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Re: [Amps] Plate Impedance, ARRL

To: craxd1@verizon.net, amps@contesting.com
Subject: Re: [Amps] Plate Impedance, ARRL
From: TexasRF@aol.com
Date: Wed, 12 Apr 2006 13:33:32 EDT
List-post: <mailto:amps@contesting.com>
 
Will and all,
as you have determined, the k=1.8 is an approximation that will keep you  
inside the ballpark at least.
 
Each tube type has it's own characteristics that will cause a shift up or  
down from k=1.8. For example, if you examine the 8877 data sheet constant  
current curves: You will see a sharp upturn in the grid current when the plate  
potential is about 600vdc. At this point, the plate voltage swing will run  
approximately plate voltage under load minus 600vdc. If we are using 4000v 
plate  
voltage, the swing is then 3400vdc. 
 
Using the earlier RCA info from this week, and assuming 1A average plate  
current the plate load impedance would be (3400 X 2) /3 or 2266.7 ohms. Using  
the K factor method, the required K would have to be 1.76 for the same  
impedance.
 
If we decide to use 2500v plate voltage, still at 1A plate current, the  
plate voltage swing would be 2500 - 600 = 1900vdc. The resulting plate load  
impedance in this case then would be (1900 X 2) / 3 =1266.7 ohms. To reach this 
 
value with the K factor method, the required K would have to be 1.97.
 
If you use a tetrode, the limiting factor for minimum plate voltage is  
screen grid current. The knee of the curve for this is very close to the screen 
 
voltage used. If you raise the screen voltage from a typical voltage to the  
maximum allowed, the plate voltage swing is reduced. With a lower plate  
voltage 
swing, a lower plate load impedance is needed. This implies that the K  factor 
would need to be higher to reach the required impedance. Of coarse with  
higher screen voltage comes higher plate current so an additional reduction in  
plate load impedance is needed and another, even different, K factor.
 
So, you can see, K is very much an approximation. Using the actual tube  
curves would seem much more precise. Having said that, the actual difference in 
 
loaded Q by using k=1.8 vs 1.97 is less than one.
 
73,
Gerald K5GW
 
 
 
 
 
In a message dated 4/12/2006 11:03:29 A.M. Central Daylight Time,  
craxd1@verizon.net writes:

Jim,

I read this over again last night, and it didn't mention  anything about 
triodes only. The main reason I posted this was to show the  differences, plus 
find where the illusive 1.8 came from in print. So far, the  only book I've 
seen 
1.8 listed in was Bill Orrs Handbook. I don't have any  newer ARRL handbooks 
past the 90's as they seemed to be the same old thing,  over and over, with not 
that much new.

Here's the thing. I use 1.8 just  like evryone else because it does get you 
there. I calculate the plate current  though the same way it was shown in this 
volume of the ARRL Handbook, by  efficiency. I seen in the old RCA Radiotron 
handbook where it said the plate  current for class AB could be as much as 3 
times. How they come up with this,  I don't know as efficiency is efficiency. 
Everything I've read says AB is  around 60% efficient, not less. To my opinion, 
there's a lot of  mis-information out there as compared to what I've seen work 
in the real  world. I would like to find out though where the factor of 1.8 
was first  mentioned.

Best,

Will


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

On 4/11/06 at 9:04 PM jkearman@att.net  wrote:

>From: "Will Matney" <craxd1@verizon.net>
>>  Class AB, K = 1.5
>
>My understanding is that this value of K  applies to triodes, where Ep can
>swing nearly to zero. For tetrodes, Ep  cannot swing below the screen
>voltage. IIRC, this has the effect of  increasing K. 
>
>It's useful to consider the consequences of  slight errors in
>component-value selection. Assuming your variable  controls (plate tuning
>and loading) have enough range to get a close  match, the negative
>consequence would be a Q different from what you  calculated. But if you
>give yourself enough range in tuning and loading  Cs, you should be able to
>tune for _best linearity_ (more important  than best efficiency) and still
>get enough Q to reduce harmonics below  FCC requirements. 
>
>If you calculate a range of plate loads by  varying K from 1.5 - 1.8, and
>then calculate tank circuit values based  on a Q range of 12-15, you should
>come up with tuning and loading cap  values that will do the job. 
>
>73,
>
>Jim,  KR1S
>http://kr1s.kearman.com/
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