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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