"o pere o" wrote in message
...
On 11/15/2010 03:49 PM, amdx wrote:
On the webpage http://www.crystal-radio.eu/engev.htm the author diagrams
a
crystal radio with Rp as the losses in L1 and C1. Then he goes further
and
loads L1C1 with an antenna, this reduces Q by 1/2.
He then adds we can calculate the value of Rp as follows: Rp=2.pi.f.L.Q
(Q
Unloaded).
My question; Why does he do all further calculations with Rp rather
than
1/2Rp, since the Q of L1C1 is
reduced to 1/2 by the antenna loading.
MikeK


From what I read, I guess:

1. He works out the optimum match between antenna and LC tank.
2. She investigates the optimum match between LC tank and load completely
unrelated to 1.

OTOH, I would be glad to find a reference on how to accurately derive the
proposed equivalent circuit of the diode detector at low signal levels.

Pere

Ya, that's what it looks like to me, I sent him an email, but it looks like
he is
not responding to any, his store is closed also, at least for now.

Here's a bunch of information about diode detectors in crystal radios.
I think the first one will answer your question, but Ben Tongue has great
info too.

http://www.crystal-radio.eu/endiodes.htm

http://www.crystal-radio.eu/endetunittest1.htm

http://www.bentongue.com/xtalset/4opd_xfr/4opd_xfr.html

http://www.bentongue.com/xtalset/6XlStSPS/6XlStSPS.html

http://www.bentongue.com/xtalset/7diodeCv/7diodeCv.html

http://www.bentongue.com/xtalset/8DetVIWG/8DetVIWG.html

http://www.bentongue.com/xtalset/10npddec/10npddec.html

http://www.bentongue.com/xtalset/17Is_n/17Is_n.html

http://www.bentongue.com/xtalset/28SqLDtL/28SqLDtL.html

Ben's index page; http://www.bentongue.com/xtalset/xtalset.html

Thanks, MikeK