I'm self learning electronics and trying to wrap my head around its characteristics. I was looking at this source. Regarding the active and saturation region graph interpretation on that page, I would like to confirm if my thinking is correct. From that graph, it states that the ratio between i_c and i_b is inconsistent at V_ce < 0.7. Beyond the saturation region, the relationship between i_c and i_b is now constant. We call that beta (β). The breakdown region is when the voltage V_ce reaches (or goes beyond) the voltage V_cc. From that link: "Suppose VBB and VCE are set to produce a current iB > 0 and VCE=0 V. Then both BE and BC junctions are forward biased. The saturation region corresponds to the case where both junctions are forward biased." Under what condition would V_ce equal to 0V while in the saturated region besides when i_b = 0 and i_c = 0? Or is it just saying that the starting position of V_ce=0V and it is expected for i_b to increase? I gathered that the cutoff region occurs beyond the midpoint of 0 volts and where V_ce=V_cc. I don't understand the concept behind the cutoff region. I would like a clearer explanation on what is the cutoff region. It says the cutoff region is when i_b = 0, but what is i_c? I think I get tripped up over the different values of i_b in that graph and its relationship with the cutoff region, whether those values are considered in the cutoff region or not. Another question: I was doing a few [URL="http://www.wisc-online.com/Objects/ViewObject.aspx?ID=SSE6104]practice problems[/URL] I found online on transistors where they say to assume that i_c = i_e (or at least, approximately equal). When I was doing those problems I initially made the mistake of calculating i_c by doing i_c = V_cc/R_c, thinking that i_c would equate to the current flowing from V_cc when in actuality, i_c = i_e. I learned that I had to find the current flowing through i_e to get a correct value for i_c. Why is this the case? I would have thought because the base acts as a switch by opening up contact between the collector and the emitter that the current from the emitter, i_e would equal to the current flowing to the collector, i_c based on using V_cc to calculate for i_c using ohm's law. I also figured that at the linear region of an npn transistor, the emitter-base region goes from forward bias (sat region configuration) to reversed bias: V_Rc + V_c = V_cc. Also, V_Re + V_be = V_b where V_be = 0.7 V. Therefore the voltage drop from the start of the emitter to ground is equal to the voltage V_bb because of forward bias from the base to the emitter. Thank you everyone.