Brian Howie wrote:


|
|
!
-----!-----
| |
| |
L C
| |
| |
R R
| |
| |
------------
|
|
|


I ran it through Spice ( laziness) - It doesn't resonate. Intuitively
you think it should have a low Q resonance at 1.6MHz , but it doesn't
Nice one.

73 Brian GM4DIJ



The trick is to make

R = sqrt(L/C)

then the impedance is real everywhere. You can use any old values for L:
and C, as long as you make R=sqrt(L/C);


That equation is obviously know from transmission lines too..

On Sun, 18 Nov 2007 19:12:48 +0000, Dave wrote:

Brian Howie wrote:


|
|
!
-----!-----
| |
| |
L C
| |
| |
R R
| |
| |
------------
|
|
|


I ran it through Spice ( laziness) - It doesn't resonate. Intuitively
you think it should have a low Q resonance at 1.6MHz , but it doesn't
Nice one.

73 Brian GM4DIJ



The trick is to make

R = sqrt(L/C)

then the impedance is real everywhere. You can use any old values for L:
and C, as long as you make R=sqrt(L/C);


That equation is obviously know from transmission lines too..

Hi Dave,

Perhaps, but not in this round.

sqrt(L/C) 1000

Besides, resonance is slightly above 1MHz.

73's
Richard Clark, KB7QHC

Richard Clark wrote:

Hi Dave,

Perhaps, but not in this round.

sqrt(L/C) 1000

Besides, resonance is slightly above 1MHz.

73's
Richard Clark, KB7QHC

For a minute, I thought he had abandoned the "new math" and gone over to
the "dark side." (or, "new-new math!") ;-)
JS

Richard Clark wrote:
On Sun, 18 Nov 2007 19:12:48 +0000, Dave wrote:

Brian Howie wrote:

|
|
!
-----!-----
| |
| |
L C
| |
| |
R R
| |
| |
------------
|
|
|

I ran it through Spice ( laziness) - It doesn't resonate. Intuitively
you think it should have a low Q resonance at 1.6MHz , but it doesn't
Nice one.

73 Brian GM4DIJ


The trick is to make

R = sqrt(L/C)

then the impedance is real everywhere. You can use any old values for L:
and C, as long as you make R=sqrt(L/C);


That equation is obviously know from transmission lines too..

Hi Dave,

Perhaps, but not in this round.

sqrt(L/C) 1000

Besides, resonance is slightly above 1MHz.

73's
Richard Clark, KB7QHC

My calculator needs fixing. When I divide 100 uH by 100 pF and take
the square root, I end up with the number 1000. Where did I go wrong?
73,
Tom Donaly, KA6RUH

Tom Donaly wrote:
My calculator needs fixing. When I divide 100 uH by 100 pF and take
the square root, I end up with the number 1000. Where did I go wrong?

The actual formula is 1/[2pi*SQRT(L*C)]
--
73, Cecil http://www.w5dxp.com

On 18 Nov, 14:11, Cecil Moore wrote:
Tom Donaly wrote:
My calculator needs fixing. When I divide 100 uH by 100 pF and take
the square root, I end up with the number 1000. Where did I go wrong?

The actual formula is 1/[2pi*SQRT(L*C)]
--
73, Cecil http://www.w5dxp.com

I see that somebody intimated a 1000 ohm resistive impedance.
My antenna on 160 is about half of that! The question I have now is
how can we relate the radiation with respect to that high
resistive impedance?
Regards
Art

Art wrote:
"The question I now have is how can we relate the radiation with respact
to that high resistive impedance?"

Efficiency = radiation resistance / radiation resistance + loss
resistance

Best regards, Richard Harrison, KB5WZI

On 18 Nov, 14:41, (Richard Harrison) wrote:
Art wrote:

"The question I now have is how can we relate the radiation with respact
to that high resistive impedance?"

Efficiency = radiation resistance / radiation resistance + loss
resistance

Best regards, Richard Harrison, KB5WZI

Nothing spectacular about that Richard, or are you relating to
something I missed?
Ofcourse nothing is real with the circuit that David provided because
the
capacitor is not real without a bypass resistance!
Art

Cecil Moore wrote:
Tom Donaly wrote:
My calculator needs fixing. When I divide 100 uH by 100 pF and take
the square root, I end up with the number 1000. Where did I go wrong?

The actual formula is 1/[2pi*SQRT(L*C)]

Not in this case. All you have to do, Cecil, is take your formula,
above, find the frequency you think is the resonant frequency, and
then use it to find the impedance across the circuit. Now, try some
other frequency. You can prove that this circuit can be replaced by
a 1000 ohm resistor for all frequencies, using network analysis, but
that's a little more difficult.
73,
Tom Donaly, KA6RUH

Tom Donaly wrote:
You can prove that this circuit can be replaced by
a 1000 ohm resistor for all frequencies, using network analysis, but
that's a little more difficult.

That's pretty interesting and not intuitively obvious.
--
73, Cecil http://www.w5dxp.com