Those tubes will be rather used in class B.
Without grid-current they put out 1200W at 3,5KV anode voltage,
with a total of 13mA grid current it will be about 1600W.
On using the formula Ep/x*Ip one should consider that Ep
will be smaller then the supply voltage, as the voltage swing
should not move below the screeen voltage, in this case Ep will
be 2,9KV for that calculation
73
Peter
-----Original Message-----
From: amps-bounces@contesting.com [mailto:amps-bounces@contesting.com] On
Behalf Of David C. Hallam
Sent: Mittwoch, 16. November 2005 02:05
To: Joe Subich, W4TV; 'Phil Clements'; Amps@contesting.com
Subject: Re: [Amps] Anode Impedance of 4-400A
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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> Amps@contesting.com
> http://lists.contesting.com/mailman/listinfo/amps
>
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