Transcript Anisotropy and Magnetization Reversal

Anisotropy, Reversal and Micro-magnetics
1. Magnetic anisotropy
(a) Magnetic crystalline anisotropy
(b) Single ion anisotropy and atom pairs model
(c) Exchange energy (anisotropy)
(d) Interface anisotropy
(e) Interlayer anti-ferromagnetic coupling
2. Magnetization reversal
(a) H parallel and normal the anisotropy axis, respectively
(b) Coherent rotation (Stoner-Wohlfarth model)
(c) Spin torque (Current induced switching)
(d) For votex
3. Micromagnetics
dynamic simulation; solving LLG equation
Magnetocrystalline anisotropy
Crystal structure showing easy and hard magnetization direction for Fe (a),
Ni (b), and Co (c), above. Respective magnetization curves, below.
The Defination of Field Ha
A quantitative measure of the strength of the
magnetocrystalline anisotropy is the field, Ha,
needed to saturate the magnetization in the hard
direction.
The energy per unit volume needed to saturate
a material in a particular direction is given by a
generation:
The uniaxial anisotropy in Co,Ku = 1400 x 7000/2
Oe emu/cm3 =4.9 x 106 erg/cm3.
How is µL coupled to the lattice ?
If the local crystal field seen by an atom is of low symmetry
and if the bonding electrons of that atom have an asymmetric
charge distribution (Lz ≠ 0), then the atomic orbits interact
anisotropically with the crystal field. In other words, certain
orientation for the bonding electron charge distribution are
energetically preferred.
The coupling of the spin part of the magnetic moment to the
electronic orbital shape and orientation (spin-orbit coupling) on
a given atom generates the crystalline anisotropy
Physical Origin of
Magnetocrystalline anisotropy
Simple representation of the role of orbital angular
momentum <Lz> and crystalline electric field in determining the strength of magnetic anisotropy.
Uniaxial Anisotropy
Careful analysis of the magnetization-orientation curves
indicates that for most purpose it is sufficient to keep only
the first three terms:
where Kuo is independent of the orientation of M. Ku1>0
implies an easy axis.
Uniaxial Anisotropy
(1)Pt/Co or Pd/Co multilayers from interface
(2)CoCr films from shape
(3)Single crystal Co in c axis from (magneto-crystal anisotropy)
(4)MnBi (hcp structure)
(5)Amorphous GdCo film
(6)FeNi film
Single-Ion Model of Magnetic Anisotropy
dε
dγ
In a cubic crystal field, the orbital states of 3d
electrons are split into two groups: one is the
triply degenerate dε orbits and the other the
doubly one d γ.
Energy levels of dεand d dγ electrons in
(a) octahedral and (b) tetrahedral sites.
Table: The ground state and degeneracy of transition metal ions
d electrons for
Fe2+ in octahedral site.
Co2+ ions
Oxygen ions
Cations
Distribution of surrounding ions about
the octahedral site of spinel structure.
Conclusion :
(1) As for the Fe2+ ion, the sixth electron should occupy
the lowest singlet, so that the ground state is degenerate.
(2) Co2+ ion has seven electrons, so that the last one
should occupy the doublet. In such a case the orbit has
the freedom to change its state in plane which is normal
to the trigonal axis, so that it has an angular momentum
parallel to the trigonal axis.
Since this angular momentum is fixed in direction, it
tends to align the spin magnetic moment parallel to the
trigonal axis through the spin-orbit interaction.
Slonczewski expalain the stronger anisotropy of Co2+ relative the Fe2+ ions
in spinel ferrites ( in Magnetism Vol.3, G.Rado and H.Suhl,eds.)
Single ion model:
Ku = 2αJ J(J-1/2)A2<r2>,
Where A2 is the uniaxial anisotropy
of the crystal field around 4f
electrons, αJ Steven’ factor, J total
anglar momentum quantum
numbee and <r2> the average of
the square of the orbital radius of
4f electrons.
Perpendicular anisotropy energy per RE
atom substitution in Gd19Co81films prepared
by RF sputtering (Suzuki at el., IEEE Trans.
Magn. 23(1987)2275.
over the nearest-neighbor ions j.
Y.J.Wang and W.Kleemann PRB 44(1991)5132.
References (single ion anisotropy)
(1) J.J.Rhyne 1972 Magnetic Properties Rare earth matals ed by R.J.elliott
p156
(2) Z.S.Shan, D.J.Sellmayer, S.S.Jaswal, Y.J.Wang, and J.X.Shen,
Magnetism of rare-earth tansition metal nanoscale multilayers,
Phys.Rev.Lett., 63(1989)449;
(3) Y. Suzuki and N. Ohta, Single ion model for magneto-striction in rare-earth
transition metal amorphous films, J.Appl.Phys., 63(1988)3633;
(4) Y.J.Wang and W.Kleemann,
Magnetization and perpendicular anisotropy in Tb/Fe multilayer films,
Phys.Rev.B, 44 (1991)5132.
Exchange Anisotropy
Co particle 2r=20nm
Schematic representation of effect of exchange coupling on M-H loop
for a material with antiferromagnetic (A) surface layer and a soft ferromagnetic layer (F). The anisotropy field is defined on a hard-axis loop,
right ( Meiklejohn and Bean, Phys. Rev. 102(1956)3047 ).
FeMn
NiFe
weak-antiferromagnete
strong-antiferromagnet
Above, the interfacial moment configuration in zero field. Below, left, the
weak-antiferromagnete limit, moments of both films respond in unison
to field. Below, right, in the strong-antiferromagnet limit, the A moment
far from the interface maintain their orientation. (Mauri JAP 62(1987)3047)
NiFe/FeMn
In the weak-antiferromagnet
limit,
KA tA ＜＜ J,
tA ≦ j / KA= tAc,
For FeMn system, tAc ≈ 5 0
(A) for j ≈ 0.1 mJ/m2 and KA
≈ 2x104 mJ/m3.
Exchange field and coecivity as function of FeMn
Thickness (Mauri JAP 62(1987)3047).
Mauri et al., (JAP 62(1987)3047) derived an expression for
M-H loop of the soft film in the exchange-coupled regime,
(tA>tAc)
θ
There are stable solution at θ=0 and π
corresponding to ± MF.
Hex along z direction
Carcia et al., APL 47(1985)178
Interface anisotropy
200 C
50 C
Total anisotropy Kt for evaporated (111) texturized polycrystalline
Co/Pd multilayers versus thickness t of Co films.
Co/Pt Multilatersa
Magnetic hysteresis loops at 20 oC.
(Si substrate)
Effective anisotropy times Co thickness versus cobalt
thickness for [Co/Pt] multilayers (Engle PRL 67(1990)1910).
The effective anisotropy energy measured for a film of
thickness d may be described as
,
(1)
or writing as
(2)
=
Keff d = 2ks + (kV -2πMs2)d
(3)
Surface Magnetic Anisotropy ?
•The reduced symmetry at the surface (Neel 1954);
•The ratio of Lz2 / (Lx2 + Ly2) is increased near the surface
•Interface anisotropy (LS coupling)
[1]J.G.Gay and Roy Richter, PRL 56(1986)2728, [2] G.H.O. Daalderop et al.,
PRB 41(1990)11919, [3] D.S.Wang et al., PRL 70(1993)869.
Interlayer AF coupling
Grunberg et al., PRL 57(1986)2442
Fe/Cr/Fe
Fe/Au/Fe
Fig.2 Spectra from Cr 8 and Au 20 with Bo along
the easy axis. The arrow indicate the suggested
magnetization direction on the two Fe layers where
Bo is supposed to point up. Observed spin-wave
propagation then is along a horizontal line.
Oscillation Exchange Coupling
Field needed to saturate the magnetization at 4.2 K versus Cr thickness
for Si(111) / 100ACr / [20AFe / tCr Cr ]n /50A Cr, deposited at T=40oC (
solid circle, N=30); at T=125oC (open circle, N=20) (Parkin PRL 64
(1990)2304).
Interlayer exchange coupling strength J12 for coupling of
Ni80Co20 layers through a Ru spacer layer. The solid line
corresponds to a fit to the data of RKKY form. Parkin et al.,
PRB 44(1991)7131.
Parkin et al., PRL 66(1991)2152
Bruno, Chappert PRL 67(1991)1602
The spin polarization of the conduction electrons gives rise to an indirect exchange
interaction Hij = J(Rij) S i·S j . The interlayer coupling is obtained by summing Hij
over all the pairs ij, i and j running respectively on F1 and F2.
Co/Au(111)/Co
Dependence of the exchange coupling J between Co layers vs the thickness tAu
of the Au(111) interlayer. Line: theoretical fit of experimental data to RKKY model,
with I33.8 erg/cm2, Λ=4.5 AL, Ψ=0.11 rad, tc=5AL and m*/m=0.16
PRL 71(1993)3023
RKKY theory
Fermi spanning vector
Magnetization Process
The magnetization process describes
the response of material to applied field.
(1) What does an M-H curve look like ?
(2) why ?
For uniaxial anisotropy and domain walls
are parallel to the easy axis
Application of a field H transverse to the EA results in rotation of the
domain magnetization but no wall motion. Wall motion appears as H
is parallel to the EA.
Hard-Axis Magnetization
The energy density
(1)
(For zero torque condition)
(2)
(For stability condition)
θ= 0 for H > 2 Ku / Ms
θ the angle between H and M
θ= π for H <
(Ku >0 )
-2 Ku / Ms
(Ku <0)
The other solution from eq.1 is given by
(2)
This is the equation of motion for the magnetization
in field below saturation -2Ku/Ms <H < 2Ku/Ms
Eq.(2) may be written as
HaMscosθ= MsH
Using cosθ=m=M/Ms , eq.3 gives
m=h,
( h=H/Ha)
(3)
m = h,
( m = M/Ms ; h = H/Ha )
It is the general equatiuon for the magnetization
processs with the field applied in hard direction for
an uniaxial material,
M-H loop for hard
axis magnetization
process
M-H loop for easy-axis magnetization process
Stoner-Wohlfarth Model
The free energy
f = -Kucos2 (θ- θo)+ HMscosθ
Minimizing with respect to θ, giving
Coordinate system for
magnetization reversal
process in single-domain
particle.
Kusin2 (θ- θo) – HMssin θ=0
Kusin2 (θ- θo) – HoMsSin θ=0
(1)
∂2E/ ∂ θ2 =0 giving,
2KuCos (θ- θo) - Ho MsCos θ=0
(2)
Eq.(1) and (2) can be written as
sin2(θ- θo) = psinθ
(3)
cos (θ- θo) = (p/2)cosθ
(4)
with p=Ho Ms/Ku
From eq.(3) and (4) we obtain
(5)
Using Eq.(3-5) one gets
(6)
The relationship between p and θo
Sin2θo=(1/p2) [(4-p2)/3]3/2
p
θo is the angle between H
and the easy axis; p=Ho Ms/Ku.
θo =45o, Ho =Ku/Ms; θo =0 or 90o, Ho =2Ku/Ms
Stoner Wohlfarth model of coherent rotation
Hc [2Ku/Ms]
M/Ms
H [2Ku/Ms]
o
Wall motion coecivity Hc
H
The change of wall energy
per unit area is
∂εw /∂ s =2IsHcos θ
θ is the angle between H and Is
Ho={1/ (2Iscos θ) } (∂εw/ ∂s)max
(1)
If the change of wall energy arises from interior stress
max
(2)
here δ is the wall thick. Substitution of (2) into (1) getting,
When ι
For common magnet, Homax =200 Oe.
≈ δ
(λ≈10-5, Is=1T, σo=100
KG /mm2.)
Ho max= πλσo/2Iscosθ
Dependence of the coercive force on the magnitude for
of internal stress nickel (a) hard-drawn in various
stress; (b) a hard-worked Ni specimen which was annealed to release the internal stress.
Coecivity from domain wall pinning
Geometry of medium showing defect region 2 and host
material in regions 1 and 3
*Friedburg and Pauil
PRL 34(1975)1234.
The total energy related to the 180o wall movement
E=∫[Ai (dθ/dx)2 + ki sin2θ – HMi cos θ] dx
(1)
where i=1,2 and 3 in the region 1,2 and 3 respectively;
Minimizing the total energy and obtaining the Euler equation
-2 Ai (d2θ/dx2 ) + 2ki sinθcos θ +HMi sin θ =0
(2)
Integrating the Eq.(2) yields the three nonlinear equations
- Ai (dθ/dx)2 + ki sin2θ - HMi cosθ = ci
where ci is an integral constant
(3)
σ = A(Əθ/ Əz)2dz
δσ = 2A [ƏθƏ δθ/ƏzƏz ]dz
= -2A(Ə 2θ/Əz2) δθdz
+2A(Əθ/Əz)δz
=  -2A (Ə 2θ/Əz2)
The boundary condition in the homogeneous regions is
θ(-∞)=0, θ(+∞)=π, (d θ/dx)x= -∞=0, (d θ/dx)x=∞=0, (4)
substituting (4) into (3), one obtains
- A (dθ/dx)2 - k sin2θ - HMs cosθ + HMs= 0
(5)
- A (dθ/dx)2 - k sin2θ - HMs cosθ - HMs= 0
(6)
- A’ (dθ/dx)2 - k’ sin2θ- HM2 cosθ = c2;
(7)
they are, respectively, Euler equations in region of (1), (3)
and (2).
Using the continuous condition at boundary,
A dθ1.2 / dx = A’ d θ1.2 / dx,
one gets
(cosθ1 +ha/2b)2 – (cosθ2 +ha/2b)2 = 2h/b
(8)
x2
D=
 dx
= A’/(AK)1/2 [(1-b)sin2θ2 -h(1-a)cosθ2
x1
+ bsin2θ1 -hacosθ1+h ] -1/2 dθ2
(9)
with h= HMs / K, a=1- M2A’/MsA, b=1- A’K’/AK.
When a, b and D are determined, namely the defects are determined, we can obtain
a set of solution of h, θ1 and θ2, among them there must be a set’s h which is
maximun hmax.
Solution 1
When the applied field small so as to h<1, then eq.9 become
D≈ [A’/(AK)1/2][ -cos2 θ2 + cos2 θ1]/[sin2 θ1sinθ1]
From eq.(8), cos2 θ1 ≈cos2 θ2+2h/b
(10) .
(11).
Substituting (11) into (10), we get
Hc = [(KD)/(Msδo)](A/A’-K/K’)(sin2 θ1 cosθ2)max
=(2K / Ms)( D / 32/3δo)(A/A’- K/K’)
(12)
Solution 2
If the thickness of the wall is much less than the thickness of the
defect, δo << D. In such a case, the reversal can be performed in
the region 2. Therefore, the contribution from the region 3 can be
ignored for θ2 =π. We get
Hc = (2K/Ms)(1-pq)[(1-(mp)1/2)2/(1-mp)2]
where p=A’/A, q=K’/K, m=M2/Ms
We see that Hc is not related to D when δo << D
(13)
w (=D/δdw);
(1) when w<<1, hc increases lineally with w
(2) when w is larger, hc is saturated;
(3) When F<<1 or E<<1, larger hc can be
obtained
The normalized Hcvs w (=D/δdw)
The normalized Hc vs wall energy
of the defect, E=A2K2/A1K1
Example-1
Nd2Fe14B
4πMs=16 KG, A≈10-6 erg/cm, K1=5x107 erg/cm3 ,
δo ≈40 A
The Nd2Fe14B grains, 5-10 µm are separated by non magnetic phase
of Nd1.1Fe4B4 and rich-Nd (0.1-1 µm ) and satisfied for w=D/δo, E=F ≈0;
If let F be 0.01, then
hc=1.6
or
Hc=hcK1M1= 40 KOe
Example-2
For amorphous alloy, W (D/δo) >>1 and F ≈1, E larger;
from fig, hc ≈0.06 is obtained.
Assuming K1 ≈100erg/cm3 and Ms=1200 emu/cm3, then
Hc=0.006 Oe
It is in agreement with experiment.
SST
1. L.Berger PRB 54(1996)9353
2. J.C.Slonczewski 3M 159(1996)1
Spin toque transfer
Tsoi et al PRL 80(1998)4281
Samples (large):
(Co/Cu)N MLs N=20-50
tCo=1.5nm, tCu=2.0-2.2nm
capped with 1.6 nm Au
Measured at 4.2 K and
Jc≈ 109 A/cm2
At negative polarity, current
flows from Ag tip into MLs.
H is perpendicular to MLS.
H
1200A Cu /100A Co /60A Cu/
25A Co /150A Cu /30A Pt/
600A Au
The device diameter
is 130± 30 nm
Katine et al PRL 84(2000)3149.
At room temperature
J=1.0x108 A/cm2
Fig.2 (a) dv/dI of a pillars device exhibits hysteretic jumps as
the current is swept; (b) Zero-bias MR hysteresis loop for the
same sample (Katine PRL 84(2000)3149).
g = [-4+(1+p)3(3 + s1 · s2)/4p3/2]-1
For I > Ic+ =αγ eS [ Heff (0) + 2πM] / g(0) from parallel to anti-parallel
For below Ic- = αγ eS [Heff (0) -2πM] / g(π) from anti-parallel to parallel
The spin-transfer model predicts that increasing H should make both
Ic+ and Ic- more positive.
Some degree of spin polarization
along the instantaneous axis parallel
to the vector s1 of local ferromagnetic
polarization in F1 will be present in the
electrons impinging on F2.
spin polarization along the axis parallel
to the vector ML of local ferromagnetic
polarization in will be present in the
electrons impinging on MR.
S1,2 = (Ieg/c)S1,2 x (S1 x S2)
g = [ -4+(1+p)3(3+S1·S2)/4p3/2]
(J.C.Slonczwski 3M 159(1996)L1)
The last term incorporates the spin-transfer effects.
The prefactor αI depends on the spin polarization
current p and the angle between the free and
pinned layers (Özyilmaz et al PRL 91(2003)067203).
Single-crystal multilayer mad by MBE;
25nm(Ga1-xMnx)As/500nm(InyGa1-y)As/100nmGaAs
on (001) semi-insulating GaAs substrate.
Sample A: x=0.05, y=0.15; B : x=0.038, y=0.23
Tc=90K for A, 110K for B
Fig 1, A micrograph and a schematic drawing of the device.
(a) a 20 µm wide channel with three pairs of Hall probes
separated by 15 µm was defined by photolithograhy
and wet etching.
(b) to reduce the Hc of two regions, 7-8 nm and 3nm of the surface
layers of regions 2 and 3, respectively, were removed
. A domain wall was prepared at the boundary
of regions 1 and 2, and its position after application
of a current pulse was monitoredby RHall=VHall/l,
using a small probe current I. Voltage VHall is measured
at the Hall probe. Magneto-optical Kerr microscopy was used
Fig 2, The hysteresis loop of regions 1,2
and 3 of sample A measured by RHall at
83K, and the temperature and the current
dependence of the averaged Hall resis-
tance, RHall.
(a) Hc shows that Hc(1)>Hc(3)>Hc(2), as
designed. No dependence of hysteresis on
current direction was observed at I=5µA.
(b) Temperature dependence of RHall measured
using I=5µA, 1.5-s current pulses; open circies
Indicate RHall for region 1, closed triangles for 2,
and open downward triangles for 3.
(c) Current dependence of RHall measured 1.5-s
current pulses at 82K.
Fig 3, The effect of successive alternating
negative and positive current pulses on
RHall for regions 1,2 and 3 of sample A.
The amplitude of the pulse is 350 µA and
its width 100ms.
At t=0, the domain wall is at the boundary of
regions 1 and 2; when a negative current
pulse is applied (t=30s), M direction in region
2 is reversed. The M direction in region 2
can be switched back.
initial state
After I=-300 µA After I=+ 300 µA
Fig 4, MOKE images of sample A using 546 nm light at 80K.
R.P.Cowburn et al., (Uni Cambridge) PRL 83(1999)1042.
Reversal in votex
Motion Equation for Magnetic vector
There is angle moment G corresponding to M and the
relation between them
M = -γ G,
(1)
under the action of Heff, M is acted by a torque,
L = M x Heff.
(2)
Due to the torque,
dG/dt = M x Heff,
(3)
substituting (1) to (3), we obtain
dM/dt = -γ M x Heff,
(4)
This is the Motion Equation for Magnetic vector
He
t2=t1+dt.
-L”
-L’
t2
M’
t1
o
M”
At t1, the position of M is represented
by M’. Under the action of He, M’ is
affected by a torque L’=M’ x He and M’
moves to the position of M” at t2=t1+dt.
As such a manner, M moves with a
constant length and a continuous variant
direction.
G’
G”
This behaviors of M is called as the
precession motion of M around He.
dM/dt = -γ M x Heff, which is a motion equation without
any damping. When the damping term in the equation exits,
M is quickly parallel to the direction of He.
The expression of the damping term:
(1) Landau-Lifshith form
(2) Gilbert form
Td = (-αγ/M) x M x ( M x Heff),
Td = (α/M) M x dM / dt
Micromagnetics-Dynamic
Simulation
(1) The film is divided into nx ∙ ny regular elements,
(2) Determining all the field on each element
(3) Solving Landau-Lifshith-Gilbert equation
Two dimension
Magnetic thin film modelded in two-dimensional
approximation. The film is divided into nx x ny elements for the simulation.
ΔM < 1.0 x10-7 G; The sum
torque T <102 erg/cc
Computation flow diagram for solving the magnetization
In the magnetic film.
(1) To set the parameters: Ms=1400 G, Ku=1x106 erg/cm3,
random anisotropy from 0-180o, A=2x10-6 erg/cm, d=20nm, α=1
(2) mx(i,j)=m sinΘcos β, my(i,j)=, mz(i,j)= dmx(i,j)=0, dmy(i,j)=0
dmz(i,j)=0
(3) mx(i,j)=mx(i,j)+dmx(i,j), dmy(i,j), dmz(i,j)
(4) ∑ mx(i,j)/N [i from 1 to 20, j from 1 to 20]
(5) t from 10-3 to 1 sec, dt=10-12
(6) To calculate Hk(i,j), Hd(i,j), Hex(i,j), Hap(i,j)
(7) To solving the LLG eqation
(8) To calculate torque and dm
Micromagnetics-dynamic simulation
Cross-tie wall in thin Permalloy film: simulated (a and b) and observed (c)
Nakatani et al., Japanese JAP 28(1989)2485.
Hysterisis Loop Simulation
(an example Co/Ru/Co and Co/Ru/Co/Ru/Co Films)
Co
Ru
Ru
Ru
Co
Co
Co
Co
Wang YJ et al., JAP 89(2001)6994;91(2002)9241; 94(2003)525.
Landau-Lifshitz-Gilbert Equation
The other fields
(1) Random anisotropy field : ha = ( m ∙ e ) ∙ hK ,
m = M/Ms , and e denotes the unit vector along the
easy axis in the cell;
(2) Exchange energy fild:
hex =
(3) Demagnetizing field (dipole-dipole interaction)
hmagi = - ∑ (1/rij3) [3(mj∙ rij)/rij –mj]
(4) The applied field
happ = h ∙ m
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