PHYSICAL REVIEW B VOLUME 35, NUMBER 9 15 MARCH 1987-II

Hot electrons in one dimension. II. Backscattering

G. D. Mahan and G. S. CanrightSolid State Division, Oak Ridge iVational Laboratory, Oak Ridge, Tennessee 37830

and Department of Physics, The University of Tennessee, Knoxville, Tennessee 37996(Received 14 October 1986)

An exact analytical solution is given to the Boltzmann equation in one dimension for a particlescattering from optical phonons in a constant electric field. The solution extends our earlier theoryto now include both forward and backward scattering of the particle. The drift velocity saturates athigh field, and the saturation velocity is found to be temperature independent.

INTRODUCTIONIn an earlier paper we introduced a new method of solv-

ing the Boltzmann equation for electron flow in one di-mension. The solution is exact, is valid for an arbitrarypotential profile V(x), and includes processes such as in-elastic scattering of the electron by optical phonons. Thisfirst paper only treated forward scattering of the electronsby the phonons. Here we discuss the solution which in-cludes both forward and backward scattering and providea detailed discussion of the case of constant electric field.

Electron or hole flow in semiconductors has traditional-ly been treated by considering the quasiequilibrium pro-cesses of drift in an electric field or diffusion in a concen-tration gradient. The miniaturization of integrated cir-cuit devices has reduced the spatial dimensions while re-taining the same voltages. Thus the electric fields haveincreased to the point where they are quite large, and thecurrent flow is often ballistic and nonequilibrium. Thissituation is called the "hot-electron prob1em. "

The current flow is rather two dimensional in mostmeasurements of hot electrons. However, recently therehave been quite a few measurements in layered semicon-ductors where the flow is primarily one dimensional.Our ultimate goal is to model realistically these interestingdevices. The present calculation should be regarded as astep in that direction.

Other theoretical approaches to the hot-electron prob-lem are primarily numerical. ' ' Both Monte Car-lo' ' and digital solutions ' '-' to the transport equa-tion are very popular. The present approach is unique inthat we proceed analytically. Other analytical methodsare limited to constant field geometries, ' ' whereas ourmethod is useful for any profile V(x) of potential energy.Most other solutions make further approximations suchas the relaxation time approximation. ' ' ' We do notneed to make this approximation, but instead solve analyt-ically the Boltzmann equation while retaining the fu11 in-elastic nature of the scattering process.

Our theory easily includes elastic scattering from im-purities and acoustical phonons, as well as inelastic

scattering from optical phonons. Here we focus on the in-elastic scattering from optical phonons, which is the mostimportant process by which hot electrons lose energy.Impact ionization is another important inelastic processfor very energetic electrons in silicon, and eventually weplan to include this process as well ~

Van Kampen argued in a historic paper against thevalidity of linear response in transport theory. His obser-vation was that the linear response of the macroscopicsystem did not guarantee linear response in the micro-scopic motions of the individual particles. Counter argu-ments against this view have been presented. Thepresent calculation is also unique in that we are calculat-ing the transport properties without assuming linearresponse. They are being done in a manner acceptable tovan Kampen's criteria of not assuming microscopiclinearity. We follow the motions of the particles down thefield, and do not linearize the equations of motion. Theresults we find at small fields are in perfect agreementwith linear response theory, and provide another coun-terexample to van Kampen's arguments.

The general method of solution follows that of paper I.'An analytical solution to the Boltzmann equation is ob-tained which is valid for a region of space x~ ~x &xj.+&.The boundary points xl occur when the potential V(x)equals multiples of the optical-phonon energy co. Thehard part of the theory is to match at xz the solutionswhich are found on two sides of this boundary point.This matching requires some boundary conditions whichare the subtle aspects of the theory. They are discussedextensively in the boundary conditions section, where theexact boundary conditions are presented for the two casesof zero field (equilibrium) and constant field. Theboundary conditions for an arbitrary field:-(x)=d V(x)/dx are still unknown.

CJENERAL SOLUTIONIn one dimension the variables which describe the elec-

tron distribution f(x,E,p) are position x, energy E, andthe direction of motion p =+1. The Boltzmann equationfor optical-phonon scattering is'

BV c) /vff(x,E,p)= [ —(1+a)f(x,E,p)[N +(N +1)e(E—co)]

+Np[f (x, E —co, p)+af(x, E —co, —p)]+(Np+ 1)[f(x, E+co, p)+af( x, E +co, —Itt )]},

35 4365 1987 The American Physical Society

4366 G. D. MAHAN AND Cr. S. CANRIGHT 35

where Xp is the phonon occupation number, lp is themean free path for making the optical phonon, and a isthe ratio of backward-to-forward scattering rates. In I wesolved the case that a=0, while here we solve the casethat e = 1. The energy variable is replaced byE= E+lco +V(x), where co is the optical phonon energy,0&v&co, and 3 is the integer index which specifies themultiples of co needed to get to E. The distribution func-tion in this notation is written as f ( x, l, lj, ), and the e

dependence is suppressed. Then it is convenient to intro-duce the symmetric F(x, l) and antisymmetric G(x, l) dis-tribution functions

F=f(x, l, +)+f(x, l, —),6=f (x, l, + ) f(x, l, ——) .

The static Boltzmann equation is solved in terms of a di-mensionless distance z=2x/lp, where lp is the mean freepath for electron scattering by optical phonons. We

G(z, O)

G(z, 1)G(z)= G(z, 2)

G(z, 3)

F(z,O)

F(z, 1)F(z)= F(z, 2)

F(z, 3)

The Boltzmann equation is

d Cr(z)/dz = —MF,d F(z)/dz = —M'Cs .

The matrix M has tridiagonal form, while M' is diagonal.The solution to this pair of differential equations wasshown in I to be

showed in I that the Boltzmann equation can be reducedto a pair of matrix equations. There are two column vec-tors G and F whose elements are G (z, l) and F(z, l)

G(z, l) =-e~ ae +e ' f cos(19+/)[C(9)e ' +D(9)e' ]—16/2

F (z, 1)= [b —(2XO+ 1)az]e ' —(2%0+ 1)e ' f cos(IO+ Q) [D (9)e' "—C(9)e ' ],—tG ZZ z&X -z&X

A, (9)= (2XO+ 1) I (9),I (9)= [1—cos(9)/cosh(G/2)],

tang = (2+e )e / sing —(2e + 1)cotO,

c1 —1+e~6, 0

(2)

where G =~/k& T and Xp ls the phonon occupation num-ber. The two parameters a and 6 are constants of integra-tion: a is interpreted as being proportional to the current,while b is interpreted as being proportional to the particledensity. The two functions C(9) and D(9) are arbitrary,and are determined by the initial and boundary condi-tions. The above two equations can be inverted to obtainC and D in terms of the amplitudes for the current G anddensity I'

De' +Ce ' =2 g G (z, l)e '~ cos(19+/),

analysis:

g E~e' cos(19+/) =0,

1=0

EI cos(19+P) cos(19'+P') =—5(O —9'),1=0 2

dOcos(19+&) cos(mO+&) =5& +e (1—e )

—G(1+m)/2Qe

De' —Ce ' = —2VI" QF(z, l)ele ' cos(19+$) .1

(3)&10& o

1+ exp( —G)

Further information needed to describe the general solu-tion are two sum rules —one each for G and F. They are

g G (z, 1)=ae /( 1 —e ),1=0

g F(z, l)e~ ——fb —az (2NO+ 1)]eG/( I —e G) .1=0

(4)

These two sum rules play a key role in the analysis, as dis-cussed below. Finally, below are listed three orthogonalityrelations which we derived and found useful in the

The two parameters a and b are used throughout, andit is worthwhile to discuss their physical meaning. Firstwe start from the standard definitions of particle densityn (x) and current j in terms of the distribution functionf (x,k)

n = dk xk

j =e f dk (k/m)f (x,k) .

Next change the integration variable from dk to dE andsums over @=+1

35 HOT ELECTRONS IN ONE DIMENSION. II. BACKSCATTERING 4367

n(x)= f F(x,E),vm dE2 E

j =e f dE G(x,E) .

The energy variable E is again changed to c.+leo, and theintegral dE is changed to dc plus the summation over allvalues of l.

f"dE=y f"de.

For the two integrals in (5) we find

n(x)= f dEQ2 (E+&~)' '

j=e dc. G x, l

The expression for the current is found readily by usingthe sum rule in Eq. (4) and the fact that a is independentof c.

j =eatage l(1 —e ) .

In one dimension the current j is a constant, which meansthat a does not depend upon x. The expression for theparticle density n (x) is less easy to evaluate. The summa-tion over F(x, t) does not have the same form as found inthe sum rule of Eq. (4). Nevertheless b is the parameterwhich is related to the density of the particles.

BOUNDARY CONDITIONS

The name staging was introduced in I to describe theprocess of redefining all of the amplitudes at a matchpoint. Figure 1 shows two match points zj and zj+& ~ Theoptical phonon scattering moves electrons up or down aset of energy levels separated by cu. These levels areshown by the horizontal lines in Fig. 1. The bottom levelis always labeled 0. A match point is where the potentialV(z) has varied sufficiently that a new level appears (flowto the right) or disappears (flow to left). There the levelsmust be relabeled. One of the key issues in this method ofsolution is to determine the boundary conditions at amatch point.

For example, consider that current flow is to the right

2 3AI

I

I1 I 2

v(z)

FIG. 1. The solid line shows the potential energy V(z) for aconstant electric field when the particle flow is to the right. Thehorizontal dashed lines are separated by energy intervals equalto the optical phonon energy. The vertical lines mark wherenew horizontal levels appear. These are the match points zjwhere one must apply the boundary conditions.

in Fig. 1. The amplitudes are continuous at a matchpoint. Because of the relabeling of levels, this continuityis expressed as

G(zj. +r), 1) =G(zj —rl, l —1),F(zj+g, l) =F(z~ —.rI, 1 —1), (7)

e Fo+ Q F(z, l) =bjrr e l(1 —e ),l

A =t2 (21VO+ 1) .

We subtract these two equations, and the summationterms mostly cancel because of the continuity of F across

where g is infinitesimal. For current flow to the right, wemight know all of the amplitudes to the left of the matchpoint. The amplitudes to the right of the match point arethen determined by the continuity relations given above,except for the two new amplitudes G (zj +g, 0) andF(z&+g, 0). They must be found by other arguments,which is a bit of a problem.

The initial values of the two new amplitudes, which weabbreviate as Fo and Go, are found by using the sum rules(4). They show, for example, that Go ——0. The sum ruleon G(z, l) is proportional to a, independent of z. Since ais proportional to the current, it has the same value foreach stage. The summation of all values of G to the leftof the match point is equal to the summation of all valuesto the right. Since G is continuous at the match point,the new value Go cannot contribute to the summation,and must start with zero value.

Thus we have shown that Go ——0, and there remainsonly the problem of finding Fo. This quantity has provedelusive, and me have not yet determined the generalboundary condition. The result has only been found forthe case of equilibrium (no electric field) and the case ofconstant electric field.

Although a is stage independent, the other constant bhas a different value for each stage. This assertion is easi-ly proved in equilibrium where a =0. Then the densityvaries with the potential according to n (z)=no exp[ —V(z)ikT]. Thus the density at each newstage must increase or decrease by the factor of exp(G).Denote the amplitude be and bJI as the amplitudes tothe right or left of the stage at zj; then b~z ——bz+& L. Inequilibrium one finds that

err——exp(G)bJL .

Each new stage introduces two new parameters into the Fsystem: b and the initial amplitude Fo. The sum rule (4)provides one relationship, but another one is needed forthese two parameters. Finding the second boundary condi-tion has proved to be difficult. The above condition forthe b s, while valid in equilibrium, is not valid for a non-equilibrium situation such as current flow.

The first step is to use the sum rule to provide a rela-tionship between the two unknown parameters. Just tothe left (z =zj —rj) and right (z+=z~+g) of the matchpoint the sum rule for F gives

e F(z,0)+ g F(z, 1)= (b&L —Az. )e Gl(1 —e G),

4368 G. D. MAHAN AND G. S. CANRIGHT 35

the match point. This yields the important relation

(1+e )Fo e——Fi+(b/z+Azj bj—l )e /(1 —e ),Fi F(——zj. —iI, O) =F(zi+q, 1) .

(9)

In order to find the second relationship between b and Fo,we examine the solution for a special case—a constantelectric field. Figure 2 shows a sawtooth line whichrepresents the value of the sum rule g(z) as a function of zfor constant electric field:-=d V/dx:

$(z)=(b~ —Az)e /(1 —e ) .

Here we use the convention that the value of z starts atzero at each match point; z is the distance from the lastmatch point. The constant field is assumed to extend toinfinity in both directions. The system has a steady stateflow of constant current. We assume that the functiong(z) is periodic, and repeats its value at each stage. Thedistance 5=zj —zj+ &

between stages is

b, =2'/(lo-) .

In order that g(z) be periodic, the value of b must be thesame at each stage. This is expressed as

We begin by outlining the method of solution. Filling inthe details takes the rest of this section. The technique isto derive a transfer matrix N(b, ) which describes how thevalues of gj+& are found from those of gj:

g~. +i ——N(b, ).gj+aL+bK . (13)

These values will also depend upon a and b, which eachmultiply their own vector contribution. The periodicityrequires that gj+~ ——gj is independent of stage. The sub-script j is dropped, and the above equation is solved forthe amplitudes

since we shall demonstrate that our solution has a currentwhich saturates at high field. Furthermore, the solutionfor constant field will often be used as the boundary con-dition for other cases.

For the constant field, the basic premise is that thesolution is completely periodic. The vectors F and Ghave the same values at every match point. We constructa double-column vector called gj which provides the valueof F and Ci at match point zj.

Fgj=

~jR ~jL (10)g=(I —N) '(aL+bK) . (14)

Ab, =b(l —e ),cr =elob/[2No(1+e )], (12)

which provides an expression for the conductivitycr=j/:-. This same result is derived with more rigor inthe following section.

CONSTANT ELECTRIC FIELD

Next we provide a complete solution for the case ofconstant electric field. This case is interesting in itself,

$(Z)

Equation (10) is combined with Eq. (9) to produce thefinal expression for Fo

F (1+e )=F, +Ah/(1 —e ) .

This boundary condition is valid for the case of constantelectric field. It should also be valid in the case that thefield goes to zero, where the system approaches equilibri-um. In that case, both boundary conditions (8) and (11)should be valid. In equilibrium one has that at the begin-ning of a stage Ft =b~ exp( —IG). Using this ansatz in Eq.(11) for Fo and F, gives for the current at small electricfield

The amplitudes G(z, l) and F(z, l) are now determined interms of a and b. The sum rule (4) can be evaluated,which gives a as a function of a, b, and A. This equationis then solved to give the current a as a function of thedensity b and electric field 6 '. The two sum rules eachgive an expression, and the two are identical.

The transfer matrix N and vectors L and K are con-structed using Eqs. (2) and (3). First Eq. (3) is used tofind the functions C(8) and D(g) from the amplitudes Fand G at zj. These values for C and D are used in Eq. (2)to find F and 6 at zj+& ——zj+A. The latter amplitudesconstitute gj+& except for Go and Fo for which we usethe two boundary conditions Go ——0 and (11) for Fo We.use the argument (j, l) of F and G to denote the values atzj+ q. The amplitudes at j + 1 are given by

F(j+1,0)=F(j+1,1)/(1+e )+Ah/(1 —e ),F(j +1, l+1)=(b —AA)e

+ g[n(FF)t F(j,m)

+n(FG)i G(j,m)],

G(j+1,0)=0,r

G(j+1, l+1)=E& ae + g[n(GG)t G(j,m)mZ. -1 ZJ

FKr. 2. A graph of the function g(z) =(b—Az) exp(G)/[I —exp( —6)] for the case of constant electricfield where 6 =su/k~T. Each stage must be alike for uniformcurrent flow, which means that g(z) must be a periodic function. where

+n(GF), F(j,m)]

35 HOT ELECTRONS IN ONE DIMENSION. II. BACKSCATTERING 4369

n (FF)=2e' " cosh(b, V A, )dO

Xcos(18+/) cos(m6+p),

n (FG) 2e(m —l)G/2 sjnh(Q~g)d0~Mr

X cos(16+p) cos(m 6+ /),n (GG) = n (FF),

n (GF) 2 ™—l~~~2 f ~

h(Qv g)

(15)

dent of z. We exploit this fact to evaluate this expressionat small field, which has b,~ oo. We expect that D is ofthe magnitude of 0[ exp( —b,v A, )] since it has that mag-nitude in Eq. (17) when z=b, . However, when we exam-ine this same equation for z =0, we find that

g e ' [G(0,1)—clF(0,1)]cos(ln/2. +$) 0-[ exp( —b, )],I

where b, =b, (2Np+1). The easiest way to satisfy this re-lation is term by term. Both 6 and F are expressed asequilibrium terms [ —exp( —Gl)] plus transients (5G, 5F).They are written as

Xcos(18+/)cos(m6+p) .

The set of Eqs. (15), when cast in terms of column vectorsg, provide the definitions of the matrix N and the vectorsK and L in Eq. (14). For example, the vector K has(1+e )

' for its first element, exp[ —G(l —1)] for theother elements which define F, and zero for all elementswhich give G. The transfer matrix N contains the ele-ments n (FF), n (FG), n (GG), and n (GF).

The above example has the transfer matrix going in thedirection of the particle flow, which is to the right. Wealso constructed the transfer matrix for flow to the left,which is against the direction of particle flow. Thisupstream transfer matrix has the feature that the inver-sion in (14) is not allowed. That is, it does not give sensi-ble results. Thus the method only works for the down-stream transfer matrix.

SMALL ELECTRIC FIELDS

For small electric fields =, we expect that5 = 2p2p/( lp ) diverges, while b goes to a constant and avanishes proportional to =. The factor aA goes to a con-stant. On the basis of dimensional analysis, we expect atsmall field that the current goes to an expression of theform

5G(0, 1) =El5F(0,1)+0[exp( —b.)], (18)

which we abbreviate as 5GI ——c.~5FI.Next examine the equation for C(6). At z =0 the

right-hand side is of order unity. This must also be trueat z =5, when the right-hand side is

exp(h) g e '/ [56(b„l)+e(5F(6,1)]

X cos(lvr/2+/) -0(1) .

Again it seems that this relationship is satisfied for eachterm in the series, which gives us the additional relation-ship

5G(b„l)= c(5F(b„l)+0—[ exp( —b, )] . (19)

This is the second important relation.The next step is to use the continuity of G and F at the

match point. This continuity gives the two relations

6 (z, 1)=asle '+5G(z, 1),F(z, l) =(b —zA)e '+5F(z, l) .

The equilibrium terms give zero when inserted in the defi-nitions for D and C. Thus the above series are actuallyexpressions for 56 and 5F. Thus a term-by-term solutionof the above equation is

a =Kb/5, (16) a ale +56 (b., I) =ae ( +"+5G(0, 1+1),

C(6) +zv2. g eGI/2[6( 1)I

(17)

+E,v 1 (6)F(z, l}]cos(18+/) .

These two expressions are most easily evaluated at 0=0,vr

since they vanish because P(0) =P(vr) =sr/2.

D(0)=D(1r)=C(0)=C(~)=0 .

A convenient place to evaluate this expression is at6=n /2 since there I (m. /2) = l.

The right-hand side of Eq. (17) appears to depend uponz, although it does not since the left-hand side is indepen-

where the constant H needs to be determined.The first step is to rewrite the expressions for C(6) and

D(6)

D(6) —z )j. g Gl/2[6(1

vr (6)F(z,l)] co,s(16+.P),

(20)

(b —ba)e +5F(b, , l)=be G'l+''+5F(0, 1+1) .

These two equations, along with (18) and (19), are suffi-cient to determine the four sets of unknowns: 56(0,1),56(b, , l), 5F(0,1), and 5F(1)., 1). The solution for 5G atz =0 is

6GO ———aco,

561 ——[a (Ep —e )+ epb(1 —e ) —a 1I),sp]/(1+op), (21)

56 ( 1 e—G)e —G(l —1)(a +b)/2 age —G(l —1)/2

(1») .

The other terms are also found easily.We know from the sum rule for currents that

+5Gl ——0 .I

Using Eq. (21) for 5G gives an equation which expresses ain terms of b and 6:

4370 G. D. MAHAN ANI3 G. S. CANRIGHT 35

be (1—e )lim a=

Aa )+a2

a, =Ãp(l+e ),02=(Ep —1/2sp)e (1 —e )/(1+e /2)

(22)

which is the final result. This agrees with Eq. (12) in thelimit that A~ap. The conductivity is proportional toe, lo, and inversely proportional to Xo, in agreement withthe result known from solving the Kubo formula for thecurrent-current correlation function. Note that our resultis valid for every temperature, whereas the Kubo resultsare only known asymptotically in the limit of low tem-perature.

According to Eq. (21) the amplitudes Gt and I't areboth proportional to exp( —Gl), which predicts that theelectron temperature is equal to the phonon temperature.Of course this is the correct low-field limit. The analyti-cal expression (22) for the current ceases to be valid athigh field when the electron temperature becomes largerthan the phonon temperature.

NUMERICAL RESULTS

Equation (14) was solved numerically for differentvalues of:"~ 6 ' and for enough 3 values to obtain con-vergence. The resulting values of GI and F~ were used inthe sum rules to obtain a as a function of b and h. Thisresult is plotted as the solid line in Fig. 3 at three tern-peratures. The only parameter needed is the optical-phonon energy where we used co=0.063 eV appropriatefor silicon. The vertical axis is the quantity(a/b) exp(G)/[1 —exp( —G)] which according to Eq. (6)is proportional to the average drift velocity per electron.The horizontal axis is 1/6 which is proportional to the

electric field. The average drift velocity does saturate athigh field as is known from experiment and pr'edicted byother theories. Results are shown for three different tem-peratures (100, 200, and 300 K). The saturation value ap-pears to be nearly independent of temperature and equalsunity. This result agrees with the saturation velocity forsilicon which changes only by 30% between 300 K andlow temperature. The dashed lines emanating from theorigin are the low-field asymptotic result as given by Eq.(22). The low-field results match up well with the high-field results in the vicinity of 1/6=0. l.

The high-field solution in (14) cannot be numericallyextended to small values of field. Small values of 6 ' aredifficult to calculate using Eq. (15). At small field:- thenb, =2'/(1p=) is large. The elements of the transfer ma-trix N(A) each become large because they each containcosh(h) or sinh(b, ). The inversion of this matrix in Eq.(14) becomes numerically inaccurate. Most transporttheories are easiest to solve at small fields and difficult atlarge fields. In contrast, the present technique is easiest atlarge fields and is difficult at small fields. In our theory,the electric field enters as an inverse power in an ex-ponent, which certainly demonstrates that we are notlinearizing the microscopic equations.

Figure 4 shows a graph of electron temperature versuselectric field when the phonon temperature is 100, 200,and 300 K. The electron temperature is found by plottingthe calculated I'I versus I on semilog graph paper and fit-ting the points to a straight line. The fit was exce11entsince the line was quite straight for values of 1& 2. Thefigure sho~s that the electron temperature increases with

400—

100

I

0.1

I

0.2

200

300 K

[

0.3 0.4 0.5

LLI

300

LLI

200

UJ

LLI

FIG. 3. A graph of the average drift velocity as a function ofelectric field:" at three different phonon temperatures. Thevertical scale is the quantity ( a /6 ) exp(6) /[ 1 —exp( —G) ]which is proportional to drift velocity. The horizontal scale is

'=:"lo/(2'), where lo is the mean free path for making opti-cal phonons of energy co=0.63 meV. The drift velocity satu-rates at high field to a value which is nearly independent of pho-non temperature. The solid line is from the high-field theory ofEq. (14). The dashed lines from the origin are the predictions ofthe low-field mobility in Eq. (22). The two solutions match upwell in the vicinity of 6 '=0. 1.

100—

00 0.1

[

O. Z

I

0.3 0.4 0.5

FICx. 4. The effective electron temperature as a function ofelectric field for a phonon temperature of 100, 200, and 300 K.The electron temperature appears to be less than the phonontemperature at small fields.

35 HOT ELECTRONS IN ONE DIMENSION. II. BACKSCATTERING 4371

electric field, in agreement with the results of other calcu-lations. However, at small fields the electron temperatureappears to fall below the phonon temperature. Thisfeature of our result is somewhat uncertain since it occursin the low-field regime where our method becomes inaccu-rate, as discussed above. However, we' and Lei andTing found similar behavior in earlier calculations, andthe phenomena seems to be real. This prediction disagreeswith the usual theory that T, —Tz is proportional to thesquare of the electric field at small fields. Also note thatthe reason that the low-field and high-field currenttheories match up so well in Fig. 3 is that the low-fieldtheory is valid as long as the electron temperature equalsthe phonon temperature. The high-field theory predictsthat T, =T~ in this region, so the two theories overlap.

Figure 5 shows how the amplitudes GI and FI vary as afunction of z between stages for the case that T= 300 Kand 5=5. For constant field this dependence is repeatedfor each stage.

0.8

N

(9 04

0.2

0

5 0

0.8

DISCUSSION

We have developed a technique for solving theBoltzmann equation exactly in one dimension for the in-elastic scattering of electrons by optical phonons. Herewe present the solution for a constant electric field and forthe case that the electron has both forward and backwardscattering while emitting or absorbing the phonons. Thissolution showed the typical behavior that the average driftvelocity was proportional to field at low field and saturat-ed at high field. The saturation velocity was found to benearly independent of phonon temperature. The electrontemperature was found to first decrease and then increasewith increasing electric field. An analytical expressionwas found for the low-field mobility which is valid at alltemperatures.

A key aspect of our theory is the concept of staging. Astage is a region of space over which the potential energyV(x) changes by the energy of one optical phonon. Thesolution within a stage is easy to find. The subtle aspectsof the theory are matching the solutions at the boundarypoints between stages. The matching process requires aknowledge of boundary conditions. The exact boundarycondition was presented for the case of uniform electricfield. With it we were able to present a complete solution

0.4

0.2

00 1.0 2.0 3.0 4.0 5.0

FIG. 5. The x dependence of the amplitudes for the densityFq and current G~ between two match points, where z=x/Ip.Results are shown for the case that T=300 K and 6=5.

ACKNOWLEDGMENT

Research support is gratefully acknowledged from theU. S. National Science Foundation Grant No. DMR-85-01101, from the University of Tennessee, and from theU. S. Department of Energy.

to this problem. The general boundary conditions are stillunknown for a general profile of potential energy V(x).The boundary conditions must be found before this casecan be solved.

G. D. Mahan, J. Appl. Phys. 58, 2242 (1985).~G. A. Baraff, Phys. Rev. 128, 2507 (1962); 133, A26 (1964).3S. M. Sze, The Physics of Semiconductor Deuices (Wiley, New

York, 1981).4H. U. Baranger and J. W. Wilkins, Phys. Rev. B 30, 7349

(1984).~J. Shah, A. Pinczuk, A. C. Gossard, and W. Wiegmann, Phys.

Rev. Lett. 54, 2045 (1985}.C. H. Yang, J. M. Carlson-Swindle, S. A. Lyon, and J. M.

Worlock, Phys. Rev. Lett. 55„2359 (1985).7A. F. J. Levi, A. R. Hayes, P. M. Platzman, and W. Wieg-

mann, Phys. Rev. Lett. 55, 2071 (1985).J. L. Oudar, D. Hulin, A. Migus, A. Antonetti, and F. Alexan-

dre, Phys. Rev. Lett. 55, 2074 (1985).M. Heiblum, M. I. Nathan, D. C. Thomas, and C. M.

Knoedler, Phys. Rev. Lett. 55, 2200 (1985).' R. A. Hopfel, J. Shah, D. Block, and A. C. Gossard, Appl.

Phys. Lett. 48, 148 (1986).' R. A. Hopfel, J. Shah, and A. C. Gossard, Phys. Rev. Lett.

56, 765 (1986).W. H. Knox, C. Hirlimann, D. A. B. Miller, J. Shah, D. S.Chemia, and C. V. Shank, Phys. Rev. Lett. 56, 1191 (1986).

' M. Heiblum, E. Calleja, I. M. Anderson, W. P. Dumke, C. M.Knoedler, and L. Osterling, Phys. Rev. Lett. 56, 2854 (1986).P. Lugli and D. K. Ferry, Phys. Rev. Lett. 56, 1295 (1986)~

' D. K. Ferry, Solid State Electron. 21, 115 (1978).

4372 G. D. MAHAN AND G. S. CANRIGHT 35

'~P. K. Basu, J. Appl. Phys. 48, 350 (1977).~Y. J. Park, T. W. Tang, and D. H. Navon, IEEE Trans. Elec-

tron Devices ED-30, 1110{1983).M. V. Fischetti, D. J. DiMaria, S. D. Bronson, T. N. Theis,and J. R. Kirtley, Phys. Rev. B 31, 8124 (1985).R. K. Cook and J. Frey, IEEE Trans. Electron Devices

ED-29, 970 (1982).X. L. Lei and C. S. Ting, Phys. Rev. 8 32, 1112 (1985).

~ S. A. Trugman and A. J. Taylor, Phys. Rev. B 33, 5575 (1986).~~N. Cx. van Kampen, Phys. Norv. 5, 10 (1971).

W. M. Visscher, Phys. Rev. A 10, 2461 (1974).