Question

An engineer needs a circuit to perform a logical function but he only has an integrated circuit with four logical operators with output values of 5 V and 0 V respectively. The engineer is lucky, the input signals of Va, Vb and Vc have voltage high and low of 5 V and O V respectively. He also found resistances in his drawer, all the same. The circuit he built is shown below. All of the resistances have the same value and the reference voltages of 3 V and 2 V were obtained through voltage divisors that are not shown for simplification. What logical function is being done by this circuit? Please justify the answer. 3V VB Vc-wo Intl 3 V 3 V- s Vo 2VG+ Please provide clear steps of the process and proper units. Thanks.

Answer 1

R R VA + Ve -R - 3 v VB R R Vea NS Vs R + 3v + Vr Vt vo + 3v av

Case I VA = VB=Vc=0 Vor = vy = vz = 0 since all logic circuit is a comberato circuit. If difference of input (utv) to positice then Vill output voltage will be SV atherecia It will be o Volt. Vp = 0 va 5 ve 2.5U (by voltage dicisin Rule - VDR) - 5 V V V t 2-5v Vo o Volt = low o

case I VA= 0 = VB, Vc = 5 Va = Vy = V2 = 0 Vp O VQ = 5 Vs = 5 Vr = Vt - Vo " = 5 Volt 3 righ Cease II VA=0 =Vc Vs = 5 Volt Vn 2.5V Vp o Volt 5 vy Vz 2.5 V = a 5 vo Us 2.5 V Vr 5V Vt 2.50 vo OV of & logic o

Case IV VA=0, UB = Ve=5 Va= 2.5V = vy 5 V V₂ = 2-5v by voltage dicussin Rule Vp 0 V P va 5 v = Vs 5V = ov Vr Vt = vo 5V logic high 'I' Case I VA =5, VB = Ve=0 - 2.5 V Vn- Vz Up OV O V Vo = ov, logic low 8 VQ - 5 V Vs = 2.5v kr 5V 2.5V Vt

Ceese T VA =Vc =5 VB=0 - 2.5 V Va Vz ou Vp ov vo 5V 5 = O Vs Vr Vt Vo 2 = 5 Volt , logic righ 'l' case VI VA VB = 5, Vc=0a 2.5V Vr=5 Vy = 5 Uz Up = 5V 5v Vs 2.5V Vo = ov, logic low o va z Vr 5V V 5V

Truth table VA ase VB Vc Vo Clogreal output) 1. O O >ov 99 O SU 3. O O ov su OV O su 6. OV O 7. ov O logic high flow logic high er 5V ov logic low 'o'