Download - MRAM (Magnetic random access memory)
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MRAM (Magnetic random access memory)
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Outline
• Motivation: introduction to MRAM.
• Switching of small magnetic structures: a highly nonlinear problem with large mesoscopic fluctuations.
• Current theoretical approaches.
• Problems: write reliability issues.
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An array of magnetic elements
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Schematic MRAM
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Write: 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.
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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
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Read: Tunnelling magneto resistance between ferromagnets• Miyazaki et al,
Moodera et al.• room temperature
magneto resiatance is about 30 %
• Fixed the magnetization on one side, the resistance is different between the AP and P configurations
• large resistance: 100 ohm for 10^(-4) cm^2, may save power
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Switching of magnetization of small structures
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Understanding the basic physics: different approaches
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Semi-analytic approaches
Solition solutions
Conformal Mapping
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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.
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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
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Conformal mapping
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From circle to square: Spins parallel to boundaries
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Navier Stokes equation (Yau)
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Numerical methods
• Numerical studies can be carried out by either solving the Landau-Gilbert equation numerically or by Monte Carlo simulation.
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Landau-Gilbert equation
• (1+2)dmi/d=hieffmi–(mi(mihieff)) • i is a spin label, • hieff=Hieff/Ms is the total reduced effective field
from all source; • mi=Mi/Ms, Ms is the saturation magnetization is a damping constant. =tMs is the reduced time with the
gyromagnetic ratio. • The total reduced effective field for each spin is
composed of the exchange, demagnetization and anisotropy field: Hieff=hiex+hidemg+hiani .
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Approximate results
• E=Eexch+Edip+Eanis.
• Between neighboring spins Edip<<Eexch.
• The effect of Edip is to make the spins lie in the plane and parallel to the boundaries.
• Subject to these boundary conditions, we only need to optimize the sum of the exchange and the anisotropy energies.
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Reliability problem of switching of magnetic random access
memory (MRAM)
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Fluctuation of the switching field
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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.
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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
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Experimental hysteresis curve
• J. Shi and S. Tehrani, APL 77, 1692 (2000).
• For large Hy, the hysteresis curve still exhibits nonzero magnetization at Hcx (Hy=0).
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Edge pinned domain proposed
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Hysteresis curves from computer simulations can also exhibit similar
behaviour• For nonzero Hy
switching can be a two step process. The bit is completely switched only for a sufficiently large Hx.
E
S
O
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• For finite Hy, curves with large Hsx are usually associated with an intermediate state.
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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.
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Another way recently proposed by the Motorola group: Spin flop
switchingTwo layers antiferromagnetically
coupled.
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• Memory in the green area.
• Read is with TMR with the magnet in the grey area, the same as before.
• Write is with two perpendicular wires (bottom figure) but time dependent.
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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.
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Switching boundaries
• Paper presented at the MMM meeting, 2003 by the Motorola group.
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This solves the bit selectivity but the field required is too big
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Stronger field, -135: Note the edge-pinned domain for the top layer
H
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Very similar to the edge pinned domain for the monlayer case.
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• Switching scenario involves edge pinned domain, similar to the monolayer case and very different from the coherent rotation picture.
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Coercive field dependence on interlayer exchange
• For the top curve, a whole line of bits is written.
• For real systems, there are fluctuations in the switching field, indicated by the colour lines. If these overlap, then bits can be accidentally written.
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Bit selectivity vs interlayer coupling: Magnitude of the
switching field
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Temperature dependence
• Hc (bilayer) >>Hc (single layer). Hc (bilayer) exhibits a stronger temperature dependence than the monolayer case, different from the prediction of the coherent rotation picture.
• Usually requires large current.
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Simple Physics in micromagnetics
• Alignment of neighboring spins is determined by the exchange, since it is much bigger than the other energies such as the dipolar interaction and the intrinsic anisotropy.
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Energy between spins
• H=0.5 ij=xyz,RR’ Vij(R-R’)Si(R)Sj(R’) ,• V=Vd+Ve+Va • The dipolar energy Vdij(R)=gij (1/|R|); • The exchange energy Ve=-J (R=R’+d)ij; d
denotes the nearest neighbors• Va is the crystalline anisotropy energy. It
can be uniaxial or four-fold symmetric, with the easy or hard axis aligned along specific directions.
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Optimizing the energy
• Eexch=-A dr ( S)2.• Eani=-K dr Sz
2.• Let S lie in the xz plane at an angle .• Eexch=-AS2 dr ( )2.• (Eexch+Eani)/ = AS22 -K sin =0.• 2=x
2-iy2.
• This is the imaginary time sine Gordon equation and can be exactly solved.
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Dipolar interaction
• The dipolar interaction Edipo=i,j MiaMjb[a,b/R3-3Rij,aRij,b/Rij
5]
• 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. The spins are parallel to the edges so that qM is small.
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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” boundary condition.