Simple equivalent resonant circuit confusion.

[Rene]

Hello,

I am reading a chapter in a RF Circuit Design book that talks about resonant circuits. In that chapter, the book author explains what happens when you apply a load to a resonant circuit. See attached picture.

In the picture, you can see that the author states that the resonant circuit sees RS in parallel with RL as its true load. Why is that? How does that make any sense? As far as I can see, RS and RL are in series not parallel.

I should point out that I think he is talking about what happens during resonance, not sure if this thing changes when the circuit is not in resonance.

Any help is greatly appreciated.

Thanks.

[Rerouter]

Looks almost like a therivin termination calculation to me, e.g. how you can terminate a circuit with a pull up and pull down.

[CatalinaWOW]

It might make sense to you if you redraw the circuit with the source and load on the same side of the LC pair.  Additionally recognize that the ideal signal source (the little circle with one cycle of sine) has zero resistance.  Any source resistance is represented by the Rs.  Which is effectively connected to ground through that funny little circle.

This can be confusing because the same little circle is treated as open when doing a norton equivalent (which might make the parallel connection more obvious).

[ejeffrey]

Yes, one way to look at this would be to transform the voltage source and its series resistance into its norton equivalent -- a high impedance current source in parallel with a resistor Rs.  Then, Rs is obviously in parallel with RL, so you can use the parallel resistance formula. 

[IanB]

If we say that the voltage source is ideal and has no internal impedance then R_S represents the entire source impedance. If we now replace the voltage source with a short circuit and tie all the grounds together, than R_S appears in parallel with R_L from the perspective of the LC tank circuit. Wouldn't that be the appropriate view of the situation?

[T3sl4co1l]

Yeah, all voltage sources can be redrawn as short circuits, and all current sources can be redrawn as open circuits, for purposes of dynamics.

This is because, for a given perturbation of the system, only a frequency at the same frequency as the source could have any interaction with it; for all other frequencies, it must act as a short circuit.  Unequal frequencies are orthogonal, so there's nothing to equate with, no way to interfere.

 You can imagine, if a stimulus were applied to the circuit some other way (say through a high value resistor, or a magnetically induced current*), then the circuit must respond the same way (i.e., it exhibits the same dynamic equations).

It's only a small stretch further to say: well, if it obeys the same dynamics, and the circuit obeys superposition (it does, it's linear), then it must also be true for equal frequency (and phase) as well; and so, the circuit will respond that way to the original source exactly the same as it does to another stimulus.

*Which, mind that these are simply other kinds of sources in the equivalent circuit, and therefore are still within this rule.  I can't make a non-circular argument here, because if it's a LTI source, it must fit into the theorem I'm trying to explain...

Or, to put it more directly: if the superposition theorem applies (i.e., you can solve for the V, I in a circuit by setting all but one independent sources to zero, cycling through each one at a time, recording the node/branch V/I conditions from each, then adding them all up), then the circuit must have the same dynamics whether the source is zero or nonzero.

Tim

[Ratch]

Rene,

Let's do a node analysis and see what is happening.  Vs is the voltage source and V is the node where all the components come together.  Setting up the node equations below and solving for V, we see that the resistors act as though they are in series.  That makes sense because a perfect capacitor and inductance in parallel present a infinite impedance.  Notice that at the resonant frequency, the inductive reactance (j X) and the capacitive reactance (-j X) have the same "X" value, but opposite "j" values.  This simplified the node analysis.

Ratch