[Amps] tube impedance, figuring tank circuit values

[David C. Hallam]

I think that Peter inadvertently left out a factor of 2.  According to my
RCA tube manual Rp is equal to twice the plate voltage swing divided by the
max current.

David
KC2JD

-----Original Message-----
From: amps-bounces at contesting.com [mailto:amps-bounces at contesting.com]On
Behalf Of TexasRF at aol.com
Sent: Saturday, April 01, 2006 12:27 PM
To: g3rzp at g3rzp.wanadoo.co.uk; dezrat at copper.net;
Chuck_Partain at maxtor.com; amps at contesting.com
Subject: Re: [Amps] tube impedance, figuring tank circuit values



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 at 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
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