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Re: Transformerless TC-Excited Capacitive Transformer?



Original poster: "Antonio Carlos M. de Queiroz by way of Terry Fritz <twftesla-at-qwest-dot-net>" <acmq-at-compuland-dot-com.br>

Tesla list wrote:
 >
 > Original poster: "Jolyon Vater Cox by way of Terry Fritz 
<twftesla-at-qwest-dot-net>" <jolyon-at-vatercox.freeserve.co.uk>
 >
 > Antonio,
 > I have had a look at the circuit diagram I sent the other day
 > and it seems that when C1 and L1 (which are common to both networks) are
 > included
 > the lefthand network is sixth-order bandpass filter
 > and the righthand a fouth-order lowpass filter -admittedly that does not
 > seem quite right.

It's not clear to me that it's impossible to design it to work in
this way, but I don't know how yet.

 > So how about a couple of bandpass filters connected oppositely across L1?
 > That way -when C1 and L1 are included-  won't both sides of the bridge be
 > sixth-order filters?

A bridge of identical filters, with the last elements interchanged
to create a balanced output, is easy to design. And with sixth-order
circuits, the energy transfer is faster. The networks are, however,
very complicated.

 > Would there still be problems with the ground capacitance "unbalancing" the
 > bridge or would the effects of ground capacitance be less dominant  with
 > this circuit than with the fourth-order "twin"?

For proper balancing, both outputs must have shunt capacitances, and
both final coils must have (they have anyway)

 > ...

I could't understand your drawing again. Please use only the courier
font for drawings.

Balanced networks can be derived from any single-ended
network of these types, by just exchanging the two last branches.
A schematic diagram for a 6th-order band-pass triple resonance network
with balanced output can be:

     o------+--2C1--+-L3/2-+-----+---------+----+
            |       |      |     |         |    |
            |       |      |     C5        L6   C7
            o       |      |     |         |    |
    PSU    gap    L2/2    2C4    +----+a   +----+b
            o       |      |     |    |    |
            |       |      |     L6   C7   C6
            |       |      |     |    |    |
     o------+-------+------+-----+----+----+

The element values would be the same of the single-ended version,
with the impedance level of the first sections divided by 2.
Note that it's not necessary to duplicate the L3/C4 section.
The construction can be as shown below. C5 and C7 are the
capacitances between the induction plane and L6/terminal and
between L6/terminal and ground, made identical:

                                (=====)a    (=====)b
                                   |           |
                                   L6          L6
                                   |           |
     o------+-2C1--+--L3/2--+--(=) | (=)---(=)-+-(=) <-Two influence
planes
            o      |        |      |
    PSU    gap   L2/2      2C4     |
            o      |        |      |
     o------+------+--------+------+---------o Ground

An example:
With 2C1=5 nF, L6=28.2 mH, and C5=C7~=5 pF, mode 7:8:9 with C7 being
0.516 of the total output capacitance, the element values are:

C7:        4.997351 pF
L6:    28200.000000 uH
C5:        4.998551 pF
C4:       75.311350 pF
L3:     3509.530334 uH
L2:      109.244413 uH
C1:     2500.000000 pF

The circuit transfers energy in just 4 cycles, producing a voltage gain
of ~31. A similar structure without L3 and C4 would have to operate
in mode 31:32, with large losses due to the long energy transfer time.
The elements would be (omit L3 and C4 in the drawing above):

C7:        5.046556 pF
L6:    28200.000000 uH
C5:        5.046639 pF
L2:      113.621877 uH
C1:     2500.000000 pF

 > Finally, won't the ground current or voltage (as measured by M1) be at a
 > MINIMUM for a properly balanced bridge whether it be fourth-order,
 > sixth-order or so on -can this "null" be used as an aid to setting up the
 > circuit?

The problem is that you can't measure the current that returns from the
distributed capacitances. Low-power tuning can always be done by
looking at the voltage over L2 while exciting the circuit with a slow
square wave in place of the gap, maybe looking also at the two output
voltages with probe antennas close to the terminals. After some
adjustments for symmetrical output and for a notch at the input
voltage, the system is in tune. Add some extra terminal capacitance
to account for streamer loading.

Antonio Carlos M. de Queiroz