GOOOOOOD ! for the schematic and logic.

But you went wrong with the maths! Re-read your assumptions in your post #1.

hie is 1.5KΩ, so Rin = 0.5 + 1.5 = 2KΩ.

Additional note: note that the current source given by the BJT (assumed linear behaviour) has no influence on Rin because of Ce. This is important to retain.

If there was no Ce, Re and the current source would have impact --> Additional equivalent resistor until ac ground = Re*(1+hfe). So, in this case, total Rin would be Rb + hie + Re*(1+hfe).

Hope you understood the importance of making simple drawings in electronics.

Sorry, i'm making too many mistake now because when I was replying to this thread I was taking a break from another tutorial sheet, trying to solve too many problems at once.

Yes you're right about the schematics, tbh I already should know to redraw the equivalent circuit but for some reason I always try and do it the quick way. They say insanity is repeating the same thing and expecting a different result lol.

 

LvW, where do you see the orientation of the current source in the schematic? Is it from the Vo arrow?

Efron - sorry.
I have overlooked the short circuit across RE because it is a bit uncommon to draw a resistor and to place a wire across it. But, of course, both diagrams are equivalent.