Solution 1:

To determine VBB & VBB';

IE,S2 = [VD(ON) - VBE,S2(ECL)] / RE2

= [ 0.8 - 0.75 ] / 0.15 = 0.333mA

Assume that IR = IE,S2

=> VBB = VCC - IR(RSH) - VBE(ECL)

= 0 - 0.333(0.33) - 0.75 = -0.86V

VBB' = -VEE + VBE(ECL) + IR(R) + VBE(ECL) - VBE(ECL)

= - 4.5 + 0.75 + 0.333x(0.88) + 0.75 - 0.75

= - 3.457 V

VBB = - 0.86V ; VBB' = - 3.457V

---