Transcript M262 Putting Computers Systems to Work

T325: Technologies for digital media
Second semester – 2011/2012
Tutorial 2 – Information Storage
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1
• Rotating media
• Optical media
• Solid-state memory
Outline
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ROTATING MEDIA
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Why are we still using CDs and DVDs despite
the huge competition of semiconductor
memories?
Question
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Semiconductor
memory
Rotating media
Simple construction
No moving parts
Capacity &
Reduced Sizes
(HDD)
Fast access times
Cheap and
effective
(optical media)
Semi-conductor vs. Rotating media
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• Both spellings are common
• Disk for magnetic media (Hard Disk)
• Disc for optical media (Blue-ray disc)
Disc or Disk ?
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• Magnetic media have a ferromagnetic surface on which
areas can be magnetised
• Some metals, such as iron and its alloys, may be
magnetised (permanent magnetism)
• Information may be encoded as a pattern of magnetism
• This pattern can subsequently be detected and the
information retrieved
• Non-volatile medium
• Patterns persist until the material is demagnetised or remagnetised
• Does not depend on a continual supply of power
Magnetic disks
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Magnetic disks
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• How read/write operations are performed in a
magnetic disk drive?
• a write head converts electrical signals to magnetised areas on the
disk surface, or changes an already magnetised area.
• a read head produces an electrical signal in response to a magnetic
field.
• The two heads are constructed as a single assembly and
mounted on an arm which can move radially across the
spinning disk, so that any point on the recording surface may
be reached.
• Data is recorded in concentric tracks, with each track being
written or read with the heads at a certain nominal radius
Magnetic disks
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• Reading and writing operation in
Magnetic disks rely on the relation
between electricity and magnetism
 What is this relation?
Magnetic disks
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• Any electrical conductor carrying a current is associated
with it a magnetic field
• A compass needle was deflected by a nearby wire carrying an
electric current  connection between magnetism and
electricity (Hans Christian Oersted (1777-1851))
• Two current carrying wires attracted or repelled each other
(depending on the direction of the currents) due to their
magnetic fields (André-Marie Ampère (1775-1836))
• Electric currents could be produced by a magnetic field as long
as the field was varying with time (Michael Faraday (1791-1867))
Relationship between electricity and magnetism
Experiments
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• How can we apply these experiments results on the
HDD R/W operations?
• A write head is an electromagnet constructed so as to
concentrate the magnetic field over a very small area of
the surface of the disk.
• If there is sufficient current through the write head, the
magnetic field is intense enough to magnetise this area
permanently (or until the next write).
• By varying the current as the disk rotates under the head,
a pattern of magnetism is built up along the track,
corresponding to a stream of bits.
Rotating media
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Does the converse of the operation work
to read out the stored information?
 Faraday’s law
Question
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• Answer: Yes, but there is a subtle and important
difference!
• Faraday’s law of magnetic induction: Changing
magnetic field DOES generate a current
• As the disk rotates under the head, the magnetic field it
encounters does indeed change frequently, with
consecutive areas being magnetised in either of two
directions.
• A head of this type does produce a voltage signal
• This signal represents the transitions between magnetic
states rather than the states themselves.
Rotating media
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• The presented R/W scheme is how almost all magnetic
recording was done up until the 1990s
 BUT!
• Growth of demand for larger data storage capacities
• Greater areal density
• Smaller diameter disks
 Magnetic bits become smaller (radially and along the
track)
• This constitutes a limitation for the original R/W scheme
 Smaller magnetic bits means smaller voltage signal
(difficult to be detected against the background noise)
Solution  Giant
MagnetoResistance
Limitation of the original R/W scheme
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• The effect of magneto-resistance is a change in the
electrical resistance of a conductor when it is placed in a
magnetic field
• Relates the voltage across a conductor to the current
flowing through it in accordance with Ohm’s law:
V=IxR
• R can be measured by applying a known voltage to the
conductor and measuring the current through it
Magneto-resistance
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• How do you think magnetoresistance can be
applied in the HDD read head?
• Changes in resistance are, in principle, rather easier to
detect than the tiny amounts of energy that the magnetic
bits of a high density disk can generate by induction.
• The power used for the measurement now comes from an
external voltage/Current source.
• The voltage and current can be made as large as
necessary for accurate results
• Limits: excessive I or V could overheat or damage the head
• Note: magnetoresistance is also known as Anisotropic
Magneto-Resistance (AMR)
Magnetoresistance for a read head mechanism
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• Although magnetoresistance was observed by William
Thomson (Lord Kelvin) 150 years ago, the change in
resistance of below 1% was too small to be practically useful.
• The real breakthrough came with the discovery of Giant
magnetoresistance (GMR), when huge resistance changes
(e.g. 50%) were reported.
• Discovered by Albert Fert in France and Peter Grϋnberg in
Germany, working independently.
• They jointly received the Nobel Prize in Physics in October 2007
for their discovery.
• GMR read heads were one of the first products of
nanotechnology!
Giant MagnetoResistance (GMR)
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• What is nanotechnology?
• The term nanotechnology covers a wide variety of ideas
and techniques, but its defining characteristic is that it
deals with objects with dimensions in the order of a
nanometer (10-9 m, or a millionth of a millimeter)
• Many of the techniques used in nanotechnology were
developed for the production of microelectronics
• Thousands and millions of transistors are built on silicon
chips
Nanotechnology
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• Purpose: to increase areal densities
• In conventional recording, or longitudinal magnetic
recording, the magnetic bits may be thought of as tiny bar
magnets which point along the track.
• In perpendicular recording, the bar magnets are aligned
at right angles to the disk and point in and out of the disk
surface.
Perpendicular Magnetic Recording (PMR)
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Perpendicular Magnetic Recording (PMR)
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• In Perpendicular recording, much stronger magnetic field
is required for writing than in case of longitudinal
recording
• Materials used for the recording surface should have a
much higher coercivity
• Coercivity is a measure of how difficult a material is to
magnetize or demagnetize
• High coercivity material  difficult to magnetize but
also difficult to demagnetize
Coercivity
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OPTICAL MEDIA
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• Data is written on a spiral track as a
series of ‘pits’ of various lengths
• Track pitch: Distance between
adjacent tracks
Pits and Lands 
Optical Media - Introduction
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http://www.sonic.net/~ideas/graphics/mma_cd.gif
• Four layers
A. A polycarbonate disc layer has
the data encoded by using
bumps.
B. A shiny layer reflects the laser.
C. A layer of lacquer protects the
shiny layer.
D. Artwork is screen printed on
the top of the disc
E. A laser beam reads the CD and
is reflected back to a sensor,
which converts it into
electronic data
Optical Media - Introduction
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• A laser beam is focused onto the track and reflected back
• The reflected beam is analyzed to detect the pits.
• Feedback mechanisms (servomechanisms, or ‘servos’ for short)
involving optics, mechanics and electronics ensure that the beam is
kept focused on the track and does not drift away from the track.
• They compensate effectively for minor warping or eccentricity and for
some vibration.
• The data surface is at the opposite side of a transparent
polycarbonate substrate from the laser reader.
• The laser light penetrates the transparent surface before it reaches a
point of focus, so it is spread over a relatively wide area and,
therefore, is less affected by minor scratches and dust particles.
Optical Media – Reading Data
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• For read-only CDs, the pits are actual indentations in the
data surface
• For rewritable media they are simply areas of different
reflectivity from the ‘land’ that surrounds them.
• These discs use a material that can exist in one of two
physical states or phases: crystalline (where the
molecules are arranged in regular patterns) and
amorphous (where there is a lack of order).
• The phase depends on how the material is heated and
cooled, and can be changed from one to the other and
back again by heat from the laser.
• This is called phase-change recording.
Read Only vs. Rewritable CDs
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• Blu-ray disc (BD) use blue lasers with a wavelength of
405 nm (shorter than CD and DVD)
• Shorter wavelength means:
• laser beam can be focused to a smaller spot on the data
surface.
• pits and track pitch (the spacing between tracks) could be
reduced in size, increasing the data capacity achievable.
• Data in Blu-ray is recorded in a similar way to previous
rewritable optical discs, although the details differ.
• The recording surface uses a phase-change material, and
the pits are areas of contrasting reflectivity.
Blu-ray
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Optical media
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Laser wavelength (nm)
Track pitch (mm)
900
800
700
600
500
400
300
200
100
0
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
Audio CD
DVD (Single
Layer)
HD DVD (Single Blu-ray disc
Layer)
(Single Layer)
Audio CD
DVD (Single
Layer)
HD DVD (Single Blu-ray disc
Layer)
(Single Layer)
Capacity (Gbyte)
30
25
20
15
10
5
0
Audio CD
DVD (Single HD DVD (Single Blu-ray disc
Layer)
Layer)
(Single Layer)
Optical media - Comparison
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• One important design decision
was the thickness of the
transparent layer covering the data
surface.
• 1.2 mm thickness in CDs
• 0.6 mm thickness in DVDs
• There are optical advantages to be
gained by having a much thinner
transparent cover layer.
• If the disc tilts at all, then a thick
layer could deflect and defocus the
beam.
Thickness of the transparent layer
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• In the Original CD, the pits were laid in a spiral pattern
on a flat surface
• Blu-ray discs have a physical spiral track on which the
pits are recorded.
• This track is a spiral groove
• The preformed groove is there to guide the beam so that it
deposits pits in the correct spiral pattern
Blu-ray Disc
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Optical media
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Blu-ray disc
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• Any optical disc player uses a servomechanism to stay on
track.
• This is a form of closed-loop control where the position
of the beam on the track is constantly monitored.
• This signal is fed back to the motors that position the
beam in such a way as to correct the departure.
• The alternative, open-loop control, does not use
monitoring and feedback, but simply relies on very
precise mechanics and favorable operating conditions to
keep errors to a minimum.
Servomechanism in Optical Media
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• As the disc rotates, the servomechanism ensures that the
laser follows the groove
• Servomechanism will be moving the laser from side to side
to follow the wobble.
• The rate at which it moves from side to side will be
determined by the length of the wobbles (the wobble
period – the distance between peaks) and the speed of
rotation.
• The length of the wobbles is predefined, so the equipment
can determine the speed of rotation from the rate at which
the laser is moving from side to side.
• A feedback loop is used to control the speed of rotation,
based on measuring the rate at which the servomechanism
is having to move the laser from side to side.
Optical media
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• Blu-ray data is held as a pattern of pits along a track.
• The laser beam reading the data is kept on the track and
in focus by a servomechanism.
• The track also produces a wobble signal.
• The wobble information is used in locating blocks of data.
• Small dust particles may corrupt only a few bytes, but
marks or scratches might extend for some distance along
the track and cause errors in a long sequence of data.
• When data is recorded it is encoded with an errorcorrecting code (Reed-Solomon (or RS) code).
• This allows a large proportion of the errors to be corrected.
Blue-ray Disc : Conclusion of Important
features
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SOLID STATE MEMORY
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• Flash memory development : storage capacity and cost
• Variety of usages: memory keys, memory cards, built-in
telephone memory
Introduction
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Flash memory is an example of memory that is
described as ‘solid-state’ What do you hear by Solid
in the term : “Solid State Memory” ?
• In science ‘the solid state’ means solid as distinct from
liquid, gas or plasma.
• Transistor (Solid-state electronics ) vs. thermionic valve
(electrons in a vacuum tube)
• “Solid state” vs. “Rotating Media”
• “solid” seems in contrast with the moving mechanical parts
of the hard drive.
Solid-state memory
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• Our main interest is in memory which stores digital media
files: audio, and still and moving images.
• Main memory is comprised of integrated circuits
• Two sorts of Main memory: ROM and RAM.
• ROM stands for Read-Only Memory
• ROM (and secondary storage media) are non-volatile
• ROM is used for the programs needed on start-up.
• RAM stands for random-access memory.
• Random Access: a data word in any location in the whole
memory can be accessed just as quickly as a data word in any
other location.
• RAM is volatile: all the programs and associated data stored in
it are lost when the power supply to the RAM is switched off.
Memory Basics
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• PROM is ‘programmable’ ROM.
• Non-programmable ROM has the data put into it when it is
manufactured.
• PROM, by contrast, is manufactured with no data stored in it and
can be programmed by the user (the computer manufacturer for
example)
• Special equipment is needed to program PROM, and it cannot be
changed once it is programmed (it cannot be erased).
• Why using the term programming rather than writing ?
• because it is all done at once ‘up front’, rather than writing
individual bits or bytes when needed, although for some devices
the distinction is not so clear-cut.
Programmable ROM (PROM)
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• EPROM is ‘Erasable’ PROM
• Erasable PROM can be erased and reprogrammed
• EPROM is used specifically for a device that can be
erased by exposure to ultra-violet light
• EEPROM is Electrically Erasable PROM
• EEPROM is erased with an electrical signal rather than
needing UV light.
Erasable PROM (EPROM)
Electrically EPROM (EEPROM)
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EPROM
 Erasing EPROM has to
be done to the whole of
the device (UV light is
shone onto the whole of
the ‘chip’)
EEPROM
 EEPROM can be erased
one bit or one byte at a
time.
 EEPROM is more
complex  costs more
and it is not possible to
fit as much memory on a
single integrated circuit.
EPROM vs. EEPROM
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What is the meaning of the term
flash in Flash Memory ?
Question
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• Flash memory is a variation of EEPROM
• The word ‘flash’ is used in reference to the way it is
erased, which is done either on the whole device at once
or on a block of data in the device and can be done
quickly, ‘in a flash’.
• “Offering the electrical erase capability, traditionally
featured by the expensive EEPROM, at cost and density
comparable to EPROM, Flash memories not only have
taken a big portion of their progenitor’s markets, but
in addition they have greatly expanded the fields of
application of non-volatile memories.” (Source: Bez and
Cappelletti, 2005, p. 84)
Flash Memory
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• Memory cell: Part of the memory that can store a single bit
• Cell must be able to exist in each of two states:
• one state that represents a data 0
• one state that represents a data 1
• Operations: Reading, Erasing, and Writing
• Reading a cell: detecting which of the two states it is in.
• Erasing a block of memory: setting all the cells in a block to the
same known state, that representing a data 1
• Writing to the cells (programming the memory): setting some of
the cells to the other state, that which represents a data 0.
• Since the remaining cells are already in the state representing a data 1,
there is no need to do anything to them
Flash Memory Cell
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Solid-state memory
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• Flash cells use semiconductor technology
• Constructed using the same or similar semiconductor
materials and manufacturing processes to those used in
microelectronics such as microprocessors.
• The basic building block of microelectronics is the
transistor
• In flash memory, a memory cell uses a single transistor.
In some other types of semiconductor memory, several
transistors are needed for a single memory cell
• This is one of the reasons that Flash memory allows high
storage densities
Semiconductor Technology
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• A transistor can be thought of as a controlled switch
• A signal on the gate connects the source to the drain
Refer to block 1 – pages 79 - 83 for more reading about transistors
Solid-state memory
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• To read data from a cell, a mechanism is needed to detect
whether or not there is charge (electrons) on the floating
gate
• To read data from a cell, you apply a voltage to the
control gate and see whether it succeeds in allowing
current to flow between the source and the drain.
• If it does allow current to flow -- if current is detected at the
drain  then the data on the cell is a 1;
• if it does not allow current to flow -- if no current is
detected at the drain  then the data on the cell is a 0.
Solid-state memory
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• Different variations of flash cells is customized for different
applications
• Details of the cell design,
• Configuration of the cells on the chip,
• Electronic circuits that accompany the memory cells
• In this course, two categories of flash memory is examined
• NAND flash memory
• NOR flash memory
• The simplest flash cells only distinguish between two
options: charge or no charge (at the floating gate)
• Multi-level flash is more sophisticated, distinguishing
between more than two levels.
Varieties of flash
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• Multi-level flash: allows more than one bit to be stored in
each cell, and hence a higher storage density.
• For example, suppose in a binary cell a data 1 corresponds to no
charge on the gate and a data 0 corresponds to an amount of charge
called C.
• If the system is able to add charge more precisely and also measure
how much is on the gate accurately, it might be possible to work
with four levels of charge: 0, C/3, 2C/3, C. These four levels could
be used to represent two bits of binary data, defined as, for example:
•
•
•
•
0 charge = 11
C/3 = 01
2C/3 = 00
C = 10
• Memory using this system would be described as “2 bits per cell”
memory.
Multi-Level Flash Memory
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• NAND flash is optimized for bulk storage, similar to the function of
rotating media
• NOR flash is optimized for storage of computer code.
• The NAND device has a much larger capacity than the NOR device
(16 Gbits compared with 1 Gbit).
• The random access time for the NOR flash is much shorter than for
the NAND flash (96 ns compared with 60 ms).
• NOR can program individual words, whereas NAND needs to
program a page at a time.
• NAND and NOR programming times are similar, being of the order
of tens of microseconds.
• It takes much longer to erase a block in the NOR flash than in the
NAND flash (1 s compared with 2.5 ms).
• The endurance of the NOR flash product is around 10 times that of
the NAND flash (100,000 cycles compared to 10,000 cycles).
NAND vs. NOR Flash Memory
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Solid-state memory
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• How does the random access time of this NAND flash
compare with that of a HDD that has a latency time of
4.17 ms?
Sol:
• The average random access time (latency) for the HDD
was 4.17 ms, which is a lot longer than the 60 micro
second for the flash.
• This is a ratio of
Solid-state memory
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• This figure is a simplified display of the
memory array configuration (it ignores the
use of two planes).
• Use information from the data sheet in
NAND memory Data sheet to find:
•
•
n, the number of pages per block
m, the total number of blocks.
• From these numbers, calculate (giving
separate answers for main memory and
spare memory in each case):
• the total number of bits in a page
• the total number of bits in a block
• the total number of bits in the whole
memory. (This should, of course, agree
with the total memory size of 16 Gbits of
main memory and 512 Mbits of spare
memory.)
Solid-State memory : Activity 2.9
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• Find : n, the number of pages per
block; m, the total number of blocks.
• From these numbers, calculate
(giving separate answers for main
memory and spare memory in each
case):
• the total number of bits in a
page
• the total number of bits in a
block
• the total number of bits in the
whole memory.
• The layout of memory cells in NAND flash allows
memory cells to be packed more closely together than in
NOR flash.
• According to a paper by Bez and Cappelletti (2005), if
the technology node is F, then the space required by a
NOR cell is 10F2 and the space required by a NAND cell
is 4.5F2.
• If the technology node F is 65 nm, use these formulas to
calculate:
• The cell size of flash NOR and NAND memory for binary
cells (1 bit per cell).
• The number of bits that can be stored on a chip that is
5mmx5mm.
Solid-state memory
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a) F = 65 nm = 65 x10-9 m.
• NOR cell size is 10F2, which is 10 x(65 x10-9)2m2 = 4.2 x10-14 m2.
• NAND cell size is 4.5F2, which is 4.5 x(65 ·10-9)2m2 =1.9 x10-14 m2.
b) 5 mm square is (5 x10-3)2m2 = 2.5 x10-5m2.
• The number of NOR cells that will fit in that area is
• (2.5 x10-5)/(4.2 x10-14) = 5.9 x108. That is 560 Mbits (with M in the
sense of 220 = 1048 576).
• The number of NAND cells that will fit in that area is
(2.5 x10-5)/(1.9 x10-14) = 1.3 x109. That is 1.2 Gbits (with G in the
sense of 230 = 1073 741824).
Solid-state memory
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TEST YOUR KNOWLEDGE QUESTIONS FROM PREVIOUS EXAMS
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• In _________ the magnetic bits may be thought of as tiny
bar magnets which point along the track
a. Perpendicular magnetic recording
b. Longitudinal magnetic recording
c. Classic magnetic recording
d. Static magnetic recording
e. None of the above
(Fall 2011 – MTA)
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• The order of optical media in terms of their capacity from smallest to
largest:
a. Single Layer Blu-ray disc < Audio CD < Single Layer DVD <
Single Layer HD DVD.
b. Single Layer DVD < Single Layer HD DVD < Audio CD < Single
Layer Bluray disc.
c. Single Layer DVD < Audio CD < Single Layer HD DVD < Single
Layer Bluray disc.
d. Audio CD < Single Layer DVD < Single Layer HD DVD < Single
Layer Bluray disc.
e. Single Layer DVD < Single Layer Blu-ray disc < Audio CD <
Single Layer HD DVD.
(Fall 2011 – MTA)
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• Flash memory is an example of memory that is described as
___________.
a. Solid-state
b. Vacuum tube
c. Random-access
d. Volatile
e. None of the above
• The change in the electrical resistance of a conductor when it
is placed in a magnetic field is called:
a. Electro-resistance
b. Magneto-resistance
c. Electromagnetic resistance
d. Magnetic change
e. Coercivity
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• One of the following is a feature of Blu-ray
a. Data could not be held as a pattern of pits along a track
b. The track produces a wobble signal which is used in
locating blocks of data
c. Recorded data is not encoded with an error-correcting code
d. It uses solid state structure
e. It is volatile
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• In ___________, the bar magnets are aligned at right angles to
the disk and point in and out of the disk surface
a. Conventional magnetic recording
b. Longitudinal magnetic recording
c. Perpendicular magnetic recording
d. Classical magnetic recording
e. None of the above
• Which of the following is True for a non-volatile medium
a. Patterns persist unless they are removed or changed
b. Does not depend on a continual supply of power
c. A typical example of this category is the Read Only Memory
d. All of the above are correct
e. None of the above is correct
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• Briefly explain each of the following items: PROM, EPROM,
EEPROM, Multi-Level flash memory, NAND flash memory
(Fall 2011 – MTA)
• What are the physical features of Blu-ray and HD-DVD
that gives them a higher capacity than CD and DVD?
(Fall 2011 – Final exam)
• Explain phase-change recording in Rewritable CDs
• Summarize the main differences (three at least) between
the NAND memory and the NOR memory.
• Explain in one or two sentences the principle of operation
of read and write head in a magnetic disk
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• What are the advantages and disadvantages of EPROM
and EEPROM when compared to each other?
• Rotating media (both optical and magnetic) are in huge
competition with semiconductor memories; none of these
technologies could dominate and eliminate the other
technologies. Explain one reason/advantage that keep
each technology in use despite this competition.
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• the following information is found in a memory chip data
sheet indicating the memory organization : 32768 bits x
128 pages x 4096 blocks x 2 planes. taking into
consideration that the organization is presented from the
smallest unit size (bit) to the largest unit size (plane).
Calculate the following
a. the size of a memory page in KiloBytes (4 marks)
b. the size of a memory block in Megabytes (3 marks)
c. the size of the memory in Megabytes (3 marks)
(Fall 2011 – MTA)
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• Consider the following information concerning a multilevel NAND flash memory
• The following levels of charge can be distinguished: 0,
C/15, 2C/15... 14C/15, C where C represents the full charge
of the transistor.
• The NAND memory cell size is 2 x10^-14 m^2
• The memory chip size is 25mm2
• Answer the following questions:
a. How many bits can be represented in each cell?
b. How many memory cells the memory chip contains?
c. Calculate the storage capacity of the memory chip in
Gigabytes
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