EECE 310 Exam, Spring 2008

Test 8 (Dec 3, 2007) It is given that Q1 and Q2 have kn = 2 mA / V2, Vtn = 1 V, Q3 has kp = 1 mA / V2, Vtp = -1 V, and for all transistors. Determine VB so that Q3 is at the edge of saturation. Solution: Since Q1 and Q2 are identical and carry the same current they must have the same VGS, assuming that they are both in saturation. If vGS2 = 2 V, VDS1 = 4 – 2 = 2 V, so that Q1 is in fact in saturation, since 2 > 2 – 1. It follows that mA. For Q3: , or (1) If Q3 is at the edge of saturation, VSD3 = VSG3 – , or: VSD3 = (6 – VB) – 1, or 5 – VB = VSD3 (2) Equations (1) and (2) and i1 = 1 mA are all that can be derived from the transistor equations. Since three unknowns are involved, a third equation must be derived from KCL and KVL. It is seen that: i1 = 1 = i3 + , or (3) Eliminating 2i3 between (1) and (3), . Adding to (2): = 0. Factoring, . This gives VB = 7 V, or VB = 4 V. VB = 7 cuts off Q3 and is discarded. Hence, VB = 4 V, VSD3 = 1 V and VSG3 = 2 V, which confirms that Q3 is at the edge of saturation. Moreover, VDS2 = 6-1 – 2 = 3 V, which means that Q2 is in saturation as assumed. 6 V Q1 2 k Q3 VB Q2 4 V 2 V 6 V Q1 2 k Q3 VB Q2 4 V 2 V i1 i3