Re: DRSSTC design procedure - draft

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Original poster: "Bob (R.A.) Jones" <a1accounting-at-bellsouth-dot-net> 


Hi all,


 > Original poster: "Antonio Carlos M. de Queiroz" <acmdq-at-uol-dot-com.br>
 >
 >
 >  > 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