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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