The Memory Component

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

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

Most modern computers are byte addressable, facilitating access to character data.

Logically, computer memory should be considered as an array.
The index into this array is called the address or “memory address”.

A logical view of such a byte addressable memory might be written in code as:

Const MemSize =
byte Memory[MemSize] // Indexed 0 … (MemSize – 1)

The CPU has two 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.

Primary Memory

Also called “core memory”, “store”, or “storage”.

Beginning with the MIT Whirlwind and continuing for about 30 years, the
basic technology for primary memory involved “cores” of magnetic material.

Requirements for a Memory Device

1. Random access by address, similar to use of an array. Byte addressable
memory can be considered as an array of bytes.
byte memory [N] // Address ranges from 0 to (N – 1)

2. Binary memory devices require two reliable stable states.

3. The transitions between the two stable states must occur quickly.

4. The transitions between the two stable states must not occur
spontaneously, but only in response to the proper control signals.

5. Each memory device must be physically small, so that a large number
may be placed on a single memory chip.

6. Each memory device must be relatively inexpensive to fabricate.

Varieties of Random Access Memory

There are two types of RAM

1. RAM read/write memory

2. ROM read–only memory.

The double use of the term “RAM” is just accepted. Would you say “RWM”?

Types of ROM

1. “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 same as a PROM, but that the contents can be erased
and reprogrammed by the PROM Programmer.

Memory Registers

MAR Memory Address Register
This specifies the address of the instruction or data item.

For a byte addressable memory, each byte has a distinct address.
For a word addressable memory, only the words have individual addresses.

MBR Memory Buffer Register
This holds the data read from memory or to be written to memory.
Occasionally called MDR for Memory Data Register

In a byte addressable memory, the MBR is usually 8 bits wide; that is, it
holds one byte.

In a 16–bit word addressable memory, the MBR would be 16 bits wide.

The size of the MBR is the size of an addressable item.

Memory Control Signals

Read / Write Memory must do three actions:

READ copy contents of an addressed word into the MBR

WRITE copy contents of the MBR into an addressed word

NOTHING the memory is expected to retain the contents written into
it until those contents have been rewritten.

One set of control signals Select – the memory unit is selected.
if 0 the CPU writes to memory, if 1 the CPU reads from memory.

Select		Action
0	0	Memory contents are not changed.
0	1	Memory contents are not changed.
1	0	CPU writes data to the memory.
1	1	CPU reads data from the memory.

A ROM has only one control signal: Select.

If Select = 1 for a ROM, the CPU reads data from the addressed memory slot.

Memory Timings

Memory Access Time

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

Less used, this is defined as the minimum time between two independent
memory accesses.

The Idea of an Address Space

The memory size is defined in terms of the amount of primary memory
actually installed.

The address space, determined by the size of the MAR, indicates the range
of addresses that actually can be generated.

Absent such kludges such as Expanded Memory and Extended Memory
(both obsolete, dating to about 1980), the memory size does not exceed
the size of the address space.

An N–bit MAR can address 2^N distinct memory locations, 0 … 2^N – 1

Computer	MAR bits	Address Range
PDP-11/20	16	0 to 65 535
Intel 8086	20	0 to 1 048 575
Intel Pentium	32	0 to 4 294 967 295

Memory Mapped Input / Output

Though not a memory issue, we now address the idea of memory mapped
input and output. In this scheme, we take part of the address space that
would otherwise be allocated to memory and allocate it to I/O devices.

The PDP–11 is a good example of a memory mapped device. It was a
byte addressable device, meaning that each byte had a unique address.

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 Reader 177,160 Control & Status Register
177,162 Data buffer 1
177,164 Data buffer 2

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

The Linear View of Memory

Memory may be viewed as a linear array, for example a byte–addressable memory

byte memory [N] ; // Addresses 0 .. (N – 1)

This is a perfectly good logical view, it just does not correspond to reality.

Memory Chip Organization

Consider a 4 Megabit memory chip, in which each bit is directly addressable.
Recall that 4M = 2²² = 2¹¹ · 2¹¹, and that 2¹¹ = 2, 048.

The linear view of memory, on the previous slide, calls for a 22–to–2²² decoder,
also called a 22–to–4,194,304 decoder. This is not feasible.

If we organize the memory as a two–dimensional grid of bits, then the design
calls for two 11–to–2048 decoders. This is still a stretch.

Managing Pin-Outs

Consider now the two–dimensional memory mentioned above. What pins are needed?

Pin Count
Address Lines 22 Address Lines 11
Row/Column 0 Row/Column 2
Power & Ground 2 Power & Ground 2
Data 1 Data 1
Control 3 Control 3
Total 28 Total 19

Separate row and column addresses require two cycles to specify the address.

Four–Megabyte Memory

Do we have a single four–megabyte chip or eight four–megabit memory chips?

One common solution is to have bit–oriented chips. This facilitates the
two–dimensional addressing discussed above.

For applications in which data integrity is especially important, one might add a
ninth chip to hold the parity bit. This reflects the experience that faults, when they
occur, will be localized in one chip.

Parity provides a mechanism to detect, but not correct, single bit errors.

Correction of single bit errors requires twelve memory chips. This scheme will
also detect all two–bit errors.

Memory Interleaving

Suppose a 64MB memory made up of the 4Mb chips discussed above.
We now ignore parity memory, for convenience and also because it is rarely needed.

We organize the memory into 4MB banks, each having eight of the 4MB chips.
The figure in the slide above shows such a bank.

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.

In general, an N–bit address, with N = L + M, is broken into two parts

an L–bit bank select, used to activate one of the 2^L banks of memory, and

an M–bit address that is sent to each of the memory banks.

When one of the memory banks is active, the other (2^L – 1) are inactive.

All banks receive the M–bit address, but the inactive ones do not respond to it.

High–Order Interleaving (Memory Banking)

In high-order interleaving, the most significant bits of the address select the
memory chip. The low–order bits are sent as addresses to each chip.

Bits	25 – 22	21 – 0
Use	Bank Select	Address to the chip

This idea is intuitively appealing, but suffers from performance penalties.

One problem is that consecutive addresses tend to be in the same chip.

The maximum rate of data transfer is set by the memory cycle time,
usually about 50 to 80 nanoseconds.

Low–Order Interleaving

In low–order interleaving, the least significant bits select the memory bank (module).

Bits	25 – 4	3 – 0
Use	Address to the chip	Bank Select

Notice that consecutive memory addresses are in different memory modules.

This allows memory access at much faster rates that allowed by the cycle time.

Faster Memory Chips

We can use the “2 dimensional” array approach, discussed earlier, to create a faster
memory. This is done by adding a SRAM (Static RAM) buffer onto the chip.
Consider the 4Mb (four megabit) chip discussed earlier, now with a 2Kb SRAM buffer.

In a modern scenario for reading the chip, a Row Address is passed to the chip,
followed by a number of column addresses. When the row address is received,
the entire row is copied into the SRAM buffer. Subsequent column reads come
from that buffer.

Memory Technologies: SRAM and DRAM

One major classification of computer memory is into two technologies

SRAM Static Random Access Memory

DRAM Dynamic Random Access Memory (and its variants)

SRAM is called static because it will keep its contents as long as it is powered.

DRAM is called dynamic because it tends to lose its contents, even when powered.
Special “refresh circuitry” must be provided.

Compared to DRAM, SRAM is
faster
more expensive
physically larger (fewer memory bits per square millimeter)

SDRAM is a Synchronous DRAM.
It is DRAM that is designed to work with a Synchronous Bus, one with a clock signal.

The memory bus clock is driven by the CPU system clock, but it is always slower.

SDRAM (Synchronous DRAM)

Synchronous Dynamic Random Access Memory

Suppose a 2 GHz system clock. It can easily generate the following memory
bus clock rates: 1GHz, 500 MHz, 250MHz, 125MHz, etc. Other rates are also possible.

Consider a 2 GHz CPU with 100 MHz SDRAM.
The CPU clock speed is 2 GHz = 2,000 MHz
The memory bus speed is 100 MHz.

In SDRAM, the memory transfers take place on a timing dictated by the memory
bus clock rate. This memory bus clock is always based on the system clock.

In “plain” SDRAM, the transfers all take place on the rising edge of the memory
bus clock. In DDR SDRAM (Double Data Rate Synchronous DRAM), the transfers
take place on both the rising and falling clock edges.

More on SDRAM

“Plain” SDRAM makes a transfer every cycle of the memory bus.
For a 100 MHz memory bus, we would have 100 million transfers per second.

DDR–SDRAM is Double Data Rate SDRAM

DDR–SDRAM makes two transfers for every cycle of the memory bus,
one on the rising edge of the clock cycle
one on the falling edge of the clock cycle.

For a 100 MHz memory bus, DDR–SDRAM would have 200 million transfers per second.

To this, we add wide memory buses. A typical value is a 64–bit width.

A 64–bit wide memory bus transfers 64 bits at a time. That is 8 bytes at a time.

Thus our sample DDR–SDRAM bus would transfer 1,600 million bytes per second.

This might be called 1.6 GB / second, although it more properly is 1.49 GB / second,
as 1 GB = 1, 073, 741, 824 bytes.

Byte Addressing vs. Word Addressing

The addressing capacity of a computer is dictated by the number of bits in the MAR.

Suppose the MAR (Memory Address Register Contains) N bits.
Then 2^N items can be addressed.

In a byte addressable machine, the maximum memory size is 2^N bytes.

If the machine supports longword addressing, but not byte or word addressing,
the maximum memory size is 2^N+2 bytes.

Word and longword addressable machines might have their memory sized
quoted in bytes, but they do not access individual bytes.

Example: 256 KB PDP–11/70 vs. the CDC–6600 with 256 K words (60 bits each).

The CDC–6600 could have been considered to have 1,920 KB of memory.
However, it was not byte addressable; the smallest addressable unit was a 60–bit integer.

Almost every modern computer is byte addressable to allow direct access to the
individual bytes of a character string.

A modern computer that is byte addressable can still issue both word and longword
instructions. These just reference data two bytes at a time and four bytes at a time.

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.

Note that computer architecture refers to addresses, rather than variables.

In a high–level programming language we use the term “variable” to indicate
the contents of a specific memory address.

Consider the statement Y = X

Go to the memory address associated with variable X

Get the contents

Copy the contents into the address associated with variable Y.

Big–Endian vs. Little–Endian Addressing

Address Big-Endian Little-Endian

Z 01 04

Z + 1 02 03

Z + 2 03 02

Z + 3 04 01

Example: “Core Dump” at Address 0x200

Note: Powers of 256 are 256⁰ = 1, 256¹ = 256,
256² = 65536, 256³ = 16,777,216

Suppose one has the following memory map as a result of a core dump.
The memory is byte addressable.

Address	0x200	0x201	0x202	0x203
Contents	02	04	06	08

What is the value of the 32–bit long integer stored at address 0x200?

This is stored in the four bytes at addresses 0x200, 0x201, 0x202, and 0x203.

Big Endian: The number is 0x02040608. Its decimal value is
2·256³ + 4·256² + 6·256¹ + 8·1 = 33,818,120

Little Endian: The number is 0x08060402. Its decimal value is
8·256³ + 6·256² + 4·256¹ + 2·1 = 134,611,970.

NOTE: Read the bytes backwards, not the hexadecimal digits.

Example 2: “Core Dump” at Address 0x200

Note: Powers of 256 are 256⁰ = 1, 256¹ = 256,
256² = 65536, 256³ = 16,777,216

Suppose one has the following memory map as a result of a core dump.
The memory is byte addressable.

Address	0x200	0x201	0x202	0x203
Contents	02	04	06	08

What is the value of the 16–bit integer stored at address 0x200?

This is stored in the two bytes at addresses 0x200 and 0x201.

Big Endian The value is 0x0204.
The decimal value is 2·256 + 4 = 516

Little Endian: The value is 0x0402.
The decimal value s 4·256 + 2 = 1,026

Note: The bytes at addresses 0x202 and 0x203 are not part of this
16–bit integer.