"Richard Clark" wrote in message
...
On Thu, 13 Jan 2011 19:51:21 -0600, "amdx" wrote:


"Richard Clark" wrote in message
. ..
On Thu, 13 Jan 2011 16:18:35 -0600, "amdx" wrote:


The article's focus is on matching a crystal radio tank to the antenna,
So as a general statement, the tank is a high impedance (mostly R)

Hi Mike,

There's your first mistake. Tank Z is never, ever "mostly R," or you
wouldn't be able to make the Q claim of 1000 (or even 10).

Ok, Richard that wasn't clear to me, I think at resonance the tank is
all R, but I put mostly R because I figured you would have an objection to
all R. So are you saying it is pure R at resonance?

Hi Mike,

Let's say there is absolutely no loss in the Tank (superconduction and
perfect dissipation values as it were); then we would have to ask
ourselves what happens to energy applied to this Tank at resonance? It
can never enter it, thus the Tank is, in effect, infinite in
resistance. But what about the circulating currents? The Tank is, in
effect, infinite in conductance.

Infinite Ohms & Zero Ohms simultaneously.

Is this the Z of the Tank? Is this the R of the Tank to which you are
matching? No, not even close and certainly it has nothing to do with
resonance - except the condition is a function of it being at
resonance. A low Z Tank or a high Z Tank each evidences the same
Infinite Ohms & Zero Ohms simultaneity given my initial condition of
absolute losslessness.

For the energy being applied to or drawn from the Tank, the Tank is in
parallel operation. For energy in the Tank, the Tank is in series
operation. Where is the Q in this duality? Q suffers by the nature
of what you call R. Q has two different values by this duality. One
is called "Loaded Q" and as you might guess, the second is called
"Unloaded Q." Consult Terman for the engineering design rationale for
optimal Qs as I suggested.

A lot there, but I didn't get anything out of it.
At resonance, does the tank look capacitive, inductive, resistive, or all
to the antenna you connect to it. (this makes the assumption attaching
the antenna didn't change anything, it did, let's say I adjusted back to
resonance)
And then my original question,
Are you saying it is pure R at resonance?
A yes or no will be fine, then I can try to reprocess the above that
didn't get anything out of.


When the discussion of "matching" seeks to employ R (pure resistance),
then the next step is toward a conjugate match and elaborations of
efficiency and maximum transfer of power. There is also an
alternative discussion called the Zo Match. This second match seems
to invite the same elaborations (many who post here try to force them
both into the same salad bowl and cover the illogic with dressing).

I'll bite,
I want maximum transfer of power, I'm still working with the tank as
a large pure R so I want the antenna to look like the same large R.
I realize there is still capacitance from the antenna to deal with.
Now are getting to conjugate :-)


When you offered the comment about "the tank is a high impedance
(mostly R)" it was steering the car off the cliff. Is this a Zo match
or a Conjugate match you are seeking? (I can already anticipate this
has gone over your head, as well as many readers. This and the
questions that follow are rhetorical.)

For instance, and returning to antennas (the purpose of this group's
discussion focus), you can have very high Z antennas with very low
resistance characteristics. Do you want a Zo Match, or a Conjugate
Match? Let me flip the antenna: you can have very high Z antennas
with very high resistance characteristics. Do you want a Zo Match, or
a Conjugate Match? Let's do this sideways: you can have very low Z
antennas with very low resistance characteristics. Do you want a Zo
Match, or a Conjugate Match? I could box the compass here, but the I
think I will let the reader off.


......which, then leads me to ask

"What do you really want?"

73's
Richard Clark, KB7QHC

I want to understand the use of an air variable to match an antenna
to the tank of a crystal radio, over the AMBCB frequency range.
With that, I found I need to understand the series to parallel
conversion,
which I now understand, just IS, it's not anything you do. A series RC has
a parallel RC equivalent.
I'm not sure how it can be both at the same time. But as long as that R is
transformed up, and minimally loads my tank, that's all good.
Then, I understand I still have C left that I can use as part of the C for
resonating my LC tank.
Mikek

John had some number issues with Tony's explanation, but the gist of
Tony's rational treatment should be your lesson as it provides for
your requested "why." It also implies (by my comments of the sudden
appearance of two new components) that our (Ham) tuners have been
designed to introduce the proper amounts of reactances in the proper
parallel/series relationships to enable the necessary transform
towards optimal Q and loading balance. The most elaborate of tuners
can change from Pi to T topologies, or series L parallel C (or series
C parallel L), or series LC, or parallel LC, or parallel C series L
(or parallel L series C)... and any of the other combinations I have
not enumerated (about 9 in all). Each shines for a particular
situation - you have named only one.


Because of the wavelength of the BCB antennas are usually short and
capacitive,
so simple tuning with a single series cap works. The problem begins when you
tune to the high end of the band and you have to much capacitance to get to
resonance with your tank. Then the inductor can be added parallel to the
series
antenna tuning cap. Alternately and you could increase the tank inductor
size.



It is not a trivial discussion by any means even when we are talking
about the addition of only two new components. So, your obtaining an
understanding is not going to be achieved at one sitting in front of
the "definitive" posting to a thread.

One problem of seeking the "definitive" posting is that it cannot be
born from a broken premise that article you were trying to figure out
is lame in the extreme.

Given everything you have revealed about it,
it didn't present a solution to its fantasy antenna.

I didn't rewrite the whole article here, there was a solution
with three equations, that, using the antenna finds the L with the
constrants
of highest operating frequency and lowest capacitance of your capacitor,
then a program is run that finds values for C (antenna) and C (tank) and
I'm not ready to go here yet C (load), he also tunes the diode/earphone
load for optimum with yet another capacitor.
The funs over for now go to get ready for work,
Thanks, Mikek
PS, he runs the program with another, what you call fantasy antenna,
and I agree...
73's
Richard Clark, KB7QHC

On Fri, 14 Jan 2011 07:20:20 -0600, "amdx" wrote:

I want maximum transfer of power, I'm still working with the tank as
a large pure R so I want the antenna to look like the same large R.

No you don't.

What you want is the lightest final load sufficient to drive a speaker
for a detectable sound matched to the Tank such that it does not
degrade its Q which in turn is the highest possible value for
supporting the largest amount of signal from the antenna at hand.

Am I wrong?

73's
Richard Clark, KB7QHC

"Richard Clark" wrote in message
...
On Fri, 14 Jan 2011 07:20:20 -0600, "amdx" wrote:

I want maximum transfer of power, I'm still working with the tank as
a large pure R so I want the antenna to look like the same large R.

No you don't.

What you want is the lightest final load sufficient to drive a speaker

We haven't got to the load yet! it's coming :-)

for a detectable sound matched to the Tank such that it does not
degrade its Q which in turn is the highest possible value for
supporting the largest amount of signal from the antenna at hand.

Am I wrong?

There is a chance the Q (loaded) will be high enough to limit audio
bandwidth.
So (I think) we couple more energy into the tank for more signal and this
would
lower Q for a wider bandwidth.

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, just in different order.

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.
Then I load this with 500Kohm load.
So..
The 1 megohm tank is loaded with 333,333ohms, 1meg//500k
The 1 meg antenna is loaded with 333,333ohms, 1meg//500k
The 500Kohm load is drive by 500kohms. 1meg//1meg


Scenerio 2.
1 megohm tank.
I put a 1 megohm load
I can drive the tank with a 500Kohm source,
So..
The 1 megohm tank is loaded with 333,333ohms, 500k//1meg
The 500 Kohm antenna is loaded with 500Kohms, 1meg//1meg
The 1 megohm load is driven by 500kohms. 1meg//500k



Scenerio 3.
1 megohm tankThe
I drive the tank with 2 megohm source and load it with a 2 megohm load.
So..
The 1 megohm tank is loaded with 1 Mohm, 2Mohm//2Mohm
The 2 meg antenna is loaded with 666,666 ohms, 1meg//2meg
2 megohm load is drive by 666,666 ohms. 1meg//2meg

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

This aught to be fun :-)
Mikek

Antonio - I think you slipped a decimal point. The parallel equivalent of the
series combo 58R-2947j is actually 149k+2948j.

Hi John,

I do not understand where your -2947j figure comes from. I see it appearing
nowhere in my calculations.

The antenna mentioned by Mikek has an impedance of 58R-1,072j which, according
to my spreadsheet, corresponds (at 1 MHz) to the parallel of 19862R and -1075j
(that is a 148,1 pF capacitor).

In any case, parallel -- series trasformations never result in a change of the
reactance sign; therefore it is not possible that a -2957j (negative) reactance
is transformed into a +2948j (positive) reactance.

73

Tony I0JX

On 1/14/2011 10:50 AM, Antonio Vernucci wrote:
Antonio - I think you slipped a decimal point. The parallel equivalent
of the series combo 58R-2947j is actually 149k+2948j.

Hi John,

I do not understand where your -2947j figure comes from. I see it
appearing nowhere in my calculations.


Well, it's not mine and it appears in you post as 54 pF. Isn't 58R in
series with 54 pF equal to 58-2947j? And isn't the parallel equivalent
of that equal to 149k ohms of resistance in parallel with -2948 ohms of
reactance (~54 pF)?

I'm pointing out that you slipped a decimal point or you would have seen
that 54 pF is too much it results in the parallel equivalent resistance
of 149k rather than 1.49M. Mike's figure of about 18 pF (17 pF series
combination) will do the job.

In any case, parallel -- series trasformations never result in a
change of the reactance sign; therefore it is not possible that a -2957j
(negative) reactance is transformed into a +2948j (positive) reactance.

You are correct. I allowed the sign of the suseptance to creep through.

73

Tony I0JX

Well, it's not mine and it appears in you post as 54 pF. Isn't 58R in series
with 54 pF equal to 58-2947j? And isn't the parallel equivalent of that equal
to 149k ohms of resistance in parallel with -2948 ohms of reactance (~54 pF)?

I'm pointing out that you slipped a decimal point or you would have seen that
54 pF is too much it results in the parallel equivalent resistance of 149k
rather than 1.49M. Mike's figure of about 18 pF (17 pF series combination)
will do the job.

Yes, at one o' clock in the morning, I slipped the decimal point. So, the total
antenna series capacitance should have been about 17 pF, not 54 pF. This
requires putting a 19-pF capacitance in series with the antenna, not 85 pF.

And the inductance resonating the residual parallel capacitance becomes 1,490 uH
instead of 470 uH.

Sorry for mistake!

73

Tony I0JX

On Fri, 14 Jan 2011 19:35:43 +0100, "Antonio Vernucci"
wrote:

Well, it's not mine and it appears in you post as 54 pF. Isn't 58R in series
with 54 pF equal to 58-2947j? And isn't the parallel equivalent of that equal
to 149k ohms of resistance in parallel with -2948 ohms of reactance (~54 pF)?

I'm pointing out that you slipped a decimal point or you would have seen that
54 pF is too much it results in the parallel equivalent resistance of 149k
rather than 1.49M. Mike's figure of about 18 pF (17 pF series combination)
will do the job.

Yes, at one o' clock in the morning, I slipped the decimal point. So, the total
antenna series capacitance should have been about 17 pF, not 54 pF. This
requires putting a 19-pF capacitance in series with the antenna, not 85 pF.

And the inductance resonating the residual parallel capacitance becomes 1,490 uH
instead of 470 uH.

Sorry for mistake!

73

Tony I0JX

Hasn't anyone pointed out that this a problem made for using a Smith
Chart?

(Since no one really seems capable of doing the math :-)

...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

On Jan 14, 6:41*pm, Jim Thompson To-Email-Use-The-Envelope-I...@On-My-
Web-Site.com wrote:
On Fri, 14 Jan 2011 19:35:43 +0100, "Antonio Vernucci"



wrote:
Well, it's not mine and it appears in you post as 54 pF. Isn't 58R in series
with 54 pF equal to 58-2947j? And isn't the parallel equivalent of that equal
to 149k ohms of resistance in parallel with -2948 ohms of reactance (~54 pF)?

I'm pointing out that you slipped a decimal point or you would have seen that
54 pF is too much it results in the parallel equivalent resistance of 149k
rather than 1.49M. Mike's figure of about 18 pF (17 pF series combination)
will do the job.

Yes, at one o' clock in the morning, I slipped the decimal point. So, the total
antenna series capacitance should have been about 17 pF, not 54 pF. This
requires putting a 19-pF capacitance in series with the antenna, not 85 pF.

And the inductance resonating the residual parallel capacitance becomes 1,490 uH
instead of 470 uH.

Sorry for mistake!

73

Tony I0JX

Hasn't anyone pointed out that this a problem made for using a Smith
Chart?

(Since no one really seems capable of doing the math :-)

* * * * * * * * * * * * * * * * * * * * ...Jim Thompson
--
| James E.Thompson, CTO * * * * * * * * * * * * * *| * *mens * * |
| Analog Innovations, Inc. * * * * * * * * * * * * | * * et * * *|
| Analog/Mixed-Signal ASIC's and Discrete Systems *| * *manus * *|
| Phoenix, Arizona *85048 * *Skype: Contacts Only *| * * * * * * |
| Voice480)460-2350 *Fax: Available upon request | *Brass Rat *|
| E-mail Icon athttp://www.analog-innovations.com| * *1962 * * |

I love to cook with wine. * * Sometimes I even put it in the food.

just plug it in and try it... if the volume isn't high enough get a
real radio!

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.)

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.

How does this impact design priorities?

73's
Richard Clark, KB7QHC

"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