>
>> >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?
>>
? "absolute power corrupts absolutely". .- Lord Acton.
>> > 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.
? And with the amplifier tuned for max with half drive and 750w out, the
grid current would be roughly the same, the anode swing would still be
3000vRMS, and the current through the anode C unchanged, so trying to
base the calculation on P makes no sense to me.
>
>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.
>
? Doesn't the half-cycle pulse of DC current come through the HV-RFC, or
am I missing something?
>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.
>
? Isn't the tank circulating current roughly Q times the anode current?
With a typical 10m Q or 15, tank circulating current would seem to be the
dominant factor.
>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.
? Are the DC blocking capacitor and the anode-C Not in series?
>
>> >> 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.
>
? There are two kinds of electrons?
>> >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.
? Agreed. However, you stated:
To: <amps@contesting.com>
(Date: Sun, 03 Nov 1996 07:10:43)
In article <55b1hb$du2@newsbf02.news.aol.com>, w8jitom@aol.com (W8JI Tom)
wrote:
"...... .......At 150 mA of grid current in a typical HF PA, a single
3CX800 has a
measured grid dissipation of almost 20 watts. ......."
? Since a 3CX800A7's cathode-grid driving potential is c. 40VRMS, the
grid dissipation would appear to be roughly = to 40v x 150mA=6w. It
looks like you used a grid current of almost 500mA to arrive at a figure
of "almost 20w". Why? // Who measured the grid dissipation at "almost
20w"?
- cheers, Tom.
Rich...
R. L. Measures, 805-386-3734, AG6K, www.vcnet.com/measures
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