tceic.com
>> >>

5


5-10 1K * 8 n = log21024 = 10 ϧ 8 8 5-11 32 ˛ 64KB 32 64KB = 16KW 16K 14 0000H 3FFFH PC

5-13 10241 16K8 (1) (2) 4K8 ަˉݦ˦ (1) 16 16K 168 = 128 (2) 14 4 2 12 10

1



2

4 16K8 5-15 16K8 SRAM 4K 4 A15A0(A0 ) D7D0 R/W A15 A14 A13 A12 A11 A10 A9 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 A8 0 1 0 1 0 1 0 1 A7 0 1 0 1 0 1 0 1 A6 0 1 0 1 0 1 0 1 A5 0 1 0 1 0 1 0 1 A4 0 1 0 1 0 1 0 1 A3 0 1 0 1 0 1 0 1 A2 0 1 0 1 0 1 0 1 A1 0 1 0 1 0 1 0 1 A0 0 1 0 1 0 1 0 1
4K 4K 4K 4K

A15A14 0A13A12 00011011 4 4K RAM

3

5-17 16K1 DRAM 64KB (1) (2) 0.5 CPU 1 (1) ()

4

(2) 16K1 128128 0.5128 = 64 CPU 1 64 ¡ 1 2ms, 2ms / 128 = 15.625 0.5 * 128 = 64 5-18 8 16 (A15A0) 8 (D7D0), MREQ() R/ W ( ) 08191 ROM 8192~32767 ()2K () ՛8K8 ROM16K12K84K88K8 SRAM CPU 0000H1FFFH8191(8K) 2000H7FFFH(24K) 8000H87FFH(2K) 8K8 ROM 8K8 RAM 3 2K8 RAM 1 A15 A14 A13 A12 A11 A10 A9 0 0 0 0 0 0 0 0 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 0 0 1 1 0 0 0 1 0 1 0 1 0 1 0 0 0 1 0 1 0 1 0 1 0 0 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 A8 0 1 0 1 0 1 0 1 0 1 A7 0 1 0 1 0 1 0 1 0 1 A6 0 1 0 1 0 1 0 1 0 1 A5 0 1 0 1 0 1 0 1 0 1 A4 0 1 0 1 0 1 0 1 0 1 A3 0 1 0 1 0 1 0 1 0 1 A2 0 1 0 1 0 1 0 1 0 1 A1 0 1 0 1 0 1 0 1 0 1 A0 0 1 0 1 0 1 0 1 0 1
2K 8K 8K 8K 8K

A15A14A13 3-8 2K 11

5



5-19 15KB 8KB EPROM 4K 8 7KB SRAM ৵4K4,2K 4,1K4. A15A0 D7D0R/ W MREQ 2 4K8 ROM2 4K4 RAM2 2K4 RAM2 1K4 RAM A15 0 0 0 0 0 0 A14 0 0 0 0 0 0 A13 0 0 0 0 1 1 A12 0 0 1 1 0 0 A11 0 1 0 1 0 1 A10 0 1 0 1 0 1 A9 0 1 0 1 0 1 A8 0 1 0 1 0 1
6

A7 0 1 0 1 0 1 A6 0 1 0 1 0 1 A5 0 1 0 1 0 1 A4 0 1 0 1 0 1 A3 0 1 0 1 0 1 A2 0 1 0 1 0 1 A1 0 1 0 1 0 1 A0 0 1 0 1 0 1
4K 4K 4K

0 0 0 0

0 0 0 0

1 1 1 1

1 1 1 1

0 0 1 1

0 1 0 0

0 1 0 1

0 1 0 1

0 1 0 1

0 1 0 1

0 1 0 1

0 1 0 1

0 1 0 1

0 1 0 1

0 1 0 1

0 1 0 1

2K

1K

A15A14 A13A12 00011011 4K ROM 0000H0FFFH CS 0 = Y 0 4K ROM 1000H1FFFH CS1 = Y1 4K RAM 2000H2FFFH CS 2 = Y 2 2K RAM 3000H37FFH CS 3 = A11 + Y 3 1K RAM 3800H3BFFH CS 4 = A10 + A11 + Y 3

EN

=

MREQ + A14 + A15

2K RAM A10A01K RAM A9A0

5-20 16 64KB I/O FC00 FFFFH 2164 (64K * 1) I/O
7

I/O 1111 1100 0000 0000 1111 1111 1111 1111.A15A10 1. A15A10 1

CS

= A15 * A14 * A13 * A12 * A11 * A10 +

MREQ

5-25 4MBCache 16KB 8 32 4 ( Cache 4 ) Cache (1) ڦŦ (2) Cache CPU 012399 100 ( ) 8 (3) Cache 6 Cache Cache (1) 4MB 22 84 = 32 5 4 2 s 324 = 128 Cache 16KB 16KB / 128B = 128 7 4MB / 16KB = 256 8

(2) Cache CPU 0 ħ CACHE 0 Cache 8 07 Cache 17 01299 012.12 CPU 099 1 13 ħz
8

Cache = (800 - 13) / 800 = 98.375% (3) Cache t 6t. Cache 6t Cache 6t * (1-98.375%) Cache 98.375%t, 6t * (1-98.375%) + 98.375%t = 1.08125t 6t / 1.08125t 5.5

9


:

5.doc

5 - 5-10

5.pdf

5___ 5 5-15. 16K8,...

.doc

- 5 5 1

ʦ.doc

ʦ - ʦ 2. 5.1

5.doc

5 - 5 5 1. CPU : (1)̧...

1-5.._.ppt

1-5..___1-5.. 1. ...

..doc

. - 5 5 1. CPU : (1)̧( IR...

5().doc

5()___5 , ...

()-5_.ppt

()-5___() ...

.doc

___,, 1 . X Y ,...

()___.ppt

()_____()(...

( ).pdf

36 1 1.


5.doc

5___ 5 5 1. CPU ...

( ).pdf

( ) - ......

5_.ppt

5 - 2. CP

_()3.doc

_()3 - 3 3 1 20 32 ˛, (1) ...

.pdf

- 1 1.

,.doc

, - 1. CPU : (1) ̧( IR); (2) ...

-()_.ppt

-() - ?

|
All rights reserved Powered by www.tceic.com
copyright ©right 2010-2021
zhit325@126.com