Download Quiz 2 Solutions for ECE65 (Fall 2009) - Problems 1, 2, and 3 - Prof. Farrokh Najmabadi and more Quizzes Electrical and Electronics Engineering in PDF only on Docsity!
ECE65 (Fall 2009), Quiz 2
Notes: 1. Write your answers on these three sheets.
- For each problem, 20% of points are alloacted for the correct final answer.
Problem 1. Consider NMOS circuit below with K = 0.4 mA/V^2 , Vt = 2 V, RD = 1 Kฮฉ and VDD = 12 V. Find vo when vi = 0 and 6 V. (6pt)
D iD
S
vi G
vo
VDD
R D
GS-KVL: vGS = vi DS-KVL: VDD = RDiD + vDS
A) vi = 0 V. From GS-KVL, we get vGS = vi = 0. As vGS < Vt = 2 V, NMOS is in cut-off, iD = 0, and vDS is found from DS-KVL:
DS-KVL: vo = vDS = VDD โ RDiD = 12 V
B) vi = 6 V. From GS-KVL, we get vGS = 6 V. Since vGS > Vt, NMOS is not in cut-off. Assume NMOS in saturation, then:
iD = K(vGS โ Vt)^2 = 0. 4 ร 10 โ^3 (6 โ 2)^2 = 6.4 mA DS-KVL: vDS = VDD โ RDiD = 12 โ 6. 4 ร 10 โ^3 ร 1 ร 103 = 5.6 V
Since vDS = 5. 6 > vGS โ Vt = 6 โ 2 = 4, NMOS is in saturation.
Therefore, vo = vDS = 5.8 V.
Problem 2. Find I in the circuit below with Si diodes(10 pts).
V 1 V 2
5V
โ5V
I
D D
1k
2k
I 1
ID
V 1 V 2
5V
I
1k I 1
2k
ID
โ5V
0.7V + 0.7V โ
โ
A good starting guess is that both didoes are forward biased: vD 1 = 0. 7 V , iD 1 > 0, and vD 2 = 0. 7 V , iD 2 > 0.
Replacing the diodes with their circuit model (see circuit), we see: V 1 = vD 1 = 0.7 V and
V 2 โ V 1 = vD 2 = 0. 7 โ V 2 = 0
Ohm Law: I 1 =
5 โ V 1
= 4.3 mA
Ohm Law: iD 2 =
V 2 โ (โ5)
= 2.5 mA
KCL: I = iD 1 = I 1 โ iD 2 = 4. 3 โ 2 .5 = 1.8 mA
Since both iD 1 and iD 2 are positive, our assumption of both diodes ON is justified and I = 1.8 mA.
Solution for the other three possible cases for diodesโ states are given below showing them all to be incorrect:
D1 ON, D2 OFF D1 OFF, D2 ON D1 OFF, D2 OFF
V 1 V 2
5V
I
1k I 1
2k
ID
โ5V
0.7V + โ
V 1 V 2
5V
I
1k I 1
2k
ID
โ5V
+0.7V โ
V 1 V 2
5V
I
1k I 1
2k
ID
โ5V
I 1 = iD 1 & iD 2 = 0 I 1 = iD 2 = (10 โ 0 .7)/ 3000 I 1 = 0 & iD 2 = 0 V 1 = 0.7 & V 2 = โ5 V I 1 = iD 2 = 3.1 mA V 1 = 5 & V 2 = โ5 V vD 2 = V 1 โ V 2 = 5. 7 > 0 .7 V vD 1 = V 1 = 5 โ 1000 i 1 vD 1 = V 1 = 5 > 0 .7 V D2 is NOT OFF vD 1 = 1. 9 > 0 .7 V D1 is NOT OFF D1 is NOT OFF
