K6GT wrote:
hi all,
> >Since input impedance equals drive power divided by peak
> >cathode current squared, you can see how the input impedance
> >stays fairly close throughout the operating range.
this has the correct units (ohms) at least
> >The input impedance is the ratio of input drive power to peak
> >cathode current.
and this has the units of potential (volts), not impedance!
73,
George T. Daughters, K6GT
K5PC wrote:
Your SWR does not change enough to see it because as you lower the input
power, you also lower the cathode current, maintaining about the same
input impedance. Try this...tune up your amp and apply 100 watts drive.
Note the cathode current. Reduce the drive to 50 watts. The cathode
current
should be close to 1/2 what it was at 100 watts input. Since input
impedance
equals drive power divided by peak cathode current squared, you can see
how
the input impedance stays fairly close throughout the operating range.
The
low Q of the tuned input L/C allows compensation for any variations in
the
drive/cathode current ratio.
This does not work out if you ask me.
Let's say 1.4 A at 100 watts drive and 0.7 A at 50 watts drive (half the
drive gives half the cathode current according to the snip above)
Impedance at 100 Watts drive: 100 / 1.4^2 = 50 Ohms
Impedance at 50 Watts drive: 50 / 0.7^2 = 100 Ohms
Does not stay fairly close at all. And it becomes even worse when power
goes down more. At 5 watts drive and 1/20th of the cathode current, the
calculated impedance becomes 1000 Ohms.
??????
Gerard, AA3ES
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