How can I simulate the Quiescent Point for a BJT, let's say 2N2222 in LTSpice?

I have this attached file but I can't remember how to do it so that I can verify my calcs.

Any help?

 

In the simulation command window, instead of doing a Transient simulation do a DC op pnt simulation.

Note that the bias point for that circuit is strongly dependent on the hFE gain of the transistor which typically can vary over a range of 3 to 1 or more between different transistors of the same type.
You should use a bias configuration that uses some negative feedback so the bias point only slightly varies with hFE.
One way to add some negative feedback is to connect Rb to the collector instead of to Vce.

 

Oh sorry, I thought I uploaded the correct circuit.

And for now, It's not an option to change the circuit. The purpose is actually to verify the stability of this setup.
The correct circuit is the one attached now!

 

So, do you still have a question?

 

So, do you still have a question?

Hi...

Not about that one...
However I have other questions.

I'm now testing a Colector-Base Bias and our teacher is asking us to calculate Rb so that Quiescent Point lands in half the Load Line!

Do you understand what I mean?

Rc=2k2Ω
Vcc=15V
Vbe=0.7V
β=200

I have to calculate the Load Line using the equation Vcc = Rc(Ib+Ic) + Vce <=> Vcc = Rc(Ic/hFE + Ic) + Vce.
Then we find Vce for Ic = 0 and after this we find Ic for Vce = 0.

I did this math and got the followig results for Load Line:
Ic = 0 => Vce = 15V
Vce = 0 => Ic = 6.784mA

Then, half of the Load Line is PFR(7.5V, 3.792mA)

With this values I can find Rb using the Entry net equation which I consider it to be Vcc = Rc(Ic + Ib) + RbIb + Vbe
Replacing Ib by Ic/β I find Rb = 400.96kΩ

If my calcs are correct, where can I find this point (PFR (7.5V;3.792mA)) in the attached plot???

 

But if Ic_max = 6.784mA than Icq = Ic_max/2 = 3.392mA

If my calcs are correct, where can I find this point (PFR (7.5V;3.792mA)) in the attached plot???

Simply plot a load line for Vcc = 15V and Rc= 2.2kΩ
(15V-V(col))/2200ohm

 

But if Ic_max = 6.784mA than Icq = Ic_max/2 = 3.392mA

Simply plot a load line for Vcc = 15V and Rc= 2.2kΩ
(15V-V(col))/2200ohm

I couldn't get that result... Never mind. I did it by hand!

 

But if Ic_max = 6.784mA than Icq = Ic_max/2 = 3.392mA

Simply plot a load line for Vcc = 15V and Rc= 2.2kΩ
(15V-V(col))/2200ohm

Can you show me your asc file?

 

Here you have. And do not forget to set both, right and left vertical axis at the same scale.