Why does my resistance in the LC tank circuit appear different?

Question

I created a parallel LC tank circuit in LTspice. I want to use the input impedance of the node "A" in the diagram below. For this purpose, I drew the V(a)/I(R3) cross-section, and the resonance peak appears around 20 kΩ.I don't understand why such a value is displayed.

I did not enter any parameters other than those shown in the diagram.

Answer 1

Here is the calculus I made with your circuit.

Made with Maple

If the resistors added serial (with L and C) are 10^(-12) Ohm, you can get this :

If the resistors added are 1 mOhm, you get this :

Maximum from 1 MegOhm ... to 331 kOhm ...

Here is the result with microcap v12

Answer 2

You have to remember you have have a very high Q circuit. So to hit that peak precisely you need to increase the number of point with a lot! A 100 points is way to small, at least 100000 to see a difference.

Answer 3

Set your sweep up so you can see if you have enough sampling points.
Something like:

.ac lin 1000 5.006meg 5.01meg

This will give a peak of about 500 kohms because LTspice inductors have a small loss built in to the models to add stability to the solver. Right+click on the inductor and set the series resistance = 0. Rerun the simulation and you'll get the expected 1 megohm resistance.