Or am I missing something here ?

 

This looks to be a reasonably straight-forward small-signal analysis problem. I could be wrong since I haven't actually worked through it.

First, find the DC solution. Then find the linearized small-signal resistance and inductance at that DC operating point and use that to find the AC response. The total response is then just the sum of the two.

Your DC solution looks correct, namely that the resistor current is 3 mA.

So, what is the small signal resistance and inductance at that point?

 

This looks to be a reasonably straight-forward small-signal analysis problem. I could be wrong since I haven't actually worked through it.

First, find the DC solution. Then find the linearized small-signal resistance and inductance at that DC operating point and use that to find the AC response. The total response is then just the sum of the two.

Your DC solution looks correct, namely that the resistor current is 3 mA.

So, what is the small signal resistance and inductance at that point?

Inductance in DC is 1 mH and resistance is 2k. But I wanted to use Norton Thevenin here.
Because in this task I had an option to use Norton/Thevenin here so I was curious.

 

Inductance in DC is 1 mH and resistance is 2k. But I wanted to use Norton Thevenin here.
Because in this task I had an option to use Norton/Thevenin here so I was curious.

In order to develop either a Thevenin or Norton equivalent circuit at the operating frequency, you need a linear inductance and capacitance. To do that, you need to determine the small signal value of your nonlinear components.

The inductance and resistance are not a function of frequency, but rather of current. Since it is nonlinear, you have large signal and small signal values.

I think, but am not sure, that you are referring to the small signal values. Show how you got each of those values.

 

In order to develop either a Thevenin or Norton equivalent circuit at the operating frequency, you need a linear inductance and capacitance. To do that, you need to determine the small signal value of your nonlinear components.

The inductance and resistance are not a function of frequency, but rather of current. Since it is nonlinear, you have large signal and small signal values.

I think, but am not sure, that you are referring to the small signal values. Show how you got each of those values.

For DC analizys I've calculated Jn which I think is correct (Norton equivelant).
For Thevenin I've found an error so I came up with solution that it is Norton impedance hence Rnorton = infinite.

So I already have Norton DC equivelant with impedance. And from this equivelant I can calculate dynamic resistance and dynamic inductance. This is how I did.

But I wanted to attach Norton AC, but I wasn't sure if the impedance was also infinite. Because if I had to calculate norton impedance by using calculations not by a trick like I did then I would have a lot of problems.

 

Forget about trying to combine the Thevenin (or Norton) equivalent circuit at DC with the one at AC into a single circuit. You can't if they contain nonlinear components because Thevenin/Norton equivalents are only valid for linear circuits.

What you CAN do is write all of the voltages and currents as the sum of a DC signal and a small AC signal and use superposition to SEPARATELY arrive at the DC and AC components and then add them together.

So you've gotten your DC components and you've gotten the dynamic component values at the DC operating point for your nonlinear devices.

Now draw the circuit that applies to your AC analysis.