> >It might then be advisable to add a slight safety factor to the
> >estimated current. Say fundamental current (given by the absolute
> >power ...
>
> ? ¿absolute power?
>
> > leaving the tank and the voltage at that point)
>
> ? How could power have anything to do with it? An amplifier can be
> tuned up at half power for an identical swing in anode potential. Since
> anode-C is the same, volts/ohms oughta be the same.
If you have 1500 watts leaving the tank to the load, you MUST have
1500 watts or MORE entering the tank.
If the anode swing is 3000 volts RMS, you MUST have at least one
half ampere ampere of RMS current flowing into the tank at that
end.
ALL of that current must go through the blocking capacitor, PLUS
the current charging the tube, the stray capacitances, and
associated components on the tube side of the capacitance.
That is the current you insist on ignoring, and that is what makes
your suggestion incomplete. If we use you suggested method, the
results are clearly wrong.
The peak current, since a class B tube perturbs the tank only
during half a cycle, must be more than the RMS current times
1.414. That's where harmonics come in.
> > At lower frequencies,
> >fundamental current calculated by the current driving the tank
> >would be very important.
> >
> ? Why if we can use the characteristic curves to help calculate peak AC
> anode potential?
Because we are interested in CURRENT through the blocking cap,
not current through the tube's capacitance, when we are selecting
a blocking cap.
> >> This suggests that a big GG triode on 2 must have pretty enormous RF
> >> grid currents flowing. An 8877 with 10pF Cpg and 3000 volts pk - pk
> >> plate swing will have 13.5 amps peak flowing through that capacity.
> >
> >Fortunately that current is involved only is I^2 R heating and not
> >electron kinetic energy heating of the grid.
> >
> >Rich loses sight of that fact, and thinks because the tube can
> >handle 10 or more amperes of capacitively coupled current n the
> >grid that means it can also handle 10 amperes of grid current driven by
> >an accelerating voltage of a hundred volts.
>
> ? Did Rich say that?
Yes. In a vain effort to prove only parasitics heat a grid, and not
kinetic energy from the electrons hitting the grids plating.
> >In the case of grid current from electron bombardment by cathode
> >emission, dissipation is given by E times I.
> >
> ? and would this be peak or RMS values?
It has to be the dissipation integrated over the entire RF cycle. The
waveform is complex. Eimac and other manufacturers measure
that effect and set a grid dissipation limit by watching for
secondary emission from the grid to suddenly increase.
That's much easier than doing a complex analysis of grid voltage
and current through the RF cycle.
73, Tom W8JI
w8ji@contesting.com
--
FAQ on WWW: http://www.contesting.com/ampfaq.html
Submissions: amps@contesting.com
Administrative requests: amps-REQUEST@contesting.com
Problems: owner-amps@contesting.com
Search: http://www.contesting.com/km9p/search.htm
|