[AMPS] blocking cap current

Rich Measures measures@vc.net
Tue, 20 Jul 1999 18:01:38 -0700




>
>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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