ROM Memory

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Transcript ROM Memory

Khaled A. Al-Utaibi
[email protected]
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Introduction
Memory Pin Connections
ROM Memory
RAM Memory
Memory Organization
Types of Memory Access
Data Alignment
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There are two main types of memory:
−(1) Read-Only Memory (ROM)
−(2) Random Access Memory (RAM)
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Pin connections common to all memory devices
are:
−Address Lines
−Data Lines
−Selection Control
−Read/Write Control
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See Figure 1 for ROM and RAM generic-memory
devices.
Figure 1: A pseudo-memory component illustrating the address, data, and control
connections.
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All memory devices have address inputs that
select a memory location within the memory
device.
Address lines are labeled from A0, the least
significant address input, to An-1 where n is the
total number of address pins.
For example, a memory device with 10 address
pins has its address pins labeled from A0 to A9.
The number of address pins found on a memory
device is determined by the number of memory
locations found within it.
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Today, the common memory devices have
between 1K (1024) to 1G (1,073,741,824)
memory locations, with 4G and larger memory
location devices on the horizon.
A 1K memory device has 10 address pins (A0–A9);
therefore, 10 address inputs are required to
select any of its 1024 memory locations.
It takes a 10-bit binary number (1024 different
combinations) to select any single location on a
1024-location device.
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If a memory device has 11 address connections
(A0–A11), it has 2048 (2K) internal memory
locations.
The number of memory locations can thus be
extrapolated from the number of address pins.
For example, a 4K memory device has 12 address
connections, an 8K device has 13, and so forth.
A device that contains 1M locations requires a
20-bit address (A0–A19).
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All memory devices have a set of data outputs or
input/outputs.
The device illustrated in Figure 1 has a common
set of input/output (I/O) lines.
Today, many memory devices have bidirectional
common I/O pins.
The data lines are the points at which data are
entered for storage or extracted for reading.
Data pins on memory devices are labeled D0
through D7 for an 8-bit-wide memory device.
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Each memory device has a control input –
sometimes more than one – that selects or enables
the memory device.
This type of input is most often called a Chip Select
(CS), Chip Enable (CE), or simply Select (S) input.
If the CS, CE, or S input is active (a logic 0, in this
case, because of the over bar), the memory device
performs a read or write operation;
if it is inactive (a logic 1, in this case), the memory
device cannot do a read or a write because it is
turned off or disabled.
If more than one connection is present, all must be
activated to read or write data.
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A ROM has an Output Enable (OE) or a Gate (G)
connection, which allows data to flow out of the
output data pins of the ROM.
If (OE) and the selection input (CE) are both
active, the output is enabled;
if (OE) is inactive, the output is disabled at its
high-impedance state.
The (OE) connection enables and disables a set
of three-state buffers located within the memory
device and must be active to read data.
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A RAM memory device has either one or two
control inputs.
If there is only one control input, it is often called
(R/W).
This pin selects a read operation or a write
operation only if the device is selected by the
selection input (CS).
If the RAM has two control inputs, they are
usually labeled Write Enable (WE), and Output
Enable (OE).
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The (WE) must be active to perform a memory
write.
The (OE) must be active to perform a memory
read.
When these two controls are present, they must
never both be active at the same time.
If both control inputs are inactive (logic 1s), data
are neither written nor read, and the data
connections are at their high-impedance state.
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Permanently stores data for the system.
Its contents do not change even if power is
disconnected.
The most common used ROMs are:
−Erasable Programmable ROM (EPROM)
−Electrically Erasable Programmable ROM (EEPROM)
−Flash
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The EPROM (erasable programmable read-only
memory) is commonly used when software must
be changed often.
An EPROM is programmed on a device called an
EPROM programmer.
Also erasable if exposed to high-intensity
ultraviolet light.
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EEPROM can be programmed and erased
without removing the chip from its socket.
Both byte and bulk erasure modes are possible.
EEPROMs are changed 1 byte at a time, which
makes them versatile but slow.
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Flash Memory overcomes this limitation of
EEPROM
This device uses in-circuit wiring to erase by
applying an electrical field to the entire chip or
to predetermined sections of the chip called
blocks.
It works much faster than EEPROMs because it
writes data in chunks, usually 512 bytes in size,
instead of 1 byte at a time.
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RAM stands for random access memory.
This device retain data as long as DC power is
applied.
Once the power is turned off all data stored in the
RAM will be lost.
The main difference between ROM and RAM is that
RAM is written under normal operation, whereas
ROM is programmed outside the computer and
normally is only read.
There are two main types of RAM:
− (1) Static Random Access Memory (SRAM)
− (2) Dynamic Random Access Memory (DRAM)
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RAM devices retain data as long as
DC power is applied (i.e. no special
action is required to retain data).
Static RAM (SRAM) uses a flip-flop as
the basic storage element.
A typical SRAM memory cell consists
of 6 transistors connected as shown
in the Figure 2 and its equivalent
representation in Figure 3.
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To read the data stored by the SRAM cell:
−The Row-select line is made active.
−The voltage difference between Column and Column
lines is sensed.
−A positive voltage indicates a logic 1 is stored.
−A negative voltage between these same lines indicates
a logic 0 is stored.
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To write data into the SRAM cell:
−The Row-select line is made active.
−To store a logic 1, the Column line is driven high and
the Column line is driven low
−To store a logic 0, the process is repeated, but this time
the Column line is driven low and Column high.
Figure 2: Implementation of a static RAM cell using 6 transistors.
Figure 3: Equivalent implementation of a static RAM cell using 6 transistors.
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A DRAM memory cell consists of a single transistor
and a capacitor as shown in the Figure 4.
Thus, DRAM chips are much denser and can hold
more data than SRAM in the same size package.
However, capacitors constantly leak electricity, which
requires a memory controller to refresh the DRAM
several times a second to maintain the data.
The DRAM cell can retain data for only 2 or 4 ms on
its integrated capacitor.
After 2 or 4 ms, the content of the DRAM must be
completely rewritten (refreshed).
The value stored in the cell is determined by the
charge of the capacitor (Charged = logic 1,
Discharged = logic 0)
Figure 4: Implementation of a dynamic RAM cell using 1 transistors.
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To read the data stored by the DRAM cell:
−Pre-charge the bit line to Vcc/2.
−Set the word line HIGH.
−A sense amplifier is used to determine the logic store
in the cell as follows:
 If current flows into the cell  the cell is at logic
 If current flows out of the cell  the cell is at logic
− Cell contents are destroyed
by the read!
− Hence, the bit value must be
written back after reading.
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To write data into the DRAM cell:
−Set the word line HIGH.
−To write logic 1 set the bit line HIGH
−To write logic 0 set the bit line LOW
− Set the select line LOW.
−Note that the stored charge for a 1 will eventually leak
off.
−Typical devices require each
cell to be refreshed once
every 2 or 4 ms.
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The organization of a memory chip refers to the
way in which its cells are arranged to provide
external data access.
For example, a particular chip may have a total
of 16 MB of storage.
Externally, however, these 16 Mb may be
accessed in several different ways:
−(1) 16M x 1 (Le., 16M bits)
−(2) 4M x 4 (i.e., 4M nibbles)
−(3) 2M x 8 (i.e., 2M bytes)
−(4) 1M x 16 (Le. 16M words)
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SRAMs and ROMs are typically arranged as bytewide (i.e. provide 8-bit external data access).
The organization of a memory chip is important
because it determines how many chips will be
required in a memory interface.
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Example 1: Using 64K x 8 SRAMs, determine the
minimum number of chips required to construct
a memory interface to each of the following
processors. For each interface, calculate the total
memory capacity provided.
−(a) 8088
−(b) 8086
−(c) 80486
−(d) Pentium
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Example 1: Using 64K x 8 SRAMs, determine the
minimum number of chips required to construct
a memory interface to each of the following
processors. For each interface, calculate the total
memory capacity provided.
−(a) 8088
−(b) 8086
−(c) 80486
−(d) Pentium
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Depending on the processor, the quantity of data
transferred per memory cycle can be:
− 1 byte (the 8088)
− 2 bytes (the 8086)
− 4 bytes (the 386 and 486)
− 8 bytes (the Pentium/Pro)
D7-D0
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Depending on the processor, the quantity of data
transferred per memory cycle can be:
− 1 byte (the 8088)
− 2 bytes (the 8086)
− 4 bytes (the 386 and 486)
− 8 bytes (the Pentium/Pro)
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Depending on the processor, the quantity of data
transferred per memory cycle can be:
− 1 byte (the 8088)
− 2 bytes (the 8086)
− 4 bytes (the 386 and 486)
− 8 bytes (the Pentium/Pro)
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To indicate which bits of the data bus will be
involved in the data transfer, the 80x86 processors
provide byte enable output pins:
−The 8086 (BHE)
− The 386 & 486 (BE3-BE0)
−The Pentium & Pentium Pro (BE7-BE0)
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To indicate which bits of the data bus will be
involved in the data transfer, the 80x86 processors
provide byte enable output pins:
−The 8086 (BHE)
− The 386 & 486 (BE3-BE0)
−The Pentium & Pentium Pro (BE7-BE0)
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Using these byte enable signals, the 80x86
processors can indicate that:
−(1) A single byte is to be transferred (only one byte
enable signal active).
−(2) A word is to be transferred (two byte enable signals
active).
−(3) A double-word is to be transferred (four byte enable
signals active).
−(4) A quad-word is to be transferred (all eight byte
enable signals active).
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Note that, when a memory transfer occurs,
consecutive memory locations must be accessed,
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Example 2: Assume a Pentium processor executes
the following instructions. Indicate the logic state
of the BE7-BE0 byte enables for each associated
memory access.
−(a) MOV AL,[0000]
−(b) MOV AX,[0000]
−(c) MOV EAX,[0000]
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Example 2: Assume a Pentium processor executes
the following instructions. Indicate the logic state
of the BE7-BE0 byte enables for each associated
memory access.
−(a) MOV AL,[0000]
−(b) MOV AX,[0000]
−(c) MOV EAX,[0000]
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Data is said to be aligned if all of the bytes to be
accessed are located within the same n-byte
boundary where n is the maximum number of
bytes that can be transferred per memory cycle.
For example the 386 and 486 processors can
access 1 byte, 2 bytes, and 4 bytes using BE3-BE0.
Thus, data is said to be aligned if all of the bytes to
be accessed are located within the same 4-byte
boundary.
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Four such address boundaries are possible for 386
and 486 processors as shown in the next figure.
Data items that span across two of these
boundaries are said to be misaligned and will
require that two bus cycles be performed.
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Example 3: Assume a 486 processor executes the
instruction MOV EAX, [00005]. Which byte enable
signals will be active? How many bus cycles will be
required?
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Example 3: Assume a 486 processor executes the
instruction MOV EAX, [00005]. Which byte enable
signals will be active? How many bus cycles will be
required?
− The double-word (four bytes) at address 0005-0008 is to
be accessed.
−Two bus cycles will be required:
 1st bus cycle will be run with BE1-BE3 active and transfer
the three bytes at address 0005-0007.
 2nd bus cycle will be run with only BE0 active. This will
transfer the byte at address 0008.