"K7ITM" wrote in message
...
On Jan 13, 9:05 am, "amdx" wrote:
Hi All,
Please look at this in fixed font.
I'm looking for understanding of a series to parallel conversion for
antenna matching.
I'm pretty sure I'm missing something, so point it out to me.
This is in regard to a crystal radio, so the match is for a low impedance
antenna to a high impedance tank circuit.
The antenna: R=58 ohms C=1072 ohms at 1Mhz.
The tank: L=240uh C=106pf Q = 1000
Tank Z =~1.5 Mohms
Here's my understanding of what I think I'm reading.
I put a matching capacitor in series with the antenna.
Antenna-- -----R-----C--------Match cap-----tank------ground.
and this is supposed to transform the circuit to this.
( Maybe better said, equivalent to this)

l------------l
l l
Antenna--- R C LC---Tank
l l
l------------l
^
Ground

I calculate an 18.5pf cap for the match, making the antenna look like 58R
and 17pf.

So this; Antenna-- -----58R-----270pf--------Match
cap18.5pf-----1.5Mohms------ground.
This converts to;
l-----------------l
l l
Antenna---58R 17pf 1.5M---LC Tank at
l l Resonance
l-----------------l
^
Ground

And I now have a 1.5 Mohms source feeding a 1.5 Mohm load.
The purpose of which is to cause minimal loading of the tank by the
antenna.
I don't understand how adding a series capacitor makes a parallel
conversion.
What do I misunderstand or do just need to believe the numbers.
Thanks, Mikek

Seems like it doesn't much matter whether the antenna is real or
imagined; the point is rather what's going on when you couple a low
impedance source to a parallel-resonant tank through a small
capacitance...

Perhaps it will help you to think first about a low impedance source,
let's say 50 ohms at 1.591MHz (1e7 radians/sec), that you want to
couple to an (imaginary) inductor, 2.5mH with Qu=500 at that
frequency, and maximize the energy transfer to the inductor. The
inductance and Q implies that the effective series resistance of the
inductor is 50 ohms. That means all we need to do is cancel out the
inductive reactance, and we can do that with a series capacitance--
that happens to be 4pF. That assumes the ESR of the capacitance is
zero, or essentially zero.

But what if the inductor has 1/4 as much inductance, 0.625mH, same
Qu? Then its effective series resistance is 1/4 as much, or 12.5
ohms, and to get the same energy dissipated in it, you need the
current through it to be 2 times as large as before (constant R*I^2).
The total capacitance to resonate the tank is 4 times as much: 16pF.
If you divide that into two 8pF caps, one directly across the inductor
and the other in series with the 50 ohm source, (very close to) half
the tank's circulating current flows in each capacitor. For the same
energy delivered to the coil, the voltage at the top of the inductor
must be half as much as before. Since the coupling capacitor to the
source is twice as large as it was before, the load impedance the
source sees must be the same as before (50 ohms, as required for max
energy transfer).

Carry that another step, to an inductor with 1/16 the original
inductance and the same Qu=500, and you need 64pF to resonate it and
16pF to couple to it from the low impedance source. Now you divide up
the total capacitance with 3/4 of it (48pF) directly across the
inductor and 1/4 of it to the source. 3/4 of the tank's circulating
current flows in the 48pF capacitor, and 1/4 of it in the 16pF to the
source; the source once again sees a 50 ohm load (because the voltage
at the top of the inductor is 1/4 of the original, and the capacitor
coupling to the source is 4 times as large -- with the current in the
source unchanged).

The key to understanding this coupling, to me, is the division of the
tank's circulating current between the two capacitances, one directly
across the coil and one in series with the source. The smaller the
inductance (at constant Q), the more circulating current required for
the same energy dissipation, and the smaller
_percentage_of_total_resonating_capacitance_ for coupling to the
(constant impedance) source.

Cheers,
Tom

Tom I need to print this out and work with it.
Did you change the inductor on purpose or just miss a
factor of ten from the numbers I posted?
Thanks, Mikek