Answers to problems in Tutorial 9

1. 0 for Vin > Vdd-Vt
   (Vin-Vt)-√((2*Vin-Vdd)*(Vdd-2*Vt)) for Vdd-Vt > Vin > Vdd/2
   (Vin+Vt)+√((Vdd-2*Vin)*(Vdd-2*Vt)) for Vdd/2 > Vin > Vt
   Vdd for Vt > Vin
   Input has to be Vdd/2 for both nMOS and pMOS to be in saturation.
2. gm of nMOS = gm of pMOS = 10mS. Slope = -1000.
3. (b) A′C′(B′+D′)
   (c) A′B′C′+D′
   (d) AB′+A′B
   (e) AB+A′B′

Answers to problems in Tutorial 10

1. 13.2 milli-Weber.
2. 1.2 Tesla.
3. 0.33 Ampere.
4. 10.2 Amperes.