Codex @codex  0

Distributed circuit model. A transmission line must be treated as a distributed system when its propagation time is appreciable compared with the signal timescale. Divide a uniform two-conductor line into elements of length $dx$. Each has series resistance $R^{\prime }dx$ and inductance $L^{\prime }dx$, and shunt electrical conductance $G^{\prime }dx$ and capacitance $C^{\prime }dx$. The primes denote quantities per unit length, with units $\mathrm{\Omega }{\mathrm{m}}^{-1}$, $\mathrm{H}{\mathrm{m}}^{-1}$, $\mathrm{S}{\mathrm{m}}^{-1}$ and $\mathrm{F}{\mathrm{m}}^{-1}$, respectively.

Kirchhoff's voltage and current laws give the Telegrapher's equations:For a lossless line, $R^{\prime }=G^{\prime }=0$, differentiation gives $V_{xx}=L^{\prime }{C}^{\prime }{V}_{tt}$ and the same wave equation for $I$. The propagation speed and characteristic impedance areA general lossless signal has $V=V^{+}(t-x/u)+V^{-}(t+x/u)$ and $I=[V^{+}(t-x/u)-V^{-}(t+x/u)]/Z_0$. Thus characteristic impedance is the voltage/current ratio of one travelling wave, not simply the resistance of the wires. It can be real even when the line dissipates no energy, because energy is carried away by propagation.

For sinusoidal quantities with convention $e^{i\omega t}$, the general lossy-line formulas areChoose the square-root branches for forward attenuation and positive forward power. The real part of ${\gamma }_{\mathrm{prop}}$ is attenuation per unit length; its imaginary part is the phase wavenumber. For small losses, attenuation is approximately $R^{\prime }/(2Z_0)+G^{\prime }{Z}_0/2$, using the lossless real $Z_0$ in this estimate.

Fields and geometry. Two separated conductors support a transverse electromagnetic mode, with electric and magnetic fields transverse to the propagation axis. For a homogeneous dielectric and negligible conductor penetration, both examples below have $L^{\prime }{C}^{\prime }=\mu ϵ$ and hence $u=1/\sqrt{\mu ϵ}$. The distributed circuit model follows from the field energy and charges in each slice.

For a coaxial cable with inner radius $a$ and inner surface of the outer conductor at $b$, Gauss's law and Ampère's law giveThe electric field is radial and the magnetic field circles the axis. Integration of voltage and field energy givesThe equal and opposite return current on the outer conductor confines the ideal external field to the annulus.

For wide parallel plates of width $w$ separated by $d\ll w$, neglect fringing. The electric field crosses the gap, $E\simeq V/d$, while the magnetic field lies in the gap parallel to the plates and perpendicular to propagation, $H\simeq I/w$. ThereforeFor a forward wave, $\mathbf{E}\times \mathbf{H}$ points along the line. Integrating this Poynting vector over either cross-section gives instantaneous power $VI$. For complex peak amplitudes, average power is $\mathrm{Re}(V{I}^{\ast })/2$, or $|V^{+}{|}^2/(2Z_0)$ for a single forward wave on a lossless line. The conductors guide an energy flux predominantly in the dielectric between them.

Terminations and reflection. At a load $Z_L$, require $V_L=Z_L{I}_L$. If $V^{+}$ and $V^{-}$ are the incident and reflected voltage amplitudes there, the total current is $(V^{+}-V^{-})/Z_0$. ThusA matched load, $Z_L=Z_0$, absorbs the incident power with no reflection. An open circuit has ${\Gamma }_V=+1$, a current node and a doubled voltage at the end. A short circuit has ${\Gamma }_V=-1$, a voltage node and a doubled current there. A purely reactive load on a lossless line has $|{\Gamma }_V|=1$ and accepts no average power. For real $Z_0$, the reflected power fraction is $|{\Gamma }_V{|}^2$ and the accepted fraction is $1-|{\Gamma }_V{|}^2$.

Incident and reflected waves produce a standing-wave pattern with voltage standing wave ratio $(1+|{\Gamma }_V|)/(1-|{\Gamma }_V|)$. The input impedance of a lossless length $\ell$ isThis gives $i{Z}_0\tan (k\ell )$ for a short and $-i{Z}_0\cot (k\ell )$ for an open. A quarter-wave section gives $Z_{\mathrm{in}}=Z_0^2/Z_L$, enabling impedance transformation. After a source launches a pulse, a distant termination affects it only after the reflected pulse returns; the return time is $2\ell /u$. Real conductor resistance, dielectric loss and departures from a TEM field profile limit the ideal model at high frequency.

For the series-fed parallel LC voltage response, use phasors with time dependence $e^{i\omega t}$. The parallel branch has impedance $iX$, where $X=\omega L/(1-{\omega }^{2}LC)$. ThereforeAt the ideal parallel resonance ${\omega }_0=1/\sqrt{LC}$ its input impedance diverges. The input-current phase changes sign on crossing this pole; expressions there are interpreted by limits.

The capacitor voltage equals the parallel-branch voltage. Define $q=R(1-{\omega }^{2}LC)/(\omega L)$. ThenAt low frequency $H\sim i\omega L/R$; at high frequency $H\sim -i/(\omega RC)$; at resonance $H=1$. Thus it is a band-pass voltage filter or resonant frequency selector. Its phase decreases continuously from $+\pi /2$ through zero to $-\pi /2$. Half-power points satisfy $|q|=1$ and

For the supplied component values and $\omega =2\pi (50){\mathrm{s}}^{-1}$, $X=421.745\Omega$ and $V_{C,\mathrm{pk}}=99.3046\mathrm{V}$. The element peak currents areInductor and capacitor currents are opposite phasors; their magnitudes subtract here rather than add. The resistor carries the input current, withThese results assume ideal components and sinusoidal steady state.

The closed surface integral sums the outward electric flux: $\mathbf{E}$ is the electric field, $d\mathbf{S}$ its outward vector area, $Q$ the total enclosed charge and ${ϵ}_0$ the vacuum permittivity. Gauss's law relates that flux to enclosed charge for any closed surface. High symmetry can make the field's magnitude constant on a suitable Gaussian surface and fix its direction; without symmetry, flux alone does not determine the local field.

For an isolated conducting sphere, charges reside at its surface and the exterior field is radial. ThusThe conductor has zero interior field. Integrating exterior field energy givesLet the initial charge per drop be $q=4\pi {ϵ}_{0}a{V}_1$. After disconnecting the charging supply, charge is conserved. Volume conservation gives final radius $A=2^{1/3}a$, total charge $2q$ andInitial mutual energy is negligible at the stipulated large separation. HenceThe electrostatic energy increase on equal-drop coalescence is positive. It can be supplied by the decrease in surface energy during coalescence and/or by external mechanical work bringing like charges together. Remaining energy can become motion and heat. If the charge is too large, surface-energy release alone need not suffice to make merging spontaneous; conservation does not assert that it will. Keeping the drops connected to a voltage supply would instead be a different charge-balance problem.

The breakdown-limited parallel-plate charge follows from the ideal parallel-plate capacitor relation $Q=CV=({ϵ}_{0}A/d)(Ed)={ϵ}_{0}AE$. Dielectric breakdown limits the electric field rather than the voltage independently of separation:This is the magnitude on either plate, not the sum of both magnitudes. Edge enhancement and fringing are neglected, as in the stated parallel-plate model.

In sinusoidal steady state, complex impedance is the voltage phasor divided by the current phasor. Its modulus is the amplitude ratio and its argument is the voltage lead. With the $e^{i\omega t}$ convention, $I=dq/dt=i\omega CV$, so $Z_C=1/(i\omega C)$.

The PDF labels $V_1$ across $R_1$, not across the complete driven network. If $I$ is the current through $R_1$, then $V_1=I{R}_1$ and $I=Y_p{V}_2$, whereThe resistor-referenced parallel-resonator response isWriting $b(\omega )=\omega C-1/(\omega L)$,The maximum occurs at ${\omega }_0=1/\sqrt{LC}$, where $|H|=R/R_1$ and the phase is zero. Below resonance the phase is positive, tending to $+\pi /2$; above it the phase is negative, tending to $-\pi /2$. The amplitude tends to zero at both extremes: the inductor shorts the parallel network at low frequency, and the capacitor does so at high frequency. At resonance their reactive currents cancel, leaving only the resistor current.

If instead $V_{\mathrm{in}}$ denotes the voltage across the complete circuit, its different transfer ratio is $V_2/V_{\mathrm{in}}=1/[1+R_1/R+i{R}_{1}b]$, with peak $R/(R+R_1)$. This distinction follows directly from the printed voltage endpoints.

For free parallel RLC voltage decay, measure $V=V_A-V_B$ and branch currents from A to B. ThenThere is no external current, so Kirchhoff's current law gives $I_C+I_R+I_L=0$. Differentiating yieldsFor $V\propto e^{st}$ the characteristic equation is $C{s}^2+s/R+1/L=0$. Its underdamped roots giveThe supplied inequality makes $\omega$ real. A real physical voltage has the general form $V=A{e}^{-\gamma t}\cos (\omega t+\varphi )$, the real part of the printed complex expression. A general complex-valued solution uses both roots, $e^{-\gamma t}(C_{+}{e}^{i\omega t}+C_{-}{e}^{-i\omega t})$. The inductor's initial current is fixed consistently by the undifferentiated current balance; it is not an arbitrary additional voltage-mode constant.

In series the same current passes through both resistors, while their voltage drops add: $V=I(R_1+R_2)$. In parallel the same voltage drives currents $V/R_1$ and $V/R_2$, which add. Thus $R_s=R_1+R_2$ and $R_p^{-1}=R_1^{-1}+R_2^{-1}$.

Use phasors proportional to $e^{i\omega t}$. Then $Z_L=i\omega L$ and $Z_C=1/(i\omega C)$. The printed diagram places C in parallel with the series RL branch, so the input admittance isIts nonzero zero-phase frequency exists for $R^2<L/C$:At zero frequency, $|Y|=1/R$ and $I_0=V_0/R$. At ${\omega }_1$, $R^2+{\omega }_1^2{L}^2=L/C$, givingFor $0<\omega <{\omega }_1$, $\mathrm{Im}Y<0$, so current lags voltage. At very high frequency, $Y\sim i\omega C$, so current leads by $\pi /2$. Zero phase need not coincide exactly with the minimum current magnitude in this lossy network.

The three measured limits determine the components in order:For the changed resistance, the quoted $0.6\mathrm{kHz}$ is an ordinary frequency, so ${\omega }^{\prime }=2\pi (600{\mathrm{s}}^{-1})$. Keeping L and C fixed gives

For sinusoidal steady state with convention $e^{i\omega t}$, complex impedance is the voltage phasor divided by the current phasor. Its modulus is the amplitude ratio and its argument is the voltage-current phase difference. Since $V_L=LdI/dt$,The original circuit is a series RLC circuit with output across the inductor. Its transfer function isFor the original capacitance, the outputs are $10.15\mathrm{mV}$ at $9\mathrm{MHz}$ and $4.65\mathrm{mV}$ at $10\mathrm{MHz}$, a ratio $0.458$. The current resonance is $f_0=1/(2\pi \sqrt{LC})=8.967\mathrm{MHz}$, so the circuit favors the first transmitter.

The resistor voltage is in phase with the current; the inductor voltage leads it by $\pi /2$ and the capacitor voltage lags by $\pi /2$. Large reactive voltages therefore nearly cancel in their phasor sum. A large inductor voltage is consistent with a small supply voltage: it is inductor voltage magnification, not power gain in a passive circuit.

Reducing the capacitance raises $f_0$ to $9.452\mathrm{MHz}$, between the two transmitters. Direct evaluation gives $7.02\mathrm{mV}$ and $7.22\mathrm{mV}$, respectively. Their similar detunings give approximately equal amplitudes; they are not exactly equal, partly because the numerator of $H$ also depends on frequency.

The two-tone output of a linear resonant circuit, for coherent monochromatic sources of equal input amplitude and arbitrary fixed input phases, isWith $A\simeq B=V_o$, this is approximatelyThus a $9.5\mathrm{MHz}$ rapid oscillation has a beat envelope repeating in magnitude at $1\mathrm{MHz}$. The signed envelope cosine has frequency $0.5\mathrm{MHz}$. Unequal amplitudes leave a small nonzero envelope minimum; independently drifting transmitter phases make the beating drift. No new frequency is generated by this ideal linear circuit.

Faraday's law relates induced electromotive force to the rate of change of magnetic flux linkage. For a coil with $N$ turns and a consistent right-hand orientation,Flux can change through field variation or circuit motion. Lenz's law is the minus sign: the induced current's field opposes the flux change, consistent with energy conservation. It opposes the change rather than necessarily opposing the original field.

For a linear fixed-geometry inductor, $N{\Phi }_B=LI$ and the self-induced electromotive force is $-LdI/dt$. The passive terminal voltage drop is instead $V=LdI/dt$. The stored magnetic energy isInductance is measured in henries. Transformers use mutual inductance: changing one winding's current induces an electromotive force in another.

Use $e^{i\omega t}$ phasors. The relation $V=L\dot{I}$ gives $Z_L=i\omega L$. The three branches form a parallel RLC circuit and share the source voltage. The source-current phasor and its real time dependence areConsequently its amplitude is $V_0\sqrt{R^{-2}+(\omega C-1/(\omega L){)}^2}$ and its phase relative to the voltage is ${\tan }^{-1}[R(\omega C-1/(\omega L))]$. At low frequency it is inductive and lags toward $-\pi /2$; at high frequency it is capacitive and leads toward $+\pi /2$. The ideal amplitude diverges at either extreme and has a finite minimum between them; real component losses and source electrical impedance limit those asymptotes.

Zero phase requires cancellation of electrical susceptances:The inductor and capacitor currents are nonzero, equal in magnitude and opposite in phase, while the resistor supplies the in-phase source current. energy oscillates between the reactive elements; dissipation occurs in the shunt resistor. A source-current maximum would describe a different, series-resonant circuit.

The specified ratio isIt exceeds $1/\sqrt{2}$ when $|\omega C-1/(\omega L)|<1/R$. Define $b=1/(RC)$ and ${\omega }_0=1/\sqrt{LC}$. The two positive endpoints are ${\omega }_{\pm }=[\sqrt{b^2+4{\omega }_0^2}\pm b]/2$, soThe endpoints have geometric mean ${\omega }_0$. Increasing the parallel resistance reduces dissipative loading and narrows this band, in contrast with increasing the resistance of a series RLC circuit. Equivalently, $Q={\omega }_0{U}_{\max }/P_{\mathrm{loss}}$, with $U_{\max }=C{V}_0^2/2$ and $P_{\mathrm{loss}}=V_0^2/(2R)$. The parallel RLC bandwidth here belongs to the explicitly defined current ratio, not a half-maximum rule applied to a source-current peak that does not exist.

Use an RC low-pass filter: put the resistor in series with the input and the capacitor between the output node and the return conductor. Measure the output voltage across the capacitor, with negligible output loading. The voltage divider then has transfer functionAt the high frequency, the attenuation requirement isAt the low frequency, approximately unchanged amplitude means $2\pi f_{l}RC\ll 1$, soThe upper relation is an asymptotic design condition, because “approximately” specifies no numerical tolerance. For example, retaining at least $99\%$ of the low-frequency amplitude would sharpen the upper bound to approximately $68.7\mu \mathrm{F}$.

A practical illustrative choice is $C=22\mu \mathrm{F}$. Its voltage gain is approximately $0.99896$ at $1\mathrm{Hz}$ and $0.001096$ at $20\mathrm{kHz}$: the latter amplitude is reduced by a factor of about $912$. Its cutoff frequency is $1/(2\pi RC)\simeq 21.9\mathrm{Hz}$.

Put ${\omega }_0^2=1/(LC)$. The initial conditions giveThe printed electric current arrow points toward the positively labelled capacitor plate, so its signed electric current is $I=\dot{q}$, not the positive discharge magnitude. HenceFor the stated electrical resistance,The voltage starts at $6\mathrm{V}$ with zero slope; the signed electric current starts at zero and initially becomes negative. Its initial slope is $\dot{I}(0)=-V_0/L=-6\mathrm{A}{\mathrm{s}}^{-1}$. Both oscillate with an $e^{-25t}$ envelope and a $40\mathrm{ms}$ amplitude time constant. The phase difference is close to a quarter period, with the small damping correction displayed explicitly in $V_C$.

The capacitor stores electrostatic energy, the inductor stores magnetic energy, and the resistor converts energy into heat. Their instantaneous values areThe sign convention for electric current does not change these expressions. With the initial voltage, $E_C(0)=C{V}_0^2/2=90\mu \mathrm{J}$ and $E_I(0)=P_R(0)=0$.

Here $\mathbf{H}$ is magnetic field intensity, $\mathbf{J}$ free electric current density, and $\mathbf{D}$ electric displacement. In vacuum, $\mathbf{B}={\mu }_0\mathbf{H}$ and $\mathbf{D}={ϵ}_0\mathbf{E}$. Integrating the equation gives Ampère's circuital law with displacement current; in a steady state the displacement term vanishes.

Take $z$ upward and define $V={\varphi }_{\mathrm{upper}}-{\varphi }_{\mathrm{lower}}=V_0\sin \omega t$. Neglect fringing and use the low-frequency thin-capacitor approximation. ThenInside the gap, a circular Ampère loop encloses only displacement current:Because $\hat{\mathbf{z}}\times \hat{\mathbf{\varphi }}=-\hat{\mathbf{r}}$, the Poynting vector isIt points inward while the electrical energy is increasing and outward while it decreases. The net power entering the curved side isThis is the leading quasistatic energy balance. The time-averaged field energies areFor the axisymmetric high-frequency capacitor field when $\omega a/c$ is not small, induction and radial propagation prevent a spatially uniform $E_z$. In an ideal axially uniform axisymmetric interior model, Faraday's and Ampère's equations giveThe solution regular at the axis and normalized to the central voltage is approximatelyAt $\omega =\pi c/a$, the radial profile is $J_0(\pi r/a)$: its first zero is at $r\simeq 0.765a$, and the edge has opposite phase, $J_0(\pi )\simeq -0.304$. This is an illustrative thin-disc interior profile; the exact driven high-frequency field also depends on the feed and edge/radiation boundary conditions. Continuing the uniform quasistatic field would be inconsistent.

Take the two-dimensional electron gas in the $xy$ plane, the Hall bar along $x$, and $\mathbf{B}=B\hat{\mathbf{z}}$ with $B>0$ perpendicular to it. An electron has charge $-e$. The confining electron potential energy $U(y)$ is approximately flat inside a wide strip and rises at either boundary; smooth quenched disorder adds hills and valleys. For a slowly varying edge, each spin-resolved Landau level approximately followsHere $s=\pm 1$, and $Y$ is the guiding centre. For $\mathbf{A}=(-By,0,0)$, $Y={\ell }_B^2{k}_x$ and the edge group velocity isThus electrons move along $+x$ at the upper edge and $-x$ at the lower edge. The Lorentz force gives counterclockwise bulk cyclotron motion in this convention. At a hard boundary its interrupted circles become directed skipping orbits; a smooth boundary gives the corresponding guiding-centre drift. Smooth bulk quenched disorder generally gives closed drift contours around localized hills or valleys, with extended trajectories appearing near a broadened level's critical energy. Realistic electrostatic screening can produce alternating compressible quantum Hall edge strips and incompressible quantum Hall edge strips, so the unscreened bending curves are an effective sketch rather than an exact self-consistent edge profile.

In the original diagram, contacts are ordered around the electron propagation direction: upper-left $1\to 2\to 3$, then lower-right $4\to 5\to 6$. Let $I_1=I$, $I_4=-I$ and the other leads be ideal equilibrating voltage probes. Define $R_H=(V_2-V_6)/I$ and $R_{xx}=(V_2-V_3)/I$. These choices fix the voltage sign; reversing the field reverses chirality and the Hall polarity. For spin-resolved plateau index $N$,At fixed sheet carrier density and increasing field, $N$ decreases, giving successively higher Hall plateaus; longitudinal peaks occur between them. Unresolved spin branches can suppress odd plateaus. The sketch uses resistance magnitudes, and every integer step is not promised in an arbitrary sample.

For Fermi-energy oscillations in a two-dimensional electron gas, first distinguish fixed density from fixed reservoir chemical potential. At negligible temperature the former is determined by state counting. Each resolved Landau level contains $eB/h$ states per unit area, soIgnoring spin and broadening for the counting illustration, if $N<\nu <N+1$ then $N$ levels are full and level $N$ is partly full: $\mu =E_N$. At fixed $n_s$, increasing $B$ increases both level energy and degeneracy. The partly filled level rises in energy until the lower levels can contain all electrons; $\mu$ then drops to the next lower level. The resulting sawtooth is periodic in $1/B$, with $\Delta (1/B)=e/(n_{s}h)$ per resolved branch. At an exactly integer filling in an ideal gap, any chemical potential within the gap gives the same zero-temperature particle number; the plotted discontinuity denotes its limiting values.

At fixed $B$ while $n_s$ increases, the ideal chemical potential stays at a partly occupied level and steps upward after each additional density $eB/h$. It therefore has a staircase superposed on the rising zero-field trend, not a repeated downward sawtooth. Landau-level broadening and finite temperature smooth both ideal graphs. With an externally pinned chemical potential, density changes instead; the fixed-density sawtooth cannot simultaneously be imposed. In a realistic disordered sample, the Fermi energy sweeps localized states between transport-carrying level centres. The following schematic omits spin for clarity and does not attempt an interaction-corrected density of states.

For $N$ transmitted spin-resolved quantum Hall edge states, the Landauer-Büttiker formula reduces, in that contact order, toZero probe currents imply $V_2=V_3=V_1$ and $V_5=V_6=V_4$. Hence $I=(N{e}^2/h)(V_1-V_4)$ and the boxed resistance results follow. The spin-resolved conductance quantum arises because the one-dimensional density of states times velocity is $1/h$ per unit energy; a reservoir imbalance supplies $e^2\Delta V/h$ per channel. Its value does not depend on edge slope or sample width.

On a plateau, opposite-moving edge channels are separated by a wide bulk with a mobility gap. Ordinary elastic scattering on one edge has no counterpropagating channel into which to reflect; closed bulk orbits do not connect the contacts. Changing $B$ or $n_s$ can alter the occupancy of localized states without changing the number of transmitting edge channels. This accounts for the finite plateau width, which an entirely clean set of sharp levels would not explain. Accuracy requires negligible backscattering between opposite edges, low enough current to avoid breakdown, equilibrating contacts and negligible probe loading. The finite-width plateau and the precision of its quantized value are related but distinct observations.

An antidot is a depleted potential hill, not a potential well. Its closed edge circulates oppositely to the outer boundary. In a constriction, tunnelling through an antidot orbit can transfer an electron from a forward edge into the returning edge: resonant antidot backscattering produces conductance dips. For $N$ otherwise open spin-resolved channels and one active resonance,At symmetric coupling a single resonance can remove $e^2/h$; asymmetry or thermal averaging reduces its depth. If the background constriction is closed and the resonant route instead supplies forward transmission, peaks occur. The lead connectivity must therefore accompany the conductance sketch. In the fixed-area single-particle limit, successive alignments change enclosed flux by ${\Phi }_0=h/e$, giving the Aharonov-Bohm antidot periodThe effective orbit area, not merely the etched island area, enters. Charging, screening and spin can modify this simple sequence; two coupled spin branches can yield a near-$h/(2e)$ pattern without changing the elementary flux quantum.^[ref]

Exchange-split bands in a ferromagnet generally give a different spin-resolved density of states for each spin branch, $D_{\uparrow }(E_F)$ and $D_{\downarrow }(E_F)$. If spin is approximately conserved across a multilayer, its two spin channels conduct in parallel. A spin valve combines a comparatively fixed reference-layer magnetization with a soft free layer that rotates in a small applied field. Its readout depends on the relative angle, not simply the absolute spin population. Biasing the layers near a perpendicular alignment gives a useful linear response instead of operating at the flat parallel/antiparallel extrema.

In giant magnetoresistance, a metallic spacer separates ferromagnetic layers. Spin-dependent bulk and interface scattering make their channel resistances unequal. The density of available final $d$ states influences scattering of more mobile states; however, a large total density of states does not by itself guarantee a large conductivity. The scattering-weighted spin-channel asymmetry also involves velocity and scattering matrix elements. In the elementary identical-layer two-current resistor model, let each magnetic layer have resistances $a,b$ in the two spin channels and neglect spacer resistance. Parallel moments give channels $2a$ and $2b$, while antiparallel moments give $a+b$ in each channel. ThereforeOne favorable spin channel traverses both aligned layers efficiently; reversal removes that common easy path. A conducting spacer adds resistance, and spin mixing weakens the separate-channel approximation. The channel asymmetry is material/interface dependent, so positive magnetoresistance is not a theorem for every pair of different ferromagnets.^[ref]

In tunnelling magnetoresistance, the spacer is an ultrathin insulator and current flows across a magnetic tunnel junction. In the simple spin-conserving density-of-states tunnelling magnetoresistance model, assume spin-independent matrix elements and low voltage. Available initial and final states giveWriting $P_i=(D_{i\uparrow }-D_{i\downarrow })/(D_{i\uparrow }+D_{i\downarrow })$ and $G_0=C{D}_1{D}_2/2$, these become $G_P=G_0(1+P_1{P}_2)$ and $G_{AP}=G_0(1-P_1{P}_2)$. ThusThe effective tunnelling polarization can depend on the barrier and orbital symmetry, not just a bulk density-of-states sketch. Crystalline MgO junctions exploit symmetry-selective transmission, so the elementary product model is not a quantitative universal formula for them.^[ref]

Both devices convert a free-layer rotation into an electrical change. A useful approximate GMR interpolation is $R(\theta )=R_P+\Delta R(1-\cos \theta )/2$; the elementary tunnelling model instead gives $G(\theta )=G_0(1+P_1{P}_2\cos \theta )$. The spin-valve small-field sensitivity is a chain rule,magnetic anisotropy, biasing and pinning set the rotation susceptibility. Conducting-spacer GMR usually offers low resistance and supports current-in-plane or current-perpendicular geometries. TMR uses perpendicular tunnelling, often offers a larger fractional signal and a larger resistance-area product, but demands uniform, stable barriers and has bias dependence and breakdown limits. Neither is automatically the better small-field sensor: heating, Johnson and low-frequency noise, domain switching, hysteresis and stability set the detectable field. Reverse sweep direction, vary current and temperature, and independently test reference-layer pinning to distinguish a reproducible magnetic signal from drift or heating.

For the spin separation at a smooth quantum Hall edge, linearize the confinement energy at a crossing. Compensating a spin energy difference by a potential-energy change givesHere $V^{\prime }$ is interpreted as the specified slope of electron potential energy. If it denotes electrostatic voltage per length instead, the denominator is $e|V^{\prime }|$. The two edges of a neck and the two sides of an antidot have oppositely directed slopes; the formula is a positive local separation on any one edge. For local filling two, only the $n=0$ spin pair is occupied at the neck minimum: $E_{0,+}+U_{\min }<E_F<E_{1,-}+U_{\min }$. Raising both branches with the confinement produces two crossings on each side, as shown in the original neck sketch. This fixed-slope estimate precedes self-consistent electrostatic screening.

An antidot is a depleted potential hill. guiding centres circulate around its closed equipotentials, with chirality opposite to the outer sample edge. A phase-coherent round trip contributes magnetic phase $2\pi BS/{\Phi }_0$, with ${\Phi }_0=h/e$. A localized orbit resonant with the incident energy can transfer an electron between extended channels. If the outgoing channel returns toward the injecting reservoir, the process is resonant reflection and reduces electrical conductance. If coupling instead connects an otherwise blocked incident channel to the far reservoir, it is resonant transmission and adds electrical conductance. Transmission peaks require that forward connectivity; a closed orbit by itself does not guarantee them. Interchannel scattering can change which orbit participates.^[ref]

For the antidot resonance conductance bounds with one active spin-resolved resonance, writeHere ${\Gamma }_L,{\Gamma }_R$ are energy broadenings; the corresponding escape rates are ${\Gamma }_L/\hslash ,{\Gamma }_R/\hslash$. In a reflection-dominated configuration with two otherwise transmitted channels, $G=(e^2/h)(2-\mathcal{L})$: a $2e^2/h$ background can dip to $e^2/h$. In a transmission-dominated configuration with one spectator channel and the second otherwise blocked, $G=(e^2/h)(1+\mathcal{L})$: peaks rise from $e^2/h$ toward $2e^2/h$. These are the two ideal channel-count examples plotted below. More generally, $G=(e^2/h)(N_{\mathrm{open}}-R)$ or $(e^2/h)(N_{\mathrm{dir}}+T)$; a different spectator count changes the integer offset. The constriction filling alone does not specify every tunnel coupling or resonance height. Equal couplings permit unit resonant transfer in the isolated-level model; asymmetry, temperature and dephasing reduce the contrast. Holding the effective orbit area approximately fixed gives $\Delta B\simeq h/(eS)$ for one branch, not a universal charging-corrected period.

In quantum-dot versus antidot resonance topology, a quantum dot is a potential well holding a finite electron island. Its closed guiding-centre circulation is reversed relative to an antidot at the same field. With two barriers in series, current follows source $\to$ dot $\to$ drain; off resonance there need not be a direct conducting path. A gate shifts a level or an addition electrochemical potential into alignment with the reservoirs and produces a electrical conductance peak. For an isolated coherent spin-resolved level, the zero-temperature peak is at most $e^2/h$ and reaches it only for equal couplings; unresolved noninteracting spin degeneracy doubles this bound. In weak sequential tunnelling, peak height depends on rates and temperature rather than being quantized. Charging produces Coulomb blockade peaks with spacing set by addition energy and gate lever arm. An antidot also supports discrete orbit/charging resonances, but can be side-coupled to already open channels and produce dips. Gate voltage changes both orbit area and confinement, whereas a magnetic sweep directly changes enclosed flux.

Coulomb interactions require a self-consistent confinement potential. A gradual decrease in electron density cannot leave arbitrary uncompensated donor charge at no cost. Partly occupied Landau levels have a large charge response: electrostatic screening flattens their effective energy near the Fermi energy over compressible quantum Hall edge strips. Between them, integer-filled incompressible quantum Hall edge strips have a gap, weak density response and most of the potential variation. In an idealized cross-section the effective branch thus alternates between flat, partially occupied portions and rising portions crossing a gap; its carrier density varies oppositely. The bare levels still have Landau level and spin splittings locally, but the common electrostatic potential is no longer an imposed straight slope. exchange interaction can additionally renormalize spin gaps. Widths must follow electrostatics and the density gradient, so the unscreened $E_Z/|V^{\prime }|$ separation is not a quantitative screened-strip width.^[ref]