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Re: [Amps] Plate impedance formula

To: "Amps@contesting.com" <Amps@contesting.com>
Subject: Re: [Amps] Plate impedance formula
From: "Will Matney" <craxd1@ezwv.com>
Date: Sun, 06 Feb 2005 10:52:04 -0500
List-post: <mailto:amps@contesting.com>
George,

Well, it's according to what the total plate current is. Let's say you had one tube drawing 100 mA. Then you add another drawing the same. That would then be 200 mA which then would divide the resistance in half. However, if you had the same two tubes and cut them back so each would draw 50 mA, it would be the same as 1 drawing a 100 mA. I should add that those values are for peak plate current and peak voltage.

I also seen another ham in Europe using a modified formula with the k factor of something like 1.87 or 1.89. He claimed it got it closer to the true values when calculating the tank circuit for a class AB1 amp. The published K factor is 1.8 in most books and the RCA tube books.

Best,

Will



On Sun, 06 Feb 2005 07:18:29 -0500, George KB2Z <Thermionic_Emission@earthlink.net> wrote:

Will, For the way the output tank sees it, in a common 2 tube amp, would that value double, half, or stay the same? Thanks, George


At 12:00 AM 2/6/05 -0500, you wrote:


It's been a few days back, I didn't go back to find the post but someone
wanted to know the formula for finding the plate impedance on a tube. For
whomever it was, as I didn't see it ever get answered, I'll post it below.
I almost overlooked it myself and forgot about it.


RP = EP / 1.8 X IP

1.8 = K factor for class AB1 (Changes with ZSAC)
RP = Plate Resistance
EP = Plate Voltage
IP = Plate Current

Will



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