Solution to problem 3 :

(i) The critical voltages are :

VIL = 1.4V

VIH = 1.7V

VOH = 4.3V

VOL = 0.5V

The critical voltages are the same as that of ALSTTL.

(ii) To calculate ICC(OH) :

IRB(OH) = {VCC - VBE(FA) - VCE(HARD)} / RB

= { 5 - 0.7 - 0.5 } / 10K = 0.38mA

IRBP1 = { VCC - 2(VSBD(ON)) } / RBP1

= { 5 - 2(0.3) } / 50K = 0.088mA

ICC(OH) = IRB(OH) + IRBP1 = 0.468 mA

(iii) To determine ICC(OL) :

IRB(OL) = VCC - 3(VBE(HARD)) / RB = 5 - 3(0.8) / 10K = 0.26mA

IRBP1 = 0.088mA

IRC(OL) = VCC - VCES(HARD) -VBE(HARD) / RC

= 5 - 0.5 - 0.8 / 2 = 1.85 mA

ICC(OL) = IRB(OL) + IRBP1 + IRC(OL) = 2.198 mA

(iii) the average power dissipation is :

Pavg = { ICC(OL) + ICC(OH) } x VCC / 2 = 6.665mW

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