Current-perpendicular-to-plane magnetoresistance

The measuring electric current crosses the layer planes. This geometry motivates the serial-layer paths in the two-current resistor model. A quantitative material calculation can also require interface electrical resistances, spin relaxation and spin accumulation; the circuit below intentionally isolates a smaller set of assumptions.

Consider two identical magnetic layers separated by a nonmagnetic layer, between reservoirs that impose the same terminal electric potential difference on both spin channels. Neglect spin-flip scattering and assume each layer obeys Ohm's law. In a magnetic layer, define positive channel electrical resistances $a,b>0$. Let the spacer contribute $r\ge 0$ to each channel. These are circuit parameters, not measured iron or chromium resistivities.

For parallel layer magnetizations, one channel meets $a,r,a$ and the other $b,r,b$. For opposite layer magnetizations, each meets one $a$, one $b$, and the spacer. The two channels share the same terminal electric potential difference and run in parallel. Writing $x\parallel y=xy/(x+y)$ givesThe inequality is algebra, given the specified network. It shows how a difference between spin-channel responses can yield a difference between magnetic configurations even though all the constituent resistors are ohmic.

For chosen $a=1\Omega$, $b=4\Omega$, $r=1\Omega$, the parallel channels have total electrical resistances $3$ and $9\Omega$, while the opposite-alignment channels each have $6\Omega$. Thus $R_P=2.25\Omega$ and $R_{AP}=3.00\Omega$. At a common terminal electric potential difference $1\mathrm{V}$, the total electric currents are $4/9\mathrm{A}$ and $1/3\mathrm{A}$, respectively.

A quoted percentage needs its denominator. Define the resistance increase relative to the low-resistance state by $m=(R_{AP}-R_P)/R_P$. The decrease relative to the high-resistance state is instead $d=(R_{AP}-R_P)/R_{AP}$. For $R_P,R_{AP}>0$,The chosen example has $m=1/3$ and $d=1/4$: a $33.3\%$ increase and a $25\%$ decrease describe the same pair of resistances. A paper reporting a field-induced drop relative to its zero-field value need not be using the same percentage convention as a plot normalized by $R_P$.

Define a dimensionless channel contrast $p=(b-a)/(a+b)$ and spacer ratio $s=r/(a+b)$. Positive $a,b$ give $|p|<1$, and substitution gives the displayed expression. Equal channel responses, $p=0$, eliminate the model's magnetic-configuration signal. Increasing the common spacer resistance at fixed $a,b$ increases the denominator and decreases $m$.

This is a controlled comparison within the model. In an actual multilayer, changing a spacer thickness can change the magnetic coupling and the interfaces as well as electrical transport. Spin-flip scattering also mixes the channels that the calculation kept independent.

Let the first layer have channel resistances $a_1,b_1>0$ and the second $a_2,b_2>0$, with a common spacer $r\ge 0$. Define $\Sigma =a_1+b_1+a_2+b_2+2r$. The same circuit rules giveSubtracting their numerators gives $(a_1-b_1)(a_2-b_2)$. The sign therefore depends on whether the two layers favor the same global spin channel. With $(a_1,b_1)=(1,4)\Omega$, $(a_2,b_2)=(4,1)\Omega$ and $r=1\Omega$, the values reverse: $R_P=3.00\Omega$, $R_{AP}=2.25\Omega$. The model's ratio $m$ is then $-1/4$. Positive magnetoresistance was a consequence of the identical-layer assumptions, not a theorem about every two-channel network.