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@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@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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