[AMPS] Re: Drive Level in amplifiers

[Rich Measures]

>>>One other point on drive level:
>>>
>>>If the point were true that input impedance varies with drive, then in 
>>>order to have an amplifier system that is operating properly, one must 
>>>always operate it at maximum drive.  
>>>If not operated then at maximum 
>>>drive, the matching network is not properly attenuating harmonics, not 
>>>providing the proper "flywheel" affect, etc.
>>
>>- The flywheel makes things mo betta, and it's always there.  
>
>Again though, my point on matching network impedances still stands.  If 
>at a low drive level my tube impedance is say 150 Ohms and at high drive 
>levels it drops to say 75 Ohms then a matching network design to match a 
>150 Ohm impedance to a 50 Ohm impedance will NOT work properly with a 
>load of 75 Ohms.

Not perfect we can live with if it works good enough.  
>
>In fact, the input impedance of a tuned circuit designed to match 50 Ohms 
>on its input to some output impedance may actually look MUCH worse with a 
>different impedance on it's output.  And you won't get proper power 
>transfer either.  Optimal power transfer to a a load (such as a tube) 
>occurs when the impedance that load sees looking back toward the source 
>is the complex conjugate of it's impedance.  For example if the tube has 
>an impedance of 125-j200 Ohms, then for proper power transfer from 
>matching network to tube, the tube wants the matching network to have an 
>output impedance of 125+j200 Ohms.  If the matching network therefore is 
>designed for an output impedance of 125+j200 Ohms, but the tube changes 
>to something like 80-j130 Ohms, optimal power transfer will not exist.  
>And if the power isn't transferred to the tube, where does it go?  It 
>goes BACK to the exciter and hence appears as an SWR.
>
Energy is stored in the tuned input flywheel during the positive half of 
the drive signal when the cathode is not drawing current.  The stored 
energy is delivered during the following negative half of the drive 
signal when the 4-1000A anode current peaks at c. 2.2A.  However, the 
picture is even more complex than this.  More on that later.  
.   
>So just having inductors and capacitors on your circuit doesn't 
>necessarily make things always better.

True.  The designer needs to optimize values for minimal trade-offs.  If 
g-g tuned input Q is too high, bandwidth will be too narrow.  If Q is too 
low, swr will be too high.  
>>.  
>>>It is not always beneficial to run an amplifier at maximum drive.  
>>
>>During ssb modulation, the drive varies widely during each spoken word.  
>>However, max. peak drive and max. anode-I is essential to correct tank 
>>adjustment.  
>
>What does correct tank adjustment have to do with the input impedance?
>
In a grounded-grid amplifier, the output circuitry is effectively in 
series, and out of phase, with the input circuitry.  Thus, all of the 
current in the output circuitry passes through the cathode input 
circuitry.  Therefore, there is interaction.  .  On paper, a 
grounded-grid amplifier looks simple, however, what takes place inside 
one of the beasts is not.   Not only is the exciter trying to drive a 
diode-like cathode, the cathode is connected to the output tank circuit 
via the path of anode current, all of which comes from the cathode.   
>>
>>Please consider that:  Below a certain drive level, zero grid current 
>>flows.  At max. drive, grid current flows.
>
>I would say that this depends on how the tube is biased, but for argument 
>sake, let's assume you are right.

No tube is biased in the positive grid V region for g-g operation.  
>
>However, what does the flow of grid current have to do with tube 
>impedance?  

If an unknown R draws no current at say one volt, is the resistance high 
or low? 

>Maybe I am showing some ignorance here, but again, I haven't 
>seen this.

First off,  series RC and series RL loads have impedance, Z.  The cathode 
represents a parallel RC load.   Parallel loads do have Z, they have 
admittance, Y.  .However, Y may be converted to a series-equivalent Z, 
and visa versa, but the process takes a bit of work.  {page 27 of March, 
1989 *QST*} 
.  Grid current is one of the two components of cathode current.  Drive 
is applied between the cathode and the grid.  The current that flows in 
the grid has some influence on input admittance.  The out of phase 
current that flows in the anode has some influence on input admittance.   
 
>
>I guess we will see eventually who is correct here.  If I am able to take 
>a noise bridge, measure the tube impedance 

With a few uW from a noise-bridge driving the cathode, one part of the 
input admittance is going to look like 27pF in parallel with an R which 
does not draw current -- i.e., infinite ohms.  The other part of the 
admittance contains current which is out of phase with the input.  This 
is goes beyond my knowledge.  .  .  However, I know what works.  .   .  
- later, Jon


cheers
Rich...

R. L. Measures, 805-386-3734, AG6K   


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