Solution 4:

(i) To determine IIL:

IIL = [VCC - VBE,I(SAT) - VCE,O'(SAT)]/RB

= [ 5 -0.8 - 0.2 ] / 3 = 4/3 = 1.33 mA

(ii) To determine IOL :

IOL = IC,O =ßF.IB,O

IB,O = IE,S - IRD

IRD = VBE,O(SAT)/RD = 0.8/1 = 0.8mA

IE,S = IC,S + IB,S

IC,S = IRC = [VCC - VCE,S(SAT) - VBE,O(SAT)] / RC

= [5 - 0.2 - 0.8]/1 = 4 mA

IB,S = IC,I = ( ßR + 1 ) IRB

IRB = [5 - 0.7 - 2 x 0.8 ] / 3 = 0.9 mA

IB,S = ( 0.3 + 1 ) x 0.9 = 1.17 mA

IE,S = IB,S + IC,S = 1.17 + 4 = 5.17 mA

IB,O = IE,S - IRD = 5.17 - 0.8 = 4.37 mA

IOL = IC,O = ßF. IB,O = 50 x 4.37 = 218.5 mA

(iii) The maximum fan-out N = IOL / IIL

N = 218.5/1.33 = 164 ( rounded off)