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========= SUPPLEMENTARY MATERIAL FOR STORAGE IN EPAPS ========
J. L. Reed
Manhattan LCD, R&D Dept., 1007 Brielle Ave., Oviedo, Florida 32765, USA
Unabridged mathematical development of governing equations for the conduction
coupled Tesla transformer
FIG.1. Schematic of conduction-coupled Tesla transformer. (a) is supply of conditioned
power, C1 is primary capacitor, L1 is primary inductor, (SG) is spark-gap switch, (b) is
conductor feeding base of secondary inductor L2, C2 is self-capacitance of L2, and (c) is
high-voltage output terminal. L1 is the common branch between the primary and
secondary loops. I1 and I2 are the currents in the circuit loops. The circuit is constructed
with a very high Q and resistances are negligible. R1 and R2 of the circuit loops are zero.
The circuit of FIG.1, with the specification of appropriate circuit element settings,
is theoretically capable of achieving, increasingly large voltage gains over corresponding
increasingly large time intervals. If  is the resonant angular frequency of loop 1, and if
r 1   is the ratio of the two characteristic frequencies of the coupled circuit, then an
upper limit voltage gain (capacitor 1 to capacitor 2) of 1/  , associated with a beat
frequency  within an envelope frequency of   / 2 , may be established for a
continuous range of circuit settings with the restriction that the quantity  must be small,
with its value determined by ε ≈ (L1/L2)1/2 = (C2/C1)1/2 (ε is the reciprocal of the voltage
gain). The circuit analysis for this xfmr assumes L1C1 = L2C2. R1 and R2 are shown in the
loop equations, to demonstrate how they would enter the analysis for investigators
wanting to develop the damping parameters for the new oscillator. In the present analysis
they are set to zero for a simplified freely oscillating circuit. Finding the “frequency
equation,” and the time histories of the primary and secondary voltages and currents are
presented as follows:
The Kirchhoff voltage drop equation for loop 1, assuming an initial charge Q0
and initial voltage V0  Q0 / C1 on capacitor 1, may be written as follows:
t
V0 
1
I1dt  L1 ( I1  I2 )  R1 ( I1  I 2 ) ,
C1 0
or, upon differentiation,
1
(1)
I1
 L1 ( I1  I2 )  R1 ( I1  I2 )  0 .
C1
(1a)
Similarly, for loop 2:
t
1
I 2dt  L1 ( I1  I2 )  L2 I2  R2 I 2  0
C2 0
(2)
and, upon differentiation,
I2
 L1 ( I1  I2 )  L2 I2  R2 I2  0 .
C2
Set the resistances to zero and add (1a) and (2a):
I1 I 2

 L2 I2  0 ,
C1 C2
so that
C
I1   1 I 2  C1L2 I2 .
C2
Substitute into (1a):
I
C
 2  L2 I2  L1 ( 1 I2  C1L2I2  I2 )  0 .
C2
C2
(2a)
(3)
(4)
Eliminate the minus signs. Let 1  L1C1 and 2  L2C2 . Then (4) may be written
C
I 2   2 I2  1I2   1 2I2  2  1I2  0 ,
C1
or
I 2  [ 2  (1 
C2
) 1] I2   1 2I2  0
C1
(5)
The characteristic frequency equation for characteristic function ei t is then
1  [ 2  (1 
C2
) 1 ] 2   1 2 4  0
C1
(6)
and so
2 
 2  (1 
C2
C
)1  [ 2  (1  2 )1 ]2  4 12
C1
C1
.
2 12
(7)
Refer now to the material set forth in the Appendix. Using the Appendix
definitions in this supplemental material (in terms common to Tesla xfmr analysis, r is
the “frequency ratio” and s2 is the “tuning ratio”)
2
r

1

s2 
1 L1C1
.

2 L2C2
(8)
The Appendix derived result
C2 1
 [ (r )   ( s )]
C1 s
[  ( x)  x 
1
],
x
(9)
implying
1
sr
r
(9a)
1  sr  r 2
(9b)
0  sr  1  r 2  1 ,
(9c)
or
or
Eq. (7) may be divided in the numerator and denominator by 2 to obtain
2 
1  s 2  s[ (r )   ( s )]  {1  s 2  s[ (r )   ( s )]}2  4s 2
.
2 1
(10)
But s (s)  s(s  1/ s)  1  s 2 , so that
r  1 / r  (r  1 / r ) sr /  1
2   (r )   2 (r )  4
.

s

s


s
/
r

2 1
2 1
2 
1

(11)
The initial current for this circuit is zero, at the time of the switch closing.
Therefore, an appropriate current solution formulation may be written
I1  A1 sin t  B1 sin t
I 2  A2 sin t  B2 sin t .
(12)
Substituting these equations Eq. (1a) and requiring the coefficients of sin t and
sin t to vanish, there results, using Eq. (11),
A1   1( A12  A22 )  0
or
A2  A1(1 
1
 12
)  A1(1 
1
).
sr
and
B1   1(B12  B22 )  0
3
(13a)
or
B2  B1 (1 
1
 12
)  B1 (1  r / s ) .
(13b)
Substituting Eq. (12) into Eq. (1), but now evaluating at time zero, there comes:
Q0   1( A1  B1  A2  B2 )
  1 ( A1  A2 )   1 ( B1  B2 )
  1 A1 / sr   1 B1r / s ,
using Eqs. (13). Using Eq. (11), the last result can be written
Q0  Q0
sr
1
.
Similarly, substitute Eq. (12) into Eq. (2) and evaluate at time zero to obtain:
Q  2
or, with Q  0 at time zero,
L1  
( I1  I 2 )  2 I2  0 ,
L2
L
I1  (1  2 ) I2 ,
L1
or
A1  B1  (1 
L2
)( A2  B2 ) .
L1
Subsituting from Eqs. (13), the last equation may be written
A1 [1  (1 
L2
1
L
)(1  )]  B1 [1  (1  2 )(1  r / s)]  0 ,
L1
sr
L1
or, using the first of Eqs. (8),
1 L
1
r L
A1[  2 (r  )]  B1[  2 (1  r / s )]  0
s L1
s
s L1
Multiply by L1 / L2 and use, from the Appendix, the result
4
(14)
L1
 s[ (r )   ( s)]
L2
(15)
to obtain
1
A1{[ (r )   ( s)]  (r  )}  B1{r[ (r )   ( s)]  (1  r / s)}  0
s
or
A1[( r 
1
1
1
1
1
 s  )  (r  )]  B1{r[r   s  ]  (1  r / s )}  0
r
s
s
r
s
or
(1/ r  s) A1  r (r  s) B1  0.
(16)
This and Eq. (14) are two equation in two unknowns that may be written in matrix form
as
1/ r  s r (r  s)  A1   0 

     ,
r  B1   Q0 
 1
so that
 A1 
1  r  r (r  s ))  0 
  

 
2
 B1  1  r   1 1 / r  s  Q0 

Q0   r (r  s) 

 ,
1  r 2  1/ r  s 
or
A1 
r (r  s)Q0
r2 1
Eqs. (13) then become
A2  A1(1 
( sr  1)Q0
.
r (r 2  1)
(17)
1
(r  s)( sr  1)
)
Q0
sr
s(r 2  1)
(18a)
and
B2  B1 (1  r / s)  
B1 
(r  s)( sr  1)
sr (r 2  1)
Q0   A2 / r
(18b)
t
1
The voltage on capacitor 2 is V2 
I 2 dt , which may also be written, using Eq. (2),
C2 0
V2  L1( I1  I2 )  L2 I2 .
But from Eq. (1), the first term on the right is also the voltage on capacitor 1, so that
5
(19)
t
1
V1  L1 ( I1  I2 )  V0   I1dt .
C1 0
(20)
Therefore, using Eq. (12) and the last equation, Eq. (19) may be written
V2  V1  L2 ( A2 cos t  B2 cos t ) ,
(21)
where
t
V1  V0 
1
I1dt
C1 0
t
1
 V0   ( A1 sin t  B1 sin t )dt
C1 0
 V0 
 V0 
1
C1
[
A1
(cos t  1)  B1 (cos t  1)]
r
[use (8)]
Q0
[r (r  s) cos t  ( sr  1) cos t  (r 2  1)]
2
C1 r (r  1)

[use (17)]
V0
[r (r  s) cos t  (sr  1) cos t ] [use (11), (14)]
r 1
V
 2 0 [(r 2  1) cos t  (sr  1)(cos t  cos t )]
r 1
sr  1
 V0 cos t  V0 2 (cos t  cos t ) ,
r 1
2
(22)
and the voltage on the primary capacitor can be written as
V1  V0 cos  t  2V0
 r 1 s   r 1 s 
sin 
 t  sin 
t,
r   2
r 
r 1  2
sr  1
2
(22a)
where
ω = 1/(λ1)1/2 = 1/(L1C1)1/2.
Noting that
Q0 L1  Q0 L1C1 / C1  V0 1
and
Q0 L2  Q0 L1L2 / L1 
6
V01
s[ (r )   (s)]
(23)
(24)

V0 1
V0 1
V0 1r
,


1
1
1
(
r

s
)(
sr

1
)
s(r   s  ) s(r  s)(1  )
r
s
sr
(25)
the right member of Eq. (21) may be evaluated as follows:
 L2 ( A2 cos t  B2 cos t )
  L2 ( A2 cos t 
s
1
 (r cos t  cos t )
r
r
  L2 A2
  L2
A2
 cos t )
r
[using (18b)]
[using (11), (23)]
(r  s)( sr  1)
1
1
Q0
 (r cos t  cos t )
2
sr
r
r 1
  L2
(r  s)( sr  1)
1
Q0 2 (r cos t  cos t )
2
r
r 1

V0
r 1
2
(r 2 cos t  cos t )
[using (18a)]
[using (14)]
[using (25)].
(26)
Now add Eqs. (22) and (26) to obtain Eq. (21), and various expressions for V2(t):
V2  V1  L2 ( A2 cos t  B2 cos t )
V2 
V0
V
[r (r  s) cos t  (sr  1) cos t ]  2 0 (r 2 cos t  cos t )
r 1
r 1
2
 V0
sr
s
( cos sr t  cos  t ) ,
r
r 1
(27)
 r 1 s   r 1 s 
2sr
sin 
 t  sin 
t.
2
r   2
r 
r 1  2
(28)
2
or
V2  V0
 2V0
 r 1 s   r 1 s 
sr  1  r  1 s   r  1 s  2V0

 sin 


sin

t

t
sin
 t  sin 
t,
r   2
r  r 2  1  2
r   2
r 
r 2  1  2
or, from Eq. (22a)
7
V2  V1  V0 cos t 
 r 1 s   r 1 s 
sin 
 t  sin 
t.
r   2
r 
r 1  2
2V0
2
(29)
From Eq. (9c), there follows
0
sr  1
 1,
r2 1
(30)
so that, from Eq. (22a), it may be determined that V1  V0 cos t  2V0 . Letting
V1 = V0 cos (ω+ t) = 2V0ψ
Eq. (29), giving the voltage of the self-capacitance C2, may be written
V2  2V0  
 r 1 s   r 1 s 
sin 
 t  sin 
t,
r   2
r 
r 1  2
2V0
2
(31)
where   1 .
The voltage gain of this circuit may be arbitrarily set by choosing some very small
positive value  such that r 1   and then evaluate the circuit resonance frequency
1
L1C1
ratio, s 
, which, according to Eq. (9a), must satisfy the condition  s  r , or
r
L2C2
1   s  1  .
This last inequality is satisfied for any given value of  if s  1 (ie. L1C1=L2C2, the
resonant condition); so choose this value. Now from the Appendix, with s  1 ,
L1 C2
1

 V (r )  V (1)  r   2  1    (1     2  ...)  2   2 ,
L2 C1
r
or

L1
C2
,

L2
C1
(32)
and so
Gain  G 
1


L2
C1
,

L1
C2
(33)
With s  1 (resonance condition) and  from Eq. (32) very small, Eq. (31) becomes
(ignoring the first term as an approximation) the time history of voltage in the secondary
circuit
V2 ≈ GVo sin(ωt) sin(ωt/2G)
(34)
With
8
  1/ L1C1
paper).
(the time history of the secondary current is derived in the text of the
(35)
The carrier frequency is given by
f c = ω/(2π) Hz.
(35a)
f b = ω/(4πG),
(35b)
The lower beat frequency is given by
which results in the achievement of extremum quarter-wave voltages at time intervals of
t1 
4
G
sec.

(35c)
In this quarter-envelop time, the carrier frequency generates
fb t 1 
4
G
cycles.
2
(35d)
APPENDIX
Here is Eq. (7), repeated, with 1  L1C1 and  2  L2C2 :
2 
1  (1 
L1
L
)2  [1  (1  1 )2 ]2  412
L2
L2
.
212
(1)
Let the frequency ratio be r   /  . Then
1  (1 
r2 
Let b  C2 / C1 . Then
L1
L
)2  [1  (1  1 )2 ]2  412
L2
L2
L
L
1  (1  1 )2  [1  (1  1 )2 ]2  412
L2
L2
.
L1 L1C1 C2 1


b  b ,
L2 L2C2 C1  2
where now    1/ 2 . Divide the numerator and denominator of Eq. (2) by 2 to get
9
(2)
r 
2
  (1   b)  [   (1   b)]2  4
  (1   b)  [   (1   b)]2  4
,
or, with c  1  b  1  C2 / C1 ,
r2 
1   c  [1   c]2  4 
1   c  [1   c]  4 
2
.
Let x  1   c and solve for x:
r 
2
x  x2  4
x  x2  4
r 2 x  r 2 x 2  4  x  x 2  4
(r 2  1) x  (r 2  1) x2  4
(r 2  1)2 x2  (r 2  1)2 x2  4(r 2  1)2 
(r 2  1)2   r 2 x2
1
(r  ) 2   x 2
r
1
x  (r  )   1   c ,
r
so that
1
c  [( r  )   1] /   1  b ,
r
and
b
C2

C1
1
1
(r  )  (  
)
r


,
or, in summary, letting   s 2 and defining
V ( x)  x 
1
,
x
(3)
there comes
C2 1
 [V (r )  V ( s )]
C1 s
(4)
where
r


s
Also, using Eq. (5) with Eq. (4),
10
L1C1
.
L2C2
(5)
L1
 s[V (r )  V ( s )] .
L2
Notice that x 
1
has an absolute minimum at x  1 .
x
11
(6)