So let's simplify the original circuit again. I've replaced the diodes this time with voltage sources to approximate ideal diodes in the forward active region. And look at the results. The current gain nearly equals β, but the current is also near enough to saturation. So we're turning the LED fully on but using very little base current to do so. However, we don't have to tweak anything to optimize this; using a different transistor (with a unique β) and you'll still see a similar current thru V1. Using a 2N2222 yielded 2.606mA, a difference of about 10uA. Changing the forward voltage drop of V1 and V2 will only adjust the "saturating" current thru V1. It's like the circuit is self adjusting. Let's look at the characteristic curve and load line:
So the Y-intercept of the load line is the theoretical max current of the circuit and is given by Imax = (Vcc-V1)/R1 = 3.3mA. The X-intercept is the maximum voltage that can be dropped across Q1 for the given circuit: Vmax = Vcc-V1 = 3.3V. I put the cross hair as close as possible to the intercept point, and sure enough, we see our 2.6mA from the simulated circuit. While our operating point is somewhat close to the saturation region, the gain indicates that there is very little saturation occuring. Again, changing the transistor or voltages of V2 & V3 will give similar results.
At this point I've done a few examples and a bunch of math and analysis, but why again? Well initially, our simulated LED's were behaving either like the resistors or like the voltage sources depending on the particular part that was used. The question is why do they operate differently?
