

1 1 0xFF 1 0x42 0xFF 0x42

1 1 0xFF 1 0xFF 0x42

1 1 0xFF 0xFF 1 0x42

1 1 0xFF 0xFF 0x42 1 0x42

1 1 0x42 0xFF 1 0x42

0x42 0xFF 1 0x42

If an output is not given, then assume that it is 0.
State Diagram D: R 3 out 1 = in Done , w C: 2 B: A: No Transfer Reset If an output is not given, then assume that it is 0. [ Figure 6.11 from the textbook ]

Encoding #1: A=00, B=01, C=10, D=11 (Uses Two Flip-Flops)

y1 y2 y1 y2 y1 y2 Present Next state state Outputs A 00 1 B 01 10 C 11
1 B 01 10 C 11 D w = 0 w = 1 y2y1 Y2Y1 Y2Y1 R1out R1in R2out R2in R3out R3in Done

Encoding #3: A=0001, B=0010, C=0100, D=1000 (One-Hot Encoding – Uses Four Flip-Flops)

Let's Complete the Circuit Diagram
R1out = R2in = y3 R2out = R3in = y2 R1in = R3out = Done = y4 D Q Y 2 1 w Clock y 3 4 Y1 = w y1 + y4 Y2 = w y1 Y3 = y2 Y4 = y3

Implementing the States with a Counter
The implementation with the counter and the 2-to-4-decoder used in Section 7.2 does something similar, but is much simpler.

(Load): I0 Extern Rin = X Done (Move): I1 Rin = X Rout = Y (Add): I2 Rout = X Ain Gin AddSub = 0 Gout (Sub): I3 AddSub = 1 [ Table 7.2 from the textbook ]

Derivation of the Control Inputs
See pages 434 and 435 in the textbook

Arithmetic Logic Unit (ALU)

Arithmetic Logic Unit (ALU)
Arithmetic Logic Unit (ALU) computes arithmetic or logic functions Example: A four-function ALU has two selection bits S1 S0 (also called OpCode) to specify the function 00 (ADD), 01 (SUB), 10 (AND), 11 (OR) Then the following set up will work S1 S0 Function ADD 1 SUB AND OR 00 01 10 11 ADD A B MUX A SUB A B A L U AND A B Result Result B OR A B Symbol S1 S0 S1 S0 (OpCode)

An Alternative Design of Four-Function ALU
The previous design is not very efficient as it uses an adder and a subtractor circuit We can design an add/subtract unit as discussed earlier Then we can design a logical unit (AND and OR) separately Then select appropriate output as result What are the control signals, Add/Sub, Select0, and Select1? Add/Sub 1 ADD/SUB A B MUX S1 S0 Function ADD 1 SUB AND OR AND A B 1 MUX Result OR A B Select0 Select1

A Digital System that Implements a Simple Processor
For this ALU, A+B if AddSub=1 A-B if AddSub=0 [ Figure 7.9 from the textbook ]

Examples of Some Famous Microprocessors

Intel's 4004 Chip [

Technical specifications
Maximum clock speed was 740 kHz Instruction cycle time: 10.8 µs (8 clock cycles / instruction cycle) Instruction execution time 1 or 2 instruction cycles (10.8 or 21.6 µs), to instructions per second Built using 2,300 transistors [

Technical specifications
Separate program and data storage. The 4004, with its need to keep pin count down, used a single multiplexed 4-bit bus for transferring: 12-bit addresses 8-bit instructions 4-bit data words Instruction set contained 46 instructions (of which 41 were 8 bits wide and 5 were 16 bits wide) Register set contained 16 registers of 4 bits each Internal subroutine stack, 3 levels deep. [

Intel's 8086 Chip [

Simplified block diagram of
Intel 8088 (a variant of 8086); 1=main registers; 2=segment registers and IP; 3=address adder; 4=internal address bus; 5=instruction queue; 6=control unit (very simplified!); 7=bus interface; 8=internal databus; 9=ALU; 10/11/12=external address/ data/control bus. [

Questions?

THE END

Instructor: Alexander Stoytchev

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