Re: SSTC theory

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

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

Tesla list wrote:
 >
 > Original poster: "Bob \(R.A.\) Jones" <a1accounting-at-bellsouth-dot-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 its
 > 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