Solution 2:

(i) To calculate IE:

IE = [ VBB' - VBE(ECL) - (-VEE) ] / RE

= [ -3.457 - 0.75 + 4.5 ] / 0.32

= 0.916 mA

VIL = VBB - 0.05 = - 0.86 - 0.05 = -0.91V

VIH = VBB + 0.05 = -0.86 + 0.05 = -0.81V

(ii) VB,O = -2R (I1) - VD(ON) ; I1 = - [ VB,O + VD,L1(ON) ] / 2R

Also from the circuit;

I2 = - VB,O / R ; But IE = I1 + I2

IE = I1 + I2 = - [ VB,O + VD(ON)] / 2R - [ VB,O / R ]

Rearranging & solving for VB,O yields;

VB,O = [ -2R(IE) /3 ]+ [ VDL(ON)/3 ]

= [ -2(0.630)(0.916) / 3 ] + 0.75 / 3

= -0.134 V

VOR(OL) = VB,O - VBE,BO(ECL)

= [ - 0.134 - 0.75 ]

= -0.884 V

Now to determine the voltage at the NOR output ;

VB,N = VB,O + VDL1(ON) + I1(R)

Substituting for I1 yields;

VB,N = VB,O + VD,L1(ON) - { [ VB,O + VD,L1(ON) ] x R} /2R

VB,N = [ VB,O /2 ] + [ VD,L1(ON) / 2 ]

= [- 0.134 / 2 ] + [ 0.75 / 2 ]

= 0.308 V

Hence the NOR output voltage is;

VNOR(OH) = VB,N - VBE(ECL)

= 0.308 - 0.75 = - 0.442 V