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Midterm 2 Exam for ECE65 (Winter 2009) - Transistor and Diode Circuits - Prof. Farrokh Naj, Exams of Electrical and Electronics Engineering

University of California - San DiegoElectrical and Electronics Engineering
Prof. Farrokh Najmabadi

The midterm 2 exam for the ece65 (winter 2009) course focusing on transistor and diode circuits. The exam includes three problems dealing with finding the state of transistors, analyzing diode circuits, and determining ib, ic, vbe, vce, and the state of a si transistor.

Typology: Exams

Pre 2010

Uploaded on 03/28/2010

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ECE65 (Winter 2009), Midterm 2
Name:
Notes: 1. Write your answers on these three sheets.
2. For each problem, 20% of points are alloacted for the correct final answer.
Problem 1. Find the state of this transistor if vGS > vt. (5pts)
By KVL: vDS =vGS .
Since vGS > vt, the NMOS is ON (should be either in active or ohmic). Furthermore,
since vDS =vGS > vGS โˆ’vt, NMOS should be in active region.
pf3

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ECE65 (Winter 2009), Midterm 2

Name:

Notes:2. For each problem, 20% of points are alloacted for the correct 1. Write your answers on these three sheets. final answer.

Problem 1. Find the state of this transistor if vGS > vt. (5pts)

By KVL: vDS = vGS. Sincesince vvGS > vt, the NMOS is ON (should be either in active or ohmic). Furthermore, DS =^ vGS > vGS โˆ’^ vt, NMOS should be in active region.

Problem 2. In the circuit below with Si diodes, find i for โˆ’ 20 โ‰ค vs โ‰ค 20 V (13pts)

V S โˆ’+

i^ i

S

iD

5k

D1D2 5k

KCL: iD 1 = iD 2 = iD KCL: is = iD + i KVL: vs = 5 ร— 103 is + vD 1 + vD 2 KVL: vD 1 + vD 2 = 5 ร— 103 i

Because the two didoes are in series,(i iD 1 = iD 2 = iD, the two diodes are either ON D >= 0) or OFF (iD = 0). For BOTH diodes being ON:equations we get: vD 1 = vD 2 = vฮณ = 0.7 V, iD > 0. Substituting in the circuit

vs = 5 ร— 103 is + vD 1 + vD 2 โ†’ vs = 5 ร— 103 is + 1. 4 โ†’ is = 2 ร— 10 โˆ’^4 (vs โˆ’ 1 .4) vD 1 + vD 2 = 5 ร— 103 i โ†’ 1 .4 = 5 ร— 103 i โ†’ i = 0.28 mA is = iD + i โ†’ iD = is โˆ’ 0. 28 ร— 10 โˆ’^3 = 2 ร— 10 โˆ’^4 (vs โˆ’ 1 .4) โˆ’ 2. 8 ร— 10 โˆ’^4 iD > 0 โ†’ 2 ร— 10 โˆ’^4 (vs โˆ’ 1 .4) โˆ’ 2. 8 ร— 10 โˆ’^4 > 0 โ†’ vs > 2 .8 V For BOTH diodes being OFF: v iD = 0 and vD 1 < vฮณ = 0.7 V and vD 2 < vฮณ = 0.7 V (and D 1 +^ vD 2 <^1 .4 V). Substituting in the circuit equations we get: is = iD + i โ†’ is = i vs = 5 ร— 103 is + vD 1 + vD 2 = 5 ร— 103 is + 5 ร— 103 i = 10^4 i โ†’ i = 10โˆ’^4 vs vD 1 + vD 2 = 5 ร— 103 i = 0. 5 vs vD 1 + vD 2 < 1. 4 โ†’ 0. 5 vs < 1. 4 โ†’ vs < 2 .8 V Therefore, forFor 2. 8 < v โˆ’ 20 โ‰ค vs < 2 .8 V, both diodes are OFF i = 10โˆ’^4 vs. s โ‰ค^ 20 V, both diodes are ON and^ i^ = 0.28 mA.