>
>Peter sez:
>> That turns out to be frighteningly high on 10 metres - take say 3000v pk
>> and 20pF, that's over 10 amps peak. I must admit that I've never actually
>> thought about that.
>>
>> As far as phase shift at the fundamental is concerned, I would say that's
>> pretty negligible. A 1000pF blocker is only 88 ohms at 1.8MHz. For a low Z
>> circuit (6 paralleled sweep tubes) you might want something a bit bigger
>> on 160.
>
>The phase shift could be an issue on ten meters, where series L
>and shunt C forms a delay line.
>
? How could it with a typical L of 200 or so nH and 5 to 10 pF?
>> The second harmonic component of a 180 degree conduction angle plate
>> current is 6dB down on the fundamental current. I think one has to treat
>> the tube as a current source here, so that the harmonic current into the
>> tank circuit is in phase with the fundamental current. Since the plate to
>> ground reactance is one half at the second harmonic, the amount of
>> harmonic current flowing into the tank will alter depending on frequency
>> i.e. the ratio of the capacity of the pi tank tuning C to plate-to-ground
>> C. This suggests that there will be a frequency where the two effects add
>> to give maximum current through the capacitor, although my feeling is that
>> they won't get much above the peak capacitor current caused by the plate
>> to ground capacity at the top end of the frequency range.
>
>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.
> added to
>circulating current (given by ALL capacitances at the tube end of the
blocker)
? We measured total C, Tom.
>plus a small additional safety margin added for
>harmonic currents.
>
>At ten meters, where the tube-end of things has several times less
>reactance than the tank's input impedance, tube capacitance will
>dominate the blocking cap current.
? Most of the amplifier designs I am familar with use way more Tune C
at 28MHz than the anode-grid capacitance. If this was not case, reaching
29.7MHz would be less than likely.
> 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?
>> 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?
>
>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?
>In the other case, it is simply I^2 R losses of the material in the
>grid itself with the current transfer handled by displacement
>currents rather than electron bombardment.
>
>When you calculate cuurent, the power leaving the tank must be
>included or else on lower frequencies there is a very large error.
? On "lower frequencies" does RMS anode potential decrease?
- cheers, Tom
Rich...
R. L. Measures, 805-386-3734, AG6K, www.vcnet.com/measures
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