MRAM (Magnetic random access memory)

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.

An array of magnetic elements

Schematic MRAM

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.

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

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

Switching of magnetization of small structures

Understanding the basic physics: different approaches

Semi-analytic approaches

Solition solutions

Conformal Mapping

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

Conformal mapping

From circle to square: Spins parallel to boundaries

Navier Stokes equation (Yau)

Numerical methods

• Numerical studies can be carried out by either solving the Landau-Gilbert equation numerically or by Monte Carlo simulation.

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 .

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.

Reliability problem of switching of magnetic random access

memory (MRAM)

Fluctuation of the switching field

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

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).

Edge pinned domain proposed

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

• For finite Hy, curves with large Hsx are usually associated with an intermediate state.

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

switchingTwo layers antiferromagnetically

coupled.

• 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.

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.

Switching boundaries

• Paper presented at the MMM meeting, 2003 by the Motorola group.

This solves the bit selectivity but the field required is too big

Stronger field, -135: Note the edge-pinned domain for the top layer

H

Very similar to the edge pinned domain for the monlayer case.

• Switching scenario involves edge pinned domain, similar to the monolayer case and very different from the coherent rotation picture.

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.

Bit selectivity vs interlayer coupling: Magnitude of the

switching field

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.

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.

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.

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.

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.

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.