" I have an external RF transistor amplifier (AB1) using 2 MRF455
transistors in common emitter mode. I have a resistive T pad with an
approximate impedance of 50 ohms on the input side of the transistors.
The SWR on the input of the amp shows 1.3:1. When I use a 2 foot
jumper between the radio and the amp the amp puts out about 100 watts
max, but when I switch to a 9 foot jumper the amp puts out about 40
watts max. The receive stays the same no matter which jumper is used.
Why does changing the length of the jumper between the radio and the
amp make such a drastic change in how much the amp puts out?
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If this is a 2m transmission system 2ft of coax equals 610 mm. Assuming a
coax velocity factor of 0.67 ,the electrical length is 1/ 0.67 * 610 equals
915mm which is about 'half a wavelength'.
Half a wave lenght of transmission line is an 'impedance repeater'
Hence if tx output- and amplifier input impedance are about the same ,there
is max power transfer.

Nine feet of coax for the above does NOT equal a multiple of half
wavelenghts hence the reduced power input to the amplifier and subsequent
lower amplifier output.

Just try to reduce the nine feet coax cable to about 8 ft which would be
about 4 half wavelengths . The result should be an amplifier output 40
Watts

You can readily make multiples of (electrical )half wavelength by using an
MFJ259B antenna analyer or even better a spectrum analyser..

An electrical half wave (or multiple half waves ) 'shorted' stub has a LOW
impedance at the other end .

Take a piece of coax which is slightly longer than a multiple of
(electrical) half wavelengths.
Fit a suitable connector at one end and connect to impedance measuring
instrument . Now penetrate the coax with a needle at the other end ,
observe the impedance.
The needle shorts the coax centre conductor with the braid.
Change frequency up or down finding the nearest freq where impedance is
minimal .
By penetrating the coax nearer the connected end you can find the length
where the impedance is minimal at the desired freq. That length is a
multiple of elctrical half wavelengths , and hence the optimal length of
your lead.

If you don't like the 'needle operation' , you can make an electrical
quarter wave '0pen' stub .
Again you start with a length of coax slightly longer , check the freq for
which the stub shows minimal impedance and subsequently cut tiny bits off
the stub until the instrument shows minimal impedance at the desired freq.

The lead you need between transceiver and amplifier is to be an EVEN
multiple of the remaining length of coax

I have done the above to optimise power transfer between a vintage Yaesu
FT290 ,2m transceiver (2.5W) and a 30 Watts power amplifier.

Note : The characteristics of open-ended and shorted transmission line
sections is well described in the ARRL Extra Class License Manual ( my copy
is from 1990)

Frank GM0CSZ / KN6WH