[SOLVED] Response of a parallel RLC circuit to an AC current?

[koushikr_in]

Hi,

I simulated a simple RLC parallel circuit using LTspice. The circuit diagram and results are attached. A current source (1 nA) connected across parallel RLC. In the AC response, I expect all the current (1nA) to flow through resistor (63 kohms). But I find only 180 pA flowing through the resistor at resonance. Why is this so?


Thanks
Koushik

 

[keith1200rs]

Probably because your frequency step is too large.

A couple of other points - the current you are using is very small and the default absolute voltage & current tolerances of Spice could cause problems. Simply use 1A for the current source.

Also, if you are trying to simulate reality you should put the resistor in series with the inductor.

Keith

 

[LvW]

Check the inductor model: Is it free of losses?

 

[keith1200rs]

Check the inductor model: Is it free of losses?

A good point.

Mind you, a poor choice of frequency step will give exactly that result.



Keith.

 

[LvW]

Yes Keith, you also have picked up a good point. Therefore, I always recommend to highlight the calculated points in the display in order to see if the resolution is good enough. On the other hand - everybody should see that something is wrong when he obeserves "edges" and other sharp transitions in the display. Something like that will never happen in reality (exceptions?).
LvW

 

[koushikr_in]

Hi everyone,

Thanks for your reply. I increased the resolution of my spice (10000 points per decade) and current source changed to 1A. I still find the current through the 68 k resistor less than 500 mA. This is independent of whether I add a resistor in series with the inductor or not. Why is it so? The inductor and the capacitor are ideal elements.

 

[FvM]

Atthough the inductor "loss" resistance has been suggested in the discussion by contributors, you should be clear about it's effect to drain part of the source current from the parallel resistor in resonance to feed the losses. Both of your plots have the 13 ohm series resistor. But with ideal L and C, this effect can't be seen. Thus I didn't understand which case you are describing. At least, you don't show a plot where it can be seen. My simple guess is, that you confused the simulation cases. Otherwise show the case, where the resistor current is considerably smaller in resonance, with ideal L and C elements.

 

[LvW]

... current source changed to 1A. I still find the current through the 68 k resistor less than 500 mA.

One of (possibly) several reasons for this "funny" behaviour may be the following:
For AC analyses, in LTSPICE the AC value is specified with it´s amplitude - in contrast to all other simulation programs.
You should check what parameter is displayed on the screen (rms or something else?).
Another hint: display the current through L||C together with the current through R.

 

[FvM]

You should check what parameter is displayed on the screen (rms or something else?).

LTspice displays magnitude rather than rms values. Because the shown plot has an axis extend to negative values, it's implicitely clear, that the plot style is "cartesian" (real/imaginary amplitude) with imaginary part deselected. So in the "ideal" resonance case, the current is equal to source magnitude. And that's what I see, using the originally posted circuit, with an appropriate frequency step.

 

[koushikr_in]

Hi all,

As suggested, I tried removing the resistor in series with the inductor. Now the simulation shows the entire current (1 A) flowing through 63 k at resonance. I guess the resistor in series to the inductor contributes to the loss by decreasing the effective Q.