DRSSTC Design Procedure - Draft Discussion
Edited/Updated: October 24, 2004
First Post.
Date : Fri, 03 Sep 2004 08:36:29 -0600. Subject : DRSSTC design procedure - draft
Original poster: "Steve Conner"
Hi all. After much messing around with simulations and theory, I think
I've finally come up with a usable design procedure for DRSSTCs. It has been
tested by simulation but not by building a real coil. So I'd be grateful if any
other
DRSSTC builders could give me their opinion. If you compare your measured
primary current, spark length, etc. with the predictions that my method makes,
we can see if it's accurate.
This method is based on the transient behaviour of the coil and uses nothing
more complex than conservation of energy. The simulated coils that I designed
with it seemed to transit quite happily into a steady state after breakout, with
the primary current stabilising around the design value. They seemed quite
insensitive to streamer load impedance, although very low impedances caused
dangerously high primary current.
This article doesn't cover driver circuits and power supplies as there is a lot
of information on them already.
1: Decide what length of spark you want to produce.
2: Use Freau's equation (spark length=1.7*sqrt(power)) to calculate the power
you will need to do this. Then divide it by two because DRSSTCs are twice as
efficient :)
3: From the power, and knowing the breakrate which is usually 100 or 120,
calculate the bang energy.
4: Now the fun starts. In a DRSSTC the primary and secondary resonators ring up
together. When breakout occurs they both empty into the streamer load together.
The split of energy between the primary and secondary is approximately (1-k):k.
So use the tightest coupling possible, we will assume that you managed to get
k=0.33. Knowing this we can size our primary and secondary for energy storage.
5: We now know enough to design the resonator. At breakout time, 0.33 of the
total energy is in the resonator. Use this to calculate the topload voltage, and
choose the toroid so that its breakout voltage is about the same as the voltage
you worked out. You may have to iterate here, as changing the toroid changes the
capacitance. Also check that the voltage doesn't add up to more than about 1.1kV
per mm of resonator (to avoid hassles with flashover)
6: Now it's time to do the primary. 0.66 of the energy is in the primary, so use
E=0.5*L*I^2 (where I is the maximum current you dare to put through your IGBT's)
to give the required primary inductance. Now choose the primary
capacitance to resonate this at the frequency of the secondary, and add 10% more
capacitance for luck. Finally, choose the voltage rating of the capacitor by
using E=0.5*C*V^2 to find V.
7: Light blue touchpaper and retire. Steve C.
Date : Fri, 03 Sep 2004 20:55:12 -0600. Subject : RE: DRSSTC design procedure draft
Original poster: "McCauley,
Daniel H"
5: We now know enough to design the resonator. At breakout time, 0.33 of the
total energy is in the resonator. Use this to calculate the topload voltage, and
choose the toroid so that its breakout voltage is about the same as the voltage
you worked out. You may have to iterate here, as changing the toroid changes the
capacitance. Also check that the voltage doesn't add up to more than about 1.1kV
per mm of resonator (to avoid hassles with flashover).
Sounds interesting Steve. I'll send you a copy of my data (PDF), and you can
hopefully compare the data. I don't have much time to do it myself.
Also, i think selection of toroid really isn't based on the break-out
voltage. Remember, we almost always use break-out points and by doing this, it
drastically changes the actual break-out voltage. Dan.
Date : Mon, 06 Sep 2004 16:32:05 -0600. Subject : Re: DRSSTC design procedure draft
Original poster: "Antonio Carlos
M. de Queiroz"
Tesla list wrote:
> Original poster: "Steve Conner"
I would like to see a numerical example for comparison. I will try a numerical
example with some of my ideas. Initially we agree:
> 1: Decide what length of spark you want to produce.
1 meter, or 39".
> 2: Use Freau's equation (spark length=1.7*sqrt(power)) to calculate the power you will need to do this. Then divide it
> by two because DRSSTC's are twice as efficient :)
Power = (39/1.7)^2/2 = 263 W
> 3: From the power, and knowing the breakrate which is usually 100 or 120, calculate the bang energy.
Energy = 263/120 = 2.2 J
> 4: Now the fun starts. In a DRSSTC the primary and secondary resonators ring up together. When breakout occurs
>they both empty into the streamer load together. The split of energy between the primary and secondary is
> approximately (1-k):k. So use the tightest coupling possible, we will assume that you managed to get k=0.33.
>Knowing this we can size our primary and secondary for energy storage.
Now we diverge. I can΄t find a good solution with high k. Trying a solution
with the fastest possible energy transfer (excitation at the arithmetic mean of
the resonances): I will use a large toroid, with, say, 50 pF of capacitance. I
will assume that the energy in the self-capacitance of the secondary coil is
included in this too, and that it also contributes for the bang energy, what is
probably not true. The output voltage has to reach:
Vmax = sqrt(2.2/(0.5*50e-12) = 297 kV. I will assume that the input voltage is
310 V (220 V rectified), and that the driver can switch, say, 200 A. I need a
voltage gain of 297000/(310*4/pi) = 752.
My design procedure would be:
1) I choose the mode, that is the ratio of the two resonances and the driving
frequency, based on the k that I want to use. The best solutions are ratios of
successive odd integers, with sinusoidal (or square wave from a bridge) input.
Mode 31:33:35 results in k=0.12 and energy transfer in 8.25 cycles. (The
formula for k is a bit complicated to list here, but is implemented in my sstcd
program.)
2) I know that the voltage gain is proportional to sqrt(Ca/Cb). (For mode
x:x+2:x+4, x and odd integer, it is: Av=sqrt(Ca/Cb)*sqrt(x*(x+4))/2 ). The
maximum input current decreases if the inductances increase (probably a square
root relation too, but I didn't verify). With all the design formulas
implemented in my sstcd program (I still didn't write a page with the
formulas...), I just have to adjust Ca until the voltage gain is high enough and
then adjust Lb so the current is ok.
In the example, I get the final element values:
Ca= 105.0000000000 nF
La= 28.9927583937 uH
Cb= 50.0000000000 pF
Lb= 60.0000000000 mH
kab= 0.1205497549
That results in:
Output frequencies: 86478.14, 92057.37, 97636.61 Hz
Voltage gain= 754.7350528497
Bang energy (square wave input): 2.2185676457 J
Maximum VCa (V)=-3242.78978 (0.55207 J) at 43.44887 us
Maximum ILa (A)= 196.61827 (0.56041 J) at 46.05492 us
Maximum VCb (V)=297670.20709 (2.21519 J) at 89.61579 us
Maximum ILb (A)= 8.57035 (2.20353 J) at 92.31185 us
It is possible to design a system with higher k, that would transfer energy
completely in less cycles, but with lower voltage gain, if excited at the
central frequency, and then excite it at one of the resonances to obtain an
arbitrarily large gain (in the absence of losses). But there is always a design
as the one above that requires less input current to transfer the same energy in
the same number of cycles (or I think so...).
Antonio Carlos M. de Queiroz.
Date : Tue, 07 Sep 2004 12:56:05 -0600. Subject : Re: DRSSTC design procedure draft
Original poster: "Antonio Carlos
M. de Queiroz"
Tesla list wrote:
>
> Original poster: "Steve Conner"
> I believe that the energy in the secondary self-capacitance _does_
> contribute, and what's more, any energy in the primary tank at the time of
> breakout contributes too. I actually think this is the secret of the
> DRSSTC's efficiency. In Steve Ward's successful self-resonant designs, the
> primary tank is full of energy at breakout. Jimmy H. originally tuned his in
> a similar way to what you suggest, to get energy "transfer" in the classical
> sense, but he found the spark output improved a lot when he retuned it to drive at one of the resonances.
The energy in the primary can only contribute to the output if the driver
continues to run after breakout, and through many cycles because of the high Q
of the system. I am considering a system adjusted so the driver stops when the
primary is empty, much as in a conventional Tesla coil.
> >Mode 31:33:35 results in k=0.12 and energy transfer in 8.25 cycles.
> >(The formula for k is a bit complicated to list here, but is implemented in my sstcd program.)
>
> This is very interesting. The system I've been playing with had k=0.11, but
> C1=12.5nF and L1=42uH. I tried exciting it at various frequencies but the
> lower resonance seemed to give the best results. When tuned in this way it
> took about 30 cycles to produce a decent spark. The peak primary current was
> about 400A, the bang energy something like 5-6 joules and the spark output was 36".
> It hit 36" regularly and would have probably gone even further, if it hadn't exploded on the first run.
Because you are driving the system at one of the resonances. The input current
then grows along with the output voltage. Surely works, but is hard for the
driver.
> >there is always a design as the one above that requires less input current to transfer the same energy in the same
> >number of cycles (or I think so...).
>
> According to your method then, I should be able to get the same spark output with 200A peak current in 8.25 cycles?
>I need to check this out. My coiling stuff is all dismantled at the moment, but I can run a PSpice simulation of the two
>coils side by side.
The simulations show this. But note the large primary capacitor that was
required. With 12.5 nF the mode would have to be higher, and the resulting k
maybe too low for practical implementation. It's possible to design the system
so any voltage is reached in any number of cycles, but the element values can
become impractical if a too low or too high mode is used (too low impedances,
too low frequency, or too low k). The restrictions on k are about the same for
a conventional Tesla coil. Something between 0.1 and 0.2 easily result in
viable designs. Low k produces larger voltage gain, but precise tuning may be
difficult. High k decreases the voltage gain, but this can be compensated with a
larger input capacitance. If the current grows excessively, larger inductances
bring it down. (The input current is inversely proportional to the square root
of the inductances, for fixed capacitances. The driving frequency too.)
> If the improvement that this design method gives is as drastic as you suggest, then we could find a way of living with
>the hard switching that is an inevitable consequence if you don't drive the system at one of the resonances.
Driving at the central frequency results in practically perfect soft switching.
The same at the resonances. Antonio Carlos M. de Queiroz.
Date : Tue, 07 Sep 2004 06:36:28 -0600. Subject : RE: DRSSTC design procedure draft
Original poster: "Steve Conner"
>I will use a large toroid, with, say, 50 pF of capacitance. I will assume that the energy in the self-capacitance of the
>secondary coil is included in this too, and that it also contributes for the bang energy, what is probably not true.
I believe that the energy in the secondary self-capacitance _does_ contribute,
and what's more, any energy in the primary tank at the time of breakout
contributes too. I actually think this is the secret of the DRSSTC's efficiency.
In Steve Ward's successful self-resonant designs, the primary tank is full of
energy at breakout. Jimmy H. originally tuned his in a similar way to what you
suggest, to get energy "transfer" in the classical sense, but he found the spark
output improved a lot when he retuned it to drive at one of the resonances. Mode
31:33:35 results in k=0.12 and energy transfer in 8.25 cycles.
>(The formula for k is a bit
complicated to list here, but is implemented in my sstcd program.)
This is very interesting. The system I've been playing with had k=0.11, but
C1=12.5nF and L1=42uH. I tried exciting it at various frequencies but the lower
resonance seemed to give the best results. When tuned in this way it took about
30 cycles to produce a decent spark. The peak primary current was about 400A,
the bang energy something like 5-6 joules and the spark output was 36".
It hit 36" regularly and would have probably gone even further, if it hadn't
exploded on the first run.
>there is always a design as the one above that requires less input current to transfer the same energy in the same
>number of cycles (or I think so...).
According to your method then, I should be able to get the same spark output
with 200A peak current in 8.25 cycles? I need to check this out. My coiling
stuff is all dismantled at the moment, but I can run a PSpice simulation of the
two coils side by side.
If the improvement that this design method gives is as drastic as you suggest,
then we could find a way of living with the hard switching that is an inevitable
consequence if you don't drive the system at one of the resonances. Steve C.
Date : Fri, 10 Sep 2004 18:08:17 -0600. Subject : RE: DRSSTC design procedure draft
Original
poster: "Steve Conner"
>Mode 11:13:15:
>http://www.coe.ufrj.br/~acmq/tesla/dr111315.gif
I'm a bit worried by this. On the 4th half cycle of primary current, ILa peaks
as Vin passes through zero. This means that the inverter has to hard switch
nearly its whole rated current.
I can't make out whether the same thing happens in the higher modes but I would
expect it to. Steve C.
Date : Fri, 10 Sep 2004 07:02:57 -0600. Subject : Re: DRSSTC design procedure draft
Original
poster: "Antonio Carlos M. de Queiroz"
Tesla list wrote:
>
> Original poster: "Bob (R.A.) Jones"
<a1accounting@bellsouth.net>
> I notice that the output energy is approximately 4 x the max energy in the
> primary C (Ca).
>
> Is this generally true for the configuration/mode your suggesting?
Very interesting observation. The ratio of maximum energies in Cb and Ca really
appears to converge to 4 as the mode gets higher. This happens for cosinusoidal
or sinusoidal excitation, and depends only on the mode. It's possible to obtain
more by using modes as x:x+1:x+4, but these modes result in hard switching. Low
modes also result in greater ratio, up to 12.8 for mode 1:2:3 (cosinusoidal
input), but this mode results in too high k.
In the other hand, excitation at the resonances easily results in more energy in
the primary than in the secondary capacitance. This relation gives a quick
estimate for the maximum input current, given the bang size:
If the ratio of energies in Cb and Ca is 4:
0.5*Ca*Vamax^2 = Ebang/4
Vamax = sqrt(Ebang/(2*Ca))
Iinmax =~ Ca*2*pi*Frequency*Vamax
Testing: The default design in the sstcd program is:
Mode: 18:19:20
Ca= 5.0000000000 nF
La= 86.0126111111 uH
Cb= 15.0000000000 pF
Lb= 28.2000000000 mH
kab= 0.1046489272
Output frequencies: 231829.58, 244709.00, 257588.42 Hz
Maximum VCa (V)= -1700.52069 (0.00723 J) at t=19.40194 us
Maximum ILa (A)= 13.02961 (0.00730 J) at t=98.05981 us
Maximum VCb (V)= -62289.94357 (0.02910 J) at t=38.81388 us
Maximum ILb (A)= 1.43719 (0.02912 J) at t=37.79378 us
Using the formula:
Vamax = sqrt(0.02910/(2*5e-9))= 1706 V
Iinmax = 5e-9*2*pi*244709*1706 = 13.12 A
Very good agreement. The relation allows then an estimation of the voltage on
the primary capacitor and of the maximum input current, from the bang energy and
from the primary capacitance. Actually, this maximum current is the minimum
required, if the system is perfectly tuned. Any mistuning (including operation
at the resonances) increases it.
I will see if I can obtain the exact expression for the factor. Antonio Carlos
M. de Queiroz.
Date : Fri, 10 Sep 2004 07:03:17 -0600. Subject : Re: DRSSTC design procedure draft
Original
poster: "Antonio Carlos M. de Queiroz"
Tesla list wrote:
> Original poster: "Bob (R.A.) Jones"
<a1accounting@bellsouth.net>
> I notice that the output energy is approximately 4 x the max energy in the primary C (Ca).
More about this. I made two simulations of normalized systems, with Ca=Cb=Lb=1,
where
it's clear what happens. During the first half of the energy transfer, the
primary and the secondary accumulate equal amounts of energy, totalizing one
half of the final energy. During the second half of the transient, the primary
transfers its energy to the secondary, and another quarter part goes directly
from the driver to the secondary.
Mode 11:13:15:
http://www.coe.ufrj.br/~acmq/tesla/dr111315.gif
Mode 51:53:55:
http://www.coe.ufrj.br/~acmq/tesla/dr515355.gif
The primary voltage and current don't have component at the driving frequency.
The beats are between the two resonances only. If the energy is not spent, and
the driver continues to operate, all the energy goes back to the power supply.
Something more complex occurs with modes where the excitation is dislocated away
from the arithmetic mean of the resonances. The first case dislocating downwards
are like these:
Mode 23:25:31
http://www.coe.ufrj.br/~acmq/tesla/dr232531.gif
Mode 103:105:111
http://www.coe.ufrj.br/~acmq/tesla/dr103105111.gif
Two beats appear before the energy transfer is complete.
Modes as 103:105:115 produce three beats:
http://www.coe.ufrj.br/~acmq/tesla/dr103105115.gif
Antonio Carlos M. de Queiroz.
Date : Fri, 10 Sep 2004 18:07:52 -0600. Subject : RE: DRSSTC design procedure draft
Original
poster: "Steve Conner"
>Any mistuning (including operation at the
>resonances) increases it.
Hmm. So operation at the resonances is now mistuning?
I'm sure when Jimmy Hynes made his original coil, he explored much the same
region of parameter space that Antonio is recommending now. I can't find
accurate data on it, but I think he had C1=0.6uF, L1=13uH and k12 roughly 0.1,
and used an excitation frequency at the geometric mean of the two resonant
frequencies.
However, the experimental results showed that it worked better when retuned to
operate at one of the resonances.
I do realise that this could have been an accident, and there may have been an
even more optimal design nearby in parameter space that he just missed. Looking
at Antonio's work, it's tempting to suggest that if Jimmy had only made some
slight changes to align his system with a particular set of magic mode numbers,
the performance would have improved drastically.
But my own personal belief is that Antonio's theory gives over-optimistic
results because it doesn't take streamer loading into account. Anyway, there's
no way we can model this yet as we don't have an accurate dynamic streamer
loading model. So again more experimental work is called for... It shouldn't be
too hard to make a DRSSTC that can be assembled in "Ward Mode" or "De Queiroz
Mode" and a comparison done.
I would be happy to try this. Maybe Antonio could suggest a design that would
turn my OLTC II resonator into an optimal DRSSTC?
Data and constraints for OLTC II:
Secondary inductance: 225mH
(Secondary + toroid) capacitance: 31pF
Resonant frequency: 62 kHz
Breakout voltage of toroid: ~600kV
Flashover voltage of resonator: ~770kV
Power source: H-bridge powered by 600V DC
Maximum I2t of power source: 600A rms for 1 millisecond (=360 x 10^6 A^2.s)
Tank capacitor: I have about 40 1uF 1000V caps that can be assembled in any
combination you like. (capacity 20J)
Desired bang energy: As high as possible given the above constraints. The design
method I published results in about 21J. The secondary can only hold about 6J
before breaking out, but the remaining energy is fed from the inverter and
primary tank straight into the discharge while it is alight.
L1=?
C1=?
k12=?
Steve C.
Date : Fri, 10 Sep 2004 23:15:06 -0600. Subject : Re: DRSSTC design procedure draft
Original
poster: "Antonio Carlos M. de Queiroz"
Tesla list wrote:
>
> Original poster: "Steve Conner"
> >Any mistuning (including operation at the resonances) increases it.
>
> Hmm. So operation at the resonances is now mistuning?
If the system is designed to operate between the resonances and the driving
frequency is changed, initially the output voltage drops and the input current
increases, but by small amounts. After a certain displacement, a second "beat"
appears after the "normal" beat of the output voltage, that reaches higher
voltage, but the input current almost doubles. With more displacement a third
beat appears, with still higher voltage and current. This goes on (with hard
switching) until a resonance is reached, where the output voltage and the input
current grow continuously and the switching is soft again. My design then
creates a local maximum for the voltage gain, with an absolute minimum for the
input current.
> I'm sure when Jimmy Hynes made his original coil, he explored much the same
> region of parameter space that Antonio is recommending now. I can't find
> accurate data on it, but I think he had C1=0.6uF, L1=13uH and k12 roughly
> 0.1, and used an excitation frequency at the geometric mean of the two resonant frequencies.
The geometrical mean is good too, specially when a resistive load is being
driven.
> However, the experimental results showed that it worked better when retuned to operate at one of the resonances.
Probably because his driver could switch larger currents.
> I do realise that this could have been an accident, and there may have been
> an even more optimal design nearby in parameter space that he just missed.
> Looking at Antonio's work, it's tempting to suggest that if Jimmy had only
> made some slight changes to align his system with a particular set of magic
> mode numbers, the performance would have improved drastically.
If the system is designed with the driver operating at its maximum safe current
by my method, no other tuning will produce better results (I think so far). If
the driver can allow more current, operation at one of the resonances can
produce greater output. But then it's a question of redesigning the system for
the greater current to obtain even greater output.
> But my own personal belief is that Antonio's theory gives over-optimistic
> results because it doesn't take streamer loading into account. Anyway,
> there's no way we can model this yet as we don't have an accurate dynamic
> streamer loading model. So again more experimental work is called for... It
> shouldn't be too hard to make a DRSSTC that can be assembled in "Ward Mode"
> or "De Queiroz Mode" and a comparison done.
My previous design method, based on an impedance matching network, produces
practically the same element values of the lossless design, so resistive loads
are not a big problem. Capacitive loads would cause mistuning, but then the idea
of using a pll to adjust the input
frequency may help.
> I would be happy to try this. Maybe Antonio could suggest a design
that
> would turn my OLTC II resonator into an optimal DRSSTC?
Let's see:
> Data and constraints for OLTC II:
> Secondary inductance: 225mH
> (Secondary + toroid) capacitance: 31pF
> Resonant frequency: 62 kHz
> Breakout voltage of toroid: ~600kV
> Flashover voltage of resonator: ~770kV
>
> Power source: H-bridge powered by 600V DC
> Maximum I2t of power source: 600A rms for 1 millisecond (=360 x 10^6 A^2.s)
>
> Tank capacitor: I have about 40 1uF 1000V caps that can be assembled in any
> combination you like. (capacity 20J)
>
> Desired bang energy: As high as possible given the above constraints. The
> design method I published results in about 21J. The secondary can only hold
> about 6J before breaking out, but the remaining energy is fed from the
> inverter and primary tank straight into the discharge while it is alight.
>
> L1=?
> C1=?
> k12=?
With the driver feeding the streamers, maybe better to use the impedance
matching design. The formulas are in: http://www.coe.ufrj.br/~acmq/tesla/sstc.html
With Lb=225 mH and Cb=31 pF, the input frequency shall be 60263 Hz.
I will assume:
600 V of peak input voltage (square wave).
600 A of maximum output current at steady state.
6 J at the output capacitance, at steady state.
This results in:
Ca: 209247.3768533974 pF
La: 33.6725524037 uH
Lb: 224997.1794122552 uH
kab: 0.1003695810
Cb: 31.0000000000 pF
Optimal streamer load: Rb=844514 Ohms
Input impedance at steady state: Ra=1.2732395447 Ohms
This design reaches steady state in 0.3 ms (18 cycles), after some
"overshoot" on the specifications:
Maximum VCa (V)= 9840.90575 ( 10.13212 J) at 124.41244 us
Maximum ILa (A)= -779.23138 ( 10.22301 J) at 128.46285 us
Maximum VCb (V)= 723602.44281 ( 8.11581 J) at 186.61866 us
Maximum ILb (A)= 8.51576 ( 8.15819 J) at 191.01910 us
This leaves some time to continue to push energy into the streamers.
The same design, but with 8 J and 41 pF at the output capacitance (a lot of
capacitive streamer loading):
Excitation at 52401 Hz
Ca: 276746.6471913645 pF
La: 33.7819780991 uH
Lb: 224997.2741393207 uH
kab: 0.1152414634
Cb: 41.0000000000 pF
Rb: 638535.0921930474 Ohms
Ra: 1.2732395447 Ohms
More primary capacitance and an adjustment in k.
Estimating the energy going to the streamers (my simulator doesn't calculate
this yet). At each cycle of 52401 Hz (19 us), at steady state, the input energy
is 0.5*1.27*600^2*19e-6= 4.34 J.
In just 5 cycles you have 21.7 Joules.
I recommend running some simulations to see if I didn't make a mistake.
Antonio Carlos M. de Queiroz.
Date : Fri, 10 Sep 2004 23:14:09 -0600. Subject : Re: DRSSTC design procedure draft
Original
poster: "Antonio Carlos M. de Queiroz"
Tesla list wrote:
>
> Original poster: "Steve Conner"
>
> >Mode 11:13:15:
> >http://www.coe.ufrj.br/~acmq/tesla/dr111315.gif
>
> I'm a bit worried by this. On the 4th half cycle of primary current, ILa
> peaks as Vin passes through zero. This means that the inverter has to hard switch nearly its whole rated current.
Ila, the input current, is the yellow line. It crosses zero almost
simultaneously with Vin in all the zero crossings of Vin, except the last before
the energy transfer is complete (at 2 seconds in that normalized simulation).
The same happens in higher modes.
Antonio Carlos M. de Queiroz.
Date : Mon, 13 Sep 2004 07:52:55 -0600. Subject : RE: DRSSTC design procedure draft
Original poster: "Steve Conner"
>In just 5 cycles you have 21.7 Joules.
>I recommend running some simulations to see if I didn't make a mistake...
No, I can believe this :) The energies that can be delivered by the "impedance
matching" type of idea are quite spectacular. I've written a parametric
simulation in PSpice, and I'll plug your design and try simulating it.
I was also working on a design of my own. My "empirical" design method (have a
guess based on the results of other peoples' experiments, then do 10 parametric
simulations around that point, pick the best one, repeat) "converged" on the
following solution
Ca=0.2uF
La=42uH (it would also have worked with 36uH with minimum performance
change, however 33uH would shift the resonance to the upper pole)
k=0.315
This gives a "Ward mode" design that self-resonates at its lower resonant
frequency. The peak primary current was around 800-1000A depending on streamer
loading, and the total energy delivered is worryingly high. Over a 300
microsecond burst I found that it would deliver between 35 and 60J depending on
what assumptions I made for streamer length. However the toroid voltage never
rises above 600kV.
The primary current is considerably more than 600A, but I'm somewhat bound to
the 0.2uF primary capacitance (3 strings of 14 caps) I can't get less without
changing to 2 strings and I don't really want to do that.
If we just multiplied everything by 600/1000 we would get about the same 20J as
your method yielded. However it seems like (as you said) mine would need more
cycles at 600A to achieve this energy, in spite of having tighter coupling?
My streamer loading model is:
Load capacitance 25pF per metre
Load resistance, 103000/(streamer length in metres)
I assume that a streamer of the final length suddenly appears when the
toroid reaches something like its breakout voltage.
Steve C.
Date : Fri, 17 Sep 2004 18:23:12 -0600. Subject : RE: DRSSTC design procedure draft
Original
poster: "Steve Conner"
>I recommend running some simulations to see if I didn't make a mistake...
Here's my first simulation, it is a transient run using the exact parameters
that Antonio posted.
http://scopeboy.com/tesla/drsstc/antonio_drsstc.gif
It seems to work exactly as Antonio said, delivering 23 Joules. And the
switching is very good, there are no hard switching events.
Now I need to investigate its sensitivity to streamer loading, which is a job
for another day.
It performs very poorly when I run it with the streamer load model I use, which
is 25*L pF of cap in series with 103000/L ohms of resistance (L is streamer
length in metres) The primary current goes over 2000A. But this is totally
different to what Antonio developed it with - 638k in parallel with 10pF, for
what (according to the delivered energy trace on my sim) would be a 2 metre
streamer. So that would be 5L pF in parallel with 1276/L kOhms.
Antonio, where did you get the 638k//10pF figure from? Do you think it is
realistic? Do you think _my_ model is realistic? Does anyone else have a view?
Thanks, Steve C.
Date : Sat, 18 Sep 2004 06:34:39 -0600. Subject : Re: DRSSTC design procedure draft
Original
poster: "Antonio Carlos M. de Queiroz"
Tesla list wrote:
>
> Original poster: "Steve Conner"
>
> >I recommend running some simulations to see if I didn't make a mistake...
>
> Here's my first simulation, it is a transient run using the exact parameters that Antonio posted.
This, I presume:
Ca: 276746.6471913645 pF
La: 33.7819780991 uH
Lb: 224997.2741393207 uH
kab: 0.1152414634
Cb: 41.0000000000 pF
Rb: 638535.0921930474 Ohms
Ra: 1.2732395447 Ohms
This design puts 6 J at Cb at steady state, not 8 as I had said.
>
>
http://scopeboy.com/tesla/drsstc/antonio_drsstc.gif
>
> It seems to work exactly as Antonio said, delivering 23 Joules. And the
> switching is very good, there are no hard switching events.
Why are your current decreasing so fast? In my simulation it goes to the
designed 600 A peak and stays there.
> Now I need to investigate its sensitivity to streamer loading, which is a job for another day.
>
> It performs very poorly when I run it with the streamer load model I use,
> which is 25*L pF of cap in series with 103000/L ohms of resistance (L is
> streamer length in metres) The primary current goes over 2000A. But this is
> totally different to what Antonio developed it with - 638k in parallel with
> 10pF, for what (according to the delivered energy trace on my sim) would be
> a 2 metre streamer. So that would be 5L pF in parallel with 1276/L kOhms.
>
> Antonio, where did you get the 638k//10pF figure from? Do you think it is
> realistic? Do you think _my_ model is realistic? Does anyone else have a view?
The 10 pF was somewhat arbitrary. I should have used more capacitance for a
streamer with 1 meter. The 639 k is the resistance that if connected to the
output reflects 1.27 Ohms at the input, or 600 A with a square wave with 600 V
of peak voltage (or 4/pi*600 V of peak
voltage on the fundamental harmonic). In that design method this is just at what
loading level you choose the input impedance to be "maximally resistive".
The equivalent series load to Cp=10 pF and Rp=639 k at f=52401 Hz would be:
Q=2*pi*f*R*C=2.1
Cs=Cp/(1+1/Q^2)=12 pF
Rs=Rp/(Q^2+1)=144 kOhms
If you use Cs=25 pF and Rs=103000 Ohms of load, the equivalent parallel load
would be:
Q=1/(2*pi*f*Cs*Rs)=1.18
Cp=Cs/(1+1/Q^2)=14.6 pF
Rp=Rs*(1+Q^2)=246 kOhms
With this load, supposing the same 41 pF of load capacitance (decrease the top
load) to keep the same input frequency (52401 Hz), the maximum energy in the
load capacitance is just 2.3 Joules to keep the 600 A of input current. This
happens because the voltage gain is the square root of the ratio of output and
input resistances. There is no way around this.
To obtain long sparks, I imagine that the best, without huge input currents,
would be to store all the energy at the output capacitance before breakout, as
done in a conventional Tesla coil, and turn off the driver as soon as the input
current exceeds a certain limit.
This would fall into my lossless design.
Trying 20 J with Ca=0.28 uF, Lb=225 mH, and Cb=45 pF, I get,
with mode 31:33:35:
Ca= 280.0000000000 nF
La= 36.6939598420 uH
Cb= 45.0000000000 pF
Lb= 225.0000000000 mH
kab= 0.1205497549
Output frequencies: 47072.74, 50109.69, 53146.64 Hz
This results in, with square wave input at 50109.69 Hz:
Maximum VCa (V)= 6276.84957 ( 5.51584 J) at 578.69787 us
Maximum ILa (A)= 552.82838 ( 5.60719 J) at 573.89739 us
Maximum VCb (V)= 991803.80546 ( 22.13268 J) at 164.57646 us
Maximum ILb (A)= -13.99169 ( 22.02384 J) at 498.84988 us
If the terminal breaks down at a good fraction of 1 MV, I
think that this may work. Note that the element values are
almost the same of the first design. Nothing is very critical.
Antonio Carlos M. de Queiroz.
Date : Sat, 18 Sep 2004 23:04:41 -0600. Subject : Re: DRSSTC design procedure draft
Original
poster: "Bob (R.A.) Jones"
<a1accounting@bellsouth.net>
Hi, Some more intersting equations after AQ intro.
> Original poster: "Antonio Carlos M. de Queiroz"
<acmdq@uol.com.br>
>
> Very interesting observation. The ratio of maximum energies in Cb
> and Ca really appears to converge to 4 as the mode gets higher.
> This happens for cosinusoidal or sinusoidal excitation, and depends
snip
> only on the mode.
> in more energy in the primary than in the secondary capacitance.
> This relation gives a quick estimate for the maximum input current,
> given the bang size:
>
> If the ratio of energies in Cb and Ca is 4:
> 0.5*Ca*Vamax^2 = Ebang/4
> Vamax = sqrt(Ebang/(2*Ca))
> Iinmax =~ Ca*2*pi*Frequency*Vamax
>
> Testing: The default design in the sstcd program is:
> Mode: 18:19:20
> Ca= 5.0000000000 nF
> La= 86.0126111111 uH
> Cb= 15.0000000000 pF
> Lb= 28.2000000000 mH
> kab= 0.1046489272
> Output frequencies: 231829.58, 244709.00, 257588.42 Hz
> Maximum VCa (V)= -1700.52069 (0.00723 J) at t=19.40194 us
> Maximum ILa (A)= 13.02961 (0.00730 J) at t=98.05981 us
> Maximum VCb (V)= -62289.94357 (0.02910 J) at t=38.81388 us
> Maximum ILb (A)= 1.43719 (0.02912 J) at t=37.79378 us
>
> Using the formula:
> Vamax = sqrt(0.02910/(2*5e-9))= 1706 V
> Iinmax = 5e-9*2*pi*244709*1706 = 13.12 A
> Very good agreement. The relation allows then an estimation of the
> voltage on the primary capacitor and of the maximum input current,
> from the bang energy and from the primary capacitance.
> Actually, this maximum current is the minimum required, if the system
> is perfectly tuned. Any mistuning (including operation at the resonances) increases it.
>
Voltage gain is given by Vg = Vin^2* (8/Pi*k)*sqrt(La/Lb)
Note: inversely proportional to k!!!
Also approximately Voltage gain Vg = Vin (8/Pi*k)*sqrt(Ca/Cb)
Hence approx bang energy Eb = Vin^2*32*Ca/(Pi*k)^2 or Vin^2*1945/*k^2
and max E for Ca is 1/4 of this.
So now here is my design strategy:
>From your break rate and power supply determine required bang energy or cost
of Ca
>From max switching frequency and peak current determine the mode and frequency
then determine the k something near 0.1 say from AQ's tables then determine Ca
from Ca = (Eb*(Pi*k)^2)/32*Vin^2
Finally the hard part select a top load that has the suitable C and break out
voltage for the bang energy Ok yes you need say 10% equations/tables for C and
break out voltage of the
top load, a fiddle factor for roughness and breakout voltage to peak factor
(Someone suggested 0.3 for the last one).
At this point a suitable coil length for the expected spark length could be
selected along with a diameter assuming H/D=5
Determine the total C using your favorite method.
Then determine La and Lb and your ready to start building.
Anyone want to start on WinSSTC?
Note all the above is assumes no break out and driving at the soft switching
frequency as described previously in this thread. ie the strategy is to design a
soft switching configuration that uses the max switching power and max wall plug
power to pump up the top load to a required voltage/bang energy which then
breaks out to form a streamer. i.e. I don't believe at present the load
characteristics after break are know sufficiently to have a reliable stratigy.
Though I know that the hard part as described above is a long way from
definitive.
Bob.
Date : Sun, 19 Sep 2004 21:09:38 -0600. Subject : Re: DRSSTC design procedure draft
Original
poster: "Antonio Carlos M. de Queiroz"
Tesla list wrote:
>
> Original poster: "Bob (R.A.) Jones"
> Some more intersting equations after AQ intro.
> Voltage gain is given by Vg = Vin^2* (8/Pi*k)*sqrt(La/Lb)
> Note: inversely proportional to k!!!
Yes. In that mode of operation more cycles in the energy transfer are required
for higher gain, and the coupling coefficient must be reduced for this. The same
happens with the impedance matching technique.
But there is something strange in your equation. The voltage gain can't depend
on Vin. Pi doesn't appear in the calculation, unless the input frequency in Hz
is present too.
> Also approximately Voltage gain Vg = Vin (8/Pi*k)*sqrt(Ca/Cb)
Something like this. But again Vin can't appear.
> Hence approx bang energy Eb = Vin^2*32*Ca/(Pi*k)^2 or Vin^2*1945/*k^2
Now the dimension is correct. My present equations don't make k appear
explicitly, so it's difficult to verify now. But you can check with the numbers
from the sstcd program, that are correct.
> and max E for Ca is 1/4 of this.
I am verifying this "4" relation. So far only for the cosinusoidal input case,
that is simpler. It's never exact, and the exact relation for any mode is quite
complicated. But it tends
to 4 for the regular modes, where the three frequencies are evenly spaced.
Antonio Carlos M. de Queiroz.
Date : Mon, 20 Sep 2004 07:55:46 -0600. Subject : RE: DRSSTC design procedure draft
Original
poster: "Steve Conner"
>Why are your current decreasing so fast? In my simulation it goes to the designed 600 A peak and stays there.
This is meant to be representative of a real DRSSTC. So the system starts up
unloaded, then the streamer load is applied at 200us (around the time when
breakout would occur) and finally the drive is turned off at 300us. It's
probably the missing streamer load that makes the result different. When I ran
the simulation for longer I saw a notch and the primary current increased again.
>The equivalent series load to Cp=10 pF and Rp=639 k at f=52401 Hz would
>be:
>Cs=12 pF
>Rs=144 kOhms
I calculated it as Cs=26pF and Rs=117k. One of us is wrong (however it's
probably me)
>the maximum energy in the load capacitance is just 2.3 Joules to keep the 600 A of input current. This happens
>because the voltage gain is the square root of the ratio of output and input resistances. There is no way around this.
I'm afraid I don't understand this. I'm a long way from being intuitively
comfortable with the "TC as a filter" design approach. When I need a filter in
one of my circuits at work, I just follow a recipe published by someone else :P
>I imagine that the best, without huge input currents, would be to store all the energy at the output capacitance
>before breakout... Note that the element values are almost the same of the first design. Nothing is very critical.
Well you would certainly think so. But the performance of the new DRSSTCs is
only possible if most of the energy is fed directly from the driver to the
streamer load. For instance on Steve Ward's ISSTC II, he gets something like 40J
bang energy using a very thin toroid (I think it's a 4" minor diameter) The
numbers don't add up.
Steve C.
Date : Mon, 20 Sep 2004 21:44:00 -0600. Subject : Re: DRSSTC design procedure draft
Original
poster: "Bob (R.A.) Jones"
Hi all,
> Original poster: "Antonio Carlos M. de Queiroz"
>
> > Voltage gain is given by Vg = Vin^2* (8/Pi*k)*sqrt(La/Lb) Note: inversely proportional to k!!!
>
> Yes. In that mode of operation more cycles in the energy transfer are required for higher gain, and the coupling
>coefficient must be reduced for this. The same happens with the impedance matching technique.
>
> But there is something strange in your equation. The voltage gain can't depend on Vin.
Yes your right the above equation is for the peak output voltage. Oops I just
noticed I had the L ratios upside down too. The following is correct, I hope,
with a few more brackets for clarity
Voltage gain(pk out V/square wave input V) = (8/(Pi*k))*sqrt(Lb/La) or
(8/(Pi*k))*sqrt(Ca/Cb)
>Pi doesn't appear in the calculation, unless the input frequency in Hz is present too.
A factor of 4/pi is required to determine the peak amplitude of the fundamental
of a square wave.
It assumes that (La*s +1/Ca*s+Ra)(Lb*s+1/Cb*s+Rb) << Lab*s, where s
Laplace operator. This is a reasonable assumption when driving at the mid
frequency and given reasonable Q's (Ra*Rb<<(Lab*s)^2). Then transfer function
reduces to 1/(Lab*Cb*s^2) substitute k*sqr(La*Lb) for Lab
Then assuming the La*Ca=Lb*Cb (only required to be closer than the split
frequencies) substitute (La*Ca*Lb*Cb) for s^2 and simplify. At start up there
is the steady state response and the transient response which are initially
equal and opposite and hence cancel. After a number of cycles they sum at the
output doubling the steady state response (similar to the x4 thing with Ca and
Cb). So x 2 then x 4/pi for the square wave drive factor.
For the second version we can substitute the ratio of the L's with the ratio of
the C's.
QED.
It also assumes validity of the lumped model for the secondary. Which as we
know (I hope) does not accurately predict the output voltage with small top
loads and does not include the transients from the higher modes of the
secondary( not significant with low k's).
I have already checked against your posted example as follows:
600v 600 A square wave input
Ca=280.0000000000 nF 600v 600 A
La= 36.6939598420 uH
Cb= 45.0000000000 pF
Lb= 225.0000000000 mH
kab= 0.1205497549
Output frequencies: 47072.74, 50109.69, 53146.64 Hz
This results in, with square wave input at 50109.69 Hz
Maximum VCa (V)= 6276.84957 ( 5.51584 J) at 578.69787 us
Maximum ILa (A)= 552.82838 ( 5.60719 J) at 573.89739 us
Maximum VCb (V)= 991803.80546 ( 22.13268 J) at 164.57646 us
Maximum ILb (A)= -13.99169 ( 22.02384 J) at 498.84988 us
sim result = 991,803.80546,
Vin*(8/Pi*k)*sqrt(La/Lb)=992,600.
or Vin*8/Pi*k)*sqrt(Ca/Cb)=999,800.
Better than 1% which is at least one order of magnitude smaller than the
inherent error in a lumped model of secondary that attempts to calculate output
voltage with small top load.
The bang energy accuracy is similar. Again this is the unloaded/prebreak out
condition.
Its also useful to note that Vmax(ca)= Vin(4/k*pi) = 6,338V. Bob.
Date : Wed, 22 Sep 2004 07:41:35 -0600. Subject : Re: DRSSTC design procedure draft
Original
poster: "Antonio Carlos M. de Queiroz"
<acmdq@uol.com.br>
Tesla list wrote:
>
> Original poster: "Bob (R.A.) Jones"
<a1accounting@bellsouth.net>
> Voltage gain(pk out V/square wave input V) =
(8/(Pi*k))*sqrt(Lb/La)
This is the -exact- expression for the mid-frequency case.
> or (8/(Pi*k))*sqrt(Ca/Cb)
Close.
> A factor of 4/pi is required to determine the peak amplitude of the fundamental of a square wave.
Ok. I forgot this.
> It assumes that (La*s +1/Ca*s+Ra)(Lb*s+1/Cb*s+Rb) << Lab*s, where s Laplace operator.
> This is a reasonable assumption when driving at the mid frequency and given reasonable Q's (Ra*Rb<<(Lab*s)^2).
> Then transfer function reduces to 1/(Lab*Cb*s^2) substitute k*sqr(La*Lb) for Lab
> Then assuming the La*Ca=Lb*Cb (only required to be closer than the split
> frequencies) substitute (La*Ca*Lb*Cb) for s^2 and simplfy.
Why to include Ra?
> At start up there is the steady state response and the transient response
> which are initially equal and opposite and hence cancell. After a number of
> cycles they sum at the output doubling the steady state response (similar to
> the x4 thing with Ca and Cb). So x 2 then x 4/pi for the square wave drive factor.
I would say that this happens because the transformer is tuned at both sides to
the same frequency, and the turns ratio is n=sqrt(Lb/La)/k in the model:
. 1:n
. o--Ca--La(1-k^2)--+ +---+---+
. ) ( | |
. ) ( Lb Cb
. ) ( | |
. o-----------------+ +---+---+
> For the second version we can substitute the ratio of the L's with the ratio of the C's.
But it happens that my design doesn't result in this !? Using the tuning
relation CaLa(1-k^2)=LbCb. The gain as function of the capacitances would be: Av=(8/pi)*sqrt(Ca/Cb)*sqrt(1/kab^2-1)
And indeed it is, exactly, but only for sinusoidal input. The other formula
works with cosinusoidal input too. Curious.
> Its also useful to note that Vmax(ca)= Vin(4/k*pi) = 6,338V
Works well too. Antonio Carlos M. de Queiroz.
