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Re: [Amps] Measuring tank coil inductance

To: Ian White <gm3sek@ifwtech.co.uk>
Subject: Re: [Amps] Measuring tank coil inductance
From: Vic K2VCO <k2vco.vic@gmail.com>
Date: Fri, 26 Apr 2013 08:49:29 -0700
List-post: <amps@contesting.com">mailto:amps@contesting.com>
Yes, I stand corrected. The word ESTIMATE was used! A lot.

I was just surprised at how different the value from the load line was, since I've seen the formula for estimating the impedance using the k factor in numerous places.

What I did was, first of all, to extrapolate the Eimac constant current curve graph to 3.2 kV (it only goes up to 3 kV). I felt justified in doing this because the ZSAC at -60v on the grid and 325v on the screen was flat from 500 to 3000 plate volts.

Then I followed G4AXX and selected a minimum point of plate voltage at 750v, corresponding to a grid voltage of -25v, because this corresponds to a maximum screen current just under 50 mA, which is the trip point for the overcurrent circuit of the screen.

Here is G4AXX's example of load lines for 2.5, 2.75 and 3 kV:
http://www.granta.g4axx.com/Linear_Design_notes/loadline4.gif
3.2 kV is represented by the right edge of the graph. If you draw another line from point A to where the extrapolated curve that represents a ZSAC of 250 mA hits that edge, you get a load line whose slope represents 1815 ohms.

So why did I use 3.2 kV instead of a lower voltage? Because the transformer I have can produce either 3.2kV or 2.65 kV under load and I felt that it would be pushing things to try to get a clean 1500 watts at the lower voltage. For once I am trying to build an amplifier that will be safe to use on SSB as well as CW, just in case I should want to!

On 4/26/2013 7:44 AM, Ian White wrote:
K2VCO wrote:
I am using the GM3SEK spreadsheet to determine the LC values. I have a
simple application on my iPhone called "E-Formulas" which quickly
solves
the equation for resonance for l, c, or f.

One interesting thing that I noticed: the spreadsheet says that you can
determine the tube load impedance by the formula Rl = Ep/(Ip*k) where k
is
1.5 - 1.7 for class AB.
Ahem... it certainly doesn't say "determine" the load impedance because
you can't do that. The wording I used throughout was "ESTIMATE the load
resistance".

The method using "k" can only possibly give a rough estimate because
tube characteristics are far too complex to be condensed into a single
"magic number".

Here is the full text (with ESTIMATE changed into capitals):

"There are various methods to ESTIMATE the load resistance RL.

"The most accurate method involves a load line which is drawn over the
tube's characteristic curves.  One end of the load line is at supply
voltage E and zero-signal anode current Io. The other end of the load
line is at peak instantaneous anode current Ip and a minimum anode
voltage Vo (typically 5-15% of E).  Calculate the slope of the load line
= (E-Vo)/(Ip-Io) and enter it on line line 35 of the spreadsheet."

For tetrodes I should have added another requirement: that the anode
must at all times remain significantly more positive than the screen, to
avoid high peak screen currents and accompanying nonlinearity.

"Alternatively, the spreadsheet offers two numerical methods of
ESTIMATING RL, using a factor K which depends on the class of operation
of the tube (class A, AB1, AB2, B or C). To use this method, enter E and
the maximum DC anode current Ia on lines 22 and 23, and your estimate of
K on line 24.  Then you can enter either the desired power output W on
line 29 or your estimate of efficiency on line 32.

"Each of the three methods - the method using the load line, or the two
methods using K - will typically give a different value for RL. You must
then CHOOSE YOUR OWN BEST ESTIMATE, and enter it on line 35."

In other words, you have to CHOOSE a suitable load resistance. The
spreadsheet can not determine it for you.

Vic continues:
Using the load line for the AB1 4CX1000A at 3.2 kV and
800 mA came out to 1815 ohms, which corresponds to a k of about 2.2.

I'm not sure what you mean by "The load line at 3.2kV and 800mA came out
to 1815 ohms"? There is a lot of missing information about the way that
you chose to construct that particular line, and the specific locations
of each end.


73 from Ian GM3SEK



--
Vic, K2VCO
Fresno CA
http://www.qsl.net/k2vco/

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