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

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

"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

"amdx" wrote in message
...
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

I received a reply to my question from the webpage author.

"In the formule Rp=2.pi.f.L.Q , I use the Q of the coil in the right-side
part of circuit diagram 2.
This Q is halve the value of the coil Q in the left-side part of circuit
diagram 2.

I consider the right side part of diagram 2 to be a new unloaded circuit,
with it own unloaded Q.
With unloaded I mean here "not loaded with diode or loudspeaker".

Maybe one might say; it's not unloaded because the antenne is connected"
But as you see in circuit diagram 2, in the right side part, there is no
antenne anymore.
It's just a new LC circuit with it's own Q value, wich we are gonna load
with diode and speaker.

The calculated value of Rp is the value which the diode sees, in the
situation with the antenna connected"

On 11/15/2010 07:22 PM, amdx wrote:
"o pere 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



Thanks! I have also found that the Agilent datasheets provide a lot
of info, with some equivalent models in close agreement with your
links'.

Pere

"o pere o" wrote in message
...
On 11/15/2010 07:22 PM, amdx wrote:
"o pere 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



Thanks! I have also found that the Agilent datasheets provide a lot
of info, with some equivalent models in close agreement with your
links'.

Pere
Hi Pere,
Found this page, has a lot of info, some is a rehash of Ben's info
but there is a lot here.
http://www.klimaco.com/HAMRADIOPAGES/xtal_how_to.htm
MikeK