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Re: [Amps] Anode Impedance of 4-400A

To: "Joe Subich, W4TV" <k4ik@subich.com>,"'Phil Clements'" <philc@texascellnet.com>, <Amps@contesting.com>
Subject: Re: [Amps] Anode Impedance of 4-400A
From: "David C. Hallam" <dhallam@rapidsys.com>
Reply-to: dhallam@rapidsys.com
Date: Tue, 15 Nov 2005 20:04:39 -0500
List-post: <mailto:amps@contesting.com>
I think we are nit picking here since the original request was for an
approximation of Zp.

However just to keep things going, I think a multiplier of 1.55/1.6 more
closely approximates class A operation.  While AB1 indicates G1 will never
be driven positive, in practice we generally come fairly close looking for
that last watt.  All the calculations I have done with the constant current
curves for power tubes in AB1 (admittedly this is not vast numbers of
calculations), the calculated Zp from the curves equates to a multiplier of
1.8/1.9.  This is what you would expect if you drive G1 close to 0 volts.

David
KC2JD

-----Original Message-----
From: amps-bounces@contesting.com [mailto:amps-bounces@contesting.com]On
Behalf Of Joe Subich, W4TV
Sent: Tuesday, November 15, 2005 7:22 PM
To: 'Phil Clements'; Amps@contesting.com
Subject: Re: [Amps] Anode Impedance of 4-400A



Phil,

What class of operation, A, AB1, AB2, B or C and what conduction
angle if class C?  The 1.8 factor is an averaging factor based
on conduction angle.  For example, class A will be Ep/Ip while a
single tube in class B will be Ep/(2*Ip) because the peak currents
are twice the DC average due to the 180 degree conduction angle.

A typical tetrode amplifier will have a lower multiplication
factor since it will generally be operated in class AB1 (no
grid current, longer conduction angle), while most triodes in
grounded grid will have a higher multiplication factor due to
the lower conduction angle (closer to class B).

To simply use 1.8 is close but if the 4-400A is being used as
a tetrode and grid driven, 1.8 will probably give too low a
plate impedance ... the multiplier for AB1 is closer to 1.55
or 1.6.

73,

    ... Joe, W4TV



> -----Original Message-----
> From: amps-bounces@contesting.com
> [mailto:amps-bounces@contesting.com] On Behalf Of Phil Clements
> Sent: Tuesday, November 15, 2005 6:40 PM
> To: 'Patti'; amps@contesting.com
> Subject: Re: [Amps] Anode Impedance of 4-400A
>
>
> The load impedance of any tube or group of tubes is anode
> voltage under load divided by total anode current times 1.8.
> It matters not what the type or brand of the tube(s) are. A
> single 8877 amp that has 4000 volts on the anode and an anode
> current of 1 amp will have a plate load impedance of 2222 ohms.
>
> (((73)))
> Phil, K5PC
>
>
>  Subject: [Amps] Anode Impedance of 4-400A
> >
> > Does anyone know the 'rough' load impedance of a pair of
> 4-400A's with
> > 3500V on the anode?
> >
> > In some of EIMAC literature (W6SAI AS-1 Publication), there
> are tables,
> > which list it as 5000 Ohms.  Is this for one tube, or two?
> >
> > The 1982 ARRL Handbook contains a design with a pair.  The
> PI network
> > values listed there, equate to a load of 5000 Ohms, with a
> Q of about 12.
> >
> >
> > I am just trying to cross a few items off of my list.
> > If someone can contact me off list, I would appreciate it.
> >
> > Thanks
> > Eric Thielking
>
>
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