[ §1 · tank ]

Primary design: impedance, turns and detuning

Nobody sets the primary current. It comes out of a division, and the number you actually choose is the impedance of the tank.

CommonGnd4 August 20269 min read

Nobody sets the primary current. It comes out of a division, and the number you actually get to choose is the impedance of the tank.

The one free parameter

The frequency is set by the secondary, so the product L·C is fixed and the two can only move as a pair. What is left is:

Z = sqrt(L/C)
L = Z/(2·pi·f)
C = 1/(2·pi·f·Z)

At a fixed frequency the primary current goes as C, the power goes as C, and the voltage across the tank capacitor does not depend on C at all. Higher impedance, less current. That is what "high impedance primary" means, and Gao puts it as a large primary inductance with a small resonant capacitor, generally low capacitance to raise the impedance.

For a QCW the motive is direct: the ontime is tens of times longer than an ordinary DRSSTC's, and the switches have to carry that current for milliseconds rather than microseconds.

Where the current comes from

I_pk = (4/pi) · V_bus / R_total

The 4/pi is the fundamental of the bridge's square wave. R_total is the copper, the switches, the bank's ESR, and the resistance reflected back from the secondary and the arc, and it is that last term that decides the answer.

The rate of rise is not the limit either. dI/dt = 2V/(pi·L) is around ten amps a microsecond on a typical tank, so the current reaches its working value in five to seven cycles, well under a per cent of a bang.

Two regimes, and almost everything else depends on which you are in

That documented case — 300 A calculated, 50 A measured — is usually filed as an arithmetic mistake. It is more likely a machine in the other regime, and the distinction is worth more than the story:

OCD-limited
  current reached the
  threshold; protection
  is what cuts it

impedance-limited
  current stopped below
  the threshold; the
  bridge gave all it had

The boundary comes out of the same division as everything else on this page. To reach the threshold at all:

R_refl < (4/pi)·V_bus/I_OCD
           - R_copper

At 325 V and a 160 A detector that is 2.59 Ω before subtracting the losses that are not the arc — the copper, the switches, the bank's ESR, which on this machine's own published figures come to roughly 0.3 Ω. So the reflected resistance has to stay under about 2.3 Ω, and past that the detector is decoration.

R_refl    current
1.0 Ω      319 A   OCD-limited
2.3 Ω      160 A   the boundary
4.0 Ω       96 A   impedance-limited
11.9 Ω      34 A   impedance-limited

And the sign of the impedance question flips between them

Which is the reason to care, because two competent answers to "should the primary be high impedance" point in opposite directions and both are right.

Uspring, on HVF 1928, argues against:

power transfer from the primary to the secondary tank will reduce primary current… in a high impedance primary, the current might be reduced to levels far below the bridge's capability. That limits input power and consequently arc length.

And the arithmetic argues for — but only under a detector that is actually reached. Power goes as I²·R_refl, and R_refl goes as the primary's inductance at the same coupling, so at a fixed threshold the power rises in proportion to the inductance:

L     I same arc   P    arc
x1        1.00    1.00  1.50 m
x1.5      0.82    1.50   1.72
x2        0.71    2.00   1.89

Arc length goes as the cube root of energy, so 50 per cent more power buys about 14 per cent more arc. That column is derived, not measured, and it is worth saying that it inherits every assumption in L ∝ E^⅓.

Both statements are true and they describe different machines. A high impedance primary gives you more arc if you are limited by current and less if you are limited by impedance. No amount of reasoning settles which one you have.

The measurement that settles it

Peak primary current on the first bangs

read peak I_pri, first flashes

reached the OCD threshold
  -> current-limited
  -> more turns help

stopped well short of it
  -> impedance-limited
  -> more turns make it worse

One number decides all of these at once:

Whether to add primary turns at all, which is the question above.
Cone or cylinder. A cylinder runs at lower current, which is easier on the bridge, on the copper and on the bank's RMS rating; a cone puts its top turn further down the secondary and lowers the flashover risk. The electrical argument and the geometric one genuinely disagree, and the current is what says which one you can afford to lose.
Whether the bank's RMS overload is real, since it scales with the same current.
Whether the detector threshold is reachable, which is the boundary above.
Whether P ∝ I²·R_refl describes this coil at all.

Turns, detuning, and one happy coincidence

More turns means more inductance, which means higher impedance and less current, and at the same time a lower primary frequency. Those are two separate jobs and adding turns does both: you detune downwards and unload the switches and the bank in the same movement.

How far to detune is not a fixed percentage. The rule is that the amount of detuning should equal the amount your arc detunes you, and the difference matters. Primary turns go as 1/(1-d), so detuning 25 per cent rather than 15 is about thirteen per cent more of them — on a ten-turn primary, more than a whole turn. Make your taps generous in both directions, because until you have measured the detuning you do not know the number.

And the figure is the secondary's movement rather than the pole's, and it is not defined without an arc length beside it: one worked machine detunes 16.7 per cent under a 1.5 m arc, and a percentage carried over from somebody else's coil is carrying their arc with it. Which movement is which, and the trajectory the percentages come off, are on one page so that they can be replaced in one place when somebody finally measures them — and one figure that used to be quoted around here has already been withdrawn for resting on the wrong capacitance.

Coupling

An ordinary DRSSTC lives at k = 0.13 to 0.17, which is what JavaTC will suggest. A QCW lives at 0.30 to 0.50.

Raising k does not lengthen the arc. The reflected resistance goes as k squared, so tightening the coupling into an unchanged tank cuts the current and the power. The builders with the longest arcs sit at 0.15 on enormous coils; the builders with the highest arc-to-secondary ratios sit near 0.5 on tiny ones. Coupling buys the ratio. Power buys the length, and there is a whole article about that.

What coupling does buy is tolerance to detuning: the poles spread further apart as k rises, and each is stiffer against what the arc does to it. The cost is that a higher k eats the detuning budget at the lower pole and pushes the upper one further from what the bridge can do.

The test that the model is describing your coil: the current should fall as one over k squared. If it does not, stop and find out why before going further.

The ceiling on coupling is set by racing sparks, not by any calculation. They have been caught at k = 0.185.

Building it

Copper tube or wide strip. At the working frequency the skin effect keeps the current on the surface, so most of a round conductor's cross-section is not doing anything.
It melts at high bangs per second. Long bursts heat the copper seriously. Air and cross-section, both taken seriously.
Water cooling works with ordinary water, but arrange a pressure relief. There is a known case of a hose exploding with steam directly into the bridge.
A metal frame under the primary is a shorted turn. Ground each rail at one end only.

The primary stands beside the secondary on the department diagrams rather than lying inside the bridge, because the two being on one axis is the only way they are coupled at all.

primarycouplingtuning