Solution to problem 2 :

(i) To determine Icc(OH) :

When one or either of the inputs are low, the output is high :

(a) Current flow through IRB :

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

= { 5 - 0.7 - 0.5 } / 37 k = 0.102mA

Icc(OH) = IRB(OH) = 0.102mA

 

(ii) To determine Icc(OL) :

When both the inputs are high the output is low :

(a) Current through IRB :

IRB(OL) = { VCC - 3 x VBE(HARD) } / RB = { 5 - 3 x (0.8) } / 37K = 0.072 mA

(b) current through IRCS :

IRCS(OL) = { VCC - VCESB(HARD) - 2 x VBE(HARD) } / RCS

= { 5 - 0.5 - 2 x (0.8) } / 50 = 0.058 mA

(c) Current through IRC :

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

= { 5 - 0.5 - 0.8 } / 14 K = 0.2643mA

ICC(OL) = IRB(OL) + IRCS(OL) + IRC(OL) = 0.3925mA

(iii) The average power dissipation taking 50 % output high & 50 % output low :

Pavg = { ICC(OL) + ICC(OH) } x Vcc / 2 = { 0.3925 + 0.102 } x Vcc/ 2 = 1.236 mW