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Pete said:
I suppose its possible that it could draw enough current to
heat the cathode with no heater volts because of dissipation in
the
resistance of the cathode, but I doubt it.
Then Tom said:
The main heating of tube elements (other than the filament) is
by
kinetic energy from electrons. Since the electrons are leaving
the
cathode, they are removing energy with their departure.
<snip>
Isn't it simple? E times I gives the dissipation, where E is the
potential difference between the electron source and the target
(ie the cathode and anode) and I is the target current (ie anode
current).
That's why we calculate the anode or grid dissipation as E*I
integrated over time, and we don't have to worry about it
heating
the cathode!
<snip>
Sorry Tom, I think that your theory is wrong.
Kinetic energy of electrons reaching the anode (or screen grid,
for that matter) comes from the electric field generated by the anode
voltage, and NOT from the thermal energy generated by the filament. You
are implying that the cathode is cooling due to electron emission and
this is not correct. Heating the cathode generate a "cloud" of electrons
around it by exciting the outermost electron layer of thorium (or other
emissive metal) atoms. This electrons have just enough kinetic energy to
break free of nucleus attraction and they are floating around the
cathode until an electric field is applied. Then, electrons are
accelerated towards the anode, gaining kinetic energy from the anode
power supply.
I am sorry that this explanation is in plain terms and not in
quantum mechanics mathematical formulas, which do not translate well
through my E-mailer. And please remember that a guy named Einstein
received a Nobel prize just for that theory ;-)
For Pete:
I'm not betting either that this hypothetical cathode will be
heated by dissipation, mostly due to the fact that the current path on
the cathode surface is short and wide. Most likely, the emissive
material will flake out or (as Tom pointed), in case of direct heated
cathode, the current will be limited.
By the way: what kind of beer do you prefer ?
73 de Mike, YO3CTK
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<UL><P><FONT COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">Pete =
said:</FONT>
<BR>
<BR><FONT COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">I suppose its =
possible that it could draw enough current to</FONT>
<BR><FONT COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">heat the =
cathode with no heater volts because of dissipation in the</FONT>
<BR><FONT COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">resistance of the =
cathode, but I doubt it.</FONT>
<BR>
<BR><U><FONT COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">Then Tom =
said:</FONT></U>
<BR>
<BR><FONT COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">The main heating of =
tube elements (other than the filament) is by </FONT>
<BR><FONT COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">kinetic energy from =
electrons. Since the electrons are leaving the </FONT>
<BR><FONT COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">cathode, they are =
removing energy with their departure.</FONT>
<BR>
<BR><U><FONT COLOR=3D"#000000" SIZE=3D2 =
FACE=3D"Arial"><snip></FONT></U>
<BR>
<BR><FONT COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">Isn't it simple? E =
times I gives the dissipation, where E is the </FONT>
<BR><FONT COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">potential =
difference between the electron source and the target </FONT>
<BR><FONT COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">(ie the cathode and =
anode) and I is the target current (ie anode </FONT>
<BR><FONT COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">current).</FONT>
<BR>
<BR><FONT COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">That's why we =
calculate the anode or grid dissipation as E*I </FONT>
<BR><FONT COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">integrated over =
time, and we don't have to worry about it heating </FONT>
<BR><FONT COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">the cathode!</FONT>
<BR>
<BR><U><FONT COLOR=3D"#000000" SIZE=3D2 =
FACE=3D"Arial"><snip></FONT></U>
<BR>
<BR><U><FONT COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">Sorry Tom, I =
think that your theory is wrong.</FONT></U>
<BR>
<BR><U><FONT COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">Kinetic energy =
of electrons reaching the anode (or screen grid, for that matter) comes =
from the electric field generated by the anode v</FONT></U><U><FONT =
COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">oltage</FONT></U><U><FONT =
COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">, and NOT from the thermal =
energy generated by the filament. You are implying that the cathode is =
cooling due to electron emission</FONT></U><U><FONT COLOR=3D"#000000" =
SIZE=3D2 FACE=3D"Arial"> and this is not correct. Heating the cathode =
</FONT></U><U><FONT COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">generate =
a "cloud" of electrons around it by exciting </FONT></U><U><FONT =
COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">the </FONT></U><U><FONT =
COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">outermost </FONT></U><U><FONT =
COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">electron</FONT></U><U><FONT =
COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial"></FONT></U><U> <FONT =
COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">layer </FONT></U><U><FONT =
COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">of </FONT></U><U><FONT =
COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">thorium (or other emissive =
metal) atoms. This electrons</FONT></U><U><FONT COLOR=3D"#000000" =
SIZE=3D2 FACE=3D"Arial"> have just enough kinetic energy to break free =
of nucleus attraction</FONT></U><U><FONT COLOR=3D"#000000" SIZE=3D2 =
FACE=3D"Arial"> and they are floating around the cathode until an =
electric field is applied</FONT></U><U><FONT COLOR=3D"#000000" SIZE=3D2 =
FACE=3D"Arial">. Then, electrons are accelerated towards the anode, =
gaining kinetic energy from the anode power supply.</FONT></U></P>
<BR>
<P><U><FONT COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">I am sorry that =
this explanation is in </FONT></U><U><FONT COLOR=3D"#000000" SIZE=3D2 =
FACE=3D"Arial">plain terms and not in quantum mechanics mathematical =
formulas</FONT></U><U><FONT COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">, =
which do not translate well through my E-mailer</FONT></U><U><FONT =
COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">. And please remember that a =
guy named Einstein received a Nobel prize just for that theory =
;-)</FONT></U></P>
<BR>
<P><U><FONT COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">For =
Pete:</FONT></U>
<BR>
<BR><U><FONT COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">I'm not betting =
either that this hypothetical cathode</FONT></U><U><FONT =
COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial"> will be heated by =
dissipation, mostly due to the fact that</FONT></U><U><FONT =
COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial"> the current path =
</FONT></U><U><FONT COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">on the =
cathode surface </FONT></U><U><FONT COLOR=3D"#000000" SIZE=3D2 =
FACE=3D"Arial">is short and wide. Most likely, the emissive material =
will flake out or (as Tom pointed)</FONT></U><U><FONT COLOR=3D"#000000" =
SIZE=3D2 FACE=3D"Arial">,</FONT></U><U><FONT COLOR=3D"#000000" SIZE=3D2 =
FACE=3D"Arial"> in case o</FONT></U><U><FONT COLOR=3D"#000000" SIZE=3D2 =
FACE=3D"Arial">f</FONT></U><U><FONT COLOR=3D"#000000" SIZE=3D2 =
FACE=3D"Arial"> direct</FONT></U><U><FONT COLOR=3D"#000000" SIZE=3D2 =
FACE=3D"Arial"></FONT></U><U> <FONT COLOR=3D"#000000" SIZE=3D2 =
FACE=3D"Arial">heated</FONT></U><U><FONT COLOR=3D"#000000" SIZE=3D2 =
FACE=3D"Arial"> cathode</FONT></U><U><FONT COLOR=3D"#000000" SIZE=3D2 =
FACE=3D"Arial">, </FONT></U><U><FONT COLOR=3D"#000000" SIZE=3D2 =
FACE=3D"Arial">the current will be limited.</FONT></U></P>
<BR>
<P><U><FONT COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">By the way: what =
kind of beer do you prefer ?</FONT></U>
<BR>
<BR><U><FONT COLOR=3D"#000000" SIZE=3D2 FACE=3D"Arial">73 de Mike, =
YO3CTK</FONT></U>
<BR>
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