Table Of ContentA dc voltage step-up transformer based on a bi-layer ν = 1 quantum Hall system
B. I. Halperin1, Ady Stern2, S. M. Girvin3
1Department of Physics, Harvard University, Cambridge, MA 02138
2Department of Condensed Matter Physics, Weizmann Institute of Science, Rehovot 76100, Israel
3 Sloane Physics Laboratory, Yale University, P.O. Box 208120, New Haven, CT 06520-8120
3
0 Abilayerelectron systeminastrongmagneticfieldatlowtemperatures,withtotalLandaulevel
0 fillingfactorν =1,canenterastronglycoupledphase,knownasthe(111)phaseorthequantumHall
2 pseudospin-ferromagnet. In this phase there is a large quantized Hall drag resistivity between the
n layers. Weconsider here structures where regions of (111) phase are separated by regions in which
a one of the layers is depleted by means of a gate, and various of the regions are connected together
J by wired contacts. We note that with suitable designs, one can create a dc step-up transformer
3 where the output voltage is larger than the input, and we show how to analyze the current flows
2 and voltages in such devices.
] PACSnumbers:
l
l
a
h I. INTRODUCTION
-
s
e Following earlier theoretical predictions1,2,3,4,5, recent
m experimentshaverevealed6 auniquebehaviorofcoupled
. electronic transport in a bilayer electronic system in the
t
a quantum Hall regime, when the two layers have a total
m
Landau level filling factor ν = 1 (each layer separately
- beingatν ≈1/2). Iftheseparationbetweenthelayersis
d
sufficiently small, relative to the distance between elec-
n
trons in a layer, the system can enter a strongly coupled
o
c state at low temperatures known as the (111) or quan- FIG. 1: One version of the transformer, with N = 3 stages.
[ tum Hall pseudospin-ferromagnet phase2,8,9,10. It was Horizontalstripesindicateregionswheretheupperlayerisoc-
cupied;shading indicates regions where thelower layeris oc-
predicted for this phase, that if there is no tunneling be-
1 cupied. Areaswithbothstripesandshadinghavebothlayers
tween the layers, and a current I is driven in one of the
v occupied,withthesysteminthestronglycoupled(111)phase
2 layers(the“activelayer”),withnonetcurrentflowingin at total Landau-level-filling ν = 1. The upper layer, divided
4 the other(“passive”)layer,then the voltagedropshould into strips connected in parallel, is used as the primary. If a
4 beidenticalinthetwolayers. Inthelimitofzerotemper- currentI flowsineachstrip,thenavoltageV =NI h/e2 is
1 2 1
1 ature, this voltage drop should be purely perpendicular inducedinthesecondarylayer,providednocurrentisdrawn.
0 to the current, and equal to I h in each layer. Experi-
3 e2
ments have confirmed this quantization of the Hall drag
0
resistance with an accuracy of order 10−3. two layers, either because the barrier is too high, or be-
/
at The properties of the (111) phase reflect a novel form cause tunneling has been supressed by application of a
m of interlayer phase coherence, which may be understood parallel magnetic field.
as a kind of superfludity in the difference of the electric In 1965 Ivar Giaver realized a 1:1 dc transformer us-
-
d currents in the two layers,and which shorts out any dif- ing flux flow resistance in magnetically coupled super-
n ferencesintheelectricfieldswithinthetwolayers1,2,3,4,5. conducting layers.19 In QHE bilayers the layer coupling
o
The coherent state has a broken symmetry which leads is of electrostatic origin but the transformer action can
c
toaGoldstonecollectivemode11 andtoagiantzero-bias be viewed within the composite boson picture as arising
:
v anomaly in the interlayer tunneling spectrum12,13,14,15, fromtheflowofChern-Simonsfluxattachedtotheparti-
i
X which have both been observed experimentally in Eisen- cles. In the case of a superconductor, flux flow is mostly
stein’s group16,17. perpendicular to the current flow while in the present
r
a InthisworkweusethisequalityoftheHallvoltagebe- case the Chern-Simons flux is attached to the particles
tweenthetwolayerstoshowthataproperlyconstructed themselvesandflowswiththem, inducing avoltagedrop
bi-layer system, incorporatingregions of the (111) phase inthesecondaryperpendicularto the currentflowinthe
separated by regions where one of the layers is depleted, primary.5
may serve as a dc voltage step-up transformer.18 More The proof of concept is most simply seen in the ge-
generally, we show how to analyze the current flows and ometry of Fig. [1], where the upper active layer is used
voltages in devices made up of alternating regions con- as the primary circuit, with N primary strips in paral-
taining the (111) phase and regions where one or the lel. A time-independent voltage V will lead to a time-
1
other layer is depleted by a top or bottom gate. We as- independent current I in each of the strips, and this
1
sume throughout that there is no tunneling between the current is independent of N. The voltage across the
2
to“hotspots”,isaconsequenceoftheclassicalKirchoff’s
laws for non-uniform systems in a magnetic field.
II. RESISTANCES AND CURRENT FLOWS
Let us define R as the ratio V /I for an N-stage
N 1 1
device of the type shown in Fig. [2], when no current is
FIG. 2: Alternate version of thetransformer. Here thelower drawnfromthe secondary. InthelimitwhereN is large,
layer is used as the primary, while the upper layer, divided we can ignore end effects, and write R = NR∗, where
N
into strips connected in series, is used as thesecondary. R∗ is a constant, assuming that all intermediate stages
∗
areidenticaltoeachother. ThevalueofR canbecalcu-
latedbyconsideringaninfinite periodicsystem,withthe
passive layer, which acts as a secondary circuit, will be unit cell shown in Fig. [3]. The primary layer occupies a
V =NI h/e2whennocurrentisdrawn. Clearly,wewill region of width w running parallel to the x-axis, which
2 1
have V larger than V if N is sufficiently large. As we we label 0 < y < w. The secondary layer is depleted in
2 1
show below, the output impedance of the transformer is a short region, of length Ld, which we take here to be
non-zerosothatifanon-zerocurrentI2isdrawnfromthe the region −Ld/2 < x < Ld/2. In the remainder of the
secondary, the secondary voltage V2 will decrease, and unitcell,oftotallengthLc,bothlayersareoccupied,and
the primary voltage V will increase, for fixed I . Nev- the system is in the (111) phase. We have assumed that
1 1
ertheless, if the current drawn is sufficiently small, the the voltage tabs attached to the secondary layer shown
secondary voltage will remain larger than the primary. inFig.[2]arenarrow,sotheydonotperturbthecurrent
Another possible geometry, having the strips con- flow when no net current is drawn from the voltage con-
nectedin seriesratherthanparallel,is illustratedin Fig. tacts. Thus thereis nocurrentflowineitherlayeracross
[2]. A time-independent current I flows in the primary the boundaries at y =0 and y =w.
1
layer, denoted layer 1. This current is driven by a bat- We assume that in the parts of the sample where one
tery with voltage V . The voltage V measured in the of the layers is depleted, the other layer (whose filling
1 2
secondary circuit, when no current is drawn, is equal factor is approximately 1/2) is in a compressible state,
to NI h/e2 for an N-stage device. (Again, the voltage characterizedby a resistivity tensor ρ1 with
1
will be reduced when finite current is drawn from the
ρ1 = −ρ1 ≡ν−1 ≈2 (1)
secondary.) In this case, the voltage V1 in the primary yx xy 1
circuit is also proportional to N. However, we show be- ρ1 = ρ1 ≡ǫ≪1 (2)
xx yy
low that with proper design, the voltage drop V will be
1
smaller than V , so that voltage gain is achieved. By We have chosen the magnetic field direction along the
2
connectingtogetherseveraldevices,withthesecondaries positive z-axis, so that ρ1 is positive, and we use units
yx
in series and the primaries in parallel, one can obtain an where h/e2 =1 for intermediate steps of the calculation.
arbitrarily large multiplication factor for the voltage. Inthe coherent(111)region,wherebothlayersareoccu-
The structures we discuss are inherently non-uniform, pied,thereisnoresistivitytoaflowofananti-symmetric
since one of the layersis depleted in parts of the sample. current. For a symmetric current the Hall resistivity is
As a consequence,the analysis of currentflows and volt- quantized ρcoh = −ρcoh = 1. The diagonal resistivity
yx xy
age drops is non-trivial. Below, we shall first carry out vanishesrapidlyatlowtemperatures andwetakeit here
such an analysis for the case where no current is drawn to be a negligibly small positive quantity. We carry out
from the secondary, and then consider the case where our analysis of the current flow patterns assuming that
I 6=0. We will mostly analyze the device shown in Fig. thediagonalresistivitiesineachofthetwophasesdonot
2
[2], and discuss the device in Fig. [1] towards the end of fluctuate with position. However, our results for the op-
the paper. We confine ourselves to the situation where erationof the device as a dc voltagestep-up transformer
the dimensions of the transformerare large compared to are independent of that assumption.
anyrelevantmicroscopiclength,includingthemean-free- Let jα(r) and φα(r) be the current density and the
pathofanycharge-carriers. Wecanthenusemacroscopic potentialinlayerα,andEα =−∇φα betheelectricfield
conductivity laws and Kirchoff’s equations to determine in layer α. In the geometry of Fig. [2] no current flows
thecurrentflowsandvoltagedropsineachlayer. Itisim- in the secondary layer, i.e., j2 = 0. This is obviously
portantto distinguish the classicalandquantum aspects true in the regions where this layer is depleted, but in
ofourcalculation: thestrikingtransportpropertiesofthe fact holds also in the coherent (111) regions. In these
(111) phase, particularly the existence of a Hall voltage regions, the superfluidity shorts out any anti-symmetric
in a layer where no current is flowing, are a consequence electric fields, so E1 = E2, and the potentials in the
of the quantum Hall effect. In contrast,the non-uniform two layersdiffer only by a constant. The anti-symmetric
current distribution we find in some of the regimes we currentinthe coherentregion,j1−j2 ,isasupercurrent,
consider, and particularly the confinement of dissipation soitscurlanddivergencebothvanish20. Infact,thesame
3
L
d
V(x,y) = 0, at x= , (7)
2
result from the vanishing longitudinal resistivity at the
(111) region. The value of V is proportional to I , and
0 1
should be chosen such that
w
dyj1 =I , (8)
x 1
Z
0
where the current density is given by
ν−1−1 ǫ
j1(r)= 1 zˆ×∇V − ∇V .
ǫ2+ 1−ν−1 2 ǫ2+ 1−ν−1 2
1 1
(cid:0) (cid:1) (cid:0) (cid:1) (9)
Theintegralin(8)maybe takenatanyconvenientvalue
FIG. 3: Schematic unit cell of device shown in Fig 2 lies be- of x. Having defined the equation and boundary condi-
tween vertical dotted lines at x = ±Lc/2. Shaded region tionsforV(r),wenoticethatV(r)isthepotentialgener-
containsthe(111)phase,wherebothlayersareoccupied;un-
atedbyacapacitorsubjecttoavoltageV andtounusual
shadedregionbetweenlinesatx=±L /2hastheupperlayer 0
d boundary conditions at the edges.
depleted. Arrows suggest the flow pattern of an “extra” in-
As we now show, there are three different regimes for
homogeneous current, resulting from the y-direction current
this problem, according to the ratio of the two dimen-
inthedepletedregion. Tothismustbeaddedauniform cur-
sionless parameters in our definition of the problem, the
rent in the x-direction, so that the total current is I . The
inhomogeneouscurrentissmallcomparedtotheunifor1mcur- aspect ratio Ld, and the longitudinal resistivity ǫ. The
w
rentin regime (i),Ld/w≪ǫ,whereǫh/e2 is thelongitudinal regimes are: (i) Ld/w ≪ ǫ; (ii) ǫ ≪ Ld/w ≪ 1/ǫ and
resistivity in thedepleted region. (iii) 1/ǫ≪L /w.
d
holds for the total (symmetric) current: the divergence A. Regime (i): L /w≪ǫ
∇·(j1+j2) = 0 due to current conservation, and the d
curlvanishessince∇×E1 =0andtheresistivitiesinthe
We start with the first regime. When L /w → 0 the
(111)phaseareindependentofposition. Thusthevalues d
systemisverywide andthe edgesmaybe neglected. Far
of both j1,j2 in the coherent regions are determined by
fromthe edges j1, E1 are independent ofposition. They
the boundary conditions: their normal component must x y
arethenequaltoI /w,bothinthe(111)regionandinthe
vanish at the top and bottom edges, y = 0 and y = w, 1
depletedregion,while∂ V vanishes. Inthe (111)region,
whilethenormalcomponentofj2 mustvanishalsoatthe y
the electric field is purely perpendicular to the current,
boundaries between the coherent and depleted regions.
but this is not true in the depleted region, where the
Thus,thecurrentj2 vanisheseverywhere,andnocurrent
longitudinal resistivity is non-zero. Substituting these
flows in the secondary layer.
values of j1, E1 in the relation E=ρ1j we obtain,
The current distribution j1 in the active layer may be x y
analyzed by solving Kirchoff’s equations. It is useful to I I 1−ν−1
define a “reduced potential” V(r) by j1 = ν ǫ 1 −E1 = 1 1 , (10)
y 1(cid:20) w x(cid:21) w ǫ
∇φ1 =∇V −zˆ×j1. (3) I 1−ν−1
E1 = 1 ǫ− 1 . (11)
V(r) is the potential corresponding to a current density x w (cid:26) ν ǫ (cid:27)
1
j1(r) in a system where the coherent regions are super-
fluids for both symmetric and anti-symmetric currents, and V0 = Ex1d. These results apply throughout the de-
and where the depleted regions have a Hall resistivity of pleted region, except for small regions, with dimensions
ν1−1−1 rather than ν1−1. Within the depleted region of order Ld, close to the top and bottom edges, at y =0
andy =w. (See SubsectionIIC,below.) The integrated
∇2V =0, (4) current I1 =L j1 leaves the depleted region near y =0,
y d y
and four boundary conditions should be imposed. The and spreads out into the (111) region, where it eventu-
first two are ally flows up towards the upper edge and back into the
depleted region near y = w. This extra current flow is
ǫ∂xV +(1−ν1−1)∂yV =0, at y =0 and y =w, shown by the arrows in Fig. [3]. The integrated current
(5) flowacrossthemidliney =w/2inthe(111)region,must
which assure that the current at the edges is parallel to be exactly equal and opposite to the integrated vertical
the edges. The other two, current I1 in the depleted region, as there can be no net
y
L current flow in the y-direction. Since a current density
V(x,y) = V0, at x=− 2d , (6) jy at a point in the (111) region must be driven by a
4
Hall electric field in the x-direction, we see that E1dx
x
along the midline of a (111) region, say from thRe point
x = L /2 to the point x = L −L /2, must be equal to
d c d
I1. Adding in the contribution from the field E in the
y x
depleted region,we see that the total voltagedrop along
the midline of a unit cell, say from the point x=−L /2
d
to x=L −L /2 is equal to I R∗, with
c d 1
FIG. 4: Network model when there is no dissipation in the
L
R∗ = d (1−ν−1)2+ǫ2 . (12) interior of the sample. Lines with arrows are bonds, with
wǫ 1 orientationasdescribedinthetext. Doublelinesaremetallic
(cid:2) (cid:3)
∗ leads. NumbersincirclesdenotetheHallconductivityσxy (in
We see that R canbe made arbitrarilysmallby using a unitsofe2/h)withineachregion. Potentialsonthebonds,in
sample with a large width w, and by making the length theprimary layer 1, and on the leads, are denoted by φa,φb,
L of the depleted region as small as possible. etc. DissipationoccursatnodeslabeledAandatthecontacts
d
Sincethereisnocurrentflowacrosstheedgesaty =0 to theleads.
and y = w, we see that E1 and E2 both vanish along
x x
these edges in the (111) regions. The potentials φα are
this figure.) The boundaries between different Hall re-
therefore constants along each of these edges, in a given
gions,orbetweenaHallregionandthe vacuum,arerep-
(111) region, and the potential difference between y =0
resentedby bondsinthemodel. Eachbondisassigneda
and y = w in a given layer is just the Hall voltage, I .
1 directional arrow, oriented so that the Hall conductance
The difference in potential between layer 1 and layer 2
σ = 1/ρ of the region on the left-hand side of the
xy yx
is an arbitrary constant that has no effect on the cur-
bond is algebraically larger than the Hall conductance
rent flows or voltage drops along the layers. Thus, if the
on the right. (An insulating region is assigned a Hall
edges of the secondary layer are connected together as
conductance σ = 0.) For each bond µ, we denote the
xy
shown in Fig. [2], and no current is drawn from the sec-
absolute value of the difference in these Hall conductivi-
ondary circuit, we obtain V = NI h/e2 as claimed in
2 1 ties by σ >0. Note that if one were to reverse the sign
µ
the introduction.
of the magnetic field, the signs of the Hall conductances
wouldchange,as wouldthe directions ofthe arrows,but
σ would be unchanged.
µ
B. Regime (ii): ǫ≪Ld/w≪ǫ−1 Let φµ denote the potential φ1 on bond µ, which is a
constant along the length of the bond. In the network
Naively one might expect the analysis above to hold model, the bond carries a current, iµ = σµφµ, and an
as long as Ld ≪ w, but in fact this is not the case. As energy flux iµφµ, with positive signs denoting transport
often happens atthe interfaceofmaterials withdifferent in the direction of the arrow. Although the current flow
Hall resistivities in the regime of strong magnetic fields, in the original problem is not actually confined to the
thecurrentdistributioncanbecomeveryinhomogeneous, edges and boundaries, the details of the flow are irrel-
when the Hall angle is large, and this can have a major evant to a computation of the voltage drop along the
effect on the voltage drop (see, e.g.,21, and references edges. The voltagedifference between any two points on
therein). However, the analysis again becomes simple the boundaries of a given Hall region is completely de-
in regime (ii), where ǫ is small compared to both L /w terminedby the totalHallcurrentcrossinga line joining
d
andw/L . We canunderstandthis regimeby takingthe the two points, if the diagonal resistivity is zero. Thus
d
limit ǫ→0 withL andw fixed. In this case the bound- there is no error introduced by associating the currents
d
ary conditions (5) - (7) imply that the reduced potential with the boundaries.
V(x,y)isaconstantalongeachboundaryofthedepleted Nodes in the network, where three bonds come to-
region,with the exceptionof two “hot spots”,at corners gether,representthemeetingpointofthreeHallregions,
where two boundaries meet. At these corners, there is Current conservation requires that the current leaving a
a discontinuity in V, or more accurately, a very rapid node must equal the current entering, while energy con-
change of the potential, on a length scale of order ǫL . servation dictates that the energy flux leaving the node
d
Inthe limit ǫ→0the resultingdivergenceofthe electric be equal to or smaller than the energy entering. There
field leads to a finite totalcurrentcrossingthe boundary are two types of nodes. When there are two bonds with
between the (111) region and the depleted region in the arrowspointingintothenode(labeledµ=1,2),andone
corner, with a finite amount of dissipation. Everywhere pointing out (labeled µ = 3), then the potentials on the
other than at the corner hot spots the electric fields re- incoming bonds are arbitrary, and the potential on the
main finite, and thus there is no dissipation other than outgoing bond is determined by current conservation:
at these corners.
φ =(φ σ +φ σ )/σ , (13)
3 1 1 2 2 3
The voltage drop in regime (ii) can be calculated us-
ing a simple networkmodel, illustrated in Fig. (4). (The with σ = σ +σ . On the other hand, for a node with
3 1 2
hot spots where dissipation takes place are marked A in oneincomingbondandtwooutgoingbonds, the require-
5
mentsofcurrentandenergyconservationdictatethatthe The total current flowing in the x-direction can be
potentials on the outgoing bonds be equal to the poten- found most conveniently by evaluating the integral (8)
tial on the incoming bond. at the center line, x = 0. The first term in (16) gives
In addition to the nodes described above, we include rise to a “bulk” contribution , Ibulk = (V ǫw/L )[(1−
x 0 d
additional junctions to represent an ohmic contact with ν−1)2+ǫ2]−2 ,whichisthesameasthatobtainedearlier
1
a metallic lead. For an ideal contact, the condition is in region (i). The second term in (16) gives rise to an
that the potential on the bond leaving the contactis the “edge current”, concentrated in regions of order L near
d
sameasthevoltageinthemetalliclead. Thepotentialon the upper and lower edges, and falling off exponentially
theincomingbondisarbitrary. (Foranon-idealcontact, away from these edges. As we are considering the sit-
one may include a series resistance which leads to an uation where w ≫ L , the edge current is independent
d
additionalvoltagedropifthereisnetcurrentflowinginto of w, and simply adds to the bulk current. In the limit
or out of the lead.) ǫ≪1, the edge current is derived entirely from the first
Using the above rules, we see that the there are no term in (9) (i.e., the Hall term) and it leads to a total
potential differences among the three bonds connected contribution Iedge = V /(ν−1−1), which is the same as
x 0 1
to each of the nodes labeled B in Fig. [4]. However, the result in regime (ii).22 Thus we find, for ǫ ≪ 1 and
there are voltage drops at the nodes marked A. If the L ≪w, including the crossover region between regimes
d
net current in the x−direction is I , then we must have (i)and(ii),theresistanceR∗ ofadepletedregionisgiven
1
I = ν (φ −φ ) = (φ −φ ) = ν (φ −φ ), etc. Then by
1 1 a b c b 1 c d
we find φ −φ =φ −φ =I R∗, where
a c b d 1 1 1 ǫw
= + (17)
R∗ =(ν−1−1). (14) R∗ ν−1−1 L ((1−ν−1)2+ǫ2)
1 1 d 1
Although the edge current near the center line x = 0
is spread out over a region of height ≈L , we note that
C. Cross-over between regimes (i) and (ii) d
very close to the interfaces, at x = ±L /2, the current
d
is concentrated in a smaller interval, of height ≈ǫL , at
d
Comparing Eqs. (12) and (14) , we see that the two
the hot-spot corner.
∗
expressions for R become equal when the aspect ratio
L /w is of order ǫ, suggesting that this is indeed the
d
boundary between regimes (i) and (ii), as claimed. In D. Regime (iii) ǫ−1 ≪L /w
∗ d
fact, one can obtain an exact expression for R that is
validthroughoutthiscrossoverregime. Todothis,letus
It is clear that the length-independent expression for
consider the problem where both ǫ and the aspect ratio ∗
R given by (14) must break down, if L is sufficiently
Ld/w are very small, but with arbitrary ratio between large. When L /w >ǫ−1(regime (iii)), thde resistance of
them. It is convenient to think of L as fixed, with w d
d each stage is proportional to the longitudinal resistivity
very large and ǫ very small. As discussed above for the
of the depleted region, and is given by
case of regime (ii), the reduced potential V on the lower
edge (y = 0) will be a constant, equal to V0, except for R∗ =ǫLd/w. (18)
a small interval, of order ǫL/d near the right corner, at
Thisregimeisofnointerest,ifwewanttocreateadevice
x = L /2, while on the upper edge (y = w) we have
d ∗
with small R .
V =0 except for a small interval near the left corner, at
The crossover between regimes (ii) and (iii) may also
x=−L /2.
d
be solved analytically by considering a depleted region
We may solve Eqs. (4,6,7) by writing
wherebothǫandw/L areverysmall,butwitharbitrary
d
xV0 ratio between them. This problem may be solved by a
V(x,y) = V − (15)
0 L similar method to that used for the crossover between
d
+ A e−kny+B ekny sink (x+L /2), regimes (i) and (ii). In fact the two problems may be
n n n d relatedby aduality transformation,inwhichthe electric
Xn (cid:0) (cid:1)
fields and the currents are interchanged, and the spatial
wThheerceoneffiisciseunmtsmAend,oBvnerapreosditeitveermininteegderbsyatnhdekvnal=uesπLndof. cnooowrdiisntahteastatrheerroetsaistteadncbeyR90∗ disegthreeess.umThoef fithnealrreessuultlst
V at y =0 and y = L. We see that as long as n ≪ǫ−1, given by (14) and (18) .
one has
V
An ≈ − 0(−1)n E. Total resistance RN
πn
V
Bn ≈ − 0e−knw (16) Eqs. (12,14,18) give R∗, the resistance for each of the
πn
(N−1)intermediatedepletedregions,inthethreeaspect-
For n > ǫ−1, the coefficients fall off more rapidly than ratio regimes. To this we must add the resistance aris-
1/n. ing from the ohmic contacts to the depleted end tabs in
6
′
Fig. 2. If the lengths L of the end tabs are such that If the secondary circuit is closed by a load resistance R,
d
′
theaspectratioL /wisintheintermediateregime,large we find
d
compared to ǫ but small compared to ǫ−1, then the re- V BR
2
sistance of the end tabs may analyzedusing the network = . (23)
V AR+AD+BC
1
model, as illustratedin Fig.[4]. We readilyfind that the
combined added resistance of the two end tabs is equal
to2ν1−1−1. Thisresistanceiscomposedofν1−1,thetwo- G. Structure in Fig. [1]
terminal resistance of the depleted system, and ν−1−1,
1
the resistance associated with the interface between the
Finally we consider the structure shown in Fig. [1],
depletedand(111)parts. Thuswefindatotalresistance
which we may analyze in a manner similar to the above.
in layer 1 of
If the aspect ratios of the depleted regions are all in the
intermediate regime ǫ < L /w < ǫ−1, and we define I
R =[(N −1)R∗+(2ν−1−1)]h ≈[(N −1)R∗+3]h as the current in each primdary strip, then we obtain th1e
N 1 e2 e2 following results for the constants in Eqs. (20-23): A ≈
(19)
3h/e2,C = h/e2,B = Nh/e2,D ≈ (N + 2)h/e2. In
[Havingarrivedatthe presentationofourfinalformulas,
this case we obtain a voltage ratio V /V ≈ N/3, which
wenowrestorethefactorofh/e2.]Foradevicecontaining 2 1
exceeds unity provided N ≥ 4, For large N, the output
twovoltagetabsinregime(ii)andone(111)region(N =
impedance is lowerthanthat ofthe geometryin Fig.[2].
1), this gives a resistance R ≈3h/e2.
1
We note then that in order to have V > V , with the
2 1
device in Fig. [2], we must choose the aspect ratio L /w
d III. ADDITIONAL REMARKS
of the intermediate depleted regions to be smaller than
ǫ. For technical reasons, it may be difficult to attach an
′ Throughout our analysis, we have assumed that the
ohmiccontacttoaveryshortendtabwhoselengthL is
d dimensions of the system are large enough for us to use
smallerthanǫw. Ontheotherhand,itshouldbepossible
macroscopic constitutive relations for the current and
to fabricate a device where the lengths L of the inter-
d voltage in each region. For the geometry of Fig. 2, the
mediatedepletedregionsareveryshort,byusingnarrow
∗ most critical requirement for our analysis is that the
wiresastopgates. AlthoughtheadditionalresistanceR
length L of the depleted regionmust be largerthan the
for each intermediate stage is small compared to h/e2 in d
meanfreepathofquasiparticlesinthesinglelayerphase,
this case,the totalresistance cannotbe smaller thanthe
sothatamacroscopicresistivitymaybeused. Ingeneral,
value ≈ 3h/e2 for a single stage device. From (19), we
for a large but finite system, there will be corrections to
seethatinordertohavealargervoltageinthesecondary
the macroscopic equations arising from the boundaries
thanintheprimary,evenforverysmallvaluesofL /ǫw,
d between regions which could either increase or decrease
the number of stages N must be ≥4.
thetotalresistance. However,theboundarycontribution
to the resistance should decrease as the reciprocalof the
length of the boundary, so that the boundary contribu-
F. Effect of finite current in secondary tion becomes negligible in the macroscopic limit.
We now consider what happens if there is a finite cur-
IV. CONCLUSIONS
rent I drawn from the secondary. The linearity of the
2
circuit means that we can write
Insummary,inthispaperwehaveconsideredtwopos-
sible geometriesfor a quantum Hallbilayer device which
V = AI +CI (20)
1 1 2
should be able to act as a dc transformer with voltage
V = BI −DI (21)
2 1 2 gain,and analyzedthe currentflow patterns in these ge-
ometries. In particular, our analysis permits us to cal-
where A = RN and B = Nh/e2 as given above, and C culate the voltage differences between any two points on
and D are constants to be determined now. If we set the edge of the sample, in either layer, given the total
I1 = 0 with I2 6= 0, we have effectively interchanged the current flow in each layer. The methods we have used
roles of the primary and secondary layers. We see from for this analysis are applicable more generally, to com-
this that C = Nh/e2. Also, if the tabs for the contacts posite systems where all components of the systems are
to layer 2 have an aspect ratio between ǫ and ǫ−1, we characterized by a longitudinal resistivity much smaller
see that D will be N times the resistance of the primary thantheHallresistivity. Finally,experimentstotestour
layer of a single stage device: D ≈3Nh/e2. The output proposals for a dc voltage step-up transformer, and to
impedance of the secondary is appropriately defined as measure the voltage drops in various geometries, would
helpstrengthenourunderstandingofthenovelinterlayer
∂V2 correlations and phase coherence found in strongly cou-
Z =− =D. (22)
(cid:18)∂I (cid:19) pled ν =1 bi-layer systems.
2 I1
7
Acknowledgments manuscript. This work was supported by NSF DMR-
0196503(SMG), DMR-0233773(BIH), the US-Israel Bi-
The authors are grateful to James Eisenstein for national Science Foundation (AS and BIH), and the Is-
several helpful discussions, including comments on the rael Science Foundation (AS).
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