I would like to clarify following detail I still not understand so far on the explanation what's the proper purpose of the coupling capacitor C1 in following receiver circuit is (found in http://makearadio.com/misc-stuff/regennotes.php ) but there are no explanations what finally the job of this coupling capacitor is):

enter image description here

The answers I got to my original question on this problem I asked a week ago, suggest in unison that this coupling capacitor C1 has an important effect on the Q factor of the L1-C2-tank, which in responsible for selection of desired frequency.

Since the goal of this L1-C2-tank is to be selective, it's Q should be relatively big, and seemingly (at least as far as I understood the answers in the linked discussion) the Q of the L1-C2-tank depends on this coupling capacitor C1.

I not understand it. Why the tank's Q depends on coupling capacitor? If we separate the L1-C2-tank from the rest of this circuit then it's just a parallel LC-circuit:

(note that the LC-tank in the first picture above there is no resistance R explicitely depicted, that's just for sake of brevity, a real coil has always a small resistant component R, which we should take into account if we want to calculate the Q of this LC-tank.

Now comes the point which confuses me: For a parallel LC-tank seemingly there is a closed formula for Q (see eg https://en.wikipedia.org/wiki/Q_factor#RLC_circuits). And the problem is that it seemingly not(!) depends on the coupling capacitor C1, instead only on L1 (+ it's resistant conponent R menstioned before) and C2? So inly on component which belong to the LC-tank, not external components like the cap C1.

Problem: So in summary we deal with two statements:

(I) (based on answers from original question ) "The Q factor of the L1-C2-tank depends explicitely on capacitor C1 as well"

(II) (based on the linked formula from wikipedia) "The Q of this LC-tank depends only on C2, L1, R

Well, obviously these two statements contradict to each other. What is the problem here and how to resolve it? Did I quote one of them in wrong way?


1 Answer 1


The antenna can be modeled as a resistor in series with a capacitor or inductor depending on its length.

If the antenna resistance is coupled too strongly to L1/C2 it will reduce the Q of the tank.

Radiation resistance

  • \$\begingroup\$ Yes, good note, I updated the image by the model circuit of the antenna & added a resistence component to the coil L1 since it's naturally to assume that every coil has real resistence; I guess that in the linked page where I found this circuit the author was to lazy to add this R and just assumed that it should be clear to the reader. \$\endgroup\$ Nov 21, 2021 at 15:53
  • \$\begingroup\$ But I think your answer missing slightly the point (I update the question to emphasise more precisely the problem) My question was about how to resolve the apparent contradiction of the two statements: (I) that the Q of the L1-C2-tank depends explicitely on capacity C1, which is not a part of the L-tank (that's what several answers on the linked question also suggest) AND (II) the explicite formula from wiki (en.wikipedia.org/wiki/Q_factor#RLC_circuits) for parallel LCR-circuits (which is of course our L1-C2-tank). \$\endgroup\$ Nov 21, 2021 at 15:54
  • \$\begingroup\$ So seemingly (I) and (II) contradict to each other, since (I) suggests that Q of L1-C2-tank depends on external components (ie the capacitor C1), while (II) says no, the tank's Q depends only on components, which belong to the tank, namely R, C2 and L1. And that's exactly what confuses me. Do you see how this seemingly contradiction between (I) and (II) can be resolved? \$\endgroup\$ Nov 21, 2021 at 15:54
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    \$\begingroup\$ @katalaveino neither. Wikipedia gives the correct formula for the Q of the LC tank, but the Q of the LC tank doesn't matter, the Q of the whole system matters. \$\endgroup\$
    – hobbs
    Nov 21, 2021 at 16:07
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    \$\begingroup\$ @katalaveino it doesn't have dependencies, and no they don't imply that, it's just what you incorrectly inferred. Slow down and read what I already wrote. \$\endgroup\$
    – hobbs
    Nov 21, 2021 at 16:57

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