Hi!
I was doing some calculations regarding the plate current in my Class AB (almost Class B) home brew GS-35b linear amplifier.
I am running the tube with 3000V on the plate, in grounded grid configuration... and I was curious as to how much peak current the tube has to produce in order to obtain my target peak output power of 1500 watts.
See the attached JPEG. [I know this stuff is not new, but it was interesting to calculate]
If I want the rf output to be 1500 watts, and if I want the peak plate voltage swing that develops across the output pi network at the output frequency (e.g. 3.8 MHz) to be 2500 volts (i.e., 83% of the full plate voltage), then I need a peak current at the fundamental frequency of 1.2 amperes. That is, the r.f. output power is the peak voltage at the fundamental frequency x the peak current at the fundamental frequency /2. The factor of 2 accounts for the use of peak values rather than rms values for the voltage and the current.
Looking at the attached JPEG, it follows that the peak plate current (not the component of the peak plate current at the fundamental frequency, but, rather, the peak of the Class B current waveform) is 2.4 amps.
When I look at the published curves for the GS-35b, I find that a single GS-35b can't deliver 2.4 amps
.
Therefore, it is not surprising that I can't get 1500 watts of peak power output on modulation peaks!
I'm going to switch to a pair of GS-35b's in parallel, to obtain the needed 2.4 amps of peak plate current.
As an aside, with 2.4 amps of peak plate current, and (therefore) 1.2 amps of peak plate current at the fundamental frequency... the tube is running at an average current (what the plate current meter reads) of 0.76 amps.
Stu
