Regarding Rp of an LC

On the webpage

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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
Reply to
amdx
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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

Reply to
o pere o

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.

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Ben's index page;

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Thanks, MikeK

Reply to
amdx

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"

Reply to
amdx

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

Reply to
o pere o

Hi Pere, Found this page, has a lot of info, some is a rehash of Ben's info but there is a lot here.

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MikeK

Reply to
amdx

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