Computer MemoryBasic Concepts

Lecture for CPSC 5155Edward Bosworth, Ph.D.

Computer Science DepartmentColumbus State University

The Memory Component

• The memory stores the instructions and data for an executing program.

• At this level, we are considering memory as a unit without structure. Cache memory will be covered fully in a future lecture.

Memory Addressability

• Memory is characterized by the smallest addressable unit:Byte addressablethe smallest unit is an 8–bit byte.Word addressablethe smallest unit is a word, usually 16 or 32 bits.

• Almost all modern computers use byte addressability to simplifies processing of character data.

Memory Control

• The CPU has 2 registers dedicated to handling memory.

• The MAR (Memory Address Register) holds the address being accessed.

• The MBR (Memory Buffer Register) holds the data being written to the memory or being read from the memory. This is sometimes called the Memory Data Register.

Memory Control Signals

• Read / Write Memory must do three actions:1. READcopy contents of an addressed word

into the MBR2. WRITE copy contents of the MBR into the

location being addresses.3. NOTHING the memory is expected to

retain the contents written into it until those contents have been rewritten.

One Option for Memory Control

• The CPU can issue one of two control signals.These are usually asserted high.

• READ WRITE

Action

0 0 Nothing. The memory is not active.0 1 The CPU writes to memory1 0 The CPU reads from memory

1 1 This is a problem, which must be solved by a design decision.

Asserting Control Signals

• A control signal is a binary signal with only two values: 0 (logic low) and 1 (logic high).

• When a control signal is to activate some circuit element, it is said to be asserted.

• An active high signal is asserted when its value is 1. It is denoted by the signal name.

• An active low signal is asserted when its value is 0. It is denoted by appending a “#” to its name, as in Select#.

Alternate Notation for Active Low

• A signal asserted low might be denoted as

• The first notation is older, and harder to depict using word processors.

• A two-value control signal enables one of two actions depending on its value. A signal to indicate either read or write might be denoted

Modern Memory Control

• Most modern designs avoid the READ and WRITE control signals, using Select# and R/W#

• If R/W# = 0, a memory write is called for.If R/W# = 1, a memory read is called for.

Select# R/W# Action1 0 Memory contents are not changed.

1 1 Memory contents are not changed.

0 0 CPU writes data to the memory.

0 1 CPU reads data from the memory.

(Classical) Memory Timings

Memory Access Time is defined in terms of reading from memory. It is the time between the address becoming stable in the MAR and the data becoming available in the MBR.

Memory cycle time is the minimum time interval between two independent memory accesses.

Synchronous Memory Timings• Synchronous memory, typically SDRAM, is rated

by the speed of the memory bus.• The speed is normally quoted in megahertz, as

in 166 MHz or 250 MHz, or some other value.• A 250 MHz memory can be attached to a 250

MHz synchronous memory bus, and transfer data at the rate of 250 million transfers/sec.

• This is one transfer every 4 nanoseconds.• The data transfer rate depends on the width of

the data bus, often as high as 64 bits.

RAM and ROM

• Technically, the term “RAM” stands for “random access memory” with no further connotations.

• In common usage, the term RAM refers to memory that is “read/write”; the CPU can both read from and write to RAM.

• The term “ROM” stands for “read only memory”. In standard operations, the CPU cannot change the values stored in ROM.

Varieties of ROM1. “Plain ROM” The contents of the memory

are set at manufacture and cannot be changed without destroying the chip.

2. PROM The contents of the chip are set by a special device called a “PROM Programmer”. Once programmed the contents are fixed.

3. EPROM is same as a PROM, but that the contents can be erased and reprogrammed by the PROM Programmer device.

Memory Mapped Input / Output• The old PDP–11/20 supported a 16–bit address space. This supported

addresses in the range 0 through 65,535 or 0 through 0177777 in octal.• Addresses 0 though 61,439 were reserved for physical memory.

In octal these addresses are given by 0 through 167,777.• Addresses 61,440 through 65,535 (octal 170,000 through 177,777)

were reserved for registers associated with Input / Output devices.

• Examples: CR11 Card Reader177,160 Control & Status Register177,162 Data buffer 1177,164 Data buffer 2

• Reading from address 0177162 would access the card reader data buffer.

Memory Mapped I/O in the MIPS

The Linear View of Memory

• This logical view is not easily implemented.

Standard Memory Organization• Modern computer memory is organized as a

collection of modules connected to a bus.• Each module comprises 8 or 9 data chips.

Module Organization

• A standard module will have 8 data chips, one for each bit in the Memory Buffer.

• Assigning one bit per chip is more efficient.

Memory Chip Organization

• As an example, here is a 4 megabit chip. It is organized as a 2-D array.

Two Options for Addressing

• The option at left has 28 pins, the option at right has 19 pins. The signals RAS# and CAS# indicate the meaning of the 11-bit address.

Memory Interleaving

• Suppose a 64MB memory made up of the 4Mb chips discussed above.

• We organize the memory into 4MB banks.• The memory thus has 16 banks, each of 4MB.• 16 = 24 4 bits to select the bank

4M = 222 22 bits address to each chip• Not surprisingly, 64M = 226.

Two Interleaving Options

• The type of interleaving dictates how the memory address is divided.

• Low-order interleaving

• High-order interleaving (banking)

Bits 25 – 4 3 – 0

Use Address to the chip Bank Select

Bits 25 – 22 21 – 0

Use Bank Select Address to the chip

Speed Up Access by Interleaving Memory

• Suppose an 8–way low–order interleaved memory. The chip timings are:80 nanosecond cycle time, and40 nanosecond access time.

• Each chip has the following timing diagram.

Timings for the Interleaved Memory

Faster Memory Chips

• After the row address is asserted, a number of column reads may proceed.

SDRAM

• SDRAM is synchronous dynamic RAM.• In SDRAM, the memory transfers take place on a

timing dictated by the memory bus clock rate. In “plain” SDRAM, the transfers all take place on the rising edge of the bus clock.

• In DDR SDRAM (Double Data Rate), the transfers take place on both the rising and falling clock edges.

A Synchronous Bus Timing Diagram

SDRAM with a Wide Bus

• DDR–SDRAM makes two transfers for every cycle of the memory bus, 1 on the rising edge of the clock cycle

1 on the falling edge of the clock cycle.• For a 100 MHz memory bus, DDR–SDRAM would have

200 million transfers per second.• Now consider a 64–bit data bus, which can transfer

64 bits (8 bytes) at a time.• Thus our sample DDR–SDRAM bus would transfer

1,600 million bytes per second.• This is 1.49 GB / second, as 1 GB = 232 bytes.

Word Addressing in a Byte Addressable Machine

• Each 8–bit byte has a distinct address.• A 16-bit word at address Z contains bytes at

addresses Z and Z + 1.• A 32-bit word at address Z contains bytes at

addresses Z, Z + 1, Z + 2, and Z + 3.• Question: How is the value stored in a 32-bit

register stored in computer memory?

Big–Endian vs. Little–Endian

• Address Big-Endian Little-Endian• Z 01 04• Z + 1 02 03• Z + 2 03 02• Z + 3 04 01

Big–Endian vs. Little–Endian

Example: “Core Dump” at Address 0x200

• Here is a sample memory map.

• What is the 32-bit integer stored at address 0x200?• Big Endian: The number is 0x02040608.

Its decimal value is22563 + 42562 + 62561 + 81 = 33,818,120

• Little Endian: The number is 0x08060402. Its decimal value is82563 + 62562 + 42561 + 21 = 134,611,970.

Address 0x200 0x201 0x202 0x203Contents 02 04 06 08

Another “Core Dump”

• Here is the same memory map.

• The 16–bit integer stored at address 0x200 is stored in the two bytes at addresses 0x200 and 0x201.

• Big Endian: The value is 0x0204.The decimal value is 2256 + 4 = 516

• Little Endian: The value is 0x0402.The decimal value s 4256 + 2 = 1,026

Address 0x200 0x201 0x202 0x203Contents 02 04 06 08

Evolution of Modern MemoryYear Cost per MB

in US $Actual component Speed

nsec.Type

  Size (KB) Cost  

1957 411,041,792.00 0.0098 392.00 10,000 transistors1959 67,947,725.00 0.0098 64.80 10,000 vacuum tubes1965 2,642,412.00 0.0098 2.52 2,000 core

1970 734,003.00 0.0098 0.70 770 core

1975 49,920.00 4 159.00 ?? static RAM1981 4,479.00 64 279.95 ?? dynamic RAM1985 300.00 2,048 599.00 ?? DRAM1990 46.00 1,024 45.50 80 SIMM1996 5.25 8,192 42.00 70 72 pin SIMM2001 15¢ 128 MB 18.89 133 MHz DIMM2006 7.3¢ 2,048 MB 148.99 667 MHz DIMM DDR22008 1.0¢ 4,096 MB 39.99 800 MHz DIMM DDR22010 1.22¢ 8,192 MB 99.99 1333 MHz DIMM DDR2