.step oct PsySc 1 64 2 would make the parameter PsySc take on values of 1, 1.414, 2, 2.828, 4,..., 45.26, 64. The 2 means 2 steps per octave, starting at 1, ending at 64.

.step dec PsySc 1 1000 2 would make the parameter PsySc take on values of 1, 3.1622, 10, 31.622, 100, 316.22, 1000. The 2 means 2 steps per decade, starting at 1, ending at 1000.

When you plot, you can elect to make either the X-axis or Y-axis log or linear.

Once the plot is up, you can elect to add none, one, or two cursors.

The main point I wanted to show you is that a .DC analysis is much more useful than a .TRAN analysis in this case...

Ok, 3 questions:

1 - Where does those values of 1.414, 2.828, etc comes from???

2 - Where does those values of 3.1622, 31.622, etc comes from???

3 - Also I would like to know if there is any tip that tells us when to use what type of simulation!

 

How's that?

Well, assume that the npn is off in this case -> Ib = 0. The current going through the two resistors would be 12v/(10k+400)=1.2ma. At that current, Vbe = 400 * 1.2ma = 500ma, not sufficient to turn on the transistor -> our assumption about the transistor being off is correct.

Isn't the voltage at the upper terminal of Rb2 the voltage divider between Rb1 and Rb2 using Vcc voltage??? Which is 400*12/10400 = 0.462V

Yes.

Then Ib = (Vbb-Vbe)/Rbb = (11.88-0.66)/9.901 = 1.133mA which matches LTSpice value!

Not sure how you got that but there is no way it is correct,

 

1 - Where does those values of 1.414, 2.828, etc comes from???

1.414^2 = 2; ...

2 - Where does those values of 3.1622, 31.622, etc comes from???

1.31622^4 = 3;

3 - Also I would like to know if there is any tip that tells us when to use what type of simulation!

If you know where spice's limitations are, you are an expert in it.

 

Ok, 3 questions:

1 - Where does those values of 1.414, 2.828, etc comes from???

Octave means value doubles: 1,2,4,8,...,16
Two steps per octave means the intermediate step must be √2 = 1.414 because √2*√2 = 2. If there were 3 steps per octave, the step would be cube-root of 2, and so on.

2 - Where does those values of 3.1622, 31.622, etc comes from???

Decade means value increases ten-fold: 1,10,100,1000, etc
Two steps per decade means the intermediate step must be √10 = 3.16122 because √10*√10 = 10. If there were 3 steps per octave, the step would be cube-root of 10, and so on.

3 - Also I would like to know if there is any tip that tells us when to use what type of simulation!

If you are interested in transient (time-dependent) behavior, use .TRAN
If you are interested in bias solutions, parameter sweeps, transfer curves, sensitivity curves, use .DC
If you are interested in frequency-dependent behavior, Bode plots, gain/phase plots, filter characteristics, use .AC

 

.....

Not sure how you got that but there is no way it is correct,

I've used the simplified circuit after having replaced Rb1 and Rb2 by it's equivalent: Rb1*Rb2/(Rb1+Rb2) = 10KΩ*1MΩ/(10KΩ + 1MΩ) = 9.901KΩ

And also Vbb = 1MΩ*12/(10KΩ + 1MΩ) = 11.88v

Then I used the Base net equation

Vbb - Rb*Ib - Vbe = 0
<=> Ib = (Vbb - Vbe)/Rb = 11.88 - 0.66/9.901 = 1.13mA

 

I've used the simplified circuit

You are solving for the case where Rb2=1M.

We were talking about the case where Rb2=400Ohm. Here is the exchange:

How's that? Isn't the voltage at the upper terminal of Rb2 the voltage divider between Rb1 and Rb2 using Vcc voltage??? Which is 400*12/10400 = 0.462V

Then Ib = (Vbb-Vbe)/Rbb = (11.88-0.66)/9.901 = 1.133mA which matches LTSpice value!

Try to solve for Ib in that case.