If you assume,
(1) perfect transistors (no leakage or hFE variation with IC and VC)
(2) that both transistors have a constant VBE of 0.6V
from Kirchhoff, an equation for the circuit is:
18V - (18- [0.6 + 0.6V) = (iB * 3.3 * 10^6) + (iB * hFEp * 390)
Which reduces to,
16.8 = (iB * 3.3 * 10^6) + (iB * hFEp * 390)
Which can be transformed to,
16.8 = iB * (3.3 * 10^6) + iB * (hFEp * 390)
Thus,
16.8/iB = (3.3 * 10^6) + (hFEp * 390)
Thus,
iB = 16.8/ (3.3 * 10^6) + (hFEp * 390)
Where:
iB : base current of the first transistor in Amps
3.3 * 10^6 : the 3.3M resistor from the 18V supply line to the base of the first transistor
hFEp : product of the hFE of the first transitor and the hFE of the second transistor
390; 390 Ohm resistor in the emitter of the second transistor
Which, if my maths is correct, shows that iB is dependent on the only variable in the circuit, hFEc. This can also be determined by an initial analysis of the circuit.
From this you can calculate the output voltage (voltage across the 390 Ohm resistor) for any hFEc and thus any combination of gains for the first and second transistors.
spec
(crossed posts)