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Re: [Amps] Plate Impedance

To: dhallam@rapidsys.com, craxd1@verizon.net, amps@contesting.com
Subject: Re: [Amps] Plate Impedance
From: TexasRF@aol.com
Date: Mon, 10 Apr 2006 20:51:53 EDT
List-post: <mailto:amps@contesting.com>
 
David is correct; in the RCA TT5 manual, page 53, step 16, it shows Ibmax  in 
the equation.
 
73/k5gw
 
 
In a message dated 4/10/2006 7:10:12 P.M. Central Standard Time,  
dhallam@rapidsys.com writes:

Will,

That is a typo.  Read the section thoroughly and you  will see that you are
to use Ibmax in the  formula.

David
KC2JD

-----Original Message-----
From:  amps-bounces@contesting.com [mailto:amps-bounces@contesting.com]On
Behalf  Of Will Matney
Sent: Monday, April 10, 2006 7:23 PM
To:  amps@contesting.com
Subject: Re: [Amps] Plate  Impedance


Gerald,

I went ahead and brought the RCA  transmitting tube manual TT-5 here to quote
what it says.

Effective  RF load resistance;

2 x (Eb - Ebmin) / Ibo

Ibo = Zero signal  plate current

"Ibo should be between 1/6 and 1/10 of Ibmax found on the  typical plate
characteristics graph".

"Determine peak plate current  (Ibmax) for zero bias (Ec1 = 0) at or slightly
below the knee of the  zero-bias curve for the value of Ec2".



*********** REPLY  SEPARATOR  ***********

On 4/10/06 at 7:10 PM Will Matney  wrote:

>Gerald,
>
>I just came in to check me e-mail and  seen your post. I was wrong on that
>being peak plate current. That  could have been something I read on the net
>as I was reading there on  several websites also. When I made the post, I
>didn't have the books  handy and was trying to recall what I read
>somewhere. The current that  is used is the Ibo as Peter G3RZP said after I
>looked in the RCA TT  book. On tonight, I'll post what it says but the
>current used is  anywhere from 1/6 to 1/10 of Ibmax as I recall. This was
>in the RCA  transmitting tube book that I finally dug out. The other I
>posted was  from the Radiotron Designers Handbook, 3rd edition, 1945.
>
>"I  seem to remember my RCA TT4 handbook listed the load impedance as  the
>plate swing divided by Ib0, the quiescent plate current. A  typo,
>obviously".
>
>Best,
>
>Will
>
>
>
>
>***********  REPLY SEPARATOR  ***********
>
>On 4/10/06 at 5:03 PM  TexasRF@aol.com wrote:
>
>>Jeff, 5400 ohms plate load flies  into the face of what we have been
>>calculating using the K factor  stuff.
>>
>>3000v/1A/1.8 is equal to 1666 ohms so one of  your numbers is flawed.
>>
>>Using the other method, 2700v X  2 /3.14 equals 1720 ohms which is much
>>closer (certainly close  enough) to the usual calculation  result.
>>
>>73,
>>Gerald  K5GW
>>
>>
>>
>>In a message dated  4/10/2006 2:32:14 P.M. Central Standard Time,
>>Xmitters@aol.com  writes:
>>
>>In a  message dated 4/10/06 9:27:25 AM  Central Daylight Time,
>>amps-request@contesting.com  writes:
>>
>><< All,
>>
>>I  started  reading through the RCA Radiotron Handbook this  evening
>looking
>>for where  the factor of 1.8 is listed  for calculating the plate  impedance
>>of
>>a
>>class  AB amplifier, and  I cannot find it. The edition I have was  the
>>older
>>one
>>back in the 40's with the black  cover. The red one was from the 50's and
>>had
>>more in it  if I recall, but I don't have it. I was wanting to find   where
>>it
>>gives this factor, and the ones for Class A,  AB1, AB2, B, and C.  Bill
>>Orrs
>>handbook gives the  factor of 1.8 on his Pi tank values table,   but
>doesn't
>>mention
>>anywhere in the text where it  came from. I also looked  in a  RCA
>receiving
>>tube
>>book and could not find it  there either, or I am  overlooking it
>>somewhere.  I
>>could have sworn it was in the Radiotron  Handbook. The  Handbook does say
>>that
>>the plate impedance is twice the  peak  voltage swing times the peak anode
>>current. What I'm  wanting to know is  where are these factors located in
>>print,  as I'd
>>like to read the whole  texts concerning this? Any  help
>>would be deeply appreciated.  Thanks to all in  advance.
>>
>>Best,
>>
>>Will
>>>>
>>
>>Will:
>>
>>I  do a great deal of mathematical analyses on  high power RF  amplifiers.
>>My
>>favorite resource is the Eimac Care  and  Feeding of Power Grid Tubes
>>available  on
>>Eimac's web site. I think  Richardson Electronics may also  have hard
>>copies
>>available. You also need  the  clear plastic overlay and a set of  constant
>>current
>>curves for the  tube you are  using. Then you can calculate some pretty
>>close
>>values  for  your desired parameters. This is the most  accurate
>>mathematical
>>model that  I'm aware of.  Some may considered as too tedious, and that's
>>fine.  It
>>is  still the most accurate method of modeling on  paper.
>>
>>My second choice  is the mathematical steps  described in the RCA
>>Transmitting
>>Tubes book   number TT-5. This book is available from many web  resources.
>>There
>>is a  step by step procedure in  there for calculation all of the operating
>>
>>parameters.  both RF and DC. The calculations rely on a table of  "K"
>>factors
>>that
>>are dependent on the plate  current conduction angle for the  class  of
>>service
>>desired. This is a fairly straight forward  mathematical  process.
>>
>>To answer your specific  question, the input impedance as seen  by the  tube
>>of
>>the tank network is RF plate voltage swing  divided by the  peak
>>fundamental
>>component of the  plate current. If you don't get this  number "right"  the
>>effect
>>is distortion, crappy efficiency and maybe  even  instability.
>>
>>Let me give you an example.  Let's say that we are  designing a class B
>>amplifier and we  know the peak plate current is one  amp. Class  B
>suggests
>>180 degree
>>plate current. Therefore  the DC plate  current is going to be the  peak
>>plate
>>current divided by 3.1416. The peak   fundamental plate current is going to
>>be
>>half of the  peak plate current or  . 5 amp.  The tube curves are used  to
>>find
>>an
>>appropriate minimum   instantaneous plate voltage based on the  desired
>>linearity
>>and the best  value to use is  tube dependent. As a generalization,  make
>>this
>>minimum  plate voltage  (ebmin)
>>equal to ten percent of your DC plate voltage. Say   your plate voltage is
>>3000
>>volts, so ebmin is then 300  volts. the plate  swing is therefore 3000 -
>
>>300  =
>>2700 volts. So the impedance the tank  needs to present to  the tube is
>>2700/.5
>>= 5400 Ohms. The power output  BTW  is plate swing times Peak  Fundamental
>>Plate
>>Current times 0.5 and this   assumes the average power of a CW  signal.
>>
>>
>>The problem with the RCA  K  values and any other constant for that matter,
>>is
>>that  they ignore the  tube characteristics. Furthermore, often times  the
>>basis
>>for which a  multiplying constant is  derived is not always known. The RCA
>>K
>>values and   the procedures in Bill Orr's famous works, all assume  that
>the
>>plate
>>current is going to be a perfect  sinusoid over the portion of a cycle  for
>
>>which
>>it conducts. This is not an accurate  assumption and fortunately for
>>approximation purposes, is usually  good enough. The advantage of the K
>>values and  Bill
>>Orr's calculation process is it gives you a reasonable   starting point for
>>a
>>design. There is always going to be  some "lab work"  so there is a
>>personal
>>balance  everyone must make individually as to how  much time is going  to
>>be
>>spent
>>calculating and how much time is  going to  be spent constructing  and
>>optimizing.
>>
>>The first question to  ask  regarding any mathematical model is "how much
>>accuracy do  I really want or  need? The usual response is "as much  as
>>possible". If
>>you have infinite  research  dollars and infinite time, this is a
>>reasonable
>>answer.  Who  really has that advantage? Obviously there is a  practical
>>answer
>>to
>>this   question.
>>
>>
>>I should point out that the  example I gave you only  applies to a class  B
>>and
>>any other class of service is going to command  a  different set of K
>>values.
>>My
>>example  was to illustrate the basic process.  When I do an analysis on  a
>>broadcast transmitter, I like to start off with  the RCA K  values to get
>>an
>>Idea for
>>what I'm dealing  with. I then follow  up using the Eimac method, a  drawn
>>Operating Line and the accompanying  calculations to get  the final,
>>usually
>
>>more
>>accurate  results. The Eimac  method is restricted BTW, in that  the
>voltage
>>waveform on the grid must be  the same kind  of waveform on the plate.
>IOW,
>>you cannot
>>drive a  tube (with  a resonant plate tank network) with a square wave  with
>>a
>>sinusoidal signal  on the plate, and expect  meaningful results with the
>>Eimac
>>Tube  Performance  Computer. It does not work.
>>
>>
>>This is  probably a more  lengthy response than you were looking fore,  but
>>I
>>hope it helps   you.
>>
>>Jeff Glass, BSEE CSRE
>>Chief  Engineer
>>WNIU WNIJ
>>Northern  Illinois   University
>>WB9ETG
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>>
>>
>>
>>
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>
>
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