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An approximate calculation of the axial potential in electrostatic lenses
NUCLEAR
INSTRUMENTS
AND METHODS
70 0969) 3~7-3~9; © N O R T H - H O L L A N D
PUBLISHING
CO.
AN A P P R O X I M A T E C A L C U L A T I O N OF T H E AXIAL P O T E N T I A L IN E L E C T R O S T A T I C LENSES G. BOSI Universitd de Moncton, Moncton, N.-B., Canada Received 14 January 1969 An approximation method yielding an accurate picture of the fieldform in the neighbourhood o f the symmetry axis in lens systems is described. Some simple problems are solved and formulae are compared with results from already known approaches.
1. Introduction Some recent papers l'z) were devoted to the attempt of carrying out an approximate description of the scalar potential of axially symmetric fields. Provided a symmetry plane is present, our approximate method yields reliable results in the neighbourhood of this plane and at suitable distances from the symmetry axis. We have also derived 3) an improved method removing the uncertainty about the values the parameter a should be given and yielding a more accurate picture of the fieldform in the vicinity of the median plane. The scalar potential turns out to be correct to within z3-terms, while the new formulae show that taking a = r, ref. 2) is just a particular solution of the equation defining the right function a(r), which is responsible for enhancing the accuracy. This procedure should be quite helpful when investigating "flat" systems, where the most interesting region is centered at the symmetry plane. "Lens" systems require some different approach, for they are conceived to utilize extended regions surrounding the symmetry axis. That is why we are going to describe another attempt of evaluating axially symmetric potentials: it should give the best results in the neighbourhood of the symmetry axis and fit the boundary conditions with an absolute precision. 2. The Laplace equation Consider an axially symmetric potential satisfying the equation
V(r,z)
g:2F/Or2+(l/r)(SV/~r)+82V/Oz2 : 0,
(1)
Let us assume: p = 2-
(6)
and neglect r4-terms:
82 V/~P 2 + g)2V/~22 = ~P2(~4V/~P4)o: o.
(7)
A first order solution is obtained by solving for Vl(p,z ) :
2vd@ 2 + 2v1/ z2 = 0.
(8)
If the boundary cross-section consists of a number of straight line segments, a solution of eq. (8) fitting the given boundary conditions is found by mapping the (-plane ( ( = p + i z ) onto the upper half t-plane (t = u + iv) and regarding the potential as the imaginary coefficient of a complex function W 1 = U 1+iV1 analytic in the half plane v > 0. As the transformation c~= ((t)
(9)
is a conformal one, the complex potential and its derivatives are (-functions analytic in the interelectrode region. Let us assume:
f(u,v) = ~p4(~)4V,/~p4),,=o,
(10)
f(u,v) being continuous along with its partial derivatives of any order in the upper half t-plane. A third order solution of eq. (7) may be obtained by solving for V3(p,z):
2v3/& + 2v3/ z2 =
(11)
We will assume: V3 = V, +A,
within an electrode-bounded region. Let us expand:
62A/6p2+~2A/~zZ = lpE(O4V,/Op4)o= o.
V(r,z) = A ( z ) + r 2 B ( z ) + r 4 C ( z ) + O ( r 6 ) ,
(2)
OV/Or = 2rB(z) + 4r3C(z) + O(rS),
(3)
~2V/~rZ = 2B(z)+ 12r2C(z)+O(r4),
(4)
(12)
whence (13)
As the last equation is correct to within p4-terms, it may be transformed as follows:
whence
02~b3/Sp 2 + ~2q53/~z2 = 0,
(14)
q~3 = A - - 7 - i2 D 4"~4V ~. 1/"0 I0 4,)p=0"
(15)
with
(1/r) (~V/~r) = c?2V/~r2 - ½r2(O4V/~r 4),_ o + 0(r4). (5) 317
318
G. BOSI
Now eq. (14) may be solved in the same way as eq. (8): at the boundary, however, (~3 must coincide with _ ~p4(~4
Vll~p4),
= o.
Provided the electrode boundary is represented by the line v = 0 in the t-plane : ¢s(U,0) = -f(u,O).
(16)
Assume the existence of a real function ff3(u) making the complex function
The above problem can be solved exactly4). The rather complicated solution turns out to be closely approximatedS), as far as the axial potential is concerned, by the simple function:
V(O,z) = Votanh(1.318z/R).
(28)
Table 1 shows the amount of discrepancies between this formula and ours. TABLE 1
W3(t) = ~bz(t,0) + i~3(t),
(17)
an analytic one in the upper half t-plane. Its real part fulfills the Laplace equation in the last domain and coincides with the right side member of eq. (16) on the boundary line. Therefore the function
zlR
Vl Vo
V3IVo
0.0
0.0000
0.0000
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 2.0 3.0 4.0 5.0 6.0
0.1310 0.2576 0.3760 0.4832 0.5776 0.6588 0.7271 0.7835 0.8293 0.8662 0.9897 0.9992 0.9999 1.0000 1.0000
0.1402 0.2739 0.3959 0.5032 0.5949 0.6716 0.7350 0.7866 0.8286 0.8624 0.9850 0.9983 0.9998 0.9999 1.0000
A(p,z) = Re {W3[t(#)] } +7-~p4(O4V1/Op4)p=o, (18) is the solution we were looking for. The axial potential is then given by: V3(0,z) = Vl(0,z) -[- Re {W3[t(iz)] }.
(19)
3. Two cylinders lens Consider two coaxial cylinders of radius R separated by a negligible distance and held at different voltages: / = + V o , for r = R and z > 0 ,
V(r,z) = - V o , for r = R and z < 0. Then: ~(t) = (2~/n)R arcsin t,
(20)
Wl(t) = iVo+(Vo/n)ln[(t+l)/(t-1)],
(21)
d4W1/d( 4 = ln3(Vo/R4)t(5+t2)(l--t2) -2.
(22)
4. Single slit lens Consider an infinite conducting plane with a circular hole of radius R. Assume the plane to separate two regions of uniform field strengths:
( ~ V o - E l z , asz--*+°°, V(r,z) Vo, for z = 0 and r = R,
As t = i v for p = 0 :
(~4V1/Op4)p
= 0
=
Vo--E2z,
½n3(Vo/R`) v(5 v z) (1 + v2) - 2, (23)
as
(29)
z~-oo.
-
f(u,O) = q~3(u,0) = 0,
(24)
W3(t) = 0.
(25)
The origin of the coordinates has been taken at the center of the hole. Then:
V(r,z) = Vo - E2z + F(r,z),
Finally:
V3(O,z) = Vo{1 - (2/re) arctan [sinh (2- }zcz/R)] -1 }. (26)
where F(r,z) is a function fulfilling eq. (1) and the following boundary conditions:
Since we mapped the region defined by
V3(O,z) = Vo[a - (4/~) arctan exp ( - 2- ~nz/R)]. (27)
[
"+(E2--E1)z , as z - ~ + o o ,
--2-}R 0 onto the upper half t-plane, the last formula is only valid for z > 0 . A function coinciding with eq. (26) for positive z-values and fulfilling the antisymmetry condition 1/3(0,- z) = - I/3(0, + z) is easily found:
(30)
F(r,z)
=0, forz=0andr=R,
---~0, a s z - - - ~ - - o o .
We obtain: ((t) = 2-~-R(t + t-1),
(31)
p(u,v)= 2-~R[u+u(u2 +v2)-l], z(u,v) = 2-~R[v-v(uZ +v2)-~],
(32)
or
(33)
THE A X I A L P O T E N T I A L
IN E L E C T R O S T A T I C
TABLE 2
z/R 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0
A = 2 - ~ ( E 2 - E1)R.
½[(z/R)2+½] (l/.n)[l+(z/R)arctan(z/R)l 0.3535 0.3570 0.3674 0.3840 0.4062 0.4330 0.4636 0.4974 0.5338 0.5722 0.6123 1.060 1.541 2.031 2.524 3.020 3.517 4.015 4.513 5.012
319
LENSES
Furthermore: d 4 W 1/d~ 4 = - 2~[-384(E2 - E , ) / R3].
0.3183 0.3214 0.3308 0.3461 0.3667 0.3921 0.4215 0.4543 0.4901 0.5282 0.5683 1.023 1.511 2.006 2.504 3.002 3.502 4.001 4.501 5.001
• ts(t4+3t2+l)(t2-1)-7,
"V5(V4--3V2+I)(l+v2) -7, f ( u , 0 ) = ~ 3 ( u , 0 ) = W3(t ) = 0.
U 1
+ i F ~ = At,
(37) (38)
Then : V3(0,z) = V o - 1(E1 + E z ) z + ½(E 2 - E ~ ) ( z 2 + ½R2) ~. (39) T h e a b o v e p r o b l e m can still be s o l v e d exactly6): V(O,z) = Vo - ½(El + E2)z + (lfiz)(E2 - E,)-
• [R + z arctan(z/R)]. T a b l e 2 shows the a m o u n t this f o r m u l a a n d ours.
and =
(36)
(c~4V1/~p4)p = 0 = 2-~[384(E2 - E l ) / R33"
(34)
A b e i n g a real c o n s t a n t . A s v t e n d s to infinity, z a n d F t e n d to 2 - ~ R v a n d 2 ~ A z / R , respectively, w h e n c e :
(40)
of discrepancies between
References Wl(t)
(35)
1) G. Bosi, Nucl. Instr. and Meth. 46 (1967) 55. 2) G. Bosi, Nucl. Instr. and Meth. 49 (1967) 194. 3) To be published. 4) p. Grivet, Electron optics (Pergamon Press, 1965) p. 164. 5) F. Gray, Bell Syst. Techn. J. 18 (1939) 1. o) Ref. 4), p. 20.
INSTRUMENTS
AND METHODS
70 0969) 3~7-3~9; © N O R T H - H O L L A N D
PUBLISHING
CO.
AN A P P R O X I M A T E C A L C U L A T I O N OF T H E AXIAL P O T E N T I A L IN E L E C T R O S T A T I C LENSES G. BOSI Universitd de Moncton, Moncton, N.-B., Canada Received 14 January 1969 An approximation method yielding an accurate picture of the fieldform in the neighbourhood o f the symmetry axis in lens systems is described. Some simple problems are solved and formulae are compared with results from already known approaches.
1. Introduction Some recent papers l'z) were devoted to the attempt of carrying out an approximate description of the scalar potential of axially symmetric fields. Provided a symmetry plane is present, our approximate method yields reliable results in the neighbourhood of this plane and at suitable distances from the symmetry axis. We have also derived 3) an improved method removing the uncertainty about the values the parameter a should be given and yielding a more accurate picture of the fieldform in the vicinity of the median plane. The scalar potential turns out to be correct to within z3-terms, while the new formulae show that taking a = r, ref. 2) is just a particular solution of the equation defining the right function a(r), which is responsible for enhancing the accuracy. This procedure should be quite helpful when investigating "flat" systems, where the most interesting region is centered at the symmetry plane. "Lens" systems require some different approach, for they are conceived to utilize extended regions surrounding the symmetry axis. That is why we are going to describe another attempt of evaluating axially symmetric potentials: it should give the best results in the neighbourhood of the symmetry axis and fit the boundary conditions with an absolute precision. 2. The Laplace equation Consider an axially symmetric potential satisfying the equation
V(r,z)
g:2F/Or2+(l/r)(SV/~r)+82V/Oz2 : 0,
(1)
Let us assume: p = 2-
(6)
and neglect r4-terms:
82 V/~P 2 + g)2V/~22 = ~P2(~4V/~P4)o: o.
(7)
A first order solution is obtained by solving for Vl(p,z ) :
2vd@ 2 + 2v1/ z2 = 0.
(8)
If the boundary cross-section consists of a number of straight line segments, a solution of eq. (8) fitting the given boundary conditions is found by mapping the (-plane ( ( = p + i z ) onto the upper half t-plane (t = u + iv) and regarding the potential as the imaginary coefficient of a complex function W 1 = U 1+iV1 analytic in the half plane v > 0. As the transformation c~= ((t)
(9)
is a conformal one, the complex potential and its derivatives are (-functions analytic in the interelectrode region. Let us assume:
f(u,v) = ~p4(~)4V,/~p4),,=o,
(10)
f(u,v) being continuous along with its partial derivatives of any order in the upper half t-plane. A third order solution of eq. (7) may be obtained by solving for V3(p,z):
2v3/& + 2v3/ z2 =
(11)
We will assume: V3 = V, +A,
within an electrode-bounded region. Let us expand:
62A/6p2+~2A/~zZ = lpE(O4V,/Op4)o= o.
V(r,z) = A ( z ) + r 2 B ( z ) + r 4 C ( z ) + O ( r 6 ) ,
(2)
OV/Or = 2rB(z) + 4r3C(z) + O(rS),
(3)
~2V/~rZ = 2B(z)+ 12r2C(z)+O(r4),
(4)
(12)
whence (13)
As the last equation is correct to within p4-terms, it may be transformed as follows:
whence
02~b3/Sp 2 + ~2q53/~z2 = 0,
(14)
q~3 = A - - 7 - i2 D 4"~4V ~. 1/"0 I0 4,)p=0"
(15)
with
(1/r) (~V/~r) = c?2V/~r2 - ½r2(O4V/~r 4),_ o + 0(r4). (5) 317
318
G. BOSI
Now eq. (14) may be solved in the same way as eq. (8): at the boundary, however, (~3 must coincide with _ ~p4(~4
Vll~p4),
= o.
Provided the electrode boundary is represented by the line v = 0 in the t-plane : ¢s(U,0) = -f(u,O).
(16)
Assume the existence of a real function ff3(u) making the complex function
The above problem can be solved exactly4). The rather complicated solution turns out to be closely approximatedS), as far as the axial potential is concerned, by the simple function:
V(O,z) = Votanh(1.318z/R).
(28)
Table 1 shows the amount of discrepancies between this formula and ours. TABLE 1
W3(t) = ~bz(t,0) + i~3(t),
(17)
an analytic one in the upper half t-plane. Its real part fulfills the Laplace equation in the last domain and coincides with the right side member of eq. (16) on the boundary line. Therefore the function
zlR
Vl Vo
V3IVo
0.0
0.0000
0.0000
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 2.0 3.0 4.0 5.0 6.0
0.1310 0.2576 0.3760 0.4832 0.5776 0.6588 0.7271 0.7835 0.8293 0.8662 0.9897 0.9992 0.9999 1.0000 1.0000
0.1402 0.2739 0.3959 0.5032 0.5949 0.6716 0.7350 0.7866 0.8286 0.8624 0.9850 0.9983 0.9998 0.9999 1.0000
A(p,z) = Re {W3[t(#)] } +7-~p4(O4V1/Op4)p=o, (18) is the solution we were looking for. The axial potential is then given by: V3(0,z) = Vl(0,z) -[- Re {W3[t(iz)] }.
(19)
3. Two cylinders lens Consider two coaxial cylinders of radius R separated by a negligible distance and held at different voltages: / = + V o , for r = R and z > 0 ,
V(r,z) = - V o , for r = R and z < 0. Then: ~(t) = (2~/n)R arcsin t,
(20)
Wl(t) = iVo+(Vo/n)ln[(t+l)/(t-1)],
(21)
d4W1/d( 4 = ln3(Vo/R4)t(5+t2)(l--t2) -2.
(22)
4. Single slit lens Consider an infinite conducting plane with a circular hole of radius R. Assume the plane to separate two regions of uniform field strengths:
( ~ V o - E l z , asz--*+°°, V(r,z) Vo, for z = 0 and r = R,
As t = i v for p = 0 :
(~4V1/Op4)p
= 0
=
Vo--E2z,
½n3(Vo/R`) v(5 v z) (1 + v2) - 2, (23)
as
(29)
z~-oo.
-
f(u,O) = q~3(u,0) = 0,
(24)
W3(t) = 0.
(25)
The origin of the coordinates has been taken at the center of the hole. Then:
V(r,z) = Vo - E2z + F(r,z),
Finally:
V3(O,z) = Vo{1 - (2/re) arctan [sinh (2- }zcz/R)] -1 }. (26)
where F(r,z) is a function fulfilling eq. (1) and the following boundary conditions:
Since we mapped the region defined by
V3(O,z) = Vo[a - (4/~) arctan exp ( - 2- ~nz/R)]. (27)
[
"+(E2--E1)z , as z - ~ + o o ,
--2-}R
(30)
F(r,z)
=0, forz=0andr=R,
---~0, a s z - - - ~ - - o o .
We obtain: ((t) = 2-~-R(t + t-1),
(31)
p(u,v)= 2-~R[u+u(u2 +v2)-l], z(u,v) = 2-~R[v-v(uZ +v2)-~],
(32)
or
(33)
THE A X I A L P O T E N T I A L
IN E L E C T R O S T A T I C
TABLE 2
z/R 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0
A = 2 - ~ ( E 2 - E1)R.
½[(z/R)2+½] (l/.n)[l+(z/R)arctan(z/R)l 0.3535 0.3570 0.3674 0.3840 0.4062 0.4330 0.4636 0.4974 0.5338 0.5722 0.6123 1.060 1.541 2.031 2.524 3.020 3.517 4.015 4.513 5.012
319
LENSES
Furthermore: d 4 W 1/d~ 4 = - 2~[-384(E2 - E , ) / R3].
0.3183 0.3214 0.3308 0.3461 0.3667 0.3921 0.4215 0.4543 0.4901 0.5282 0.5683 1.023 1.511 2.006 2.504 3.002 3.502 4.001 4.501 5.001
• ts(t4+3t2+l)(t2-1)-7,
"V5(V4--3V2+I)(l+v2) -7, f ( u , 0 ) = ~ 3 ( u , 0 ) = W3(t ) = 0.
U 1
+ i F ~ = At,
(37) (38)
Then : V3(0,z) = V o - 1(E1 + E z ) z + ½(E 2 - E ~ ) ( z 2 + ½R2) ~. (39) T h e a b o v e p r o b l e m can still be s o l v e d exactly6): V(O,z) = Vo - ½(El + E2)z + (lfiz)(E2 - E,)-
• [R + z arctan(z/R)]. T a b l e 2 shows the a m o u n t this f o r m u l a a n d ours.
and =
(36)
(c~4V1/~p4)p = 0 = 2-~[384(E2 - E l ) / R33"
(34)
A b e i n g a real c o n s t a n t . A s v t e n d s to infinity, z a n d F t e n d to 2 - ~ R v a n d 2 ~ A z / R , respectively, w h e n c e :
(40)
of discrepancies between
References Wl(t)
(35)
1) G. Bosi, Nucl. Instr. and Meth. 46 (1967) 55. 2) G. Bosi, Nucl. Instr. and Meth. 49 (1967) 194. 3) To be published. 4) p. Grivet, Electron optics (Pergamon Press, 1965) p. 164. 5) F. Gray, Bell Syst. Techn. J. 18 (1939) 1. o) Ref. 4), p. 20.