Basically each electron is accelerated thru an
electrical potential (voltage) and gains kinetic energy. This energy goes
some place when the electron strikes the anode of the tube. Well, it goes
into heat. If the plate of an amplifier fed a zero impedance load. That
is, no power out and the plate voltage would not decrease with an increase
of plate current, all of the power (VxI) would go into kinetic energy and
be dissipated in the plate as heat. So,at zero efficiency all the plate
dissipation would be simply VxI. However, as an amplifier the plate voltage
swings to a lower value when the current increases and swings higher when
it deceases. This is due to the voltage drop across the load impedance
(power output). The power applied to the tube's plate circuit by the power
supply ends up being divided between the electrons' kinetic energy (plate
dissipation) and the load (output power). What does not go to the load goes
into heat.
73
Bill wa4lav
At 08:56 AM 9/18/02 -0500, Kim Elmore wrote:
>Amidst all the nostalgia over AM (my Dad, W5JHJ, still has a working Globe
>Champion with 275 W of plate-modulated AM; I vividly remember watching the
>866A mercury-vapor rectifiers glow blue and pulsate as he spoke), I've
>been wondering what the *physical* mechanism is that heats a vacuum tube
>anode as it operates. I'd appreciate enlightenment. We all take for
>granted the fact that these things get hot as they operate, and I nod my
>head properly when discussing efficiency and plate dissipation, but what,
>physically, is heating the anode?
>
>Kim Elmore, N5OP
>
> Kim Elmore, Ph.D.
> University of Oklahoma
> Cooperative Institute for Mesoscale Meteorological Studies
>"All of weather is divided into three parts: Yes, No, and Maybe. The
>greatest of these is Maybe" The original Latin appears to be garbled.
>
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