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Re: [Amps] 8877 input Z

To: xxw0qe@comcast.net, amps@contesting.com
Subject: Re: [Amps] 8877 input Z
From: TexasRF@aol.com
Date: Fri, 28 Mar 2014 15:40:53 -0400 (EDT)
List-post: <amps@contesting.com">mailto:amps@contesting.com>
Larry, you may be right. But there are no commercial 8877 amplifiers using  
that scheme and don't recall seeing any published homebrew amps using it  
either.
 
The 54 ohms and 42 pF are not in a series configuration. It is 54R shunted  
by 42 pF. The series equivalent will be different for every band because 
the  42pF will have a different xc for every band. If the parallel xc is 
different  from band to band the series equivalent circuit is going to have a 
different R  and xc from band to band. Conventional design uses an input 
network for each  band to deal with those band to band variations.
 
A Q of 3 or so is thought to be needed to preserve linearity. That would  
would require a shunt C of about 18 ohms when R=54. That is like 315 pF at 
28MHz  and doubling for every halving of frequency. At 28MHz, 42 pF is only 
about 135  ohms or a Q of about .4.
 
I will stop there since I certainly can't say I have been there and done  
that with this new design thinking.
 
73,
Gerald K5GW
 
 
 
 
 
In a message dated 3/28/2014 1:48:36 P.M. Central Daylight Time,  
xxw0qe@comcast.net writes:

If the  load is 50 ohms (using 50 just for example purposes) in parallel 
with 42pF  a series inductance matches well enough that a given 
inductance will be  good for more than 1 band.  However if you add 
another 42pF shunt  capacitor from the exciter side of the inductor and 
change the inductance  to about 91nH the circuit will match from 1.8 to 
30MHz with a maximum  exciter SWR of 1.15:1 with NO switching required at 
all.  If you want  it even lower a combination low pass and high pass "L" 
network can be done  (only 1 more component).

73,
Larry, W0QE


On 3/28/2014  11:54 AM, TexasRF@aol.com wrote:
> Hardy, the 54 ohms is a nominal  figure that will vary with rf drive 
level.
> Certainly close enough for  designing input networks. Higher drive levels
> will  result in a  lower input impedance; lower drive levels will result 
in a
>  higher  input impedance.
>   
> The 42 pF is the  capacitance between the cathode and grid primarily. So,
> you will need  to provide an inductance to cancel that out. The inductance
> will   be different for each band of course.
>   
> 73,
>  Gerald K5GW
>   
>   
>    
>   
>   
> In a message dated  3/27/2014 7:55:49 A.M. Pacific Daylight Time,
> n7rt@cox.net  writes:
>
> I am  looking for the input impedance of an 8877.  I found one source and
> that was  K6DC's (SK) article years ago  that said it was 54 ohms with no
> frequency  dependency or  reactive component. The input C is 42 pF 
according to
> the data   sheet. So I will assume I can use 54 ohms in parallel with 42 
pF.
>  Anyone have  anything different?
> 73 Hardy  N7RT
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