Larry Benko wrote:
> Just a small clarification:
>
> A 3 dB pad will increase the return loss byt 2*3dB = 6dB.
>
> A 2.0:1 SWR is equivalent to a reflection coefficient of .333 and a
> return loss of 9.542dB. Therefore it takes a 4.77dB pad to guarantee a
> input SWR of <= 2.0:1.
>
> However let's think of it like this: Assuming an original SWR of 3.0:1,
> then a 2 dB pad will produce a resulting SWR of 1.922:1. A measly 6 ft.
> piece of RG-58 coax between the transmitter and the amp will have less
> than 0.2dB of loss at 28MHz which will not improve the input SWR much at
> all. Even if the 6 foot piece of coax is made from RG-174 the loss at
> 28MHz will only be 0.3dB which still won't do much. Am I missing
> something here?
Without time to do any calculations, here's a guess: The cathode current
waveform contains a lot of harmonic energy due to it's 'spikey'
waveform. A low Q input network means does not fully isolate the cathode
from the outside world at harmonic frequencies so that the impedance at
the input connector of the amp has an effect on the current flow. 6' of
cable has a significant electrical length from low vhf upwards and will
have resonant properties at some frequencies depending on how the output
of the exciter terminates it.
6' will be a good length for a particular combination of amp, rig and
band(s). A different length might be better depending on the rig and
other factors.
Steve
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