Solutions to Homework Chapter 9-1

(~04/30/2004)

9.3

Since v₁ = - 1.6 V > than V_REF = - 2 V

Transistor Q₁ is ON and conducts all the I_EE current

Hence, I_C1 = I_EE = 5mA

V_BE2 = -v₁+V_BE1 + V_REF = 0.3 V

Hence Q₂ is almost cut-off

Hence I_C2 = 0

V_C1 = -I_c1*R_C = -5mA * 350ohm = -1.75V

V_C2 = -I_c2*R_C = -0mA* 350ohm = 0 V

9.26

Five input ECL OR gate

9.27

Four input ECL NOR gate