[Amps] 1:9 un-un for grid driven tetrode

[Peter Voelpel]

That 1:9 unun and 450 ohm resistor is obviously for use with a high power
linear tetrode amplifier.
Passive grid would not work with grid current.

Tetrodes requiring a few hundred volts of drive are for instance 4CX10000 or
4CX15000 and similar

73
Peter, DJ7WW

-----Original Message-----
From: Amps [mailto:amps-bounces at contesting.com] On Behalf Of Rob Atkinson
Sent: Sonntag, 24. Oktober 2021 13:19
To: amps at contesting.com
Subject: [Amps] 1:9 un-un for grid driven tetrode

The unun is (my opinion) a waste of time.  Just connect your grid
input network, usually a pi network for tetrodes, to the grid and
design it to do all the impedance transformation.  It can be band
switched and you'll be fine.  I don't know what tube this is but its
data sheet should give you the driving peak RF grid v.   You also
didn't specify the class of operation, but for class C the grid has to
be biased around 5 times past cutoff.   The data sheet should have
"typical operating" specs for a certain plate v. and operating class.
Divide the peak RF drive v. by the grid current given for the class of
operation.  That will give you the grid impedance.   Construct a "grid
dummy load" of carbon resistors that are close to the grid Z value and
connect them from the grid socket lug to ground.  Now you can put a
vswr analyzer on the RF input jack and futz with your input network to
get it in the ballpark to get the grid Z to 50J0.   To get the drive
needed, multiply the peak RF grid v. by grid current.   You can get
the needed grid current by dividing the peak RF v. by the grid Z.
Tetrodes usually do not require a lot of drive compared to triodes.

73
Rob
K5UJ
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