current in tube
Hi mlurye,
I don't think it is a stupid idea at all. This is only my opinion on current and the Gray Tube. With how I see the tube operation, I still don't think there is anything mystical about it...just profound.
With the tube setup, current (if you're talking "electron" current") may have the option of moving from either the LV rod to the HV rod when the HV jumps. Or from the grids to the rods when the potential jumps to the rods.
That is IF there is current moving according to how it is 'supposed' to move. In the opposite direction of the positive potential and in the same direction as the negative voltage potential flow.
In both cases, the HV is moving into a LV +...at the LV rod AND on the other side of the inductive load.
There is a difference in both of these.
The LV + potential at the low voltage rod is not even there UNTIL the diode is connected to the circuit by the commutator.
The LV + potential at the grids is ALREADY sitting there.
In both cases when the HV jumps...IF there is current...the electron current is forced to leave the + terminal of the battery. Will that easily happen? Can that happen in the short blip of an impulse of the HV discharge? Since current is so much slower than the instantaneous potential and with the increase in discharge time of the HV cap by the diode shutting off - the current is offset even more so it is highly doubtful that there can be much current at all that is causing any of these effects based on known principles.
Hi mlurye,
I don't think it is a stupid idea at all. This is only my opinion on current and the Gray Tube. With how I see the tube operation, I still don't think there is anything mystical about it...just profound.
With the tube setup, current (if you're talking "electron" current") may have the option of moving from either the LV rod to the HV rod when the HV jumps. Or from the grids to the rods when the potential jumps to the rods.
That is IF there is current moving according to how it is 'supposed' to move. In the opposite direction of the positive potential and in the same direction as the negative voltage potential flow.
In both cases, the HV is moving into a LV +...at the LV rod AND on the other side of the inductive load.
There is a difference in both of these.
The LV + potential at the low voltage rod is not even there UNTIL the diode is connected to the circuit by the commutator.
The LV + potential at the grids is ALREADY sitting there.
In both cases when the HV jumps...IF there is current...the electron current is forced to leave the + terminal of the battery. Will that easily happen? Can that happen in the short blip of an impulse of the HV discharge? Since current is so much slower than the instantaneous potential and with the increase in discharge time of the HV cap by the diode shutting off - the current is offset even more so it is highly doubtful that there can be much current at all that is causing any of these effects based on known principles.
Comment