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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