Re: DRSSTC driver tests- Dual resonance disaster

["Tesla list" <tesla@xxxxxxxxxx>]

Original poster: "Bob (R.A.) Jones" <a1accounting@xxxxxxxxxxxxx>

Hi Steve,

Original poster: "Steve Conner" <steve.conner@xxxxxxxxxxx>

 >These

>>state transimpedance (ie Iin=f(Vin, Vout, L1, L2, C1, C2, k, omega) where
>Iin, Vin, Vout are complex quantities) I want to know this because it would
>let you predict the worst case primary current in a self resonant coil,
>knowing the breakout voltage and the coil constants, and adding the
boundary
>condition that Iin and Vin are in phase.

>I know that several of the transfer functions are ill defined at the poles
>for a system with no loss resistance (for instance vout/vin tends to
>infinity, Iin/Vin tends to infinity) but I believe Vout/Iin should tend to
a
>constant at the poles. My reasoning is that Vout and Iin both tend to
>infinity, and infinity divided by infinity could perhaps be a finite number
;)

>If you (or anyone else) have the transfer function for the dual resonator
in
>pole-zero form I'd be very interested to see it. I tried deriving it myself
>but it's a long time since I've been in a maths class :-(

The transfer function simplifies if your driving at the mid frequency i.e.
the uncoupled frequency of the primary and secondary assuming they have the
same frequency. Most of the Ls and Cs cancel.

At that frequency the input current does not grow with out bound even in the
loss less case. Because you are not driving it at one of the poles. If you
assume the transient/s have die away (some how! this is supposed to be the
lossless case), then steady state input voltage is equal to Isecondary*Lm*s
so the secondary current is bounded by the input voltage which bounds the
output voltage as Isecondary flows thru Csecondary.  At that point the input
current would be zero. In practice (high Q) it will be just large enough to
supply the losses. So in the lossless case the steady state transimpedance
is some finite value divided by zero. (assuming the transient had died away
some how). If you did generate the transimpedance with the transient present
it presumable would be time depended  sinusoidaly going positive and
negative as power is first input to then fed back to the supply.

If you assume you never get past the first primary peak then the peak input
current can be calculated from the equations I posted a while back. See a
copy of one of  the posts on the thread below.
With a bit more work knowing the peaks of the input follow approximately a
sinwave you can calculate the input current for each peak of the out put
voltage if you know the mode.
What we don't apparently know is what happens after break out.  But if your
controller forces not only zero current switching but in phase drive voltage
and current then presumable if there is input current then is input power
too. That presumably would also prevent the power being fed back to the
supply so possibly several envelopes could be used to build up the output
voltage.

Robert

Re: DRSSTC design procedure - draft

Hi all,

> Original poster: "Antonio Carlos M. de Queiroz" <acmdq@xxxxxxxxxx>
>
>
>  > 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 simplfy.
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.
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