David, at the risk of over simplifying a complex issue, some thoughts:
With a low loss matching network, the tube efficiency is primarily related
to the class of operation. Class C can be 75%, class A might be 30%. As the
class of operation moves toward class C, the efficiency improves. The zero
drive idle current decreases as this happens and the linearity gets worse
and worse.
So, you can decide how much non-linearity is acceptable for the improved
efficiency.
I have built a 70cm power amplifier using the GS23B (aka 4CX1600U) and it
measured greater than 60% efficiency running class AB2. To achieve that
level of efficiency required driving the tube as hard as the grid and screen
dissipation specifications allowed. The harder the tube was run, the better
the power output and efficiency.
As the bias is increased to lower the idle current, the class of operation
moves toward class C. The associated K factor moves away from 1.5 toward
and beyond 1.8. I forget the class C k factor but it is greater than 2.0 as I
recall.
Hope this makes sense.
73,
Gerald K5GW
In a message dated 10/28/2013 10:59:52 A.M. Pacific Daylight Time,
dave@n8zfm.com writes:
I am attempting to calculate the Pi-L tank values for a GS23b HF amplifier
I am building and I am a bit confused. I know this tube is usually used at
VHF and above but OM power does use one at HF and I happen to have several
of them I would like to use. I need to somehow know the efficiency, I
Think, and either the info for this usage is missing on the data sheet for
the tube or I am not experienced enough to figure it out.
Using a spreadsheet I got from G3SEK. I am using the following values,
3000V on Anode, max of 1A DC anode current and planning on getting 1500W
output, using a K factor of 1.5. So the implied efficiency is 50% and a
plate load of 2000 Ohms, all well and good but 50% seems low and I would
think I would be closer to 60%. If I guess at 61% efficiency then the
plate load is 2440 ohms and I would at 3000v/1A be running 1830W output.
Then to play with this some, if I put in 3000V/.890A and assume the same
1500W out the plate load is 2247 ohms at 56% efficiency, that changes the
second calculation to 55% efficiency with implied output power of 1469W
and 2200 ohms.
Is it safe to assume the 2.2K plate load is close enough?
I am missing something here in knowing how to use the calculations and
spreadsheet correctly. This is not one I want to learn by making mistakes
with HV and expensive vacuum caps.
I can't seem to find anything to help me know what the efficiency
would/should be.
73,
David Trainor
N8ZFM
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