"Richard Clark" wrote in message
...
On Fri, 14 Jan 2011 10:47:50 -0600, "amdx" wrote:

So (I think) we couple more energy into the tank for more signal and this
would
lower Q for a wider bandwidth.

Hi Mike,

How? (It would be a sad day for us all if more input to a Tank
lowered its Q were true.)

My thought was, if we couple more energy from the antenna, it loads
the tank more lowering Q. The thought might need more work....

Here's a question I have brewing.

I have three circuits to put together, a source, a tank and, a load.
I have two scenerios. hmm..seems as though I have three!
For now assume they are all resistive.
These are all set up for maximum power transfer

Actually, that remains to be seen. We must first establish that we
have the same available power in all three which I will give the
arbitrary value of 1 to simplify the math. Also, you mix your source
loading in these models, so in the sense of Norton/Thevenin sources I
describe the power being supplied by a parallel current source to keep
the units uniform.

Scenerio 1.
Let's say the tank is 1 megohm.
I drive the tank with a 1meg source, so now I have 500Kohm circuit
impedance.
This presumes a parallel current source by your description of the
input and tank appearing as a 500K circuit. This is why set the
initial condition of there being a current source for all scenarios.
Then I load this with 500Kohm load.
The parallel current source then sees 250K Ohm for the same power
available to all scenarios.
Pavailable = 1 = i²·250K
i = sqrt(1/250K)
As the current does not divide evenly, then we will work to find the
power to the load through voltage sharing. Obviously, there is the
same voltage across the three components, hence:
e = i·250K = 500
Pload = e²/500K = 0.50

Scenerio 2.
1 megohm tank.
I put a 1 megohm load
I can drive the tank with a 500Kohm source,
This does not qualify either a parallel current nor series voltage
source, but as both are fungible to design with the same value
resistance, then I will proceed as before with all three resistors in
parallel to a parallel current source:
Pavailable = 1 = i²·250K
i = sqrt(1/250K)
Obviously, there is the same voltage across the three components,
hence:
e = i·250K = 500
Pload = e²/1000K = 0.25

Scenerio 3.
1 megohm tankThe
I drive the tank with 2 megohm source and load it with a 2 megohm load.
So..
This does not qualify either a parallel current nor series voltage
source, but as both are fungible to design with the same value
resistance, then I will proceed as before with all three resistors in
parallel to a parallel current source:
Pavailable = 1 = i²·500K
i = sqrt(1/500K)
Obviously, there is the same voltage across the three components,
hence:
e = i·500K = 707
Pload = e²/2000K = 0.25

I have no clue where maximum power is delivered
from the antenna to the load.

Any clues now? Barring any math or conceptual error on my part, then
by one account more power (that is one measure of success) is
delivered to the load when its resistance is lowest.


Scenerio 1 is how I have always thought about the system.
Nice to know where max power transfer is.

How does this impact design priorities?

I'm still at design highest Q tank circuit then transform antenna
to match Z of tank. That's about as I want to go for now.
Not ready to get into that diode thing again.
Unless you've been studying :-)
running for cover.....
Thanks, Mikek


73's
Richard Clark, KB7QHC