All,
As promised, I am quoting from the 1991 ARRL Handbook from a piece under
"Design Guidelines and Examples", starting at page 15-15, at the last
paragraph. This was from the design example of an 8877 tube.
"The maximum plate voltage allowable for the 8877 is specified to be 4000 V. A
plate potential of 3500 V is chosen to minimize grid current, which is well
within published guidelines. In order to reach the 2500-W input power, 714 mA
of plate current must be drawn at this plate voltage".
(Starting on page 15-16)
"The next step in the design process is to calculate the optimum plate load
resistance at this plate voltage and current for class AB operation. From the
earlier questions, Rl is calculated to be 3268 ohms. This result is approximate
since the value of K in class AB is only approximately 1.5. For ease in further
calculations, the plate resistance can be rounded off to 3500 ohms".
The first decision that has to be made is to choose input and output matching
networks. Almost any network can be used at the output to transform the 3500
ohm plate impedance down to 50 ohms, but experience shows that the pi and pi-l
networks yield the most practical component values".
The "earlier questions" calculation comes from these formulas on page 15-3 at
the top and I quote.
"The optimum tube load resistance is"
Rl = Vp / ( K x Ip )
where
Rl = the appropriate load resistance
Vp = the DC plate potential in votes
Ip = the DC plate current in amperes
K = a constant that approximates the RMS current to DC current ratio
appropriate for each class. For the different classes of operation;
Class A, K = 1.3
Class AB, K = 1.5
Class B, K = 1.57
Class C, K = 2
End quote.
That's what I read last night, and I said would post this morning. The plate
current was calculated the same way I do it by using the efficiency of 60% for
class AB. 3 times the current would blow the 60% efficiency all to hell. Also
notice they're saying that the "plate load resistance" is the same as "plate
impedance". All Q&A welcome.
Best,
Will
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