Re: Resonant charging design

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Barton B. Anderson by way of Terry Fritz <twftesla-at-qwest-dot-net>" <tesla123-at-pacbell-dot-net>

Hi Richie,

Thank you for this step by step procedure. I just read through it but
haven't began calculating. If this
works out, it will be a revolutionary step to producing long sparks and
efficient running coils. You didn't
happen to graph out 60 Hz did you? Can't wait to get started. I'm building
a new coil and will pull this
into the design. It will be interesting to see how it works out. Might be a
while before first light on
this coil however.

Take care,
Bart

Tesla list wrote:

> Original poster: "R.E.Burnett by way of Terry Fritz <twftesla-at-qwest-dot-net>"
<R.E.Burnett-at-newcastle.ac.uk>
>
> Hi guys,
>
> About a week ago,  I said that I would go through my method for
> designing the AC resonant charging circuit for a tesla coil.
>
> What I am going to explain is a way in which I propose that the
> ballast and tank capacitor values for a particular system can be
> determined at the design stage.  i.e.  You choose the rotary speed,
> and how many kW of power you want,  and use the following steps to
> obtain the required ballast inductance and tank capacitance.  As a
> bonus this method also achieves good power factor.
>
> I should say that this is very much "simulation based" at this
> stage,  as I haven't built lots of different systems to prove every
> combination.  However,  I have used Microsim extensively to verify
> my work,  and have got promising results.  What little practical
> work I have been able to do has also closely matched the predicted
> performance.
>
> Ok,  here goes......
>
> Let's say I want to design a Tesla Coil to process 10 kW of real
> power,  using a 12 kV power transformer,  and a 371 BPS asynchronous
> rotary gap.  Power feed is from 250V -at- 50Hz,  and at this stage I
> have no idea what ballast inductance or tank capacitor value would
> give good results.
>
> One possible approach would be to try many combinations of ballasts
> and capacitors in simulations until the combination that results in
> best performance is found.  This is what I did and it was _VERY_ time
> consuming !  I did this so hopefully you don't have to take this
> hit and miss approach.
>
> STEP 1: Determine what the resonant frequency of the charging
>         circuit should be for your chosen rotary firing rate (BPS.)
>
> We know that the ballast inductor at the primary side of the power
> transformer resonates with the tank capacitor at the secondary.
> There have been many references to LTR caps.  We are not concerned
> with whether the cap is larger or smaller than a particular value.
> We are only concerned with what the actual resonant frequency of the
> charging circuit is.  In this design approach there is nothing
> special about resonance at 50 or 60 Hz.
>
> For a whole range of break rates,  I tried simulating charging
> circuits with different resonant frequencies.  Particular attention
> was paid to the charging waveforms and power factor.  As I explained
> before,  it is the resonant frequency of the charging circuit which
> determines all of the timing related stuff like how quickly the cap
> charges between firings,  and the power factor.  So the resonant
> frequency should be made right to start with.
>
> The graph here...
>
> http://hot-streamer-dot-com/temp/reschrg1.gif
>
> ...displays what I think the resonant frequency of the charging
> circuit should be for different rotary break rates to get a power
> factor of 0.85
>
> In our example the rotary break rate is 371BPS so the charging
> circuit should be designed to resonate at 85.2 Hz.
>
> That ensures a PF of 0.85 at 371BPS,  and ties down the product of
> the ballast and the tank capacitor,  but there are still many
> combinations that would give the same resonant frequency.   Indeed
> all combinations which give the correct resonant frequency do give
> the same charging waveforms and good power factor !  As long as
> the product of L and C is right for the chosen BPS,  the resonant
> frequency is right and the charging circuit works correctly.
>
> Choosing various combinations of L and C only changes one thing.
> The _POWER THROUGHPUT_.
>
> STEP 2: Determine what the ballast inductance should be.
>
> If changing the ballast and tank cap values together only alters the
> power throughput,  then we can adjust these values to get whatever
> power level we want.  However if we don't know where to start,  how
> can we pick a value for one of them without guessing and trying
> something first ?
>
> Well,  a good place to start is by ballasting the transformer to
> whatever power throughput we desire.  In our example,  the desired
> power throughput is 10kW,  so we need to work out what ballast
> inductance will limit the transformer to 10kVA when its secondary
> winding is shorted.
>
> 10kW represents a current of 40 Amps at 250 Volts.  Therefore the
> ballast choke should present an impedance of 6.25 Ohms.  At our
> supply frequency of 50Hz,  that requires 19.9 mH of inductance.
>
> This figure of 19.9mH for the ballast is a "first-approximation".
> Although it draws 10kVA when the transformer is short-circuited,
> it is unlikely that this will give exactly 10kW of real power
> throughput at our chosen break rate.  Life is just not that
> easy ;-)
>
> STEP 3: Adjust the ballast inductance depending on chosen break rate.
>
> If we used the ballast value calculated above,  the real power
> throughput would be somewhat less than our desired 10kW.  The actual
> shortfall depends on the rotary break rate.
>
> The graph here...
>
> http://hot-streamer-dot-com/temp/reschrg2.gif
>
> ...shows what the real power output is as a fraction of the
> "ballasted VA"  for different rotary speeds.  Notice that low speeds
> give almost the full "ballasted VA" as real power, but higher speeds
> give progressively less real power for the same ballast setting.
>
> For our chosen rotary speed of 371 BPS,  it can be seen that the
> running Tesla Coil would process only 64% of the 10kVA that it was
> ballasted to draw with a short-circuit.   Therefore we must multiply
> the ballast inductance by 0.64 to correct for this shortfall and meet
> the desired power throughput criteria.
>
> 0.64 x 19.9mH = 12.74 mH
>
> The "corrected" ballast inductance is 12.74mH,  and this is the value
> that we must use to get our full 10kW of power throughput when running.
>
> STEP 4: Calculate the required tank capacitor value.
>
> Now that we know the ballast inductance,  and the resonant frequency
> of the charging circuit,  we can finally calculate what the tank
> capacitance should be.
>
> First we need to imagine that the ballast inductor is at the high-
> voltage side of the power transformer, instead of at the low voltage
> primary side.    This enables us to deal with the ballast inductor and
> tank capacitor like a series resonant circuit.  The equation:
>
> F = 1 / [ 2 pi sqrt (L x C) ]  can then be used as normal.
>
> In our example the power transformer steps 250 Volts up to 12000 Volts,
> so it has a turns ratio of 48.  Impedances are transformed by the
> square of the turns ratio,  so our 12.74 mH ballast inductor becomes
> 48 x 48 x 0.01274  =  29.35 H  when referred to the secondary side.
>
> Re-arranging our resonant frequency equation and plugging-in the values
> for F (85.2) and L (29.35) gives us a tank capacitance of 119 nF.
>
> And that is it.
>
> To summarise the ballast is 12.74 mH and the tank capacitor is 119nF.
> This gives us a system that processes 10kW of real power at 0.85 PF
> with our 12kV transformer and 371 BPS rotary spark gap.  (Running a
> Microsim simulation with these parameters gave me 10080 W of power,
> 11999 VA,  and therefore a PF is 0.84)
>
> There are two useful advantages of knowing that the approximate
> power factor is 0.85:
>
> 1. We can estimate the VA or current draw from the 250 Volt supply.
>         In this case the real power is 10kW,  so the VA = 10000 / 0.85
>         The VA = 11.76 kVA,  or 47 Amps from our 250 Volt supply.
>
> 2. We can calculate the optimum PFC capacitance to improve PF further.
>         With a PF of 0.85,  the VARs are roughly equal to _half_ the
>         watts.  Therefore we need 5kVAr of PFC correction capacitance
>         across the supply to achieve the absolute maximum power factor.
>         That is a capacitive current of 20 Amps at 250 volts requiring
>         255uF of capacitance at 50Hz.  This will cancel any remaining
>         reactive current and maximise power factor.
>
> Increasing or decreasing the power:
>
> Remember that all the timing type behaviour is determined by the
> resonant frequency.  If you want to double the power throughput of your
> design,  all you need to do is double the tank capacitance and halve
> the ballast inductance.  (Assuming the power transformer can cope ;-)
>
> This will process twice the power due to the reduction in impedance.
> However,  the charging waveforms,  peak voltages,  and power factor
> are not changed because the resonant frequency of the charging circuit
> has not changed !
>
> There is nothing particularly special about the design method described
> here.  All I have done is run a large number of simulations to find out
> what works best at various break rates.  Then I have condensed this data
> into two graphs showing:
>
> 1. What the resonant charging frequency should be for various BPS.
> 2. How much the ballast needs to be adjusted to get the desired power.
>
> Now you can work backwards from Watts and BPS to get ballast inductance
> and tank capacitance reasonably easily.  Think of this method as being
> like a lookup table.
>
> Congratulation if you have read this far,  I appreciate that this stuff
> is heavy going !!!  This kind of approximations and calculations are
> ideally suited to a computer program,  so I intend to produce a simple
> program to do all this stuff automatically.  It will ask for your
> intended power level in kW and your rotary speed in BPS.  It will then
> calculate the suggested ballast and capacitor values as explained above,
> but much faster than using graphs and a calculator.
>
> I hope that the above information is error free,  and has provided some
> insight into the work that I have been doing for some time.  If you
> have any questions,  criticisms or suggestions I will try to deal with
> them as quickly as possible.
>
> Regardless of whether you understand the method or you just use the
> computer program,  I hope this work will lead to bigger sparks for all ;-)
>
>                                                 Cheers,
>
>                                                 -Richie Burnett,
>                                                 (Newcastle, UK)