Let's start by looking at the circuit in a more traditional manner by using passive loads. Above I've shown the simulation results using 1K resistors instead of LED's. Of course, this circuit saturates rather dramatically. We only see a gain of 1.6. Using Ohm's law and assuming a Vb for the transitor of about .7V, we find that R2 can only draw around 700uA more than R3 before Q1 saturates. Our above simulation confirms the estimate. Why is this? Well if we remove R1 altogether (it doesn't change the circuit in any significant way) we can derive the following equations:
I(R2) = (Vcc-Vc)/R2 with ideal saturation: I(R2) = (5-0)/1000 = 5mA
I(R3) = (Vcc-Vb)/R3 with subsitutions: I(R3) = (5-.7)/1000 = 4.3mA
Current Gain = I(R2)/I(R3) = 5/4.3 = 1.163
The approximate calculations match fairly close to the simulation results. It turns out that Vb is greater than the assumed .7V diode drop due to the fairly large base current. Also, Vc cannot be pulled all the way down to 0V. Even so, the number differ only slightly. And besides, the ultimate point is that any additional current thru R2 is going to be related to the difference in voltage between Vb and Vc, essentially the diode drop from base to emitter. Typically we'd push a much smaller current through the base by using an appropriate resistor so that we can drive a large load from some no so tough part, such as a microcontroller. So by using identical resistors on the base and collector, we don't really see much advantage. But we do with the LED's.....
