CENG 241 Digital Design 1 Lecture 12

31
CENG 241 Digital Design 1 Lecture 12 Amirali Baniasadi [email protected]

description

CENG 241 Digital Design 1 Lecture 12. Amirali Baniasadi [email protected]. Memory. Memory unit: Stores binary information A collection of cells Two types of memory: RAM-Random Access Memory ROM-Read Only Memory RAM: Can read and write ROM:Programmable Logic Device ( PLD ). - PowerPoint PPT Presentation

Transcript of CENG 241 Digital Design 1 Lecture 12

Page 1: CENG 241 Digital Design 1 Lecture 12

CENG 241Digital Design 1

Lecture 12

Amirali [email protected]

Page 2: CENG 241 Digital Design 1 Lecture 12

2

Memory

Memory unit: Stores binary information A collection of cells

Two types of memory: RAM-Random Access Memory ROM-Read Only Memory

RAM: Can read and write ROM:Programmable Logic Device (PLD)

Page 3: CENG 241 Digital Design 1 Lecture 12

3

Programmable Logic Device: PLD

Programming: hardware procedure to insert bits into the configuration.

Different PLDs: ROM, Program Logic Array (PLA), Program Array Logic (PAL), Field Programmable Field Array (FPGA)

PLD may include hundreds of millions of gates

To show logic we use concise forms

Page 4: CENG 241 Digital Design 1 Lecture 12

4

Conventional and Concise Symbols

Page 5: CENG 241 Digital Design 1 Lecture 12

5

Random Access Memory

The time to transfer data in and out the device is the same

Information stored in group of bits called words.

Each word is assigned an address.

Page 6: CENG 241 Digital Design 1 Lecture 12

6

Memory Content Example

1024 memory locations: 10 bit address

16 bit data

Page 7: CENG 241 Digital Design 1 Lecture 12

7

Write and Read Operation

Write Operation: 1.Apply the binary address to address lines 2.Apply the data to the data lines 3.Activate the write input

Read Operation: 1.Apply the binary address to the address lines 2.Activate the read input

Page 8: CENG 241 Digital Design 1 Lecture 12

8

Memory Decoding

Memory Decoding: Select the memory word specified by the address

A memory with m words and n bits per word consists of m x n storage cells and decoding logic.

Page 9: CENG 241 Digital Design 1 Lecture 12

9

Memory Cell

Page 10: CENG 241 Digital Design 1 Lecture 12

10

4 x 4 RAM

Page 11: CENG 241 Digital Design 1 Lecture 12

11

Coincident Decoding

Regular decoding is costly: A decoder with k inputs and 2K outputs requires 2K AND gates with

k inputs per gate.

Total number of gates can be reduced by using two-dimensional decoding:

Basic idea: arrange memory cells in a ( as close as possible to) square configuration.

Use two k/2 input decoders instead of one k input decoder

Page 12: CENG 241 Digital Design 1 Lecture 12

12

Two-Dimensional Decoding

Instead of using a single 10 x 1024 decoderwe use two 5x32 decoders.

One decoder picks the row, one the column

Page 13: CENG 241 Digital Design 1 Lecture 12

13

Two-Dimensional Decoding

Needs 64 5-input AND gates instead of 102410-input gates.

Address is divided to two equal parts

What if impossible?

Page 14: CENG 241 Digital Design 1 Lecture 12

14

Address Multiplexing

Two types of RAM: Static RAM (SRAM) & Dynamic RAM (DRAM)

DRAM needs refreshing but has less number of transistors DRAMs have four times the density of SRAMs. DRAM is almost 4 times cheaper than SRAM. DRAM consumes less power. Since DRAM are large in size, they are arranged in two-

dimensional arrays.

Page 15: CENG 241 Digital Design 1 Lecture 12

15

Address Multiplexing

Note that the same line is used for both row and column.

Therefore address decoding is done in two steps

Page 16: CENG 241 Digital Design 1 Lecture 12

16

Read-Only Memory

Page 17: CENG 241 Digital Design 1 Lecture 12

17

32x8 ROM

Each OR gate has 32 inputs

Page 18: CENG 241 Digital Design 1 Lecture 12

18

ROM Programming

At 00000, 10110110 is stored. At 11111, 00110011 is stored.

1’s are connected ( x) 0’s are not.

Page 19: CENG 241 Digital Design 1 Lecture 12

19

Combinational Circuit Implementation

We can assume that each output bit can be considered as a Boolean function.

Combinational circuits can be used.

Example A7(I4,I3,I2,I1,I0)= Σ(0,2,3,……29)

Page 20: CENG 241 Digital Design 1 Lecture 12

20

Example 7-1

Design a circuit using a ROM that accepts a 3-bit number and generates the square.

Page 21: CENG 241 Digital Design 1 Lecture 12

21

Combinational PLDs

A combinational PLD consists of gates divided into AND array and OR array gates to provide an AND-OR sum of product implementation.

Program Logic Array (PAL): Most flexible PLD, both AND and OR arrays are programmable

Page 22: CENG 241 Digital Design 1 Lecture 12

22

Programmable Logic Array

Two differences of PLA with PROM:

1-PLA does not provide full decoding 2-PLA does not generate all minterms

Page 23: CENG 241 Digital Design 1 Lecture 12

23

Program Logic Array (PLA)

Each input goes through a buffer and an inverter

F1= AB’+AC+A’BC’

F2= (AC+BC)’

Page 24: CENG 241 Digital Design 1 Lecture 12

24

PLA Programming Table

inputs Output T C Product Term A B C F1 F2

AB’ 1 1 0 - 1 -AC 2 1 - 1 1 1BC 3 - 1 1 - 1 A’BC’ 4 0 1 0 1 -

Page 25: CENG 241 Digital Design 1 Lecture 12

25

Example 7-2

Implement the following two Boolean functions with a PLA

F1(A,B,C)= Σ (0,1,2,4) F2(A,B,C)= Σ (0,5,6,7)

Page 26: CENG 241 Digital Design 1 Lecture 12

26

Example 7-2

Page 27: CENG 241 Digital Design 1 Lecture 12

27

Program Array Logic (PAL)

PAL: PLD with a fixed OR array and programmable AND array.

Page 28: CENG 241 Digital Design 1 Lecture 12

28

Fuse Map for PAL

w(A,B,C,D)= Σ (2,12,13)x (A,B,C,D)= Σ (7,8,9,10,11,12,13,14,15)y (A,B,C,D)= Σ (0,2,3,4,5,6,7,8,10,11,15)z (A,B,C,D)= Σ (1,2,8,12,13)

w=ABC’+A’B’CD’x =A+BCDy =A’B+CD+B’D’z =ABC’+A’B’CD’+AC’D’+A’B’C’D =w+AC’D’+A’B’C’D

Has four inputs, by usingw, we reduce inputs to 3.

Page 29: CENG 241 Digital Design 1 Lecture 12

29

Fuse Map for PAL

w=ABC’+A’B’CD’

x =A+BCD

y =A’B+CD+B’D’

z =ABC’+A’B’CD’+AC’D’+A’B’C’D =w+AC’D’+A’B’C’D

Page 30: CENG 241 Digital Design 1 Lecture 12

30

2003 final exam

A 32Kx8 memory uses coincident decoding by splitting the internaldecoder into X-selection and Y-selection. a)What is the decoder size f oreach selection and how many gates and transistors are required f ordecoding the address? b)What will the decoder size and the number oftransistor and gates be if coincident decoding is not used?(12 points)

Page 31: CENG 241 Digital Design 1 Lecture 12

31

Summary

Memory & Programmable Logic