>
>Ah, another strange question from that guy in Arizona...
>The question is: how do you calculate the operating conditions for driving
>a tetrode in the supercathode drive circuit?
>I have constant current curves, all the Eimac app notes, etc.
>Allow me to tell you what I can do:
>1. Starting point...cathode and control grid tied together, screen DC
>ground, normal plate voltage.
>--> I would assume then that since the cathode and grid are tied together
>that the "grid voltage" on the constant-current curves is always
>zero...since "grid voltage" really means "grid-to-cathode voltage"...right?
yes
>2. Apply sine wave of some amplitude A to the cathode.
>--> Then the screen voltage actually is -A...since "screen voltage" is
>really "screen-to-cathode voltage"...right again?
The screen v. is the signal v.
>3. Find Ib, Ic, Isc for every 15 degrees of cathode drive (that's how the
>Chaffee analysis works, at least) from the constant current curves.
>
>Here's where I'm stuck. Again, I believe Ec=0 always. I can use the
>3/2-power law on the curves to find Ib, Ic, Isc for the changing screen
>voltage. BUT WHAT'S THE PLATE VOLTAGE?
The instantaneous anode [plate] potential depends on what R-load is
adjusted for.
>For some reason I can't figure it
>out. It isn't static. But the impedance is non-constant (just like any
>other tube analysis) so I can't just do ohm's law. It would seem I have
>too many variables...load impedance, instantaneous plate voltage...then I
>can find the currents.
The impedance of the load is constant. The critical moment of truth in a
linear amplifier occurs at the peak in anode current at the bottom of the
swing in anode potential. The optimal peak anode current can be found
by multiplying the rated anode current by three. The instantaneous
voltage at the anode when E-grid-cath = 0v depends on loading. For best
linearity, the anode should be loaded so that anode swing stays to the
right of the knee in the constant current curve. In a PL-172/8295, the
minimum peak anode potential apparently should not be allowed to go below
roughly 600v.
>At first I thought that since the plate amplification factor is so high in
>a tetrode (i.e., plate voltage has almost no effect on plate current) that
>it didn't really matter what the instantaneous plate voltage was. But then
>the formula for power output is 1/2*(Peak fundamental RF current)*(Peak
>Plate voltage swing) from the Chaffee analysis. So I need to be able to
>accurately calculate the plate swing after all...
The swing in anode potential is controlled by the person who tunes up the
amplifier. For a tetrode or pentode, correct loading (correct swing)is
indicated by screen current at max. drive. Too much Isc means too light
loading/too much swing. Too little Isc means too heavy loading/too
little swing. .
>And then again, if you want to do this to a triode, you definitely would NEED
the instantaneous
>plate voltages.
>
This is controlled by the operator who tunes the amplifier.
>Unfortunately I have very few examples to work against to verify a
>methodology...the "original" QST articles on Semi- and Super-cathode drive
>had two worked examples (4CX300A and 811A), fortunately I have those curves
>so I can get some verification.
To operate a PL-172 in supercathode drive service, approx. 600 peak
volts (at 3 peak amperes) of drive would be needed. My guess is that the
power gain would be around 4. . To operate a PL-172 in grid-driven
Class AB!, approx. 120 peak volts of drive would be needed.
>
- later, Scott
Rich...
R. L. Measures, 805-386-3734, AG6K, www.vcnet.com/measures
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