On Mon, 27 Apr 2009 13:43:41 -0700, Tim Wescott
wrote:

Paul Keinanen wrote:
On Mon, 27 Apr 2009 05:32:07 -0700 (PDT), David
wrote:

On Apr 25, 10:51 pm, Tim Wescott wrote:
David wrote:
Was just wandering if anyone has used or experimented with television
IF, Video and Detector coils, most are slug tuned coils that have a
few uh to several hundred uh, some are sheilded some are not, i have
about 500 that i bought years ago, a lot of them look very close too
the old loopstick type coils, looking for
ideas.
Thanks David
They'll have very low Q, and therefore not really be suitable for a
crystal radio, where low-Q coils fight your ability to get good
selectivity without burning up all your signal before it gets to the
headphones.

There's a whole bunch of _other_ cool things you can do with them, just
not Xtal sets.

--

Tim Wescott
Wescott Design Serviceshttp://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" was written for you.
See details athttp://www.wescottdesign.com/actfes/actfes.html
What would be a good Q range for AM broadcast & Shortwave bands.

If your _loaded_ Q is 100, the -3 dB bandwidth at 1 MHz would be 10
kHz (i.e. +/- 5 kHz from the carrier). The question is, what should
the _un_loaded Q be ?

If you have a full sized antenna, the signal strength would be
sufficient even with an unloaded Q of just 100-200.

At the middle of the HF band (10 MHz) a loaded Q of 1000 would be
required for a single station bandwidth and quite large helical
resonators would be required to get a usable unloaded Q without
damping the resonant circuit Q too much.

Paul OH3LWR

What Paul didn't mention is that the ratio between the unloaded and
loaded Q is pretty much the ratio between the energy coming from the
antenna and the energy that's wasted in the tank.

I.e. if your unloaded Q equals your loaded Q, then you're using up _all_
the energy in the tank, and there's none left over for your headphones.
If your unloaded Q is 200 and your loaded is the 100 that you'd want
for AM reception, you're using as much energy heating up your tank
circuit as you are using to drive your headphones.

I fully agree with this. With a Qu/Ql ratio 2:1 you end up with only 6
dB insertion loss.

However, the antenna capture area and hence the captured power (for a
constant field strength and hence constant power density) is inversely
proportional to the square of frequency (-6 dB/octave).

While a 1/2 wavelength dipole might be sufficient to feed a crystal
detector at 1 MHz, a 1/2 wave dipole at 10 MHz will unfortunately
produce only 1/100 the power.

Paul OH3LWR