ORIGINAL MESSAGE:
On Tue, 15 Dec 2009 17:51:16 -0500, "Gary Schafer" <garyschafer@comcast.net>
wrote:
>Why do you think that the anode voltage will swing to 2x power supply
>voltage?
>
>73
>Gary K4FMX
REPLY:
Because that's how amplifiers with resonant anode tank circuits work.
At idle, the DCV on the anode will be whatever the supply is. When driven to
full output, on the negative-going part of the cycle the instantaneous anode
voltage will approach zero. On the other half cycle, the flywheel effect of the
tank circuit will drive the anode to approximately twice the idling DCV by
coupling back through the plate blocking cap. That's how the pi-net supplies the
"missing" half cycle of the RF by virtue of the flywheel effect.
Think of it this way: When running at full power, on the pi-net side of the
blocking cap, the voltage swings from zero to the full negative supply on one
half cycle, and to full positive on the other half cycle, due to the flywheel
effect of the pi-net. This AC signal is coupled back to the tube by the blocking
cap and adds to the DCV. While it is true that the tube will not be conducting
during most of the positive swing, the voltage is still present on the anode and
the tube must handle it without arcing.
73, Bill W6WRT
_______________________________________________
Amps mailing list
Amps@contesting.com
http://lists.contesting.com/mailman/listinfo/amps
|