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Micromagnetics 101

Micromagnetics 101. Spin model: Each site has a spin S i. There is one spin at each site. The magnetization is proportional to the sum of all the spins.

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Micromagnetics 101

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  1. Micromagnetics 101

  2. 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.

  3. Exchange energy • Eexch=-JI,d 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

  4. 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

  5. Interaction with the external field • Eext=-gB 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)

  6. 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 MiaMjbiajb(1/|Ri-Rj|). • Edipo=sr¢ M( R) r¢ M(R’)/|R-R’| • If the magnetic charge qM=-r¢ M is small Edipo is small

  7. 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=-Ki Siz2

  8. Modifies Landau-Gilbert equation •  M / t - M£H + r¢Jm = -M/ +h •  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=Jr2M; Han=K M0.

  9. 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.

  10. 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.

  11. Physical understanding

  12. 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.

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

  14. 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.

  15. 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.

  16. Closure domain: Simulation vs analytic approximation • =tan-1[A tn( x', f) cn(v [1+kg2]0.5y', k1g)/ dn(v [1+kg2]0.5 y', k1g)], • kg2=[A22(1-A2)]/[2(1-A2)2-1], • k1g2=A22(1-A2)/(2(1-A2)-1), • f2=[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

  17. 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.

  18. 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.

  19. Current directions: • Current induced torque • Magnetic random access memory

  20. Nanopillar Technique (Katine, Albert, Emley) Au (10 nm) -Multilayer film deposited (thermal evaporation, sputtering) on insulating substrate Co (3 nm) Cu (6 nm) Co (40 nm) Cu (80 nm) -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 Cu Polyimide insulator -Current densities of 108 A/cm2 can be sent vertically through pillar

  21. Nonmagnetic Ferromagnet 2 Ferromagnet 1 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 Antiparallel (AP) Positive Current Parallel (P) Cornell THALES/Orsay NIST

  22. Modifies Landau-Gilbert equation • M / t - M£H + r¢Jm = -M/ +h. • The magnetization current Jm is nonzero.

  23. Charge and magnetization current • Je=-r V -e Dr n -DMr (M¢ p0) • J=- Mr ( V p0) - DM' r M - D' r (  np0) • p0=M0/|M0|; M0 is the local equilibrium magnetization, • V=-Er+W; W(r)=s d3r'  n(r')/|r-r’|

  24. 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 Hxor Hy is zero. These bits will not be turned on.

  25. 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. B A X Hy C Hx

  26. 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.

  27. Another way recently proposed by the Motorola group: Spin flop switching Electrical current required is too large at the moment

  28. 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.

  29. Magnetization is not uniform: coherent rotation model is not enough

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