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A fast solution of poisson's equation in FETs
Solid-State Electronics Vol. 37, No. 2, pp. 373-376, 1994
~
Copyright © 1994 Elsevier Science Ltd Printed in Great Britain, All rights reserved 0038-1101/94 $6.00 + 0.00
Pergamon
A FAST SOLUTION OF POISSON'S EQUATION IN FETs C. D. KOURKOUTAS l and G. J. PAPAIOANNOU 2 ~Department of Physics Chemistry and Material Technology, TEI Athens, od Ag Spyridonos, Egaleo 122 43 and 2Solid State Physics Section, Athens University, Panepistimiopolis, Zografou 157 71, Greece
(Received 7January 1993; in revised form 12 July 1993) Abstract--In the present work we propose a fast solution of the Poisson's equation in FETs. The method is valid, if the deep centers are uniformly distributed within 1 #m below the interface between the active layer and the substrate. The implicit method is convenient for the solution of the differential equation. The convergence to the final solution is accelerated significantly if the correction factor of the iterative procedure is considered with appropriate weight.
I. I N T R O D U C T I O N
The knowledge of the precise donor concentration in FETs is necessary for controlling its electric properties. As it is known, the C - V method, which is commonly used, is precise only if the dopant is uniformly distributed within several Debye lengths of the free carriers[l]. However, in the general case the capacitance measurements do not correspond to a unique donor profile. Several methods have been proposed for the approximate determination of the doping profile by correcting the C-Vmeasurements[2,3]. In recent publications[4,5] the determination of the doping profile is approximated by means of successive iterations, in which the Poisson's euqation is solved each time for an assumed profile until the obtained result is in good correlation with the measured profile. Usually, when we solve the Poisson's equation we are seeking solutions, which result in zero potential gradient at infinite, or practically at large enough distance below the contact[2,6]. This method is time consuming, as we must go more than 1/~m below the contact in order to achieve a satisfactory accuracy. In the present work we propose a fast solution of the Poisson's equation in FETs by shifting the boundary condition from infinite depth below the Schottky barrier to the interface between the active layer and the substrate abbreviating significantly the time needed for the final solution. This method is valid, if the deep centers are uniformly distributed in the substrate within few micrometers (1-2 #m) below the interface.
ity complete ionization of the acceptors and neglect the effects of deep donors and holes. The Poisson's equation is thus written: d2u(x) q2 rN+ . . _ NA-- n(x)] dx 2 - E%kTI Do~X/
(1)
where
E~(x)
u(x) = - - kT
(2)
is the reduced conduction band energy measured from the Fermi level. The meaning of all other symbols used in this work is explained in Table 1. The ionized donor concentration N~o(x) is given by: NDo(X) U~o(X) = (3) 1 + 2 exp[-- u(x) + q)o] with
EDo
EDo kT
(4)
The free electron concentration n(x) is approximated by the formula: Arce .(x) n(x) - 1 + 0.25 e -u(x) "
(5)
NDo, n
2. THEMETHOD We consider a F E T with a donor distribution Nno(X) and uniformly distributed deep donors and acceptors with concentrations N o and N A. Within the active layers x < w (Fig. 1), where the Fermi level is close to the conduction band, we assume for simplicSSE37/2-K
0
W
X
Fig. 1. Hypothetical doping concentration Noo and free electron concentration n profiles in a FET. 373
374
C.D. KOURKOUTASand G. J. PAPAIOANNOU Table I Non NO Na n p Ev EOo ED EA Eg Nc NV
kT q¢ e0
+ 4N~ In
Shallow donor concentration Deep donor concentration Acceptor concentration Free electron concentration Hole concentration Fermi energy Shallow donor energy Deep donor energy Acceptor energy Energy gap Effective density of states in the conduction band Effective density of states in the valence band Thermal energy Absolute electron charge Static dielectric constant Permittivity of free space
1 + 0.25 e x p [ - u ( w ) ] 1 + 0.25 exp[--Uinf]
+ N~[expIu(w) -- Eg] -- exp(uinf -- Eg)]} (11) where Uinf is the potential energy at infinite depth, which can be calculated from the charge-balance equation for x = oo:
Nfi + p = N A
In the substrate (x > w), where the donor concentration NDo is zero, the existing charge is due to electrons, which are diffused there from the active layer, holes in the valence band and ionized deep centers. The Poisson's equation becomes: q~ _ rN+ EeokT t D ( x ) - - N A - - ( x ) - - n ( x ) + p ( x ) ] "
(6) The positively charged deep donor concentration N~ is given by (3), (4), if we replace the index " D o " with "D". The negatively charged acceptors are given by: NA NA(X ) = (7) 1 + 2 exp[u(x) -- EA]
x =oo.
(12)
We can solve now the Poisson's equation within the active region, from x = 0 to x = w calculating the band potential u(x) by means of a numerical method. The implicit method[6] after some necessary modifications, which will be demonstrated below, is very convenient, as it converges quickly to the final solution. We fix equally spaced points:
x(j)=jAx, d2u(x) _ dx 2
+n,
j=0,
1,2 . . . . L
over the region 0 . . . w, and expand the potential energy u(x) in Taylor series retaining terms up to the second derivative. We write thus: duO')
u(j'- l)=u(jl-~d~-x
1 d2u(j) • 2
Ax + ~ x 2
zxx
(13)
and
u(j+l)=u(j)+~x)AX
+ ~15d2uO') - x 2 Ax2.
(14)
where EA
(8)
EA -- k T
From (13) and (14) we obtain: u(j-l)-2u(j)+u(j+l)=d2u(j)" ~ dx
and the hole concentration p ( x ) is given by:
p ( x ) = Nv exp[u(x) -- Eg]
(9)
zxx
2 (15)
where the second derivative in the right hand side of (15) is given by (1) for x = x(j). F o r j = L we find:
where
E,
E~ = ~ .
(10)
We multiply both sides of (6) by du/dx and integrate from w to infinite taking into account that: d2u d u _ l dx 2 dx
d (du~ 2 2 dx \ d x / I
and that the potential gradient at infinite is zero. We obtain thus the following relation between the potential energy u(w) and its gradient du(w)/dx at the interface (x = w):
u(L + 1) = u(L - 1) +' 2 du(w) ~ Ax
where the potential gradient at x ( L ) = w is given by (11). The potential u(0) at x = 0 is known and equals the sum of the built in potential and the applied reverse bias voltage. Equations (15) constitute therefore a nonlinear system of L equations and L u n k n o w n potentials u(j) at the points x ( j ) with j = 1 , 2 . . . L. To solve it we suggest a trial potential profile u')(j). If the true profile is /2(r-I-l)(j), we set:
3u(n(j) = u(~ + l)(j) _ u(,.)(j). (du(w)) 2
(16)
(17)
2q~ {'AtDIn 1 +0"5exp[u(w)--ED] = ~
+Naln
1 + 0 . 5 exp[Uin f -- ED]
1 + 2 exp[u(w) -- EA] 1 + 2 exp[uinf -- ED]
To the first approximation we write: exp[u("~U) +
6u~°)U)] =
[1 +
6u'~U)]exp[uC'~U)]. (18)
375
A fast solution of Poisson's equation in FETs Neglecting for small 5u(j) the variation in the denominator of (3) and (5) we find that the free electron concentration is: n(~+ l)(j) = [1 - 5u(~)(j)]n(V)(j)
(19)
and the charged donor concentration is N ~ (~'+')(j) = [1 + 6ua')(j)]N~o(~')(j).
(20)
Since u(0) is known, 6u(O) ~- 0. On the other hand equation (16) yields 5u(L + 1 ) = 3 u ( L - l) for A x e 0 . We obtain therefore the linear system: - ~ (")(1)6u (~)(1) + 6u (~)(2) = fl (~')(i)
6u(V)(1) - a(~)(2)fu(~)(2) + 6u(~)(3) = fl(')(2)
26u(°(L - 1) - a(~)(L)fu(°)(L ) = fl")(L ) (21) where
q2~Ax2 a')(J ") = 2 + ~
[n(V)(j) + N~)o(~')(j)]
(22)
q2 Ax 2 fl')(j) = ~ [mGo')(j) - X A - n')(j)].
(23)
and
solution up to infinite depth, where the boundary conditions du/dx = 0 and u = Uinr are satisfied. The solutions, which correspond to curves (2) and (3) are obtained by shifting the boundary condition u = um~ to finite distance d from x = 0 (d = 0.2 and 0.6/~m). As the accuracy of our calculations is better than l % , the boundary condition should be shifted at least 1 # m below the interface in order to obtain in the region x < 0 a solution differing from the corresponding to curve (1) with the same accuracy. In usual FETs the active layer thickness is 0.20-0.25/~m. Following the proposed method we abbreviate therefore significantly the solution of Poisson's equation. (In this example our solution is five times faster than the obtained by the conventional method.) In Fig. 3 we see the solution of Poisson's equation for a VPE-like donor profile in a G a A s FET, where we considered the presence of two deep centers, a donor like native defect such as the EL2 at 0.69 eV below the conduction band with concentration N D = 1.5 × 1021m -3 and a midgap acceptor with various concentrations N A up to 2 x 1022 m -3. In our calculations we set Ax = 1 nm. The bias voltage was fixed at 0.8 V.
ND, n (m -31
The solution of the system is repeated with the new potential profile u (v+ l)(/.) as the trial potential until the obtained corrections 3u(]) are sufficiently small. To initiate the iterative procedure we use the following trial potential: u(')(/) : - l n { ~ [ - ( - ~
+ ~ exp(--EDo)) + 2
ND -- NA
q) (24)
for j = l , 2 . . . . L. Equation (24) satisfies the charge-balance equation at any point x(/). The convergence to the final solution is accelerated significantly, if the two terms in parentheses in (22) are of the same order of magnitude. This can be done by multiplying one of them with a weighting factor, for example, N3o(X) by n(x)/NDo(X). Relation (22) is reduced then to:
q~ Ax [ a(v'(jl=2+~Llq
N3o(~)(j)] ,,),., NDo(J)-~n (j).
1022
(1)
(2)
(25/
3. EXAMPLES
In Fig. 2 we see the solution of the Poisson's equation for a step function doping profile in G a A s at 300 K. The numerical data are taken from [7]. The solution, which corresponds to curve (i) is obtained from the proposed method and is equivalent to the
-0.05
0
0.05
x (~m) Fig. 2. The solution of Poisson's equation for a step function doping profile in GaAs at 300K. Curve (I) is obtained from the proposed method. Curves (2) and (3) are obtained by setting u~,r at infinite distance d from x = 0. Curve (2): d = 0.6tim. Curve (3): d =0.2#m.
376
C . D . KOURI(OUTASand G. J. PAPAIOANNOU NDo, n (1023 m -3) 1.5 (1) -
1,0
m
(5) (4) (3) (2) 0.5 -
I 0.05
0.10
0.15
0.20
x (~tm) Fig. 3. Free electron concentration n at 300 K in the active region of a GaAs FET with VPE-like donor profile [curve (I)] and various acceptor concentrations. (2): N A = 2 x 1022m -3, (3): N^ = 1022m -3, (4): N A = 5 x I021m -3 and (4): N A = 0. The deep donor concentration is in each case 1.5 x 1021m -3.
Using relation (22) for the correction factor ~(~)(/') we o b t a i n a solution with 8 u ( j ) < 0.01 after m a n y iterations, as the convergence into the space charge region (x < 0 . 0 7 / t i n ) is slow (Su(V+l)/fu (v) = 0.9). If we use relation (25) instead, then the convergence becomes m u c h faster a n d the solution is o b t a i n e d after six iterations with the same accuracy.
4. S U M M A R Y
U s i n g the continuity condition (11) at the interface between the active layer a n d the substrate o f a F E T we abbreviate significantly the solution o f the Poisson's equation. The m e t h o d is valid, if the deep centers are uniformly distributed. The implicit m e t h o d for the solution of the differential e q u a t i o n a n d converges after few iterations to the final sol-
ution, if the correction terms are considered with a p p r o p r i a t e weight. REFERENCES
I. D. P. Kennedy, P. C. Murlay and W. Kleinfelder, IBM J. Res. Develop. 12, 399 (1968). 2. W. C. Johnson and P. T. Panousis, IEEE Trans. Electron Devices ED-18, 965 (1971). 3. D. P. Kennedy and P. R. O'Brien, IBM J. Res. Develop. 13, 212 (1969). 4. K. Iniewsky and C. A. T. Salana, Solid-St. Electron. 34, 309 (1991). 5. X. H. Fu and J. N. Chen, Solid-St. Electron. 35, 181 (1992). 6. Mamoru Kurata, in Numerical Analysis for Semiconductor Devices. Lexington Books, D. C. Heath and Company, Toronto, Mass. (1982). 7. D. L. Rode, in Semiconductors and Semimetals (Edited by R. K. Willardson and A. C. Beer), Vol. 10, Chap. 2. Academic Press, New York (1974).
~
Copyright © 1994 Elsevier Science Ltd Printed in Great Britain, All rights reserved 0038-1101/94 $6.00 + 0.00
Pergamon
A FAST SOLUTION OF POISSON'S EQUATION IN FETs C. D. KOURKOUTAS l and G. J. PAPAIOANNOU 2 ~Department of Physics Chemistry and Material Technology, TEI Athens, od Ag Spyridonos, Egaleo 122 43 and 2Solid State Physics Section, Athens University, Panepistimiopolis, Zografou 157 71, Greece
(Received 7January 1993; in revised form 12 July 1993) Abstract--In the present work we propose a fast solution of the Poisson's equation in FETs. The method is valid, if the deep centers are uniformly distributed within 1 #m below the interface between the active layer and the substrate. The implicit method is convenient for the solution of the differential equation. The convergence to the final solution is accelerated significantly if the correction factor of the iterative procedure is considered with appropriate weight.
I. I N T R O D U C T I O N
The knowledge of the precise donor concentration in FETs is necessary for controlling its electric properties. As it is known, the C - V method, which is commonly used, is precise only if the dopant is uniformly distributed within several Debye lengths of the free carriers[l]. However, in the general case the capacitance measurements do not correspond to a unique donor profile. Several methods have been proposed for the approximate determination of the doping profile by correcting the C-Vmeasurements[2,3]. In recent publications[4,5] the determination of the doping profile is approximated by means of successive iterations, in which the Poisson's euqation is solved each time for an assumed profile until the obtained result is in good correlation with the measured profile. Usually, when we solve the Poisson's equation we are seeking solutions, which result in zero potential gradient at infinite, or practically at large enough distance below the contact[2,6]. This method is time consuming, as we must go more than 1/~m below the contact in order to achieve a satisfactory accuracy. In the present work we propose a fast solution of the Poisson's equation in FETs by shifting the boundary condition from infinite depth below the Schottky barrier to the interface between the active layer and the substrate abbreviating significantly the time needed for the final solution. This method is valid, if the deep centers are uniformly distributed in the substrate within few micrometers (1-2 #m) below the interface.
ity complete ionization of the acceptors and neglect the effects of deep donors and holes. The Poisson's equation is thus written: d2u(x) q2 rN+ . . _ NA-- n(x)] dx 2 - E%kTI Do~X/
(1)
where
E~(x)
u(x) = - - kT
(2)
is the reduced conduction band energy measured from the Fermi level. The meaning of all other symbols used in this work is explained in Table 1. The ionized donor concentration N~o(x) is given by: NDo(X) U~o(X) = (3) 1 + 2 exp[-- u(x) + q)o] with
EDo
EDo kT
(4)
The free electron concentration n(x) is approximated by the formula: Arce .(x) n(x) - 1 + 0.25 e -u(x) "
(5)
NDo, n
2. THEMETHOD We consider a F E T with a donor distribution Nno(X) and uniformly distributed deep donors and acceptors with concentrations N o and N A. Within the active layers x < w (Fig. 1), where the Fermi level is close to the conduction band, we assume for simplicSSE37/2-K
0
W
X
Fig. 1. Hypothetical doping concentration Noo and free electron concentration n profiles in a FET. 373
374
C.D. KOURKOUTASand G. J. PAPAIOANNOU Table I Non NO Na n p Ev EOo ED EA Eg Nc NV
kT q¢ e0
+ 4N~ In
Shallow donor concentration Deep donor concentration Acceptor concentration Free electron concentration Hole concentration Fermi energy Shallow donor energy Deep donor energy Acceptor energy Energy gap Effective density of states in the conduction band Effective density of states in the valence band Thermal energy Absolute electron charge Static dielectric constant Permittivity of free space
1 + 0.25 e x p [ - u ( w ) ] 1 + 0.25 exp[--Uinf]
+ N~[expIu(w) -- Eg] -- exp(uinf -- Eg)]} (11) where Uinf is the potential energy at infinite depth, which can be calculated from the charge-balance equation for x = oo:
Nfi + p = N A
In the substrate (x > w), where the donor concentration NDo is zero, the existing charge is due to electrons, which are diffused there from the active layer, holes in the valence band and ionized deep centers. The Poisson's equation becomes: q~ _ rN+ EeokT t D ( x ) - - N A - - ( x ) - - n ( x ) + p ( x ) ] "
(6) The positively charged deep donor concentration N~ is given by (3), (4), if we replace the index " D o " with "D". The negatively charged acceptors are given by: NA NA(X ) = (7) 1 + 2 exp[u(x) -- EA]
x =oo.
(12)
We can solve now the Poisson's equation within the active region, from x = 0 to x = w calculating the band potential u(x) by means of a numerical method. The implicit method[6] after some necessary modifications, which will be demonstrated below, is very convenient, as it converges quickly to the final solution. We fix equally spaced points:
x(j)=jAx, d2u(x) _ dx 2
+n,
j=0,
1,2 . . . . L
over the region 0 . . . w, and expand the potential energy u(x) in Taylor series retaining terms up to the second derivative. We write thus: duO')
u(j'- l)=u(jl-~d~-x
1 d2u(j) • 2
Ax + ~ x 2
zxx
(13)
and
u(j+l)=u(j)+~x)AX
+ ~15d2uO') - x 2 Ax2.
(14)
where EA
(8)
EA -- k T
From (13) and (14) we obtain: u(j-l)-2u(j)+u(j+l)=d2u(j)" ~ dx
and the hole concentration p ( x ) is given by:
p ( x ) = Nv exp[u(x) -- Eg]
(9)
zxx
2 (15)
where the second derivative in the right hand side of (15) is given by (1) for x = x(j). F o r j = L we find:
where
E,
E~ = ~ .
(10)
We multiply both sides of (6) by du/dx and integrate from w to infinite taking into account that: d2u d u _ l dx 2 dx
d (du~ 2 2 dx \ d x / I
and that the potential gradient at infinite is zero. We obtain thus the following relation between the potential energy u(w) and its gradient du(w)/dx at the interface (x = w):
u(L + 1) = u(L - 1) +' 2 du(w) ~ Ax
where the potential gradient at x ( L ) = w is given by (11). The potential u(0) at x = 0 is known and equals the sum of the built in potential and the applied reverse bias voltage. Equations (15) constitute therefore a nonlinear system of L equations and L u n k n o w n potentials u(j) at the points x ( j ) with j = 1 , 2 . . . L. To solve it we suggest a trial potential profile u')(j). If the true profile is /2(r-I-l)(j), we set:
3u(n(j) = u(~ + l)(j) _ u(,.)(j). (du(w)) 2
(16)
(17)
2q~ {'AtDIn 1 +0"5exp[u(w)--ED] = ~
+Naln
1 + 0 . 5 exp[Uin f -- ED]
1 + 2 exp[u(w) -- EA] 1 + 2 exp[uinf -- ED]
To the first approximation we write: exp[u("~U) +
6u~°)U)] =
[1 +
6u'~U)]exp[uC'~U)]. (18)
375
A fast solution of Poisson's equation in FETs Neglecting for small 5u(j) the variation in the denominator of (3) and (5) we find that the free electron concentration is: n(~+ l)(j) = [1 - 5u(~)(j)]n(V)(j)
(19)
and the charged donor concentration is N ~ (~'+')(j) = [1 + 6ua')(j)]N~o(~')(j).
(20)
Since u(0) is known, 6u(O) ~- 0. On the other hand equation (16) yields 5u(L + 1 ) = 3 u ( L - l) for A x e 0 . We obtain therefore the linear system: - ~ (")(1)6u (~)(1) + 6u (~)(2) = fl (~')(i)
6u(V)(1) - a(~)(2)fu(~)(2) + 6u(~)(3) = fl(')(2)
26u(°(L - 1) - a(~)(L)fu(°)(L ) = fl")(L ) (21) where
q2~Ax2 a')(J ") = 2 + ~
[n(V)(j) + N~)o(~')(j)]
(22)
q2 Ax 2 fl')(j) = ~ [mGo')(j) - X A - n')(j)].
(23)
and
solution up to infinite depth, where the boundary conditions du/dx = 0 and u = Uinr are satisfied. The solutions, which correspond to curves (2) and (3) are obtained by shifting the boundary condition u = um~ to finite distance d from x = 0 (d = 0.2 and 0.6/~m). As the accuracy of our calculations is better than l % , the boundary condition should be shifted at least 1 # m below the interface in order to obtain in the region x < 0 a solution differing from the corresponding to curve (1) with the same accuracy. In usual FETs the active layer thickness is 0.20-0.25/~m. Following the proposed method we abbreviate therefore significantly the solution of Poisson's equation. (In this example our solution is five times faster than the obtained by the conventional method.) In Fig. 3 we see the solution of Poisson's equation for a VPE-like donor profile in a G a A s FET, where we considered the presence of two deep centers, a donor like native defect such as the EL2 at 0.69 eV below the conduction band with concentration N D = 1.5 × 1021m -3 and a midgap acceptor with various concentrations N A up to 2 x 1022 m -3. In our calculations we set Ax = 1 nm. The bias voltage was fixed at 0.8 V.
ND, n (m -31
The solution of the system is repeated with the new potential profile u (v+ l)(/.) as the trial potential until the obtained corrections 3u(]) are sufficiently small. To initiate the iterative procedure we use the following trial potential: u(')(/) : - l n { ~ [ - ( - ~
+ ~ exp(--EDo)) + 2
ND -- NA
q) (24)
for j = l , 2 . . . . L. Equation (24) satisfies the charge-balance equation at any point x(/). The convergence to the final solution is accelerated significantly, if the two terms in parentheses in (22) are of the same order of magnitude. This can be done by multiplying one of them with a weighting factor, for example, N3o(X) by n(x)/NDo(X). Relation (22) is reduced then to:
q~ Ax [ a(v'(jl=2+~Llq
N3o(~)(j)] ,,),., NDo(J)-~n (j).
1022
(1)
(2)
(25/
3. EXAMPLES
In Fig. 2 we see the solution of the Poisson's equation for a step function doping profile in G a A s at 300 K. The numerical data are taken from [7]. The solution, which corresponds to curve (i) is obtained from the proposed method and is equivalent to the
-0.05
0
0.05
x (~m) Fig. 2. The solution of Poisson's equation for a step function doping profile in GaAs at 300K. Curve (I) is obtained from the proposed method. Curves (2) and (3) are obtained by setting u~,r at infinite distance d from x = 0. Curve (2): d = 0.6tim. Curve (3): d =0.2#m.
376
C . D . KOURI(OUTASand G. J. PAPAIOANNOU NDo, n (1023 m -3) 1.5 (1) -
1,0
m
(5) (4) (3) (2) 0.5 -
I 0.05
0.10
0.15
0.20
x (~tm) Fig. 3. Free electron concentration n at 300 K in the active region of a GaAs FET with VPE-like donor profile [curve (I)] and various acceptor concentrations. (2): N A = 2 x 1022m -3, (3): N^ = 1022m -3, (4): N A = 5 x I021m -3 and (4): N A = 0. The deep donor concentration is in each case 1.5 x 1021m -3.
Using relation (22) for the correction factor ~(~)(/') we o b t a i n a solution with 8 u ( j ) < 0.01 after m a n y iterations, as the convergence into the space charge region (x < 0 . 0 7 / t i n ) is slow (Su(V+l)/fu (v) = 0.9). If we use relation (25) instead, then the convergence becomes m u c h faster a n d the solution is o b t a i n e d after six iterations with the same accuracy.
4. S U M M A R Y
U s i n g the continuity condition (11) at the interface between the active layer a n d the substrate o f a F E T we abbreviate significantly the solution o f the Poisson's equation. The m e t h o d is valid, if the deep centers are uniformly distributed. The implicit m e t h o d for the solution of the differential e q u a t i o n a n d converges after few iterations to the final sol-
ution, if the correction terms are considered with a p p r o p r i a t e weight. REFERENCES
I. D. P. Kennedy, P. C. Murlay and W. Kleinfelder, IBM J. Res. Develop. 12, 399 (1968). 2. W. C. Johnson and P. T. Panousis, IEEE Trans. Electron Devices ED-18, 965 (1971). 3. D. P. Kennedy and P. R. O'Brien, IBM J. Res. Develop. 13, 212 (1969). 4. K. Iniewsky and C. A. T. Salana, Solid-St. Electron. 34, 309 (1991). 5. X. H. Fu and J. N. Chen, Solid-St. Electron. 35, 181 (1992). 6. Mamoru Kurata, in Numerical Analysis for Semiconductor Devices. Lexington Books, D. C. Heath and Company, Toronto, Mass. (1982). 7. D. L. Rode, in Semiconductors and Semimetals (Edited by R. K. Willardson and A. C. Beer), Vol. 10, Chap. 2. Academic Press, New York (1974).