Regarding Rp of an LC
 

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

Regarding Rp of an LC
 

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

Regarding Rp of an LC
 

"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

Regarding Rp of an LC
 

"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"

Regarding Rp of an LC
 

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

Regarding Rp of an LC
 

"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