1/2002 JNM 1

With 20 bits, 1,048,576 different combinations are available. Each memory location is assigned a different combination. Each memory location is 1-byte wide. Therefore, the memory space of the 8086 consists of 1,048,576 bytes or 524,288 16-bit words.

The 8086 has a 20-bit Address Bus

1/2002 JNM 2

Problem

• The 8086 has a 20-bit address bus. Therefore, it can access 1,048,576 bytes of memory. How many bits and how many HEX digits are required to access 1M memory?

• 2N = 1M (where N is in bits)

• N = 20 bits = 20/4 = 5 HEX digits

1/2002 JNM 3

Positional Notation (Hex Digits)

1/2002 JNM 4

Problem Continued

• But each register is only 16 bits (4 HEX digits) wide. How are all of the locations accessed?

• Segmented Memory – all of memory is divided into 64-kByte segments. (4 HEX digits (16 bits) can access 64k different locations)– 216 64k

1/2002 JNM 5

Within the 1 MB of memory, the 8086 defines 4 64KB memory blocks.

Segmented Memory

7FFFF

The segment registers point to location 0 of each segment. (The base address)

DS: E000 CS: B300

SS: 7000 ES: 5D27

1/2002 JNM 6

Segment Register Defaults

1/2002 JNM 7

Segments/ Segmentation

• Segments are variable-sized areas of memory used by a program containing either code or data.

• Segmentation provides a way to isolate memory segments from each other. This permits multiple programs to run simultaneously without interfering with each other.

• A segment selector is a 16-bit value stored in a segment register.

• A logical address is a combination of a segment selector and an offset(16-bit for 8086).

1/2002 JNM 8

Logical Addresses

• Later in the course, we will get into translation of logical addresses to linear addresses (using descriptor tables and paging).

• For now, using a virtual-8086 mode, we can determine the “physical address” by adding the offset to the base. Remember with 8086 mode, there is only 1MB of memory.

1/2002 JNM 9

Determining the Physical Address from a Logical Address

• Shift the segment value left 4 bits (add a zero to the right), then add the offset

• 3D7F:023C 3DA2Ch

1/2002 JNM 10

Real-Mode Addessing

• In real mode addressing, the lowest 640K of memory is used by both the operating system and application programs.

• Video memory and hardware controllers have reserved memory locations.

• C0000h-FFFFFh are reserved for system ROM.• IVT (Interrupt Vector Table) is at first 1024 bytes

of memory (00000-03FFFh)

1/2002 JNM 11

Memory Map for an 80x86 computer running MS-DOS

1/2002 JNM 12

Memory Banks

• Even though the 8086 has a 16-bit data bus, it is byte addressable.

• Memory is divided into two 8-bit banks

BE0 BE1

Even Addresses Odd Addresses

1/2002 JNM 13

Memory Chip Organization

• Remember that memory chips are usually defined as a k x n device (k=#locations, n=number of cells per location)– Examples

• 16 x 1 (16M bits)

• 4M x 4 (4M nibbles)

• 2M x 8 ( 2M bytes)

• 1M x 16 (16 M words)

• SRAM and ROM are typically arranged x8 (byte wide)