Re: Trying to model loop antenna (why does capacitance go down when adding parallel cap?)
- From: donald@xxxxxxxxxxxxx (Don Pearce)
- Date: Tue, 26 Apr 2005 09:36:37 GMT
On Mon, 25 Apr 2005 22:27:14 +0100, John Woodgate
<jmw@xxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
>I read in sci.electronics.design that Larry Brasfield
><donotspam_larry_brasfield@xxxxxxxxxxx> wrote (in
><ZHcbe.57$636.3219@xxxxxxxxxxxxxxx>) about 'Trying to model loop antenna
>(why does capacitance go down when adding parallel cap?)', on Mon, 25
>Apr 2005:
>
>>If you measure the first resonant frequency and Q, then measure
>>inductance at a much lower frequency, you (or many folks here) should
>>be able to calculate the capacitance needed to tune to 60 KHz.
>
>The self-resonant frequency is very much higher than 60 kHz. It is
>irrelevant. I'm beginning to think that you, too, can only 'think
>complicated'.
>
>With such a small inductance tuned with something like 10 nF, the Q will
>be high enough to disregard, at least for the first try-out.
I think everybody is missing an important point here. This isn't just
a resonant circuit - it is an antenna. That means that there is a
major resistive term missing - the radiation resistance. If it is a
decent antenna, this will be the dominant term, and will degrade the
effective Q of the circuit massively. Certainly it will overpower the
loss resistance. So no, I don't think Q will be high enough to
disregard.
d
Pearce Consulting
http://www.pearce.uk.com
.
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