Micromagnetics 101

Spin model: Each site has a spin Si

• There is one spin at each site.

• The magnetization is proportional to the sum of all the spins.

• The total energy is the sum of the exchange energy Eexch, the anisotropy energy Eaniso, the dipolar energy Edipo and the interaction with the external field Eext.

Exchange energy

• Eexch=-J Si¢ Si+

• The exchange constant J aligns the spins on neighboring sites .

• If J>0 (<0), the energy of neighboring spins will be lowered if they are parallel (antiparallel). One has a ferromagnet (antiferromagnet

Magnitude of J

• kBTc/zJ¼ 0.3

• Sometimes the exchange term is written as A s d3 r |r M(r)|2.

• A is in units of erg/cm. For example, for permalloy, A= 1.3 £ 10-6 erg/cm

Interaction with the external field

• Eext=-gB H S=-HM• We have set M=B S.• H is the external field, B =e~/2mc is the Bohr

magneton (9.27£ 10-21 erg/Gauss).• g is the g factor, it depends on the material.• 1 A/m=4 times 10-3Oe (B is in units of G);

units of H• 1 Wb/m=(1/4) 1010 G cm3 ; units of M (emu)

Dipolar interaction

• The dipolar interaction is the long range magnetostatic interaction between the magnetic moments (spins).

• Edipo=i,j MiaMjb[a,b/R3-3Rij,aRij,b/Rij5]

• Edipo=i,j MiaMjbiajb(1/|Ri-Rj|).

• Edipo=s r¢ M( R) r¢ M(R’)/|R-R’|

• If the magnetic charge qM=-r¢ M is small Edipo is small

Anisotropy energy

• The anisotropy energy favors the spins pointing in some particular crystallographic direction. The magnitude is usually determined by some anisotropy constant K.

• Simplest example: uniaxial anisotropy

• Eaniso=-Ki Siz2

Modifies Landau-Gilbert equation

• M / t - M£H + r¢Jm = - M/

• is the thermal noise.

• Ordinarily the magnetization current Jm is zero.

• H is a sum of contributions from the exchange, Hex; the dipolar Hdipo, the anisotropy and the external field: H=He+ Hex + Hdipo +Han; Hex=JrM; Han=K M.

Some mathmatical challenges

• The dipolar field is long range:

different scheme has been developed to take care of this. These include using fast Fourier transforms or using the magnetostatic potential. For large systems, the implicit scheme takes a lot of memory.

Preconditioner: Just the exchange. (it is sparse.) Physically the exchange energy is usually the largest term.

Alternative approach

• Monte Carlo simulation with the Metroplois algoraithm.

• This is the same as solving the master equation: dP/dt=TP where T is the transition matrix.

Physical understanding

Three key ideas at finite temperatures:

• Nucleation

• Depinning

• Spins try to line up parallel to the edge because of the dipolar interaction. The magnetic charge is proportional to , and this is reduced.

M

Approximation

• Minimize only the exchange and the anisotropy energy with the boundary condition that the spins are parallel to the edge.

Two dimension:

• A spin is characterized by two angles and . In 2D, they usually lie in the plane in order to minimize the dipolar interaction. Thus it can be characterized by a single variable .

• The configurations are then obtained as solutions of the imaginary time Sine-Gordon equation r2+(K/J) sin=0 with the “parallel edge” b.c.

Edge domain: Simulation vs Analytic approximation.

• =tan-1 [sinh(v(y’-y’0))/(- v sinh((x’-x’0)))],

• y’=y/l, x’=x/l; the magnetic length l=[J/2K]0.5;

=1/[1+v2]0.5; v is a parameter.

Closure domain: Simulation vs analytic approximation

• =tan-1[A tn( x', f) cn(v [1+kg

2]0.5y', k1g)/ dn(v [1+kg2]0.5

y', k1g)], • kg

2=[A22(1-A2)]/[2(1-A2)2-1],• k1g

2=A22(1-A2)/(2(1-A2)-1), f

2=[A2+2(1-A2)2]/[2(1-A2)]• v2=[2(1-A2)2-1]/[1-A2].• The parameters A and can

be determined by requiring that the component of S normal to the surface boundary be zero

For Permalloy

• For an important class of magnetic material, the intrinsic anisotropy constant is very small.

• r2=0. For this case, conformal mapping ideas are applicable.

An example

• Constraint: M should be parallel to the boundary!

• For the circle, a simple solution is =tan-1y/x.

• Conformal mapping allows us to get the corresponding solution for the rectangle.

Current directions:

• Current induced torque

• Magnetic random access memory

Nanopillar Technique (Katine, Albert, Emley)

-Multilayer film deposited (thermal evaporation, sputtering) on insulating substrate

Au (10 nm)Co (3 nm)Cu (6 nm)

Co (40 nm)

Cu (80 nm)

-Current densities of 108 A/cm2 can be sent vertically through pillar

-Electron-beam lithography, ion milling form pillar structure (thicker Co layer left as extended film)

-Polyimide insulator deposited and Cu top lead connected to pillar

Polyimide insulator

Cu

Magnetic Reversal Induced by a Spin-Polarized Current

Large (~107-109 A/cm2) spin-polarized currents can controllably reverse the magnetization in small (< 200 nm) magnetic devices

Parallel

(P)

Antiparallel

(AP)

Ferromagnet 1 Ferromagnet 2

Nonmagnetic

Cornell THALES/Orsay NIST

Positive Current

Modifies Landau-Gilbert equation

• M / t - M£H + r¢Jm = - M/

• The magnetization current Jm is nonzero.

Charge and magnetization current

• Je=-r V -e Dr n -DM r (M¢ p0)

• J=- M r ( V p0) - DM' r M - D' r ( n p0)

• p0=M0/|M0|; M0 is the local equilibrium magnetization,

• V=-Er+W; W(r)=s d3r' n(r')/|r-r’|

Two perpendicular wires generate magnetic felds Hx and Hy

• Bit is set only if both Hx and Hy are present.

• For other bits addressed by only one line, either Hx or Hy is zero. These bits will not be turned on.

Coherent rotation Picture

• The switching boundaries are given by the line AC, for example, a field at X within the triangle ABC can write the bit.

• If Hx=0 or Hy=0, the bit will not be turned on.

Hx

Hy

A B

C

X

Bit selectivity problem: Very small (green) “writable” area

• Different curves are for different bits with different randomness.

• Cannot write a bit with 100 per cent confidence.

Another way recently proposed by the Motorola group: Spin flop

switchingElectrical current required is too

large at the moment

Simple picture from the coherent rotation model

• M1, M2 are the magnetizations of the two bilayers.

• The external magnetic fields are applied at -135 degree, then 180 degree then 135 degree.

Magnetization is not uniform: coherent rotation model is not

enough