Midterm Exam for ECE65 (Winter 2007) - Circuit Design with OpAmps and Diodes, Exams of Electrical and Electronics Engineering

The midterm exam for the ece65 (winter 2007) course focusing on circuit design with operational amplifiers (opamps) and diodes. The exam includes three problems: designing a high-pass filter circuit, finding the output voltage for a given input voltage, and determining the range of input voltage for which a diode is off. Detailed solutions for each problem.

Typology: Exams

Pre 2010

Uploaded on 03/28/2010

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ECE65 (Winter 2007), Midterm
Name:
Notes: 1. Write your answers on these three sheets.
2. For each problem, 20% of points are alloacted for the correct final answer.
3. Use the following information in solving or designing circuits: OpAmps are powered
by ±15 V power supplies (power supplies not shown), have a unity-gain bandwidth of
106Hz, a short-circuit current limit (maximum output current limit) of 250 mA, and a
slew rate of 1 V/µs. In circuit design, use commercial resistor and capacitor values of 1,
1.1, 1.2, 1.3, 1.5, 1.6, 1.8, 2, 2.2, 2.4, 2.7, 3., 3.3, 3.6, 3.9, 4.3, 4.7, 5.1, 5.6, 6.2, 6.8, 7.5,
8.2, 9.1 (×10nwhere nis an integer). You can also use 5 mH inductors.
Problem 1. Design a circuit with a transfer function of H(j ω) = 1
1j105 and a
minimum input impedance of 10 kΩ. (10pts)
The transfer function is for a first-order high-pass filter with ωc= 105rad/s. Since
K= 1, we can (and should) use a passive filter. Prototype of the circuit is shown below
with
vo
i
v
C
R
+
+
ωc=1
RC = 105
Zi|min =R10 kΩ
Choosing R = 10 kΩ, we get:
C=1
c
=1
104×105= 109F = 1 nF
Therefore, R= 10 kΩ and C= 1 nF (both commercial values).
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ECE65 (Winter 2007), Midterm

Name:

Notes: 1. Write your answers on these three sheets.

  1. For each problem, 20% of points are alloacted for the correct final answer.
  2. Use the following information in solving or designing circuits: OpAmps are powered by ±15 V power supplies (power supplies not shown), have a unity-gain bandwidth of 106 Hz, a short-circuit current limit (maximum output current limit) of 250 mA, and a slew rate of 1 V/μs. In circuit design, use commercial resistor and capacitor values of 1, 1.1, 1.2, 1.3, 1.5, 1.6, 1.8, 2, 2.2, 2.4, 2.7, 3., 3.3, 3.6, 3.9, 4.3, 4.7, 5.1, 5.6, 6.2, 6.8, 7.5, 8.2, 9.1 (× 10 n^ where n is an integer). You can also use 5 mH inductors.

Problem 1. Design a circuit with a transfer function of H(jω) = 1 1 − j 105 /ω and a

minimum input impedance of 10 kΩ. (10pts)

The transfer function is for a first-order high-pass filter with ωc = 10^5 rad/s. Since K = 1, we can (and should) use a passive filter. Prototype of the circuit is shown below with

vi vo

C

R

ωc =

RC

= 10^5

Zi|min = R ≥ 10 kΩ

Choosing R = 10 kΩ, we get:

C =

Rωc

104 × 105

= 10−^9 F = 1 nF

Therefore, R = 10 kΩ and C = 1 nF (both commercial values).

Problem 2. Find vo/vi (Assume OpAmp is ideal). (10pts)

vi vo

v 1 v 2

vn

vp

R

R R R

R

R −

Ideal OpAmp: ip ' in ' 0.

Non-inverting terminal of the OpAmp is grounded, thus vp = 0

We have negative feedback: vn ≈ vp. Since vp = 0, vn ≈= 0

Using node-voltage method:

Node vn vn − vi R

vn − v 1 R

→ vn − vi + vn − v 1 = 0 → v 1 = −vi

Node v 1 v 1 − vn R

v 1 − 0 R

v 1 − v 2 R

→ v 1 − vn + v 1 + v 1 − v 2 = 0 → v 2 = 3v 1 = − 3 vi

Node v 2 v 2 − v 1 R

v 2 − 0 R

v 2 − vo R

→ v 2 − v 1 + v 2 + v 2 − vo = 0 → vo = 3v 2 − v 1 = − 9 vi + vi = − 8 vi

Thus vo/vi = −8.