[Amps] Measuring tank coil inductance

Vic K2VCO k2vco.vic at gmail.com
Sun Apr 28 02:15:28 EDT 2013


Let me see if I understand correctly.

Suppose I want to have the amplifier work exactly according to spec. Then when I tune it 
with a steady carrier I want to have enough drive so that the zero-signal bias of -60 V 
goes up to about -25 V on positive peaks.  This is the point on the curve chosen for the 
minimum of the plate voltage swing. This means that 35 V peak is developed across the 50 
ohm grid resistor, which requires about 25 W peak. Therefore I should drive the amplifier 
with about 25 x 0.7 = 17.5 W measured by the usual wattmeter.  Or I could multiply the 
screen current shown on the curve at the minimum plate voltage point by 0.7 and tune for that.

Is this correct?

On 4/26/2013 9:12 AM, TexasRF at aol.com wrote:
> Vic, the peak screen current will be several times the indicated current. The trip 
> circuits have a time constant and don't respond to the peak current that comes along at 
> the rate of one per rf cycle. The screen bypass C tends to soak up much of the peak 
> current as well.
> When rf drive is adjusted for rated screen current, the net result is an 
> instataneous plate voltage minimum value quite a bit less than what you are plotting. 
> The load line will be shifted somewhat as a result. Also, the power generated is 
> somewhat higher than you are expecting as the peak plate current is increased as well.
> Just as the indicated plate current is an average over the complete rf cycle, so are the 
> screen and grid currents.
> 73,
> Gerald K5GW
> In a message dated 4/26/2013 8:49:49 A.M. Pacific Daylight Time, k2vco.vic at gmail.com writes:
>
>     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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>

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



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