On 18 Nov, 11:12, Dave wrote:
Brian Howie wrote:

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L C
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R R
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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

finally come up 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..- Hide quoted text -

- Show quoted text -

David,
After all those discussions where you have been baiting me on my
antenna
you finally come up with what I have been hitting you with.
Rember my comments where a radiator can be any shape, any elefation as
long as the
element is in equilibrium. Finally the penny has dropped with respect
to the LC ratio!
Art

Cecil, W5DXP wrote:
"Since the two resistances are equal, seems to me the resonant frequency
would be where the two reactances are equal."

Yes that`s the unity power factor point. There`s a rule that when the
circuit Q`s not less than 10, fo=1/2pi on the sq.rt. of LC. For lower
Qs, the calculation is more laborious. I sure miss my ARRL Lightning
Coil Calculator!

Best regards, Richard Harrison, KB5WZI

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

On 18 nov, 13:11, Dave wrote:
What is the resonate frequency of this network, as determined between
the top and bottom of what I have drawn?

I don't know how well the drawing will come out, but it consists of:

100 uH in series with 1000 Ohms.
100 pF in series with 1000 Ohms

The two two networks above are in parallel

i.e.

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L C
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R R
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hello Dave,

Normallly the resonant frequency of circuit is the frequency where Zin
is real. The problem with this circuit is that Z is real everywhere
and Q will be zero. So in my opinion it is useless to define a
resonant frequency for this circuit. The only other option you have
is to find the frequency where Im(current left leg) = -Im(current
right leg), 1.600 MHz.

Best regards,

Wim
PA3DJS
www.tetech.nl

On Sun, 18 Nov 2007 21:45:54 GMT, "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?

Hi Tom,

You didn't, I misread micro for nanohenry.

No resonance as specified.

73's
Richard Clark, KB7QHC

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

Cecil Moore wrote:
Dave wrote:
What is the resonate frequency of this network, as determined between
the top and bottom of what I have drawn?

I don't know how well the drawing will come out, but it consists of:

100 uH in series with 1000 Ohms.
100 pF in series with 1000 Ohms

The two two networks above are in parallel

Since the two resistances are equal, seems to me the
resonant frequency would be where the two reactances
are equal. Where the 100 uH line crosses the 100 pf
line on the reactance chart in the ARRL Handbook is
in the ballpark of 1.591549431 MHz. :-)

But it does not resonate at 1.591549431 MHz - or anywhere else for that
matter. The impedance is 1000 Ohms, purely resistive, at any frequency.

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

Dave wrote:
Roy Lewallen wrote:
Is this by any chance an exam question?

No, it is not. I was shown it by a lecturer of mine more than 10 years
ago. The result is quite interesting.

With the given values, it's a constant-impedance network. I've used one
many times in time domain circuit designs. Its impedance is a constant
real value of 1000 ohms at all frequencies. Since "resonance" implies a
single frequency (at which the reactance is zero), this circuit isn't
resonant at any frequency. The circuit is often used in time domain
applications (e.g., oscilloscopes) where it's sometimes necessary to
provide a constant impedance load but you're stuck with a capacitive
device input impedance. In that situation, the C is the input C of the
device. However, the transfer function isn't flat with frequency-- you
end up with a single pole lowpass rolloff, dictated by the R and C values.

For anyone who cares about such matters, "resonate" is a verb,
"resonant" is the adjective, and "resonance" the noun. A resonant
circuit resonates at resonance.

Roy Lewallen, W7EL