I am supposed to calculate the diode current when the switch is closed and also determine what region the transistor is working in when the LED diode is conducting current.

The way I tought to go about this problem is to first calculate the collector voltage so I can determine which region the transistor is working in.

First I need the base current:
Ib = 14.3 V÷15k = 0.95 mA

Now I can calculate the collector current using β:
Ic = Ib*β = 95 mA
I am not sure when I am allowed to use this formula because I read somewhere that this formula only stands true when the transistor is working in active region. I don't understand when I can use it and how I should know if the transistor is in active region. If someone could explain this it would be very helpfull.

Now I can calculate Vc:
Vc = 15 V - 2 V - Ic * 1.5k = -129.5 V
The 2 V voltage drop comes from Vd (diode
breakthrough voltage).

I do not how to procced from here. What does this tell me? My best assumption would be that the transistor is working in saturated region but that is just a wild guess and I don't even know how to use that information.

 

So, far so good. But now you need to think about the result you get -129.5 V and what this result tries to tell you.
And remember that in your circuit you have Vcc = +15V (no negative voltage) and II Kirchhoff's law always holds.

 

... Here is a table relating to the different transistor operating regions. It will be necessary to correctly calculate the required voltages.
https://learn.sparkfun.com/tutorials/transistors/operation-modes

 

So, far so good. But now you need to think about the result you get -129.5 V and what this result tries to tell you.
And remember that in your circuit you have Vcc = +15V (no negative voltage) and II Kirchhoff's law always holds.

That's exactly what I'm confused about. What does Vc = -129.5 tell me? Vce is less than Vces. So is the transistor in saturation or in active mode? How do I use that information to procced?

 

Is this negative voltage possible in your circuit? Can the transistor "create" negative voltage? If not, how can your transistor be in the active region?

 

Is this negative voltage possible in your circuit? Can the transistor "create" negative voltage? If not, how can your transistor be in the active region?

Transistor cannot create negative voltage. Can you explain to me how to see if the transistor is in active or saturated region and how to use this information? Sorry if I'm annoying I'm just really confused.

 

Is this help?
https://forum.allaboutcircuits.com/threads/bjt-saturation-question-s.81150/#post-577296
https://electronics.stackexchange.c...an-npn-bjt-in-saturation-region/276266#276266

 

Instead of calculating with the base current first, try calculating the collector saturated current of 8.5mA first. Then calculate the base current of almost 1mA.
Isn't it saturated?