Regarding BJT, forward active mode of operation i don't seem to get something. I will use NPN type. In the forward active regime, Vbe>= 0.75 V, so electrons flow from the emitter through the base .Ok, what i do not understand is why if Vbc <= 0.75 V there is a current Ic = Beta*Ib, how come electrons flow from emitter to collector if that part of the transistor is "closed" (Vbc <= 0.75)
simulate this circuit – Schematic created using CircuitLab
The thing you need to know is that the base is very thin/short.
So the electrons travel from emitter to base "thinking" it's a standard forward PN junction. However, like I said, the base is very small so many of the electrons end up at the collector side of the base and suddenly find themselves inside a big-#ss collector :-)
So there they all combine and form the collector current. Only few electrons make it out of the base without ending up in the collector, this explains the small base current.
So the base-collector junction itself isn't operating as if it is in reverse mode. It's not the junction causing the current, it is the base that is injecting carriers into the collector.
It's a bit unclear what you are really asking, but it seems the answer is "Because that's what transistors do". For more details, see semiconductor physics.
Also, you should be thinking about the current thru the base, not so much the B-E voltage as driving things. B to E looks like a diode to the external circuit, so will have about 700 mV drop depending on current. But, the function of voltage to current is quite steep, and current is ultimately what matters anyway.
V2 in your schematic should be a current source, not a voltage source. Let's say you set it for 100 µA and the transistor has a gain of 50. That means the collector can sink up to (100 µA)50 = 5 mA. If R1 were 300 Ω, then it would drop (5 mA)(300 Ω) = 1.5 V. The collector would therefore be at 3.5 V.
The above example was assuming the transistor had a gain of exactly 50 at the operating point we chose. Real transistor gains vary widely. The datasheet generally specifies a minimum, but no maximum. It's not unheard of for actual parts to have 10x more gain than the minimum guaranteed value.
Unpredictable gain is one reason transistors used in linear operation are biased with some feedback. The actual C-E voltage is used to adjust the bias current to keep the collector voltage within a useful range over the possible range of gains from part to part.
Forward active region biases the device so that it doesn't matter what value Vbc is so long as it's within electrical tolerances and less that 0. The Vbc voltage has nothing to do with determining current through Ic in forward active.
As for "close" I don't understand what you mean. Current flows like that because we forced it to do so. The BJT is constructed so that in an NPN with the Vbe forward biased and the Vbc reverse biased that current flows from the collect through to the emitter with a small leakage current added by the base.
