> Meanwhile back at the ranch, I
> looked up the formulas for calculating the values in a
> pi-network in the
> ARRL Handbook, and put them into a Microsoft Excel
> spreadsheet.
> Interestingly, when I did the calculations for my
> operating conditions, and
> those in the schematic, they were far apart as far as
> values are concerned;
> thus the reason for the inquiry here on the reflector. I
> will probably go
> with the values that provide the correct match for the
> operating conditions
> in my amp. Thanks for the response.
Mike,
Looking back at the posts it *seems to me* you are thinking
the design of the tank sets the load Z at the tube. That's
what I think is going on, so I'll answer that way.
It really doesn't matter much what you initially set Q at
when the tank has more than one adjustment knob, as long as
you stay within a reasonable amount. Many people
over-engineer network operating Q. The design of the tank
sets the loaded or operating Q, most of the time, is only a
rough approximation using plate tuning cap Cx/R where R is
the anode operating or load impedance presented by the
network. Pick up many books and you'll see they start that
way. They say "set Q at 12-14" and they use a rough
approximation like Ep/ 1.5*I = RL or some other very rough
approximations. Then the builder spends two weeks trying to
get within 1% of a number that might be 50% off and not mean
much in the first place.
You can drive yourself crazy, as most people do, trying to
get tank operating Q to some mystical magical number like
12. The truth is "12" is a number that was largely pulled
out of a place between the back pockets of our jeans ....and
the operating Q most people and amateur books list and use
is an approximation also.
If we design the tank for an operating Q of 12 it might wind
up being 9 or 20 in operation, and we won't notice the
difference as long as operating Q is between some wide
limits.
Here are the limits:
The lower Q limit is some value more than the square root of
the operating R of the tube loadline over the load
resistance at the output plus a margin. If the tube loadline
winds up at 5000 ohms and the load is 50, that's a
5000/50=100 and the square root is 10. We need a minimum Q
that is some reasonable value over 10, or the network will
not work as a pi. What happens if we don't do that is the
loading cap will be "mushy" and the output might peak with
minimum loading C. This comes from a network operating Q
that is too low.
The upper limit is generally where circulating currents get
so high things desolder or heat up. If we doubled Q tank
losses would approximately double, but they are so low in a
good tank it doesn't matter much anyway for overall
efficiency. After all, if we have 1% tank loss and we double
it is only 2%. That number gets lost in the other losses we
cannot control. You'd be amazed at how little power loss it
takes to heat up a component in a 1000 watt amplifier, so
the real problem is generally heating....not efficiency.
Low order harmonics change with network operating Q, but
even that isn't as bad as we might think. Even doubling
network Q won't help if something is way out of spec.
So here is my suggestion. Just look at the LOWEST possible
operating impedance you might have by approximation. Put
your favorite little conduction angle fudge factor in there
too. That would be the very lowest Ep/ x*Ip you will ever
have and the biggest conduction angle fudge factor. Say it
is 3000/1.5 * 1 = 2000. Now find the square root of the
ratio to load impedance, and add a safety factor for
headroom to that. 2000/50= 40. Sqrt 40= 6.3 and a fudge of
4. That's ten. Now design for a network Q of ten into that
load Z.
If the initial numbers came out 4000/50, you'd need an
initial Q of 9 plus 4, or 13.
Under any other condition the optimum anode operating
impedance will be higher, and the network Q will go up, so
you will always exceed the minimum requirement by at least
your initial fudge factors. Say you only run half the
current at the same conduction angle. The impedance would
approximately double and the network operating Q, when
adjusted for maximum power transfer, would also
approximately double.
Does this make sense to you?
73 Tom
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