EECE 310 Exam, Spring 2008

Test 6 (Nov 19, 2007) When vDS of an E-NMOS transistor is increased from 2 V to 4 V, with vGS maintained constant, iD is multiplied by a factor of 1.5. If Vt = 1 V, determine the valid value, or values, of vGS. Solution: The transistor cannot be in saturation in both cases, because of the increase in iD. Hence it must be in the triode region at vDS = 2 V, but could be in the triode region or possibly in saturation at vDS = 4 V. (a) Assuming that the transistor is in the triode region in both cases: , and . Dividing: , or , and vGS = 6 V. (b) Assuming that the transistor is in saturation when vDS = 4 V: , and , as before. Eliminating iD: . This gives: V. At vDS = 2 V, , which means that the transistor is in the saturation region, contrary to what was assumed to begin with. Hence should be discarded. Now we have to check if vDS = 4 V and gives operation in the saturation region as assumed. But which means that the transistor is in the triode region, contrary to what was assumed in this part. Hence, the only valid value of vGS is that obtained in (a).