Solution 1:

(i) To find VIL and Vo(IL) :

VIL = [2Vo - Vdd + Vt,P + (KN/KP).Vt,N] / [ 1 + (KN/KP)]

VIL = [2Vo - 4.75] / 2.25 -----------------------> 1

Vo = [2.25VIL + 4.75] / 2

The quadratic expression :

(KN/2)[(VIL - Vt,N)^2] =

KP[ ( Vdd - VIL + Vt,P)(Vdd - Vo) - (( Vdd - Vo)^2)/2 ]

substituting the values into the expression and reducing;

(5VIL^2) + 30VIL - 55 = [ 8Vo + 8 VoVIL - 4 Vo^2]

Substituting for Vo in the above expression and solving

for VIL;

VIL = 2.016V

Vo(IL) = 4.643V

 

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(ii) To find VIH and Vo(IH) :

VIH = [Vdd + Vt,P + (KN/KP)( Vt,N + 2Vo )] / ( 1 + (KN/KP)]

= [5 - 1 + (100/80)( 1 + 2Vo )] / [ 1 + (100/80)]

VIH = [5.25 + 2.5Vo] / 2.25------------------------------------->1

Vo = [2.25VIH - 5.25] / 2.5

The quadratic expression :

KN[ (VIH - VTN).Vo - Vo^2/2 ] =

(KP/2)[ (Vdd - VIH + Vt,p)^2]

Substituting the values into the expression & reducing;

2.5[(VIH.Vo - Vo - (Vo^2/2)] = [ 16 - 8VIH + VIH^2]

Substituting for VIH in the above equation & solving for Vo:

Vo = 0.3871V

VIH = 2.763V

(iii) To find VM:

VM = [Vdd + Vt,P + Vt,N.Sqrt(KN/KP)] / [ 1 + Sqrt(KN/KP)]

VM = [5 - 1 + 1.Sqrt(100/80)]/[1 + Sqrt(100/80)]

VM = 2.416V

(iv) To find VOH & VOL :

VOH = Vdd = 5V

VOL = gnd = 0V

 

Voltage Transfer Characteristics