This document complements this other
one, discussing possible design procedures for a "double resonance
solid-state Tesla coil", but now considering a lossless system [1]. The
objective is to design the structure below, when the driving signal is
sinusoidal (or a square wave, since this narrow bandwidth system responds
essentially only to the fundamental harmonic), requiring that after a
certain number of cycles of the input all the energy in the network is
concentrated at the output capacitance *C*_{b}. This is then equivalent to a Tesla
coil, where functionality is achieved through the zero-state response of
the network, caused by the input voltage V_{in}, instead of by the zero-input
response of the classical system, caused by an initial voltage in C_{a}, with V_{in }= 0.

The design can be more easily done by first designing the
transformerless structure below, and then converting it to the final
structure through the relations shown. w_{0}
is the "base frequency" in rad/s that multiplies *k*, *l*,
and *m* (see below). The usual design would be to specify *C*_{a},
*C*_{b}, and *L*_{b}, and compute w_{0}
and *L*_{a} from the equations. In all cases the coupling
coefficient *k*_{ab} depends only on the "mode" k:l:m.

w_{0}^{2}*C*_{b}*L*_{b}
= *L*_{2}*C*_{2}

w_{0}^{2}*C*_{a}*L*_{a}
= *C*_{1}(*L*_{1}+*L*_{2})

*k*_{ab} = (*L*_{2}/(*L*_{1}+*L*_{2}))^{1/2}

The circuit is a 4th-order network with two resonance
frequencies (pairs of poles at *j*w in
the transfer function), and excitation is at a third frequency. I will
assume that these three frequencies are in a ratio *k*:*l*:*m*
or three positive integers. I will also consider initially a normalized
design where the three frequencies are just *k*, *l*, and
*m* rad/s (w_{0} = 1). Three
possible designs that result in complete energy transfer can be
identified:

a) The input signal is a cosinusoid, and the three
frequencies are in a ratio of successive integers with odd differences (as
1:2:3, or 10:11:14). The excitation is at the central frequency,
l. Complete energy transfer occurs in *l*/2 cycles of the
input. The required element values are:

*C*_{2} = 1

*L*_{2} = 1/*l*^{2}

*L*_{1} = *l*^{2}/((*k*^{2}-*l*^{2})(*l*^{2}-*m*^{2}))

*C*_{1} = ((*l*^{2}-*m*^{2})(*k*^{2}-*l*^{2}))/(*k*^{2}*m*^{2})

b) The input signal is a sinusoid, and the three frequencies
are in a ratio of successive odd integers with double odd differences (as
1:3:5, or 21:23:29). The excitation is at the central frequency,
l. This is the most practical design.Complete energy transfer
occurs in *l*/4 cycles of the input. The element values are:

*C*_{2} = 1

*L*_{2} = (*k*-*l*+*m*)/(*klm*)

*L*_{1} = *l*(*k*-*l*+*m*)/((*k*-*l*)(*k*+*m*)^{2}(*l*-*m*))

*C*_{1} = (*l*-*m*)(*k*+*m*)^{2}(*k*-*l*)/(*km*(*k*-*l*+*m*)^{2})

c) The input signal is a sinusoid, and the three frequencies
are in a ratio of successive odd integers with double odd differences, as
in design (b). The excitation is at the upper frequency, *m*. This
design is usually impractical, but listed here too. Complete energy
transfer occurs in *l*/4 cycles of the input. The element values
are:

*C*_{2} = 1

*L*_{2} = (*k*-*l*+*m*)/(*klm*)

*L*_{1} = *-m*(*k*-*l*+*m*)/((*k*-*l*)^{2}(*k*+*m*)(*l*-*m*))

*C*_{1} = -(*l*-*m*)(*k*+*m*)(*k*-*l*)^{2}/(*kl*(*k*-*l*+*m*)^{2})

There is no solution (with positive elements) for excitation at the lower
frequency, *k*.

The voltage gains (maximum V_{out} / peak V_{in}) for the three designs can be
calculated as:

a)Av = (*C*_{a}/*C*_{b})^{1/2} 2*km*/((*k*^{2}-*l*^{2})(*l*^{2}-*m*^{2}))^{1/2}

b)Av = (*C*_{a}/*C*_{b})^{1/2} (*km*/((*l-m*)(*k-l*)))^{1/2}

c)Av = (*C*_{a}/*C*_{b})^{1/2} (*kl*/((*k+m*)(*l-m*)))^{1/2}

When the multipliers k:l:m are successive (increasing by 2 in designs (b) and (c)), these expressions reduce to:

a) Mode *k*:*k*+1:*k*+2: Av = (*C*_{a}/*C*_{b})^{1/2}
*k*(*k*+2)/(4*k*^{2}+8*k*+3)^{1/2}

b) Mode:* k*:*k*+2:*k*+4: Av = (*C*_{a}/*C*_{b})^{1/2}
(*k*(*k*+4))^{1/2}/2

c) Mode:* k*:*k*+2:*k*+4: Av = (*C*_{a}/*C*_{b})^{1/2}
*k*^{1/2}/2

Designs (a) and (b) generate voltage gains approximately proportional to
*k*. In design (c) the gain increases only with the square root of
*k*.

Note that for these circuits there is always the possibility of driving
at one of the resonances, what produces a continuously growing output,
that ideally can result in any voltage gain. But the input current grows
too at the same rate. For given input and output capacitances, and output
inductance, The design procedures (a) and (b) above always reach a given
output voltage faster, and with smaller maximum input current. The only
problem is that the coupling coefficient may become too small (the voltage
gain is inversely proportional to it), and the bandwidth of the system too
small too, but the same problem occurs in a system driven at a resonance,
if the input must be kept for many cycles. The same idea can be extended
for higher-order networks [2].

Examples:

Sinusoidal input, design (b) mode 11:13:15, L_{b}=30
mH,
C_{a}=10
nF, C_{b}=15
pF

Perfect energy transfer in 3.25 cycles.

Normalized transformerless circuit:

C_{1}=
0.0969696970

L_{1}=
0.0625000000

C_{2}=
1.0000000000

L_{2}=
0.0060606061

Final circuit:

C_{a}=
10.0000000000
nF

L_{a}=
49.3636363636
uH

C_{b}=
15.0000000000
pF

L_{b}=
30.0000000000
mH

k_{ab}=
0.2973176585

Output frequencies: 203172.34, 240112.77, 277053.19 Hz

The waveforms above are for 180 V of input peak voltage. The predicted
voltage gain is 165.8312395178.

A simulation results in:

Maximum V_{Ca}
(V) = 571.69140 ( 0.00163 J) at 20.86709 us

Maximum I_{La}
(A) = 8.54905 ( 0.00180 J) at 19.84698 us

Maximum V_{Cb}
(V) = 29849.64157 (0.00668 J) at 13.53135 us

Maximum I_{Lb}
(A) = 0.67034 ( 0.00674 J) at 14.54545 us

Maximum voltage gain obtained = 165.83134

Ratio of maximum energies in the capacitors = 4.08927

Note that C_{a}
and L_{a}
must store only about 1/4 of the output energy. This happens in all
practical cases.

It's also possible to design the system with irregular spacing of the three frequencies. If the mode is specified as 23:25:31 (double and add 1 (or -1) to k, l, and m, and then decrease l by 2), complete energy transfer occurs at a "second peak" of the transient waveform envelope, with a somewhat larger voltage gain.

Normalized transformerless circuit:

C_{1}=
0.0583556942

L_{1}=
0.0828760860

C_{2}=
1.0000000000

L_{2}=
0.0065077139

Final circuit:

C_{a}=
10.0000000000 nF

L_{a}=
36.0683367993 uH

C_{b}=
15.0000000000 pF

L_{b}=
30.0000000000 mH

k_{ab}=
0.2698266359

Calculated voltage gain= 199.0254031804

Maximum V_{Ca}
(V) = 642.53699 (0.00206 J) at 46.46265 us

Maximum I_{La}
(A) = 10.48317 (0.00198 J) at 45.50255 us

Maximum V_{Cb}
(V) = 35822.55794 (0.00962 J) at 26.12061 us

Maximum I_{Lb}
(A) = 0.80015 (0.00960 J) at 27.13471 us

Maximum voltage gain obtained = 199.01421

Ratio of maximum energies in the capacitors = 4.66239

If the central frequency is further approximated to one of the border
frequencies, the next solution produces a maximum at a "third peak", then
at a "fourth peak", and so on. These modes don't seem very practical,
however. In the example the output voltage increased by 20%, while the
maximum input current increased by 22.6%

A more practical design:

The previous designs didn't produce a high enough output voltage for a
spark producing coil. A more powerful system can be obtained by increasing
the primary capacitance and using a higher mode. Limiting the coupling
coefficient to 0.1, otherwise tuning becomes too critical, the highest
mode is 37:39:41 with design b. Complete energy transfer occurs in 9.75
cycles. To obtain 150 kV with 180 V square wave input (a half bridge
powered by the rectified 127 V power line, equivalent to 180*4/pi = 229
V), and using the same secondary coil and top load that I used in other
practical projects (L_{b}
= 28.2 mH, C_{b} = 10.4
pF), C_{a} has to be
increased to 12 nF. The final elements are then:

C_{a}=
12.0000000000 nF

L_{a}=
24.6977719183 uH

C_{b}=
10.4000000000 pF

L_{b}=
28.2000000000 mH

k_{ab}=
0.1021618888

Output frequencies: 279182.274741, 294273.208511 (excitation),
309364.142281 Hz

Maximum V

Maximum I

Maximum V

Maximum I

Maximum voltage gain obtained = -841.81267

Ratio of maximum energies in the capacitors = 3.91106

In the plot below, the driver is switched off after 10 cycles, and ideal free-wheeling diodes in the half-bridge return most of the energy to the power supply:

Note that the input current is in phase with the driving voltage. The use of primary current feedback to ensure zero-current switching on the driver results in the same waveforms, up to the instant when the output voltage is maximum. The driver shall be switched off at this instant, or, with feedback, a cycle leading to three times the input current occurs. Without feedback, the driver returns the energy to the power supply, with the same waveforms shown above. Note that the input current is inverted in the energy return phase.

Excitation at the resonance frequencies results in higher input current for a given output voltage, more cycles to reach the maximum, and input current out of phase with the driving voltage. If primary current feedback is used, the waveforms return to the case of excitation between the resonances shown.

The program drsstcd can make these calculations and simulate the results.

References:

[1] A. C. M. de Queiroz, Multiple resonance networks with incomplete energy transfer and operating with zero-state response, 2005 IEEE ISCAS, Kobe, Japan, Vol. I, pp. 236-239, May 2005. In IEEE Explore.

[2] A. C. M. de Queiroz, The triple resonance network with sinusoidal excitation, ICECS 2007, Marrakech, Morocco, pp. 1143-1146, December 2007 (includes some corrections). In IEEE Explore.

Created: 7 September 2004

Last update: 2 June 2020

Developed and Maintained by Antonio Carlos M. de Queiroz

Lamento informar que o Prof. Antonio Carlos Moreirão de Queiroz faleceu há algum tempo.

Sei que esta página é visitada constantemente. Assim, gostaria de saber se temos algum visitante (interessado) que seja da UFRJ. Se for, por favor, envie um e-mail para watanabe@coe.ufrj.br.

Comento que é impressionante ver o que Moreirão foi capaz de fazer. Ele não só projetou os circuitos, mas também fez todo o trabalho de marceneiro (melhor que muitos que já vi e eram profissionais).

Segundo Moreirão contou em uma palestra, ele só levou choque uma vez. Sem querer encostou o dedo médio em um capacitor com alta tensão que se descarregou através do dedo. A corrente ao passar por uma das articulações a danificou e doía sempre que dobrava esse dedo. Mas, segundo ele, já tinha acostumado.