[Moon-net] Gain of a tetrode (TH347) - corrected

Peter Voelpel df3kv at t-online.de
Mon Oct 1 18:01:59 CDT 2007


 Hi Günter and the group,

Generally for linear service 0,5A of idle current is just right which means
less negativ voltage on the grid.
With a given drive the bias voltage has to be overcome by the rf voltage for
full output.
So with more negative bias the rf voltage needed will be higher.
With too less drive maximum power will always be higher at higher idle
current.
Running the tube in class B or closer to class C always requires more drive
but efficiency will be higher because the load line gets longer.
With plenty of drive there will be no problem with power output, but  the
actual gain will alsways be lower then with true class AB.
At the given idle current gain will not vary much until the compression
point is reached.

73
Peter


-----Original Message-----
From: moon-net-bounces at list-serv.davidv.net
[mailto:moon-net-bounces at list-serv.davidv.net] On Behalf Of Koellner,Guenter
(NSN - DE/Germany - MiniMD)
Sent: Montag, 1. Oktober 2007 20:31
To: moon-net-bounces at list-serv.davidv.net; moonbounceboard
Subject: [Moon-net] Gain of a tetrode (TH347) - corrected


Hello,
 
back to a technical question:
 
On 23cm, my amp is using a TH347. I have about 75W drive (from a GSM
amplifier) and am getting about 950W output from it. While tuning and
optimizing, I found that the gain increases when I am increasing the idle
current, so this situation is at 300mA idle current.
 
I am now going to get more drive power, the plate current might become too
high and the plate loss too much. In order to compensate here, I may reduce
the idle current. But, how much less gain may I get when I for example
reduce idle to 50mA. Not that I now drive it with 3dB more and on the other
hand loose 3dB of gain, so that at the end the output is equal.
 
73, Günter (dl4mea)
 
 
corrected: was 300mA drive power -> 300mA idle current

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