[AMPS] Henry 8K at 1500 watts Out -- Warning Long

Jim Reid jreid@aloha.net
Tue, 8 Dec 1998 11:00:59 -1000


Jim,  KH7M wrote,  in part:

>>In fact,  with 100 watts into the 8K,  and it's 13 dB gain,  you will
>>get out around 2000 watts,  so better cut the drive in back until
>>the Bird in the 8K reads  no more than 1500 out with the key down,
>>SSB peaks.  


Rich,  AG6K responded:

>However, the tank of the Henry 8k is designed for a much lower RL -- 
>i.e., the tank is designed for 2.5a of anode current.  In order to 
>produce this much anode current, c. 400w of drive is needed.   
>>......

And,  presumably under this plate current,  the QL, the
loaded Q of the tank will be something above 10,  or so.


However,   Rich has caused me to revert to Terman,  hi.
I think the result of driving for only 1500 watts out,  is a lower
efficiency of use of the AC power operating the 8K.  I  have
heard from 8K owners who do just this,  operating both
SSB and AM (!),  who report that they receive glowing reports
about their signals audio quality,  so I would guess very
linear operating by the 8K.

Now,  I also think,  that lowering the drive to the grid,  with  the
amp designed as Rich points out,  results in a low Emax on the
grid,  and a smaller conduction angle of plate current,  but I
do not know how the Henry engineers have the bias set
up in the 8K.  My rather early copy of Terman discusses
RF power amps only in terms of Class C operation,  but
some of the same ideas apply,  I believe,  when figuring
what is going on in the circuit.

In any case,  the result from low drive is that there is a lower
voltage developed across the plate tank circuit,  and thus
Emin,  the difference between the plate supply voltage
(which in the 8K is something like 4200 volts,  I believe)
and the EL,  the maximum voltage across the tank is
much greater than the applied grid voltage,  Emax.

So there will be an impact on the loaded Q of the plate
tank since this Q depends upon the square of EL
divided by a product term which includes the plate
current squared and the term RL,  as Rich has said.

Now with lower than the 8K "design" drive of 400 watts,
per Rich,  it is not immediately clear what will really
happen to  the loaded Q of the plate tank,  as both
EL in the numerator is lower,  and the current in
the denominator is lower,  and both are squared terms.
The inductance of the tank in the denominator is constant,
but the R,  plate resistance is going to differ from that which
would exist with 2.5 amps of plate circuit current.

So,  I am not in a position to complete the analysis,
as I do not know all the particulars, nor do I have
a copy of the Eimac data sheet on the 3CX3000A7
used by Henry in the 8K.

However,  if the loaded Q drops,  the result will be,
perhaps,  an interesting tuning problem for the
amp user,  not sure about this either,  hi.  In the
text by Terman I have out here on Kauai,  he
writes that it is not uncommon for "very high
power Class C amplifiers"  for values of loaded
Q to be as low as 2 or 3!  Again,  I do not know
how far toward AB or B operation the 8K is
from a Class C triode transfer characteristic.

Terman goes on,  in a text footnote,  to offer that with
such low loaded Q values,  it will be necessary to use
a tuning procedure that causes the tank circuit to 
still offer a  resistance impedance to the tube,
even though this impedance may  not be the
maximum impedance obtainable from the load.

In  another section of his text,  Terman discussed the
behavior of parallel circuits with low Q.   He writes
that when circuit Q is low,  the curve of circuit impedance
as a function of frequency still has a shape that resembles
a resonance curve,  unless the Q approaches,  or is
less than unity.  At low circuit Q,  the maximum impedance
'no longer necessarily occurs at the frequency of series
resonance,  and the condition of unity power factor does
not necessarily occur either at the frequency of resonance
or when the impedance is a maximum.  the actual behavior
for a given Q will depends upon the division of circuit
resistance between the inductive and capacitive branches
of the tank.  Terman goes to say that what occurs in
the low Q tank of a power amplifier is going to depend
upon the distribution of resistances between the capacitive
and inductive branches,  and which branch is tuned to
effect resonance.

In the case of the Henry 8K,  the inductance of the tank
is tuned,  while the capacitance is fixed by doorknob
caps,  I think.  In this case,  Terman says that maximum
impedance and unity -power- factor  conditions will 
only coincide if all the circuit losses are concentrated
within   the capacitive branch.  I presume this would
be the case in the Henry 8K via the use of doorknobs,
and the rather large and healthy appearing conductor
of the plate tank inductor as shown on the Henry
web page.

So,  I see no harm to the 8K in operating it at low drive,
and constraining it's power output to legal limits,  where
it evidently provides very linear operation;  but at the
cost of probably very poor efficiency.  After all,  the
AC mains are probably delivering 8 or 10 kW of power
to the 8K power supply,  and with only 1500 coming out,
that is only about 15% efficiency,  hi.  But not sure the
fellow who buys and 8K is that concerned about his
power bill.  Who is to say??

Will really appreciate learning what else I ought to know 
about the situation of applying low drive to these big
amps.  I believe the result is very long life of the amp
components,  and a very nice on-the-air resulting
signal,  SSB,  CW and AM (since the amp transfer
characteristic resulting is apparently so very
linear,  a low power transceiver's AM signal
can be amplified very nicely by the 8K,  I think).

73,  Jim,  KH7M
On the Garden Island of Kauai




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