Solution to problem 3:

(i) For NOR output high :

(a) IRE = {(VBB - 0.75) + (VEE)}/RE

= ( -1.175 - 0.75 ) + 5.2/1.18 = 2.775V

(b) IRDN = VNOR(OH) - VEE/ RDN

= -0.75 + 5.2 / 3 = 1.483mA

 

(c) IRDO = 0 - IRExRcr - VBE(ECL) - VEE / Rcr

= 0 - 2.775x0.3 - 0.75 + 5.2 / 3K = 1.206 mA

INOR(HIGH) = 5.464 mA

(ii) for NOR output low :

(a) IRE = (VIH - VBE(ECL)) - VEE/RE = ( -1.225 -0.75 + 5.2 )/1.18 = 2.817mA

(b) IRDN = 0 - IRExRci - VBE(ECL) - VEE / RDN

= 0 - 2.817x0.29 - 0.75 + 5.2 / 3 = 1.21 mA

(c) IRDO = VOR(OH) - VEE / RDO = -0.75 + 5.2 / 3K = 1.483 mA

INOR(LOW) = 5.511mA

(iii) Average power dissipation :

Pavg = INOR(LOW) + INOR(HIGH) x VEE / 2

= 28.535 mW