SSTC Theory Discussion
Edited/Updated: October 23, 2004
June 22, 2004.
I have added some material, and pictures showing the frequency
and time responses of the trial design.
http://www.coe.ufrj.br/~acmq/tesla/sstc.html Antonio Carlos M. de Queiroz
June 23, 2004.
Original poster:
Marco.Denicolai@tellabs.com
Q. Hello Antonio, I find very interesting this new (at least to me) starting
point to design a TC. If I get it right, you design a Butterworth bandpass
filter with given center frequency and end up with a setup where L1*C1=L2*C2.
A.
Yes. That design reduces the network to an ideal transformer.
Q. But isn't it true that the voltage vs. frequency diagrams is the usual one?
I mean two peaks, more or less separated depending on the coupling coefficient
(if I recall)?
A. Yes, and really, the response has two peaks if the driver impedance is negligible. The two peaks would reduce to one if the impedance of the driver were identical to the designed input resistance of the network. The network can also be designed so there is only one peak when the output loading is as designed, but it seems that this design is not so good. See the plots at the end of my page. http://www.coe.ufrj.br/~acmq/tesla/sstc.html
Q. I am just wondering, isn't it so that if one wants to drive a traditional TC
with an SSTC driver (remove the RSG, use an IGBT/FET bridge), the correct drive
frequency is simply f=1/(2*PI)*SQRT(1/(L1*C1)), supposing that L1*C1=L2*C2?
A. Hummm.
Looking better at my equations I see that L1*C1=L2*C2 is valid for the prototype
filter, and that for it the driving frequency is really as you say. But observe
that in the network with transformer, Lb*Cb = (1-kab^2)*La*Ca = L1*C1. So,
in this network the primary and secondary circuits are -not- tuned to the same
frequency! The driving frequency is the resonant frequency of the secondary
circuit alone. The primary circuit is tuned to a slightly lower frequency.
Strange, but true. Antonio Carlos M. de Queiroz.
June 24, 2004. Original poster:
steve.conner@optosci.com
Q. So, in this network the primary and secondary circuits are -not-tuned to the
same frequency! The driving frequency is the resonant frequency of the
secondary circuit alone. The primary circuit is tuned to a slightly lower
frequency. Strange, but true.
A1.
Interesting. This matches with experimental results. The ISSTC people found that
the primary has to be tuned ~15% lower. I thought it was streamer loading
but...Steve C.
A2.
True for any tuned coupling
network of this configuration, such as coupling of a tuned transmitter output to
a tuned load. Of course, there's streamer loading too. Ed.
A3. June 24, 2004. Probably a combination of the two things. In the ideal case what happens is that at one side the tuning is with the "magnetizing inductance" of the transformer, and at the other with the "leakage inductance". With this, for a single frequency excitation, the reactances disappear. Antonio Carlos M. de Queiroz.
A4. June 25, 2004. I don't think I was ever convinced that streamer loading was a culprit. The typical operation mode is to ring the secondary up until it lets go and one usually designs the secondary to not let go until the bulk of primary energy has reached it (at which point there is little left in the primary to transfer). In my opinion, recent attempts to use matching theory are valid only if one wants to feed a continuous arc in CW operation. * I seriously doubt its validity when applied to either classic disruptive coils or the ISSTC which is pretty much the same thing when examining the operation of the secondary in such coils. Past experience with my disruptive coils often (if not always) showed better results with the primary tuned to what would have been the LSB generated with the tunings equal. This was referred to in the past as "offset tuning" and has appeared in early papers on TCs. I forget which ones but I have seen voltage vs tuning graphs in some of them. I still have those papers buried in a mountain at home. Malcolm.
*
Response 1 (June 25,
2004). I agree that matching theory is no good for disruptive coils. But for
the ISSTC I think it is useful. The ISSTCs built so far have small toroids with
a breakout point, and a relatively long burst length, that carries on after
breakout. Scope traces I have seen show the primary current ringing up
relatively quickly, and then levelling off at breakout.
This I think of as a "quasi-CW" ISSTC, and it can potentially have a "bang
energy" much greater than a disruptive coil with the same components, as the
inverter feeds the discharge directly.
The opposite is a "quasi-disruptive" ISSTC, with a big toroid, full-rated tank
capacitor, no breakout point, and the inverter switching off around breakout
time so it never feeds the discharge directly. It has roughly twice the bang
energy of a disruptive coil with the same components.
Whether the "quasi-CW" thing is a red herring, will have to be proved by
experiment, by measuring the tank capacitor voltage and seeing how the stored
energy compares to the bang energy. But, I think even if it turns out that all
practical ISSTCs are quasi-disruptive, the matching theory will still get you
into the right ballpark. Steve C.
Follow up to above
response.
June 28, 2004.
Original poster: m.j.watts@massey.ac.nz
Hi Steve, What I was suggesting is that in the case of the
ISSTC, matching to the *unloaded* resonator would, in my opinion, be more useful
than matching to a load. Malcolm.
Response to above post. June 28, 2004. I can't get my head around the combination of the concepts "matching" and "unloaded". Matching, to me, means adjusting a circuit for most efficient transfer of real power from a source to a load. If there is no load, then all the math goes out the window. There just isn't anything to match. **
**(
There is always something to match. As you noted below, resonator esr is at a
minimum (but never zero, even for a superconducting structure) as the radiaton
(sp?) resistance climbs and this is a function of resonator size and operating
frequency. The point is, matching to this gives a real resistance to match to
(if one does indeed want to indulge in matching) which then makes some design
procedure for the matching network possible. Such a match would give very high
output voltages given the high shunt impedance that terminates the resonator. I
think this answers the points you make below.
Your point about matching to the discharge thereby allowing the driver to boost
the bang energy on the fly is well taken. However, two things get in the way of
this plan as I see it: - the discharge impedance is ill-defined and dependent on
a number of variables, available power being one of them and I can't see how you
can design a matching network to operate correctly given this - the driver has
to scale with the planned coil output in order to cope and do useful things
which means more expense. Allowing the resonator to ring up to a desired energy
level and then cutting the driver off when a discharge has begun (might?)
relieve this burden. Maybe not given the dreaded phase reversal which would then
occur.
Antonio writes: Original poster: "Steve Conner"
<steve.conner@optosci.com>
> It doesn't :) Antonio presumably used a sine wave generator
set to the resonant frequency in his sim. But the feedback drive systems that we
use just flip their own phase whenever the resonant system tries to flip over
and push energy back. So, an ISSTC could theoretically ring up to infinity > (eep).
Interesting consideration. But if the system is trying to
return energy to the driver, and the phase of the driver is inverted to match
the current, it will extract energy from the system faster. A simulation shows
that this indeed happens, with the secondary voltage dropping twice faster. But
at the same time the driver is pushing more energy into the primary circuit, and
at the next phase reversal the output voltage is two times larger. The input
current and output voltage waveforms resemble beating waveforms, but two times
larger+ after each notch (lossless case). Antonio Carlos M. de Queiroz. +
(The input current and output voltage waveforms resemble beating waveforms, but
two times larger after each notch (lossless case). Oops... Larger by
the same amount at each beat. Antonio Carlos M. de Queiroz).
I note there are two "ifs" included in the first sentence of
the reply. There fore, the explanation in the subsequent sentences is
conditional. Tesla obviously didn't believe this would happen and I'm at a loss
to see why it should too. Tesla of course did not have the precision of modern
electronics at his disposal and his endeavors may well have been doomed to
failure given the inadequate technology of his day. Malcolm).
End of **
insert.
If you have a plan for getting round this problem, I would be _very_ interested
to see it :-o
BTW, I believe the base impedance of an unloaded resonator is just the
resistance of the wire (counting skin effect, radiation resistance etc) IOW,
it's a loss resistance, and not any impedance that has any significance as to
the spark producing properties of the coil. Matching to this makes no sense,
IMO, and is a bad idea.
Try a little thought experiment- A superconducting Tesla resonator would have a
base impedance of zero*
at resonance, so do you think it would be impossible to drive, and fail to
produce any sparks? Common sense says the opposite: it would work just like an
ordinary one, but better.
*Not
counting radiation resistance, which is going to be pretty minuscule anyway.
Steve C.
Response 2. June 25, 2004. Really, considering that:
1) The load is seriously nonlinear, and only effectively appears after breakout.
2) Most systems are not operated continuously, but in short bursts, maybe just
short enough to build up enough energy in the secondary system for breakout.
Something that shall be looked in the design is then what happens while the
output voltage is rising, in a condition of, ideally, no load. The ideal would
be to always present a resistive impedance to the driver while in this
condition. But this is precisely what happens if the load is removed from the
"matching theory" design. The input current is always in phase with the input
voltage while the output voltage is rising. If the energy is not spent, after
some cycles, the output voltage reaches a maximum (of about two times the
designed output voltage) and starts to fall. While it is falling the input
current is in opposite phase to the input voltage, returning energy to the power
supply. I will see if I can work out the details of the curious waveforms that
appear, and see if they are naturally in this way, or are forced to this way by
the matched design, that appears to work well in this aspect too. Antonio
Carlos M. de Queiroz
Response to above post.
June 28, 2004. I think matching to the base of an unloaded resonator is a most
fruitful approach in a number of respects:
- noting that a capacitor that is running down in a traditional disruptive
system still continues to pump the secondary until it is empty (given ideal k
for this to happen), I see no reason why a drive system operating in what is
effectively CW mode should ever suffer the phase reversal inherent in the
traditional endpoint of energy transfer.^
- the way appears to be open for machines to be built to generate enormous
voltages in the MV range with very high storage energies to boot with a rather
modest drive system (Greg?)
- the ISSTC is a system where shooting for a *high* secondary Q really comes
into its own.
- by carefully shaping a breakout point, one can tailor a given secondary to let
go before it destroys itself. In fact one can progressively shape it to restrain
further and further as confidence rises. A bit like opening up the gap in fact.
I expect the matching to be not-too-critical given the self-adjusting drive
frequency together with the transformation of MOhms to a few tens of Ohms by Zo^2.
Again, none of this is really that new. Sloan's CW system worked in much the
same way. Malcolm.
^ It doesn't :) Antonio presumably used a sine wave generator set to the resonant frequency in his sim. But the feedback drive systems that we use just flip their own phase whenever the resonant system tries to flip over and push energy back. So, an ISSTC could theoretically ring up to infinity (eep). Steve C.
Response to above response. June 29, 2004. Interesting consideration. But if the system is trying to return energy to the driver, and the phase of the driver is inverted to match the current, it will extract energy from the system faster. A simulation shows that this indeed happens, with the secondary voltage dropping twice faster. But at the same time the driver is pushing more energy into the primary circuit, and at the next phase reversal the output voltage is two times larger. The input current and output voltage waveforms resemble beating waveforms, but two times larger after each notch (lossless case). Antonio Carlos M. de Queiroz
Brief recap and answers: > Original poster:
m.j.watts@massey.ac.nz
> I think matching to the base of an unloaded resonator is a most fruitful
approach in a number of respects:
>
> - noting that a capacitor that is running down in a traditional disruptive
system still continues to pump the secondary until it is empty (given ideal k
for this to happen), I see no reason why a drive system operating in what is
effectively CW mode should ever suffer the phase reversal inherent in the
traditional endpoint of energy transfer.
The phase reversal appears in the system designed for a resistive load, if the
load is removed. The reason is that the excitation is not at one of the
resonances, but somewhere between them. I made some plots. See in:
http://www.coe.ufrj.br/~acmq/tesla/sstc.html. If the excitation is
moved to one of the resonances, there is no phase reversal during the output
voltage rise, but in both cases the designed maximum voltage is reached in
approximately the same number of cycles. Actually, the faster rise without load
is obtained by excitation at the central frequency, and not at the resonances.
***
> - the way appears to be open for machines to be built to generate enormous
voltages in the MV range with very high storage energies to boot with a rather
modest drive system (Greg?) - the ISSTC is a system where shooting for a *high*
secondary Q really comes into its own.
Not necessarily:
The ratio between the effective load resistance and the input resistance
determines the maximum voltage gain: Vout/Vin=sqrt(Rload/Rin). But as the
excitation frequency and the inductances can be chosen in a wide range, the Q
for a good secondary don't have to be necessarily very high. In my example, that
converts 229 V to 114.6 kV, a voltage gain of 500, the Q of the secondary is
just 8.5.
> - by carefully shaping a breakout point, one can tailor a
given secondary to let go before it destroys itself. In fact one can
progressively shape it to restrain further and further as confidence rises. A
bit like opening up the gap in fact.
>
> I expect the matching to be not-too-critical given the self-adjusting drive
frequency together with the transformation of MOhms to a few tens of Ohms by Zo^2.
>
> Again, none of this is really that new. Sloan's CW system worked in much the
same way.
Even Tesla tried to build "forced response" coils, by using fast gaps, and
Fessenden's high frequency generators for radio transmitters were just this same
thing. Antonio Carlos M. de Queiroz.
*** Response. June 30, 2004. I don't know the details of your simulation perhaps I missed something. For a single frequency excitation of a linear system at steady state the out put is also a single frequency. In the case above in does not mater whether you drive at a frequency in-between or not. i.e. the output must be a constant amplitude sine wave with he same frequency as the input at steady state if the transient has decayed.
(response to above
statement.
July 1, 2004.
I am considering a lossless circuit. The transient never decays, and adds to the
steady-state response. In the lossy cases, where a resistive load was
considered, the transient is what causes the output to ramp instead of suddenly
reaching the steady state value, what would be impossible. Antonio).
I suspect you may be observing the transient caused by the
switch on of the sine wave. I believe the transient will excite the two poles
and produce the two frequencies components of the classical beat envelope that
then exponentially decays. The drive frequency component will beat with the
two transient frequency components until they have decayed.
(response to above
statement.
July 1, 2004.
Yes. The transient is excited by the sudden application of the input sinusoidal
voltage (or square wave). Antonio).
I had thought that it may be possible to utilize the transient in such a way as
to significantly increase the final output voltage while maintaining maximum
input during the build up phase by picking the appropriate driving frequency. I
believe one contributor may have already explored this but it was not clear from
his contribution what the relationship of the drive to the two poles was.
(response to above
statement.
July 1, 2004.
At their geometrical mean. If this is not the optimum (apparently is), is quite
good. Antonio).
It would be interesting to analyze how a base current or field feedback system
oscillates and what its transient behave is.
My own interest is in zero crossing switching/feedback. Meaning that with a
series resonant primary it should be possible to use the zero current point to
control the switches and hence guarantee soft switching. At least initially the
driving voltage changes direction at the zero crossing points of the input
transient current. Such a method could be combined with peak current control
i.e. stop switching at the next zero crossing when a given peak current is
exceeded. Such a system may be less prone to self destruction and only requires
connection to the driver stage.
(response to above
statement.
July 1, 2004.
It appears possible to just look at the input current, reverting the polarity of
the input voltage at each change in direction of the current. Antonio).
It is informative to consider steady state conditions and consider steady state
matching options. However it may be more important to consider the transient
conditions particular for SSTC's that have very high maximum powers relative to
average wall plug power. In such systems a steady state conditions may never be
reached. This may also be important if soft switching is required. as even if
softswitching is achieved at steady state, will it also be soft during the start
up ???
(response to above
statement.
July 1, 2004.
The approach that I described works well too during the output voltage rise. The
driving frequency falls precisely at the geometrical mean of the two resonances,
and this forces resistive input impedance for any resistive output load,
including no load, and even during the initial transient. Antonio).
Consider the following approach. Lets assume some big beefy igfets say 600A peak
current and say an average wall plug current of 20A. Assuming direct switching
and pulsed operation with a triangular peak current profile then that's a 15 to
1 duty cycle. Then lets assume a pulse rate of 120Hz then that's an on time
of 556us. Additionally lets assume interrupter type operation as so the goal
is to reach max secondary voltage. So the matching problem is what
configuration will ramp up the driver current to 600Apk in 556us. If we assume
100kHz frequency and a 5kW wall plug power which say translates to a 30 x 6in
secondary assume a standard flat primary with a coupling round about 0.2. How
many turns on the primary and what size/rating of primary cap will get the
approximately 42J in to the system via an approximately 100kHz 310V ramping up
to 600Apk in 556us. Regards, Bob Jones.
(response to above
statement.
July 1, 2004.
Trying a design:
Imax=600 A.
Vin=310 V peak, square wave.
The input resistance is then (4/pi*310)/600 = 0.6578 Ohms.
556 us corresponds to 55.6 cycles of 100 kHz. A bandwidth around 100/55.6 =
1.7986 kHz would be adequate (not exact). A problem is that with the matched
design the output energy is function of the bandwidth, and so it can't be
specified independently of the output energy.
The required bandwidth is:
B=(16/(pi^2*sqrt(2)))*Vin^2/(Rin*Energy)= 634.6 Hz (after dividing by 2pi).
The voltage gain is then determined by the output capacitance:
n=sqrt(sqrt(2)/(Cb*B*Rin))
Assuming 40 pF of secondary+terminal+streamers capacitance:
n=3672
The resulting element values:
Ca: 10.9 nF
La: 233 uH
Lb: 63.3 mH
kab: 0.00449
Cb: 40 pF
Then required bandwidth resulted too narrow, and so the coupling between the
coils become too loose. This doesn't look as a realistic design...
To accumulate 42 J in a reasonable time, you need more input current.
Antonio Carlos M. de Queiroz).
June 30, 2004. Original poster:
srward16@hotmail.com
HI, I’ve been trying to follow along here though my electronics training (none)
sort of hinders things.
Anyway, thought some of you may be interested in some tests i did with my
original ISSTC. I decided it was time to try primary current feedback to drive
the coil. So how i went about it was put an 90:1 CT (home made) on the output
of my H-bridge driver. The CTs output drives yet another CT (60:1) to basically
reduce the current on the output... mainly so I don't overload my driver board
with a lowZ feedback source. Anyway, it works.
A pic of the setup:
http://www.hot-streamer.com/srward16/ISSTC/2004_06_29/100_0068.JPG
I took some pics with various tunings, but it seems my original tuning (used
with secondary base feedback) works the best. The following is the RF output
from the secondary coil:
http://www.hot-streamer.com/srward16/ISSTC/2004_06_29/100_0067.JPG
This is PRE-BREAKOUT. You can see that im turning the bridge off at about 7.5
divisions in that picture.
Here is what happens when i start to get breakout:
http://www.hot-streamer.com/srward16/ISSTC/2004_06_29/100_0066.JPG
Notice that the later half of the envelope looks distorted... i think this
correlates with when breakout is achieved (i can reduce the burst length to that
point and still hear just a slight crackling). So indeed the ISSTC system relys
on driving lots of *CW-like* power into each spark. As i turned it up more, it
appeared to breakout after just 6 cycles. Also, i can keep adding cycles to
lengthen the sparks, while this is obvious, it seems that this effect was more
pronounced with primary feedback than with secondary feedback (but i may need to
test this again). (Malcolm response: In some respects this is really
disappointing because it means that current is ramping up in the primary as well
as the secondary does it not? If so, it looks like there is no escape from
scaling the power supply component capabilities to match an increasing power
output (in which case who cares about feeding an established streamer CW). I had
thought it would be possible if not likely to prevent the driver from seeing
increasing currents like this. One cannot stop a base-drive system from seeing
with peak resonator base currents although these are typically far lower than
primary currents. So it looks as if in the double-tuned system, the artifacts of
double-tuning remain despite a constant feed of power into the primary (Y-N)?)
Also, this is an older shot of the RF output with secondary base feedback:
http://www.hot-streamer.com/srward16/ISSTC/waveforms/100_0074.JPG
Note that its a rather linear ring up, but as you begin to
really push the power, it looks like the predictions in Antonios paper (fast
ring up and levels off for the rest of the burst). Steve Ward. (Malcolm
response: I'd like to see what's happening in the primary circuit in that
scenario and it would be interesting to see the primary current waveforms for
the previous case also).
Response to above post. July 1, 2004. Nice clean waveforms Steve. Can't wait to see the beast at our Teslathon. Try some 4 sec time delay photos in a dark room at f3.5. They should look great. Dr. Resonance.
July 1, 2004. Original poster:
steve.conner@optosci.com
>In some respects this is really disappointing because it means that current is
ramping up in the primary as well as the secondary does it not?
Yes, this is exactly what happens. The energy in the primary circuit (1/2*Lp*Ip^2,
or 1/2*Cp*Vp^2) ends up roughly equal to the energy in the secondary circuit.
There are two differences though:
1: The energy in the primary mostly empties into the secondary after the
inverter shuts down. If we are cynical and say that half of it goes to the
secondary and the other half returns to the DC link, then we are still achieving
a bang energy 1.5 times greater than a conventional coil with the same primary
component ratings.
Response to 1. above: Right. I also noted that there was no phase reversal
present in the oscillogram of the secondary. Malcolm.
2: If the system runs in the quasi-CW mode, where the top voltage is heavily
clamped by breakout, the primary voltage and current will level off. Therefore,
the energy in the primary and secondary circuits is no longer increasing,
however the energy transferred to the discharge is still increasing.
This allows for an almost unlimited effective "bang energy" by just increasing
the inverter on-time, even though the voltage/current ratings of the primary and
secondary are modest.
Response to 2. above: Sure. Malcolm.
You can also imagine that it would allow the creation of sparks that were long
compared to the length of the resonator, without flashover, and indeed Steve
Ward's ISSTCs (and some built by others) do behave like this.
It seems that the way to promote this quasi-CW operation is to use tight
coupling (k=0.3). Steve C.
Response to above statement. The primary ringing up like that will be a function of loose coupling - i.e. energy is not transferred to the secondary as fast as it is being delivered to the primary. I figured that would be the case anyway but was still surprised to see the beating. I'm not surprised any more considering the relatively low coupling constant. Obviously, the tighter the coupling the better but I realize that only so much is practicable. On the plus side though, it would still seem that the primary L/C ratio can be usefully boosted to keep primary (and device) currents down while recognizing that voltages across the caps and coil would climb significantly. Malcolm
July 1, 2004.
Original poster:
steve.conner@optosci.com
>Then required bandwidth resulted too narrow, and so the coupling between the
coils become too loose. This doesn't look as a realistic design...
I had real trouble trying to calculate this one too. Doing it as a
quasi-disruptive design, using conservation of energy, I ended up with 100uH and
20nF for the primary components, and the primary capacitor would ring to over
44kV.
Using the L-match approach for a CW design, starting with a secondary top
impedance calculated from (power input/breakout voltage) I couldn't find a
solution at all :( If I wanted a reasonable loaded Q for the primary, the
inductance tended to infinity :((
I haven't figured out how to turn the handle on Antonio's method yet :(((
>To accumulate 42 J in a reasonable time, you need more input current.
I agree. AFAIK, Steve Ward's medium sized coil was running about 600A peak, with
a 380V DC link, for a ~10J bang energy and 1kW power input. Steve C.
July
2, 2004.
Original poster:
dhmccauley@easternvoltageresearch.com
I recently took measurements on my new ISSTC 2 coil. At 330V DC bus voltage I
was getting about 600A peak at 250us PW and 100Hz PRF. The primary capacitor
was about 0.1uF.
I also measured output arc current and measured 36A peak current to a ground
rod. Dan.
July 2, 2004. Brief recap and response.
> Original poster:
steve.conner@optosci.com
>
> >Then required bandwidth resulted too narrow, and so the coupling between the
coils become too loose. This doesn't look as a realistic design...
>
> I had real trouble trying to calculate this one too. Doing it as a
quasi-disruptive design, using conservation of energy, I ended up with 100uH and
20nF for the primary components, and the primary capacitor would ring to over
44kV.
Another problem with
too high Q systems. The voltage over the primary capacitor may rise excessively.
> Using the L-match approach for a CW design, starting with a secondary top
impedance calculated from (power input/breakout voltage) I couldn't find a
solution at all :( If I wanted a reasonable loaded Q for the primary, the
inductance tended to infinity :((
A simple L-match for
very high resistance changes results in impractical elements. In that case:
Power input (peak, at the driving frequency): 310*(4/pi)*600=237 kW Breakout
voltage: 300 kV If all the power goes to the output: Iload peak)=237/300=790 mA
Rload=300000/0.79=380 kOhms
Input resistance: 310*(4/pi)/600=0.659 Ohms
An L-match network operating at 100 kHz and converting 380 kOhms
to 0.659 Ohms:
Q: 759 !
C: 3180 pF
L: 0.796 mH
This circuit would take forever to reach the steady state.
It's better to make a double L-match, performing two impedance conversions by
the same factor:
Q: 27.5
C2 = 115 pF
L2 = 21.9 mH
(These convert 380 kOhms to 500.4 Ohms)
C1 = 87.6 nF
L1 = 28.9 uH
(These convert 500.5 Ohms to 0.695 Ohms)
Looks much better. Reaches 300 kV in 22 cycles (simulated).
The stored energy in C2 reaches 5.18 J.
> I haven't figured out how to turn the handle on Antonio's method yet :(((
Me too ;-) If the
design above looks reasonable (even with the quite too large C2), a doubly tuned
band-pass design shall be similar:
Rload=380 kOhms, Rin=0.659 Ohms
Voltage gain = sqrt(380000/0.659)=759
Only the bandwidth remains free: Using 5 kHz and the formulas for the doubly
terminated filter:
Ca: 85.5 nF
La: 29.7 uH
Lb: 21.3 mH
kab: 0.035
Cb: 119 pF
Almost the same values, and 5.3 J in Cb.
Note the low coupling. Antonio Carlos M. de Queiroz.
Brief recap and response. July 2, 2004.
>It's
better to make a double L-match, performing two impedance conversions by the
same factor:
>Q: 27.5
>C2 = 115 pF
>L2 = 21.9 mH
>(These convert 380 kOhms to 500.4 Ohms)
>C1 = 87.6 nF
>L1 = 28.9 uH
>(These convert 500.5 Ohms to 0.695 Ohms)
If I understand correctly, an inductively coupled Tesla coil also has a third
untuned stage of impedance conversion, of ratio (L2:L1)^2. Did you take this
into account in this calculation, or are you just assuming a L-match network
base-feeding the secondary? Or have I got it completely wrong? Steve C.
July 3, 2004. Response to above post & question.
In
this case there is no transformer, just two L-matches in cascade:
. o---L1---+---L2---+---o
. | |
. C1 C2 Rload
. | |
. o--------+--------+---o
C2 is the combination between the self-capacitance of L2, the
distributed capacitance of the terminal, and streamer capacitance. Quite too
large in the example. By using two stages, a high gain system can be built
without excessively high Qs. The version with a transformer can be derived from
a cascade of a high-pass L-match and a low-pass L-match, or from a band-pass
filter. The cascade above can be designed as a filter too, resulting in
"maximally resistive" input resistance, but the design is more complex than in
the bandpass case. Antonio Carlos M. de Queiroz.
July 3, 2004.
Response to Antonio and Steve C.
I don't have much of a handle on Antonio's method either. Best I can do is that
the burst time relates to the step response and the step response is related to
the bandwidth of a filter so by picking a particular filter type, a
Butterworth, then using filter theory it can be solved. In a master oscillator
configuration some insensitivity to frequency is required whether the flatness
of a Butterworth is too flat I don't now. However a feedback driven system and
in particular a drive current feedback would be more tolerant.
I now believe that during the build up the input will only appear resistive if
its driven at the center frequency and in phase with the transient. My initial
reaction to this is it is at the minimum in the voltage gain but may be
compensated for by changing the turns ratio. If its soft switching we want we
have no other option.
I think the reason my design parameters did not work well is that I had fixed
the frequency. I think frequency must be reduced so that with reasonable
coupling factors the time to the max energy is sufficiently long to get the
required energy in. Ss you suggest peak current could also be increased. I
assume its a design trade off between long lower current low frequency burst and
high current shorter higher frequency burst.
Or putting it an other way for a given peak current and bang energy the burst
length is determined which then fixes the maximum frequency. Bob Jones.
July 3,
2004. Brief recap and response.
Original poster:
"Antonio Carlos M. de Queiroz"
<acmdq@uol.com.br>
Tesla list wrote: Original poster: "Bob \(R.A.\) Jones"
a1accounting@bellsouth.net
>
> Response to Antonio and Steve C.
>
> I don't have much of a handle on Antonio's method either. Best I can do is
that the burst time relates to the step response and the step response is
related to the bandwidth of a filter so by picking a particular filter type, a
Butterworth, then using filter theory it can be solved.
Yes.
> In a master oscillator configuration some insensitivity to
frequency is required whether the flatness of a Butterworth is too flat I don't
know. However a feedback driven system and in particular a drive current
feedback would be more tolerant.
The flatness provides some insensitivity to frequency and errors in the element
values, specially in the "resistiveness" of the input impedance. A matched
design with current feedback will self-tune to the right frequency.
> I now believe that during the build up the input will only appear resistive
if its driven at the center frequency and in phase with the transient.
Yes. Specially if the load is
varying.
> My initial reaction to this is it is at the minimum in the voltage gain but
may be compensated for by changing the turns ratio. If it’s soft switching we
want we have no other option.
The central frequency can be a minimum or the maximum, depending on the load.
With light load it is a minimum. With the designed load, in the doubly
loaded design, the voltage gain is still a minimum, close to the two maxima at
its sides, but the input impedance is maximally resistive (continues to be
approximately resistive in all the bandwidth of the filter). The system can also
be designed to present a flat maximum at the central frequency, but then the
input impedance is not maximally resistive (it is just at the central
frequency). With high load, it is a maximum.
> I think the reason my design parameters did not work well is that I had fixed
the frequency. I think frequency must be reduced so that with reasonable
coupling factors the time to the max energy is sufficiently long to get the
required energy in. Ss you suggest peak current could also be increased. I
assume its a design trade off between long lower current low
frequency burst and high current shorter higher frequency burst.
>
> Or putting it an other way for a given peak current and bang energy the burst
length is determined which then fixes the maximum frequency.
Using as parameter the maximum
energy stored in the load-side capacitance after the steady state is reached, in
the doubly terminated band-pass Butterworth filter design it results as:
E=(Vin*4/pi)^2/(2^(1/2)*B*R)
where Vin is the peak value of the input square wave, B is the filter bandwidth
in rad/s, and R is the input resistance. We see then that the energy depends
only on the bandwidth, and not in the frequency (w0). But the number of cycles
required for reaching the steady state is about w0/B, and so the output energy
is proportional to the number of cycles, as expected, and inversely proportional
to the frequency.
Or, it is directly proportional to the burst time, no matter which is the
frequency.
This calculation assumes a resistive load with constant value connected all the
time, what is not very realistic, but that is close to what happens with a load
that is connected only after breakout.
Something that I still don't like is that, for resistive load, there is no
mechanism to limit the input current if the load decreases in resistance, other
than to stop the driver if the current grows excessively. The input current,
however, doesn't change much from cycle to cycle, even if the output is
short-circuited, and so a safe control is not difficult. Antonio Carlos M. de
Queiroz.
