The design suffers from a few potential problems:
(1) The Winding of the transformer and its total reactance (AC/DC) should be closely balanced. If it isn't there is no obvious adjustment to fix this as it stands.
(2) The wear in the tube should be even along with emission currents from both ends of the heater element. Typically however, tube emission varies unpredictably over the tube-life.
(3) There should be no current through the grid, and the resistance path to ground should be relatively low, to keep the bias stable and well-defined.
Now lets talk about why the circuit cannot deliver:
(1) The winding for a heater-transformer will not be wound the way a Push-Pull Output xformer is. In those, the windings and current must be balanced, to cancel the magnetization and core-saturation. A typical heater winding is a single coil, not a bifilar winding, and so, balancing it has no meaning in regard to D.C. flow. It WILL ALWAYS magnetize the transformer core, and use up precious core-flux. That is why it is better for the other windings to leave unneeded windings unused. Nothing is free.
The only issue then is to balance the D.C. flow for the purpose of even wear on the cathode emitter. But this also will fluctuate with tube-wear and is uncontrollable from outside of the tube. At best, you can have a compensating circuit which monitors D.C. idle current through both terminals, and adjusts itself.
(2) There is a real danger here that any attempts at adjustment of current-flow will exascerbate tube element wear unevenness (and also applied voltage). I leave that to your imagination.
(3) In many cases, Direct-Heated Triodes have relatively low Bias-points, and conduct some grid current during a cycle. In this case, Cathode-Bias technique may be less stable than desired.
I've made a chart to help understand the Direct Heated Triode problem:
The 811a is a good example, because the heater-voltage is significant, and the bias might reasonably be set quite low, even zero volts, or a +ve value for class A2 operation.
The very nature of a tube is such that the grid, with very little voltage, can control large current and voltage swings through the plate.
In part the reason why this works is that the grid is much closer to the cathode than the plate, which carries the High Voltage attractive force. The greater distance of the plate in turn lessens its influence.
It follows that small changes in distance between cathode and grid have the same effect as small changes in voltage on the grid. Although distances can't be changed, the effects can be understood by the analogy of distance.
For this purpose, I've redesigned the diagram for the triode,
in order to simulate the effect of the difference in voltage at each end of the cathode,
and its influence on cathode current flow through the screen,
when D.C. voltage is used instead of A.C. .
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"Could one heat the filament withThe answer is that using a split-supply would make no real difference,
a well-filtered or regulated +3.15V and -3.15V supply?"
because the effect here takes place INSIDE the tube.
Secondly, looking at the diagram, one could in theory ground the cathode in three places, at either end of the D.C. supply, or in the middle (of a split supply).
But grounding in the middle would be disastrous, for another reason:
Instead of flowing directly to ground from cathode,
a center-connection would cause the plate current to flow through
the INTERNAL RESISTANCE of the power-supply!
This internal resistance will be unknown, and unpredictable.
For one thing, this resistance is likely to vary with current, heat-effects, and frequency-sensitive factors.
Thus,
(1) There is no advantage in relation to the cathode emission in using a split-supply (i.e., +/- center-tap).-------------------------
(2) There could be a serious negative effect on the quality of the plate-current through the cathode to ground.
While a center-tap connection is not beneficial,
it indeed DOES make a difference which end of the D.C. supply is grounded!
If one grounds the +ve side, one can force the current
to flow all the way along the cathode. This will equalize both current loads and temperature across the cathode.
In the diagram above, the -ve side is grounded, exasperating the imbalance.
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