This document complements this
other one, discussing possible design procedures for a "double
resonance
solid-state Tesla coil", but now considering a
lossless system. 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 that 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.

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.

Created: 7 September 2004

Last update: 29 January 2013

Developed and Maintained by Antonio Carlos M. de
Queiroz