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Memory Organization

Memory Organization. Microprocessor and Interfacing 261313. CPU 8088 10 MHz 8/16 Bit 1M RAM. CPU 80486 66 MHz 32 Bit 8M RAM. Pentium 133 MHz 32 Bit 16M RAM. Pentium MMX 200 MHz 32 Bit 48M RAM. Pentium II 366 MHz 32 Bit 128M RAM. Pentium M 1.6 GHz 32 Bit

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Memory Organization

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  1. Memory Organization Microprocessor and Interfacing261313

  2. CPU 8088 10 MHz 8/16 Bit 1M RAM

  3. CPU 80486 66 MHz 32 Bit 8M RAM

  4. Pentium 133 MHz 32 Bit 16M RAM

  5. Pentium MMX 200 MHz 32 Bit 48M RAM

  6. Pentium II 366 MHz 32 Bit 128M RAM

  7. Pentium M 1.6 GHz 32 Bit 512M RAM

  8. Core2 Duo 2.0 GHz 32/64 Bit 4G RAM

  9. MHz RAM CPU Bits

  10. What does “Bits” mean?

  11. How many bits Is this PIC CPU?

  12. ระบบ 32 Bit กำลังเจอทางตัน?

  13. RAM ใหญ่ที่สุดที่ระบบ 32 bit มองเห็นมีค่าเป็นเท่าใด?

  14. Memory Limit of 32bit Machines = 4 Giga Bytes

  15. How to support bigger memory? 1.Upgrade to 64-bit systems 64-bit CPU 64-bit OS

  16. Limitation of moving to 64-bit systems Need 64-bit Hardware Drivers Applications need to be re-compiled with a 64-bit compiler

  17. How to support bigger memory? 2. Expand existing bits

  18. วิธีแก้ปัญหาของ Intel Physical Address Extension (PAE) 4 bits 32 bits = 64 GB

  19. PAE Support CPU with PAE (Since P-Pro) PAE Aware OS

  20. Benefits of PAE • Backward Compatible Limitations • Taking advantage of the added memorystill requires changes in the software

  21. Memory Limit of 64 Bit Machines?

  22. How big can we get?

  23. สองเรื่องน่ารู้เกี่ยวกับหน่วยความจำของPIC Microcontroller

  24. 1. When Memory Grows Beyond Anticipation(เมื่อหน่วยความจำขยายโตเกินคาด)

  25. 1.1 การอ้างอิงตำแหน่งใน Flash Memory

  26. K = 11 bit Maximum memory = 2^11 = 2K Bytes

  27. PIC 16F877 has 8K Flash Memory: How do we access all of it? 7 0 PCLATH (F# 0x0A) Bit 3-4 in PCLATH are usedfor memory access Total Memory becomes 2^13 = 8K Bytes

  28. 7 0 Creating a 13 bit memory address 13 11 10 0 OPCODE CALL or GOTO k (Literal) PCLATH 12 10 0 13 Bit Memory Address

  29. PIC Flash Memory is Divided in to 4 Pages of 2KBytes 12 10 0 2K 2K 2K 2K

  30. Fix this program

  31. Address 0x900 exceeds 11 bits 12 10 0 0x900 = 0 1 0 0 1 0 0 0 0 0 0 0 0 0x01 0x100

  32. 12 10 0 0x900 = 0 1 0 0 1 0 0 0 0 0 0 0 0 0x01 0x100

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