Peter, I can't make this information agree with practice. For example, I
have a TH347 amplifier on 23cm that runs 3000vdc at 1.8A on the plate.
Conventional design would yield a plate impedance of 925 ohms using a k=1.8
factor.
This amplifier runs class AB2.
With a screen voltage of 600v, the plate voltage swing is close to 2400v. I
have read that peak plate current runs three times average typically or 5.4A
in this case. Using your information, this would yield a plate impedance of
444 ohms.
Clearly, some key factor is missing here. Perhaps my peak current equals
three times average current factor is in error? Perhaps the plate voltage swing
should include the "missing" half cycle?
73,
Gerald K5GW
In a message dated 4/1/2006 5:04:08 A.M. Central Standard Time,
g3rzp@g3rzp.wanadoo.co.uk writes:
You have to figure on the supply volts under load.
Let's assume we're in AB1 for an example. From the tube data sheet, we can
choose the plate voltage when the grid volts are zero - usually just about the
knee in the characteristics. That tells us that the load impedance is equal
to the supply volts minus the knee volts divided by the plate current at Vg=0.
Exactly the same principle applies if running AB2 or Class B, except you
have to decide the positive grid volts that represents the peak of the cycle.
Then you have the peak plate current, and the minimum plate volts. Again, the
difference between the supply volts at that particular amount of current load
and the minimum plate volts, divided by the peak plate current, gives you the
load resistance.
The factors relating the peak plate current, conduction angle and the DC
drain on the PSU are more complex....
73
Peter G3RZP
_______________________________________________
Amps mailing list
Amps@contesting.com
http://lists.contesting.com/mailman/listinfo/amps
_______________________________________________
Amps mailing list
Amps@contesting.com
http://lists.contesting.com/mailman/listinfo/amps
|