Experiment No.6, 7, 8 and Transistor. (part3)

Experiment No.8
Vary the base resistor and measure changes in voltage and current for Vce, Vbe, Ic and Ib. Then Plot a load line.

 The following circuit is set up on a bread board: 

With LED and transistor stay the same, the experiment is to obtain a variety of results(Vbe; Vce; Ib; Ic) from a variety of Rb( ranged from 2k to 1M). Hence, plot a load line so we can determine the switching and current amplifying characteristic of this transistor.

Vbe stabilizes from 0.6-0.8 indicating a behavior similar to a forwarded rectifying diode.
 +At low Rb values like 2k or 47K or even 1k (tested); Vce are all very low and lower than Vbe; indicating that C-E junction is switched on and connected nearly as good as a conductor. This is because as Rb is low( in this case the following load line graph will show that from 2k to 47k the transistor is in the saturated region) hence the base current is large enough to eliminate the depleted margin between base n emitter, making it easier for the current from collector to emitter. 
Notice that even though the 47k is proportionally bigger than 2k but the amount of Amplified Ic is about the same (7-8mA), this tells me that 8mA might be the practical amplifying limit for this transistor. But the thing is, 47k and 2k are both near the margin of base current where it kinda breaks through the breakdown point of Vbe on a B-E diode(refer to diode forward Voltage drop break down point), allowing the best current amplified and only results in minimum voltage drops( Vce).
+At high value of Rb (220k; 330k; 1M) we don't see Vce really low anymore, they are now significantly big, and bigger than any of the diodes forward Vd( LED n B-e diode). This is because Rb is too large, its impact on eliminating the depleted region B-E is small that Ib only make the base a little bit more positive than the emitter. As a result, the amplified current is rationally smaller and it takes more Voltage drops to amplify.
+ From Rb= 47k to 1M, the Ib value seems unreadable on the multimeter, hence specific result can be calculated for need of determining the Beta value.

The Load line:

 The graph tells me that when operating @ Rb = 220k is the lowest that the transistor doesn't seem to be in the Saturated REgion anymore. So i guess where it starts is about 100k and above the transistor will not be so efficient as it starts to draw more than 0.3 to wayy more than 2.0 V on Vce. And while being at it, the Ib is quite low and it doesn't make positivity much of the base to the emitter, hence it doesn't help Ic much.
 Lets believe that in the saturated region the beta is maximum. Then: Beta= Ic/Ib.
Rb=2k; Rc=470; V=5V; Vbe=0.8V. Vrb=V-Vbe= 5- 0.8= 4.2V=> Ib=Vrb/Rb= 4.8/2000=2.4mA.
Beta is Ic/Ib= 8/2.4=3.33
 On the other hand:
 Rb=330k: Vbe=0.69V. Vrb=5-Vbe=5-0.69=4.31V=> Ib=4.31/330000= 1.3e^-5 approx~ 13 uA.
So the above statement about "in the saturated region the beta is maximum" is not entirely true. I reckon its more complicated than just Ic/Ib.