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Lecture 11-1 FPGA W e have finished combinational circuits, and learned registers. Now Lecture 11-1 FPGA W e have finished combinational circuits, and learned registers. Now

Lecture 11-1 FPGA W e have finished combinational circuits, and learned registers. Now - PowerPoint Presentation

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Lecture 11-1 FPGA W e have finished combinational circuits, and learned registers. Now - PPT Presentation

Chap 9 CH 2 Complex Programming Logic Devices X ilinx XCR3064XL CPLD Function block 16 macrocells PLA Macrocell a flip flop multiplexers IA routes signals Input of function block ID: 634626

chap function lut expansion function chap expansion lut block table logic x2x3 shannon

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Slide1

Lecture 11-1 FPGA

W

e have finished combinational circuits, and learned registers. Now are ready to see the inside of an FPGA.Slide2

Chap 9

C-H

2

Complex Programming Logic Devices

X

ilinx XCR3064XL CPLDFunction block (16 macrocells)= PLAMacrocell = a flip flop + multiplexersIA routes signals

Input of function blockSlide3

Chap 9

C-H

3

Function Block and MC

Signal from PLA ->

marcocell -> I/O pinUse CAD tool to fit the design into the PLD.Slide4

Chap 9

C-H

4

Field Programmable Gate Arrays (FPGA)

Logic cell: configurable logic blocks (CLBs)

Input/Output blocks (I/O blocks)Slide5

Chap 9

C-H

5

Configurable Logic Block

Inside a CLB: function generators (LUT), FFs, and MUXs

LUT: lookup table (truth table) is a reprogrammable ROM (16 1-bit words)Slide6

Chap 9

C-H

6

A Lookup Table (LUT)

If we want F =

abc (one minterm)1110 (and F=1) + 1111 (and F=1)Or if we want F = a’b’c’d’ + a’b’cd’ + …abcd

. (15 minterms)Require a single function generator. Program the LUT table to get what we want.

a b c d F

0 0 0 0 0

0 0 0 1 1

0 0 1 0 1

…..

1 1 1 1 1

This is a 4-variable function

generator.

Bit stored in the LUT.

To be implementedSlide7

Chap 9

C-H

7

Application of Shannon’s

Expansion Theorem

What if # of variables > 4 variables

f (x1, x2,… xn)= xi’ f

(x1, x2,..xi-1, 0, xi+1

, … xn) + xi

f

(x

1

, x

2

,..x

i-1

, 1, x

i+1

, …

x

n

)

= x

i

f

0

+ x

i

f

1

f

(a, b, c, d, e) = a’

f(0, b, c, d, e) + a f(1, b, c, d, e)

Let a = 0, what lefts are terms with b, c, d, e

2-1 MUX F = a’I

1

+ a I2Slide8

Shannon’s

Expansion

Theorem: a majority function f = x1x2 + x1x3 + x2x3= ~x1(x2x3) +

x1(x2+x3+x2x3); //let x1 = 0 for f, and let x1= 1 for f= ~x1(x2x3) + x1(x2+x3)What is the circuit using a 2-1 MUX for x1 ?Slide9

Shannon’s Expansion

Theorem: XOR

Ex: f = x1⊕ x2 ⊕ x3 = ~x1(0⊕ x2 ⊕ x3) + x1(1⊕ x2 ⊕ x3) = ~x1(x2

⊕ x3) + x1~(x2 ⊕ x3)

f

x

2

x

1

f

= x1

⊕ x2

f

x

3

x

1

x

2

0

0

0

1

1

0

1

1

0

1

1

0

0

0

0

1

1

0

1

1

1

0

0

1

x

1

x

2

x

3

f

0

0

0

0

1

1

1

1

x

2

x

3

Å

x

2

x

3

ÅSlide10

Shannon’s expansion for more than two variables

f

(x

1, x2,… xn)= xi’ f (x1, x2,..xi-1, 0, xi+1, … xn) + xi f (x1, x2,..xi-1, 1, xi+1, … x

n)= xi’

f0 + xif1

f (x1, x2,… xn)= x1’

x2’ f (0, 0,

x3… xn

)

+ x

1

x

2

f

(0,

1,

x

3

x

n

) +

x

1

x

2

f

(1,

0, x3… x

n) + x1 x2 f (1, 1, x3… xn)Say for x1 and x2; This expansion can be implemented by a 4-to-1 multiplexer where x1 and x2 are the selection signals.