3.Magnetic Materials..

Download Report

Transcript 3.Magnetic Materials..

MAGNETIC MATERIALS
SD-JIITN-PH611-MAT-SCI-2013
All magnetic phenomena
result from forces between
electric charges in motion.
In an atom, magnetic effect may arise due to:
1. Effective current loop of electrons in atomic orbit (orbital
Motion of electrons);
2. Electron spin;
3. Motion of the nuclei.
SD-JIITN-PH611-MAT-SCI-2013
FUNDAMENTAL RELATIONS
1. RELATION BETWEEN B, H and M
A magnetic field can be expressed in terms of Magnetic field
intensity (H) and Magnetic flux density. In free space, these
quantities are related as
B  0 H
(1.1)
In a magnetic material, above relation is written as
B  H
(1.2)
Here 0 = absolute permeability of free space,
 = absolute permeability of the medium and
/ 0 = r = relative permeability of the magnetic
material.
SD-JIITN-PH611-MAT-SCI-2013
MAGNETIZATION (M)
Magnetization is defined as magnetic moment per unit
volume and expressed in ampere/ meter. It is proportional to
the applied magnetic field intensity (H).
M  H
(1.3)
Here,  = r – 1 = Magnetic susceptibility (cm-3).
Let us consider
B  H
 B  0  r H  0  r H  0 H  0 H
 B  (r  1)0 H  0 H
 B  0 H  0 H
 B  0 M  0 H
 B  0 ( M  H )
(1.4)
SD-JIITN-PH611-MAT-SCI-2013
CLASSIFICATION OF MAGNETIC MATERIALS
Diamagnetic:
Examples:
Materials with –ve magnetic susceptibility .
 (Au) = - 3.6 cm-3,
 (Hg) = - 3.2 cm-3
 (H2O)) = - 0.2X10-8 cm-3)
Paramagnetic: Magnetic materials with +ve and small
magnetic susceptibility .
Examples:
 (Al) = 2.2X10-5 cm-3
 (Mn)= 98 cm-3
Ferromagnetic: Magnetic materials with +ve and very large
magnetic susceptibility .
Examples:
Normally of the order of 105 cm-3.
SD-JIITN-PH611-MAT-SCI-2013
2. A MICROSCOPIC LOOK
In an atom, magnetic effect may arise due to:
1. Effective current loop of electrons in atomic orbit (orbital
Motion of electrons);
2. Electron spin;
3. Motion of the nuclei.
SD-JIITN-PH611-MAT-SCI-2013
MAGNETIC MOMENTS AND ANGULAR MOMENTUM
1. ORBITAL MOTION
Consider a charged particle moving in a circular orbit (e.g. an
electron around a nucleus),
The magnetic moment  may be
given as
  IA
 q  r 2
qr 2

2
  


But
L  r  p  r  mv

2
 L  mrv  mr 
Therefore,
q 

L
2m

(2.1.1)
SD-JIITN-PH611-MAT-SCI-2013
For an electron orbiting around the nucleus, magnetic
moment would be given as
e 
L  
L  e
2m
2m

Where,  B 
l (l  1)   B l (l  1)
(2.1.2)
e
 9.27 x10  24 = Bohr Magneton
2m
In the equation (2.1.2)

e
 
2m
L
L
(2.1.3)
orbital gyro-magnetic
ratio, .
SD-JIITN-PH611-MAT-SCI-2013
2. ELECTRON SPIN
Electrons also have spin rotation about their own axis. As a
result they have both an angular momentum and magnetic
moment.
But for reasons that are purely quantum mechanical, the
ratio between  to S for electron spin is twice as large as it is
for a orbital motion of the spinning electron:
e 
S   S
m

(2.2.1)
3. NUCLEAR MOTION
Nuclear magnetic moment is expressed in terms of nuclear
magneton

e
n 
2m p
mp is mass of a proton.
(2.3.1)
What happens in a real atom?
In any atom, there are several electrons and some
combination of spin and orbital rotations builds up the total
magnetic moment.
The direction of the angular momentum is opposite to that of
magnetic moment.
Due to the mixture of the contribution from the orbits and spins
the ratio of  to angular momentum is neither -e/m nor –e/2m.
 e 
  gJ 
J
 2m 

Where g is known as Lande’s g-factor. It is given as
gJ  1
J ( J  1)  S ( S  1)  L( L  1)
2 J ( J  1)
SD-JIITN-PH611-MAT-SCI-2013
DIAMAGNETISM
Diamagnetism is inherent in all substances and arises out of
the effect of a magnetic field on the motion of electrons in an
atom.
Suppose an electron is revolving around the nucleus in atom,
the force, F, between electron and the nucleus is
F  m 2 r
When this atom is subjected to a magnetic field, B, electron
also experiences an additional force called Lorentz force
FL  erB
Thus when field is switched on, electron revolves with the new
frequency, , given by
F  erB  m '2 r
 m 2 r  erB  m '2 r
SD-JIITN-PH611-MAT-SCI-2013
 m 2  m '2  eB
m 2 r  erB  m '2 r
 ( 2   '2 ) 
For small B,
eB
m
 (   ' )(   ' ) 
  '  
eB
  2 
m
  
and
eB
m
  ' 2
eB
2m
Thus change in magnetic moment is
er 2
er 2 eB
  
  
2 2m
2
er 2
[  
]
2
e2r 2 B
   
4m
SD-JIITN-PH611-MAT-SCI-2013
Suppose atomic number be Z, then equation (2.9) may be
written as,
i z
e
  
2
 ri
2
i 1
4m
B
Where, summation extends over all electrons. Since core
electrons have different radii, therefore
e2 Z  r 2 
  
B
4m
If the orbit lies in x-y plane then,
 r 2  x 2    y 2 
If R represents average radius, then for spherical atom
 R 2  x 2    y 2    z 2 
For spherical symmetry,
 R2 
 x  y  z 
3
2
2
2
Therefore,
 r 2  x 2    y 2 
2
2

R

2
 r 
3
Therefore, equation (2.10) may be written as
e 2 BZ 2
  
(  R 2 )
4m 3
2
2
R
Ze B
Thus,    
4m
3
Ze 2 B 2
   
R
6m
2
If there are N atoms per unit volume, the magnetization
produced would be
NZe 2 B 2
M  N  
R
6m

0 NZe2 H
6m
R2
Susceptibility, , would be
0 NZe 2 B 2
0 M

R

B
6mB
 
NZ0e
6m
2
R2
This is the Langevin’s formula
for volume susceptibility of
diamagnetism of core electrons.
Conclusions:
Diamagnetic susceptibility
0 NZe2  R 2 
M


H
6m
1. Since   Z, bigger atoms would have larger susceptibility.
2. dia depends on internal structure of the atoms which is
temperature independent and hence the dia.
3. All electrons contribute to the diamagnetism even s
electrons.
4. All materials have diamagnetism although it may be
masked by other magnetic effects.
Example: R = 0.1 nm, N = 5x1028/ m3
4 107  5 1028  (1.6 1019 ) 2  (0.1109 ) 2
6
3
 


3

10
cm
6  9.11031
SD-JIITN-PH611-MAT-SCI-2013
PARAMAGNETISM
Paramagnetism occurs in those substances where the
individual atoms, ions or molecules posses a permanent
magnetic dipole moments.
The permanent magnetic moment results from the following
contributions:
- The spin or intrinsic moments of the electrons.
- The orbital motion of the electrons.
- The spin magnetic moment of the nucleus.
SD-JIITN-PH611-MAT-SCI-2013
Examples of paramagnetic materials:
- Metals.
- Atoms, and molecules possessing an odd number of
electrons, viz., free Na atoms, gaseous nitric oxide (NO) etc.
- Free atoms or ions with a partly filled inner shell: Transition
elements, rare earth and actinide elements. Mn2+, Gd3+, U4+
etc.
- A few compounds with an even number of electrons
including molecular oxygen.
SD-JIITN-PH611-MAT-SCI-2013
CLASSICAL THEORY OF PARAMAGNETISM
Let us consider a medium containing N magnetic dipoles per
unit volume each with moment .
In presence of magnetic field,
potential energy of magnetic
dipole
 
V    .B   B cos 
Where,  is angle between
magnetic moment and the
field.
B =0, M=0
B ≠0, M≠0
V   B (minimum) when   0
It shows that dipoles tend to line up with the field. The effect
of temperature, however, is to randomize the directions of
dipoles. The effect of these two competing processes is that
some magnetization is produced.
Suppose field B is applied along z-axis, then  is angle made
by dipole with z-axis. The probability of finding the dipole
along the  direction is
f(θ )  e

V
kT
e
μ B co s θ
kT
f() is the Boltzmann factor which indicates that dipole is more
likely to lie along the field than in any other direction.
The average value of z is given as
z
 f ( )d


 f ( )d
z
Where, integration is carried out over the solid angle, whose
element is d. The integration thus takes into account all the
possible orientations of the dipoles.
SD-JIITN-PH611-MAT-SCI-2013
Let n(θ) be the number of dipoles per unit solid angle at θ,
we have
n( )  n0 e
pE cos 
kT
The number of dipoles in a solid angle d
n0 e
pE cos 
kT
Note: Here d is
calculated as follows:
d
 n0 e
pE cos 
kT
2 sin d
Dipole moment of dipoles making angle  with the field
(along x-axis) is
p x  p cos 
Therefore, the dipole moment along the field within angle d
 n0 e
pE cos 
kT
2 sin d ( p cos  )
Now, average dipole moment (Total dipole moment divided
by total no. of dipoles) can be written as

n e
pE cos 
kT
0
p
2 sin  ( p cos  )d
0

n e
0
0
pE cos 
kT
2 sin d

n e
pE cos 
kT
0
p
2 sin  ( p cos  )d

0

n e
0

pE cos 
kT
2 sin d
p

p
pE cos 
x
kT
and
p e a  e a 1
 a

a
p e e
a
CASE 2: When a is very low (at
high temperature) i.e. a <<1
a
 L(a ) 
3
0

e
0
0
Let
e
pE cos
kT
Np 2 E
 Po 
3kT
pE
a
kT
sin  cos d
pE cos
kT
sin d
p
a
  L( a ) 
p
3
p  o E
a
 p p
3
p2
 o 
3kT
p2E
 p
3kT
1
 o 
T
Some of the quantities are replaced and the result is almost same
Substituting z =  cos and d = 2 sin d

z 
  cos  2 sin e
0

 2 sin e
B cos
kT
B cos
kT

d

  cos  sin e
0

 sin e
d
0
B cos
B cos
kT
kT
d
d
0

z 
  cos  sin e a cos d
0
Let

 sin e
a cos
d
B
kT
a
0
Let cos = x, then sin d = - dx and Limits -1 to +1
1
z 
  xeax dx
1
1
ax
e
 dx
 ea  ea 1 
 z    a
 
a
a
e  e
1
SD-JIITN-PH611-MAT-SCI-2013
coth( a )
 ea  ea 1 
1



coth(
a
)

z    a
 


a
a
e

e
a



  z  L(a)
[a 
B
kT
]
Langevin
function, L(a)
a a3 2 a5
L( a )  

 
3 45 945
In most practical situations a<<1,
therefore,
a
L(a ) 
3
a  2B
 z   
3 3kT
The magnetization is given as
N 2 B
M  N z 
3kT
Variation of L(a) with a.
( N = Number of dipoles per unit volume)
SD-JIITN-PH611-MAT-SCI-2013
N 2 B N 2 0 H

M  N z 
3kT
3kT
N0  2
M


H
3kT
This equation is known as CURIE LAW. The susceptibility is
referred as Langevin paramagnetic susceptibility. Further,
contrary to the diamagnetism, paramagnetic susceptibility is
inversely proportional to T
Above equation is written
in a simplified form as:
C

T
N0  2
where, C 
3k
Curie constant
SD-JIITN-PH611-MAT-SCI-2013
Self study:
1. Volume susceptibility ()
2. Mass susceptibility (m)
3. Molecular susceptibility (M)
Reference: Solid State Physics by S. O. Pillai
SD-JIITN-PH611-MAT-SCI-2013
QUANTUM THEORY OF PARAMAGNETISM
Recall the equation of magnetic moment of an atom, i. e.
e 
S  - S
m

e 
L  L
2m
e 
 J  -g
J
2m


Where g is the Lande’ splitting factor given as,
J ( J  1)  S ( S  1)  L( L  1)
g  1
2 J ( J  1)
  
here, J  L  S
Consider only spin,
L0 J S
S ( S  1)  S ( S  1)
 g  1
2S ( S  1)
e 
 J   S  -2
S
2m


2
e 
 S  - S
m

SD-JIITN-PH611-MAT-SCI-2013
Consider only orbital motion,
S 0
J L
J ( J  1)  S ( S  1)  L( L  1)
g  1
2 J ( J  1)
L( L  1)  L( L  1)
 g  1
1
2 L( L  1)
e 
 J   L  -1 L
2m


e 
 L  L
2m

Let N be the number of atoms or ions/ m3 of a paramagnetic
material. The magnetic moment of each atom is given as,

e 
 J  -g
J
2m
In presence of magnetic field,
J z  MJ
according to space quantization.
Where MJ = –J, -(J-1),…,0,…(J-1), J i.e. MJ will have (2J+1)
values.
PARAMAGNETIC SUSCEPTIBILITY IN QUANTUM TERMS
 j 2  g 2  B 2 J ( J  1)
g 2  B 0
N 0  J
M

N
J ( J  1) 
H
3kT
3kT
2
N0 
M


2
2
H
3kT
Np  
2

eff
B
0
where,
3kT
C

T
2

peff  g[ J ( J  1)]
C
1
2
Npeff  B 0
2
where,
C
T
2
3k
This is curie law.
Further,
peff  B   J
 peff 
J
B
Thus Peff is effective number of Bohr Magnetons. C is Curie
Constant. Obtained equation is similar to the relation obtained by
classical treatment.
SD-JIITN-PH611-MAT-SCI-2013
Calculation of peff:
1. Write electronic configuration.
Say for
6C
2
2
1s 2s 2 p
Partially filled
sub-shell
2
2. Find orbital quantum number (l) for partially filled sub-shell.
l 1
In the given case:
3. Obtain magnetic quantum number. In the given example:
In the given case:
ml  1,0,1
4. Accommodate electrons in d sub shell according to Pauli’s
exclusion principle.
For the given case:
ml
1
0
-1
ms
SD-JIITN-PH611-MAT-SCI-2013
5. Apply following three Hund’s rules to obtain ground state:
(i) Choose maximum value of S consistent with Pauli’s
exclusion principle.
ml
1
0 -1
In the given example:
ms
S   ms  2 
1
1
2
(ii) Choose maximum value of L consistent with the Pauli’s
exclusion principle and rule 1.
In the given example:
L 1
(iii) If the shell is less than half full, J = L – S and if it is more
than half full the J = L + S.
ml
1
0
-1
ms
5. Obtain J. Since, shell is less than half filled therefore,
J  L  S  1 1  0
6. Obtain g. In the given example:
gJ  1
J ( J  1)  S (S  1)  L( L  1)
2 J ( J  1)
Now calculate peff using
peff  g[ J ( J  1)]
1
2
For 6C, peff = 0, hence it does not show paramagnetism.
SD-JIITN-PH611-MAT-SCI-2013
WEISS THEORY OF PARAMAGNETISM
Langevin theory failed to explain some complicated
temperature dependence of few compressed and cooled
gases, solid salts, crystals etc. Further it does not throw light
on relationship between para and ferro magnetism.
Weiss introduced concept of internal molecular field in order to
explain observed discrepancies. According to Weiss, internal
molecular field is given as
H i  M
Where  is molecular field coefficient.
Therefore the net effective field should be
 H e  H  M
But, we know from classical treatment of paramagnetism that
a
B
M  Ms( )
(For a << 1)
(a 
)
3
kT
SD-JIITN-PH611-MAT-SCI-2013
N 2 0 H e
a
M  Ms( ) 
3
3kT
N 2  0
M 
( H  M )
3kT
N 2  0
N 2  0
 M (1 
) 
H
3kT
3kT
N 2  0 H
M 
N 2  0
3kT (1  
)
3kT
M

H
N 2  0
N 2  0


2
N  0
N 2  0
3kT (1  
)
3k (T  
)
3kT
3k
C

T  c
Curie-Weiss Law.
Paramagnetic
N 2 0
 curie point
where  c 
3k
2
N

0
and C 
3k
Curie constant
SD-JIITN-PH611-MAT-SCI-2013
FERROMAGNETIC MATERIALS
SD-JIITN-PH611-MAT-SCI-2013
FERROMAGNETIC MATERIALS
Certain metallic materials posses permanent magnetic
moment in the absence of an external field, and manifest
very large and permanent magnetization which is termed as
spontaneous magnetization.
Example: Fe, Co, Ni and some rare earth metals such as Gd.
 ~ 106 are possible for ferromagnetic materials.
Therefore, H<<M and B  0 H  0 M
 B  0 M
Spontaneous magnetization decreases as the temperature
rises and is stable only below a certain temperature known as
Curie temperature.
SD-JIITN-PH611-MAT-SCI-2013
ORIGIN:
Atomic magnetic moments of un-cancelled electron spin.
Orbital motion also contributes but its contribution is very small.
Coupling interaction causes net spin magnetic moments of
adjacent atoms to align with one another even in absence of
external field. This mutual spin alignment exists over a
relatively larger volume of the crystal called domain.
Dimension ~ 10-2 cm, No. of atom/ domain 1015 to 1017
The maximum possible
magnetization is called saturation
magnetization.
M s  N
There is also a corresponding
saturation flux density Bs (=0Ms).
SD-JIITN-PH611-MAT-SCI-2013
DOMAINS AND HYSTERESIS
What happens when magnetic field is applied to the ferromagnetic
crystal?
According to Becker, there are two independent processes
which take place and lead to magnetization when magnetic
field is applied.
1. Domain growth:
Volume of domains oriented favourably w. r. t to the field
at the expense of less favourably oriented domains.
2. Domain rotation:
Rotation of the directions of magnetization towards the
direction of the field.
ORIGIN OF DOMAINS
According to Neel, origin of domains in the ferromagnetic materials
may be understood in terms of thermodynamic principle that
IN EQUILIBRIUM, THE TOTAL ENERGY OF THE SYSTEM IS MINIMUM.
Total energy:
1. Exchange energy;
2. Magnetic energy;
3. Anisotropy energy and
4. Domain wall or Bloch wall energy.
1. Exchange Energy
 
Ee  2 J e  Si  S j
It is lowered when spins are parallel. Thus, it favours an
infinitely large domain or a single domain in the specimen.
2. Magnetic Energy
This arises because the magnetized specimen has free
poles at the ends and thus produce external field H.
Magnitude of this energy is
1
8
2
H
 dv
Value of this energy is very high and can be reduced if the
volume in which external field exists is reduced and can be
eliminated if free poles at the ends of the specimen are
absent.
3. Anisotropy energy
For bcc Fe
[100] easy direction
[110] medium direction
[111] hard direction
For Ni
[111] easy direction
[110] medium direction
[100] hard direction
The excess energy required to magnetize a specimen in a
particular direction over that required to magnetize in the
easy direction is called crystalline anisotropy energy.
4. Domain wall or Bloch wall energy
Domain wall creation involves energy which is known as domain
wall energy of Bloch energy.
EXCHANGE INTERACTION IN MAGENTIC MATERIALS
Heisenberg (1928) gave theoretical explanation for large
Weiss field in ferromagnetic materials.
Parallel arrangements of spins in ferromagnetic materials
arises due to exchange interaction in which two neighboring
spins. The exchange energy of such coupling is
Eexch  2 Js1s2
Here J = exchange integral. Its value depends upon separation
between atoms as well as overlap of electron charge cloud.
J > 0, favors parallel configuration of spins, while for J < 0,
spins favors anti-parallel.
If there are Z nearest neighbors to a central ith spin, the
exchange energy for this spin is
 
Eexch  2 J ij Si .S j  2ZJS 2
j 1
This energy must be equal to K as at , ferromagnetic
order is destroyed. Thus,
2ZJ ij S 2  K
 J ij 
K
2ZS 2
Thus
criteria
for
ferromagnetism (due to Slater)
becomes – atoms must have
unbalanced spins and the
exchange integral J must be
positive. Alloys like Mn-As, CuMn
and
Mn-Sb
show
ferromagnetism.
Idea of magnetic energy due to domain:
Based on the definition of these energies a scheme is drawn
below which helps in minimization of energy of the system:
Domain closure
Single domain
Magnetic energy
high
Domain halved
magnetic energy
reduced
Elimination of magnetic
energy by domain closure
BLOCH WALL
(Due to Bloch)
The entire change in spin
direction between domains does
not occur in one sudden jump
across a single atomic plane
rather takes place in a gradual
way extending over many atomic
planes.
Bloch Wall
Because for a given total
change in spin direction, the
exchange energy is lower
when change is distributed
over many spins than when
the change occurs abruptly.
From the Heisenberg model, exchange energy is
Eexch
 
 2 J e Si .S j
 Eexch  2 J e S 2 cos 0
Substituting cos 0  1 
0 2
2
 ...
Eexch  2 J e S (1 
2
(Where 0 is the angle
between two spins.)
0 2
2
 ...)
For small angle , the change in exchange energy when
angle between spins change from 0 to  is
Eexch  Eexch ( )  Eexch (0)
 Eexch  2 J e S 2 (1 
0 2
2
 ...)  (2 J e S 2 )
Eexch  2 J e S 2 (1 
0 2
 ...)  (2 J e S 2 )
2
 2 J e S 2  2 J e S
 Eexch  J e S 20
2

2 0
2
 ...  2 J e S 2
2
Thus exchange energy increases when two spins are rotated
by an angle from exact parallel arrangement between them.
Now suppose the total change of angle between two domains
occurs in N equal steps.
Thus the change of angle between
two neighbouring spins =
Eexch  J e S
2

2
0
N2
0
N
Eexch  J e S
2

2
0
N2
Thus total energy change in N equal steps
(Eexch )T  J e S
2

2
0
N
2
N
 ( Eexch )T  J e S
2

2 0
N
Thus total energy change decreases when N increases.
Q. Why does not the wall becomes infinitely thick.
Ans. Because of increase of the anisotropy energy. Therefore
competing claims between exchange energy and anisotropy
energy leads to an equilibrium thickness.
Exchange energy per unit area (refer to the
figure),
2

Eexch  J e S 2 0 2 (Where  = )
0
Na
Eanis  KNa
Anisotropy energy is
Where K = anisotropy constant, a = lattice constant
Thus total wall energy would be
Ew  J e S 2
0 2
Na
2

dE

  J e S 2 20 2  Ka
dN
N a
2
 KNa

dE
 0   J e S 2 20 2  Ka
dN
N a
2
For minimum E wrt N
 JeS
Thus
2
0 2
2
N a
 Ka  N  J e S
2
2
Ew  J e S
 JeS
2
0 2
Na
2
(JeS
2
2
0 2
Ka
3
 N  (JeS
 KNa
0 2
2

2
0
Ka
1
2
2
)
a
3
 K (JeS
2
0 2
Ka
3
1
2
) a
2
0 2
Ka
3
)
1
2
Ew  J e S 2
(JeS
0 2
2
2 0
Ka
 K (JeS 2
1
2
3
) a
0
0
2
2
 Ew  J e S
2
1
2
Je S
1
2
K a
 Ew 
1
2
1
2
a
Thus
Ka
 KJ e S
3
2
a2
3
1
2
) a
1
2
K a
1
2
1
2
1
2
0
1
2
J e K S0
1
2
0 2
 K Je S
3
2
0
a
1
2
J e K 12
Ew  2 S0 (
)
a
This is equation for Block wall energy.
a
Magnetization by
domain rotation
Domain growth irreversible
boundary displacements.
Domain growth reversible
boundary displacements.
Crystal
structure
FeO·Fe2O3 (Iron ferrite)
Tetrahedral site:Fe ion is surrounded by four oxygens
Octahedral site:Fe ion is surrounded by six oxygens
Applications of Magnetism
Applications of Magnetic Induction
• Tape / Hard Drive
• Tiny coil responds to change in flux as the magnetic
domains (encoding 0’s or 1’s) go by.
• Credit Card Reader
–Must swipe card
 generates changing flux
–Faster swipe  bigger signal
Capacity of the disk: Sectors and tracks
• The road map consists of
magnetic markers embedded in
the magnetic film on the surface
of the disk.
• The codes divide the surfaces of
the disk into sectors (pie slices
and tracks (concentric circles).
• These divisions organize the disk
so that data can be recorded in a
logical manner
• and accessed quickly by the
read/write heads that move back
and forth over the disk as it spins.
• The number of sector and tracks
that fit on a disk determines the
disk’s capacity.
•
A typical track is shown in yellow; a
typical sector is shown in blue. A sector
contains a fixed number of bytes -- for
example, 256 or 512. Either at the drive or
the operating system level, sectors are
often grouped together into clusters.
Basic principles of magnetic recording
1. The drive channel electronics
receive data in binary form from
the computer and converts
them into a current in the head
coil.
2. The current in the coil reverses
at each 1 and remains the same
at each 0.
3. This current interaction with the
media results in magnetization
of the media, which direction
depends on the current
direction in the coil.
Magnetic heads: MR sensors
1. Today's magnetic head typically consists
of an MR (magneto-resistive) or GMR
(giant MR) reading head and a thin-film
inductive write head.
2. MR head design is based on the ability
of metals to change their resistivity in
the presence of a magnetic field. This
effect was first found in 1857.
3. The alloy of Ni and Fe (81%/19%) is
widely used in MR heads and is called
Permalloy.
1. MR heads are suitable for extremely
high bit density and have superior
signal-to-noise ratio when compared to
inductive read heads.
Multiple platters and heads are commonly used to increase the
amount of information
• How many platters in
this hard disk? 3
• How many read/write
heads? 6
Physical organization of the disk
Little bit about RAM
• Basic Terms:
RAM: Random Access Memory (volatile in case of semiconductor RAM so we need to
reboot the system after every occasion of power-cut)
SRAM: Static RAM (Used mostly for Cache Memory, Static RAM holds information
in memory as long as the power is on. It doesn't have to be constantly
refreshed, like standard Dynamic RAM (DRAM). Static RAM is faster than
DRAM but it's more expensive and takes up more space.)
DRAM: Dynamic RAM
SDRAM: Short for Synchronous DRAM, a type of DRAM that can run at much higher clock
speeds than conventional memory. SDRAM actually synchronizes itself with the
CPU's bus and is capable of running at 133 MHz, about three times faster than
conventional FPM RAM, and about twice as fast EDO DRAM and BEDO DRAM.
DDRAM: Double Data Rate RAM. Simply, this is RAM which handles data at twice the
speed of the old SDR (Synchronous Dynamic RAM.) DDR RAM typical operates at
DDR speeds of 266MHz, 333MHz, and 400MHz (actual speeds are 133, 166 and 200
respectively.)
MRAM: Magnetic RAM
•
It has magnetic memory which is nonvolatile=>you
don’t need to reboot the system after every power
cut.
There are three technologies:
1. Hybrid semiconductor/magnetic devices
2. Magnetic tunnel junction
3. All metal spin transistors
The leakage field from the magnetic elements can be
detected by a hall effect sensor and from the sign of the
hall voltage the orientation of the memory bit can be
determined.
Other applications:
• Ground fault interrupter
• Electric guitar
http://entertainment.howstuff
works.com/electricguitar1.htm
• Magnetic detector at LRC
• Metal detector
• Generator
• …